ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 https://ncertmcq.com/ml-aggarwal-icse-solutions-for-class-10-maths/

More Exercise

Take π = \(\\ \frac { 22 }{ 7 } \), unless stated otherwise.

Ex 17.4 Class 10 Ml Aggarwal Question 1.
The adjoining figure shows a cuboidal block of wood through which a circular cylinderical hole of the biggest size is drilled. Find the volume of the wood left in the block.
Ex 17.4 Class 10 Ml Aggarwal
Solution:
Diameter of the biggest hole = 30 cm.
Radius (r) = \(\\ \frac { 30 }{ 2 } \) = 15 cm
and height (h) = 70 cm.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q1.2

Ml Aggarwal Class 10 Ex 17.4. Solutions Question 2.
The given figure shows a solid trophy made of shining glass. If one cubic centimetre of glass costs Rs 0.75, find the cost of the glass for making the trophy.
Ml Aggarwal Class 10 Ex 17.4. Solutions
Solution:
Edge of cubical part = 28 cm
and radius of cylindrical part (r) = \(\\ \frac { 28 }{ 2 } \) = 14cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q2.2

Ml Aggarwal Class 10 Solutions Mensuration Question 3.
From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.
Solution:
Edge of a cube = 14 cm
Volume = (side)³ = (14)³ = 14 × 14 × 14 cm³ = 2744 cm³
Now diameter of the cone cut out from it = 14 cm
Ml Aggarwal Class 10 Solutions Mensuration
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q3.2

Ml Aggarwal Class 10 Mensuration Question 4.
A cone of maximum volume is curved out of a block of wood of size 20 cm x 10 cm x 10 cm. Find the volume of the remaining wood.
Solution:
Size of wooden block = 20 cm × 10 cm × 10 cm
Maximum diameter of the cone = 10 cm
and height (h) = 20 cm
Ml Aggarwal Class 10 Mensuration

Ml Aggarwal Class 10 Mensuration Solutions Question 5.
16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm x 8 cm x 8 cm and then the box is filled with water. Find the volume of the water filled in the box.
Solution:
Given
Radius of each glass sphere = 2 cm
Ml Aggarwal Class 10 Mensuration Solutions

Ml Aggarwal Class 10 Solutions Chapter Mensuration Question 6.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of the wood in the entire stand, correct to 2 decimal places.
Ml Aggarwal Class 10 Solutions Chapter Mensuration
Solution:
Dimensions of cuboid = 15 cm × 10 cm × 3.5 cm
and radius of each conical depression (r) = 0.5 cm
and depth (h) = 1.4 cm
Volume of cuboid = l × b × h
Ml Aggarwal Class 10 Solutions Gst

Ml Aggarwal Class 10 Solutions Gst Question 7.
A cuboidal block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.
Solution:
Side of cuboidal = 7 cm
Diameter of hemisphere = 7 cm
and radius (r) = \(\\ \frac { 7 }{ 2 } \) cm
Ml Aggarwal Class 10 Solutions

Ml Aggarwal Class 10 Solutions Question 8.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in the given figure). If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.
Www.Ilovemaths.Com For Class 10 Icse Solutions
Solution:
Height of the cylinder = 10 cm
and radius of the base = 3.5 cm
Total surface area
= Curved surface area of cylinder + 2 × Curved surface area of hemisphere
Ml Aggarwal Class 10 Solutions Icse Gst

Www.Ilovemaths.Com For Class 10 Icse Solutions Question 9.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy.
Solution:
Total height of the toy = 15.5 cm
Radius of the base of the conical part (r) = 3.5 cm
Ml Aggarwal Class 10 Solutions Icse
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q9.2

Ml Aggarwal Class 10 Solutions Icse Gst Question 10.
A circus tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas used to make the tent.
Solution:
Radius of base of cylindrical portion of tent (r) = \(\\ \frac { 24 }{ 2 } \) = 12 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q10.2

Ml Aggarwal Class 10 Solutions Icse Question 11.
An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².
Solution:
Total height of the tent = 85 m
Height of cylindrical part (h1) = 50 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q11.2

Question 12.
From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.
Solution:
Radius of solid cylinder (r1) = 7 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q12.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q12.2

Question 13.
The given figure shows a wooden toy rocket which is in the shape of a circular cone mounted on a circular cylinder. The total height of the rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted green and the cylindrical portion red, find the area of the rocket painted with each of these colours. Also, find the volume of the wood in the rocket. Use π = 3.14 and give answers correct to 2 decimal places.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q13.1
Solution:
In the given figure,
The total height of the toy rocket = 26 cm
Diameter of cylindrical portion = 3 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q13.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q13.3

Question 14.
The given figure shows a hemisphere of radius 5 cm surmounted by a right circular cone of base radius 5 cm. Find the volume of the solid if the height of the cone is 7 cm. Give your answer correct to two places of decimal.
Solution:
Height of the conical part (h) = 7 cm
and radius of the base (r) = 5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q14.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q14.2

Question 15.
A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is \(\\ \frac { 2 }{ 3 } \) of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal
Solution:
Radius of base of hemisphere = \(\\ \frac { 7 }{ 2 } \) m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q15.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q15.3

Question 16.
A circular hall (big room) has a hemispherical roof. The greatest height is equal to the inner diameter, find the area of the floor, given that the capacity of the hall is 48510 m³.
Solution:
Let h be the greatest height
and r be the radius of the base
Then 2r = h + r ⇒ h = r
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q16.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q16.2

Question 17.
A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains \(41 \frac { 19 }{ 21 } \) m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building
Solution:
Volume of air in dome = \(41 \frac { 19 }{ 21 } \) m³
= \(\\ \frac { 880 }{ 21 } \) m³
Let radius of the dome = r m
Then height (h) = r m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q17.1

Question 18.
A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket. (Use π = 3.14).
Solution:
Height of the cylindrical part (A) = 12 cm
Diameter = 6 cm
Radius (r) = \(\\ \frac { 6 }{ 2 } \) = 3 cm
Slant height of the conical part (l) = 5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q18.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q18.2

Question 19.
A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid. (Take π = 3.14)
Solution:
Diameter = 3.5 cm
Radius (r) = \(\\ \frac { 3.5 }{ 2 } \) = 1.75 cm
Height of cylindrical part (h1) = 10 cm
and height of conical part (h2) = 6 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q19.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q19.2

Question 20.
A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.
Solution:
Height of cylindrical part = 13 cm
Radius = 5 cm
Radius of cone (r) = 5 cm
Height of cone (h) = 12 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q20.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q20.2

Question 21.
The given figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1 : 200, find
(i) the total surface area of the solid in π m².
(ii) the volume of the solid in π litres.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q21.1
Solution:
(i) In the given figure,
Height of cylindrical portion (h) = 8 cm
Radius (r) = 3 cm
Scale = 1 : 200
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q21.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 Q21.3

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11

More Exercises

Midpoint Calculator is used to find the midpoint between 2 line segments using the midpoint formula. Use our calculator to find accurate midpoints step by step.

Question 1.
Find the co-ordinates of the mid-point of the line segments joining the following pairs of points:
(i) (2, – 3), ( – 6, 7)
(ii) (5, – 11), (4, 3)
(iii) (a + 3, 5b), (2a – 1, 3b + 4)
Solution:
(i) Co-ordinates of the mid-point of (2, -3), ( -6, 7)
\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) or \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q1.2

Question 2.
The co-ordinates of two points A and B are ( – 3, 3) and (12, – 7) respectively. P is a point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of P.
Solution:
Points are A (-3, 3), B (12, -7)
Let P (x1,  y1) be the point which divides AB in the ratio of m1 : m2 i.e. 2 : 3
then co-ordinates of P will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q2.1

Question 3.
P divides the distance between A ( – 2, 1) and B (1, 4) in the ratio of 2 : 1. Calculate the co-ordinates of the point P.
Solution:
Points are A (-2, 1) and B (1, 4) and
Let P (x, y) divides AB in the ratio of m1 : m2 i.e. 2 : 1
Co-ordinates of P will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q3.1

Question 4.
(i) Find the co-ordinates of the points of trisection of the line segment joining the point (3, – 3) and (6, 9).
(ii) The line segment joining the points (3, – 4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, – 2) and \(\left( \frac { 5 }{ 3 } ,q \right) \) respectively, find the values of p and q.
Solution:
(i) Let P (x1, y1) and Q (x2, y2) be the points
which trisect the line segment joining the points
A (3, -3) and B (6, 9)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q4.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q4.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q4.3

Question 5.
(i) The line segment joining the points A (3, 2) and B (5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k.
(ii) A point P divides the line segment joining the points A (3, – 5) and B ( – 4, 8) such that \(\frac { AP }{ PB } =\frac { k }{ 1 } \) If P lies on the line x + y = 0, then find the value of k.
Solution:
(i) The point P (x, y) divides the line segment joining the points
A (3, 2) and B (5, 1) in the ratio 1 : 2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q5.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q5.3

Question 6.
Find the coordinates of the point which is three-fourth of the way from A (3, 1) to B ( – 2, 5).
Solution:
Let P be the required point, then
\(\frac { AP }{ AB } =\frac { 3 }{ 4 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q6.2

Question 7.
Point P (3, – 5) is reflected to P’ in the x- axis. Also P on reflection in the y-axis is mapped as P”.
(i) Find the co-ordinates of P’ and P”.
(ii) Compute the distance P’ P”.
(iii) Find the middle point of the line segment P’ P”.
(iv) On which co-ordinate axis does the middle point of the line segment P P” lie ?
Solution:
(i) Co-ordinates of P’, the image of P (3, -5)
when reflected in x-axis will be (3, 5)
and co-ordinates of P”, the image of P (3, -5)
when reflected in y-axis will be (-3, -5)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q7.2

Question 8.
Use graph paper for this question. Take 1 cm = 1 unit on both axes. Plot the points A(3, 0) and B(0, 4).
(i) Write down the co-ordinates of A1, the reflection of A in the y-axis.
(ii) Write down the co-ordinates of B1, the reflection of B in the x-axis.
(iii) Assign.the special name to the quadrilateral ABA1B1.
(iv) If C is the mid point is AB. Write down the co-ordinates of the point C1, the reflection of C in the origin.
(v) Assign the special name to quadrilateral ABC1B1.
Solution:
Two points A (3, 0) and B (0,4) have been plotted on the graph.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q8.1
(i)∵ A1 is the reflection of A (3, 0) in the v-axis Its co-ordinates will be ( -3, 0)
(ii)∵ B1 is the reflection of B (0, 4) in the .x-axis co-ordinates of B, will be (0, -4)
(iii) The so formed figure ABA1B1 is a rhombus.
(iv) C is the mid point of AB co-ordinates of C” will be \(\frac { AP }{ AB } =\frac { 3 }{ 4 } \)
∵ C, is the reflection of C in the origin
co-ordinates of C, will be \(\left( \frac { -3 }{ 2 } ,-2 \right) \)
(v) The name of quadrilateral ABC1B1 is a trapezium because AB is parallel to B1C1.

Question 9.
The line segment joining A ( – 3, 1) and B (5, – 4) is a diameter of a circle whose centre is C. find the co-ordinates of the point C. (1990)
Solution:
∵ C is the centre of the circle and AB is the diameter
C is the midpoint of AB.
Let co-ordinates of C (x, y)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q9.1

Question 10.
The mid-point of the line segment joining the points (3m, 6) and ( – 4, 3n) is (1, 2m – 1). Find the values of m and n.
Solution:
Let the mid-point of the line segment joining two points
A(3m, 6) and (-4, 3n) is P( 1, 2m – 1)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q10.1

Question 11.
The co-ordinates of the mid-point of the line segment PQ are (1, – 2). The co-ordinates of P are ( – 3, 2). Find the co-ordinates of Q.(1992)
Solution:
Let the co-ordinates of Q be (x, y)
co-ordinates of P are (-3, 2) and mid-point of PQ are (1, -2) then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q11.1

Question 12.
AB is a diameter of a circle with centre C ( – 2, 5). If point A is (3, – 7). Find:
(i) the length of radius AC.
(ii) the coordinates of B.
Solution:
AC = \(\sqrt { { \left( 3+2 \right) }^{ 2 }+{ \left( -7-5 \right) }^{ 2 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q12.1

Question 13.
Find the reflection (image) of the point (5, – 3) in the point ( – 1, 3).
Solution:
Let the co-ordinates of the images of the point A (5, -3) be
A1 (x, y) in the point (-1, 3) then
the point (-1, 3) will be the midpoint of AA1.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q13.1

Question 14.
The line segment joining A \(\left( -1,\frac { 5 }{ 3 } \right) \) the points B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Calculate
(i) the value of a
(ii) the co-ordinates of P. (1994)
Solution:
Let P (x, y) divides the line segment joining
the points \(\left( -1,\frac { 5 }{ 3 } \right) \), B(a, 5) in the ratio 1 : 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q14.1

Question 15.
The point P ( – 4, 1) divides the line segment joining the points A (2, – 2) and B in the ratio of 3 : 5. Find the point B.
Solution:
Let the co-ordinates of B be (x, y)
Co-ordinates of A (2, -2) and point P (-4, 1)
divides AB in the ratio of 3 : 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q15.1

Question 16.
(i) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7 ,6) ?
(ii) In what ratio does the point ( – 4, b) divide the line segment joining the points P (2, – 2), Q ( – 14, 6) ? Hence find the value of b.
Solution:
(i) Let the ratio be m1 : m2 that the point (5, 4) divides
the line segment joining the points (2, 1), (7, 6).
\(5=\frac { { m }_{ 1 }\times 7+{ m }_{ 2 }\times 2 }{ { m }_{ 1 }+{ m }_{ 2 } } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q16.1

Question 17.
The line segment joining A (2, 3) and B (6, – 5) is intercepted by the x-axis at the point K. Write the ordinate of the point k. Hence, find the ratio in which K divides AB. Also, find the coordinates of the point K.
Solution:
Let the co-ordinates of K be (x, 0) as it intersects x-axis.
Let point K divides the line segment joining the points
A (2, 3) and B (6, -5) in the ratio m1 : m2.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q17.1

Question 18.
If A ( – 4, 3) and B (8, – 6), (i) find the length of AB.
(ii) in what ratio is the line joining AB, divided by the x-axis? (2008)
Solution:
Given A (-4, 3), B (8, -6)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q18.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q18.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q18.3

Question 19.
(i) Calculate the ratio in which the line segment joining (3, 4) and( – 2, 1) is divided by the y-axis.
(ii) In what ratio does the line x – y – 2 = 0 divide the line segment joining the points (3, – 1) and (8, 9)? Also, find the coordinates of the point of division.
Solution:
(i) Let the point P divides the line segment joining the points
A (3, 4) and B (-2, 3) in the ratio of m1 : m2 and
let the co-ordinates of P be (0, y) as it intersects the y-axis
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q19.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q19.2

Question 20.
Given a line segment AB joining the points A ( – 4, 6) and B (8, – 3). Find:
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii)the length of AB.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1. So,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q20.1

Question 21.
(i) Write down the co-ordinates of the point P that divides the line joining A ( – 4, 1) and B (17,10) in the ratio 1 : 2.
(ii)Calculate the distance OP where O is the origin.
(iii)In what ratio does the y-axis divide the line AB ?
Solution:
(i) Let co-ordinate of P be (x, y) which divides the line segment joining the points
A ( -4, 1) and B(17, 10) in the ratio of 1 : 2.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q21.1

Question 22.
Calculate the length of the median through the vertex A of the triangle ABC with vertices A (7, – 3), B (5, 3) and C (3, – 1)
Solution:
Let D (x, y) be the median of ΔABC through A to BC.
∴ D will be the midpoint of BC
∴ Co-ordinates of D will be,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q22.1

Question 23.
Three consecutive vertices of a parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0). Find the fourth vertex D.
Solution:
Let O in the mid-point of AC the diagonal of ABCD
∴ Co-ordinates of O will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q23.1

Question 24.
If the points A ( – 2, – 1), B (1, 0), C (p, 3) and D (1, q) from a parallelogram ABCD, find the values of p and q.
Solution:
A (-2, -1), B (1, 0), C (p, 3) and D (1, q)
are the vertices of a parallelogram ABCD
∴ Diagonal AC and BD bisect each other at O
O is the midpoint of AC as well as BD
Let co-ordinates of O be (x, y)
When O is mid-point of AC, then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q24.1

Question 25.
If two vertices of a parallelogram are (3, 2) ( – 1, 0) and its diagonals meet at (2, – 5), find the other two vertices of the parallelogram.
Solution:
Two vertices of a ||gm ABCD are A (3, 2), B (-1, 0)
and point of intersection of its diagonals is P (2, -5)
P is mid-point of AC and BD.
Let co-ordinates of C be (x, y), then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q25.1

Question 26.
Prove that the points A ( – 5, 4), B ( – 1, – 2) and C (5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square.
Solution:
Points A (-5, 4), B (-1, -2) and C (5, 2) are given.
If these are vertices of an isosceles triangle ABC then
AB = BC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q26.1

Question 27.
Find the third vertex of a triangle if its two vertices are ( – 1, 4) and (5, 2) and mid point of one sides is (0, 3).
Solution:
Let A (-1, 4) and B (5, 2) be the two points and let D (0, 3)
be its the midpoint of AC and co-ordinates of C be (x, y).
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q27.1

Question 28.
Find the coordinates of the vertices of the triangle the middle points of whose sides are \(\left( 0,\frac { 1 }{ 2 } \right) ,\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right) and\left( \frac { 1 }{ 2 } ,0 \right) \)
Solution:
Let ABC be a ∆ in which \(D\left( 0,\frac { 1 }{ 2 } \right) ,E\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right) andF\left( \frac { 1 }{ 2 } ,0 \right) \),
the mid-points of sides AB, BC and CA respectively.
Let co-ordinates of A be (x1, y1), B (x2, y2), C (x3, y3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q28.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q28.2

Question 29.
Show by section formula that the points (3, – 2), (5, 2) and (8, 8) are collinear.
Solution:
Let the point (5, 2) divides the line joining the points (3, -2) and (8, 8)
in the ratio of m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q29.1

Question 30.
Find the value of p for which the points ( – 5, 1), (1, p) and (4, – 2) are collinear.
Solution:
Let points A (-5, 1), B (1, p) and C (4, -2)
are collinear and let point A (-5, 1) divides
BC in the ratio in m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q30.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q30.2

Question 31.
A (10, 5), B (6, – 3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM = \(\\ \frac { 1 }{ 2 } \) BC.
Solution:
Co-ordinates of L will be
\(\left( \frac { 10+6 }{ 2 } ,\frac { 5-3 }{ 2 } \right) or\left( \frac { 16 }{ 2 } ,\frac { 2 }{ 2 } \right) or(8,1)\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q31.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q31.2

Question 32.
A (2, 5), B ( – 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and.Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.
(i) Find the co-ordinates of P and Q.
(ii) Show that PQ = \(\\ \frac { 1 }{ 3 } \) BC.
Solution:
A (2, 5), B (-1, 2) and C (5, 8) are the vertices of a ∆ABC,
P and Q are points on AB
and AC respectively such that \(\frac { AP }{ PB } =\frac { AQ }{ QC } =\frac { 1 }{ 2 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q32.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q32.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q32.3

Question 33.
The mid-point of the line segment AB shown in the adjoining diagram is (4, – 3). Write down die co-ordinates of A and B.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q33.1
Solution:
A lies on x-axis and B on the y-axis.
Let co-ordinates of A be (x, 0) and of B be (0, y)
P (4, -3) is the mid-point of AB
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q33.2

Question 34.
Find the co-ordinates of the centroid of a triangle whose vertices are A ( – 1, 3), B(1, – 1) and C (5, 1) (2006)
Solution:
Co-ordinates of the centroid of a triangle,
whose vertices are (x1, y1), (x2, y2) and
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q34.1

Question 35.
Two vertices of a triangle are (3, – 5) and ( – 7, 4). Find the third vertex given that the centroid is (2, – 1).
Solution:
Let the co-ordinates of third vertices be (x, y)
and other two vertices are (3, -5) and (-7, 4)
and centroid = (2, -1).
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q35.1

Question 36.
The vertices of a triangle are A ( – 5, 3), B (p – 1) and C (6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, – 1).
Solution:
The vertices of ∆ABC are A (-5, 3), B (p, -1), C (6, q)
and the centroid of ∆ABC is O (1, -1)
co-ordinates of the centroid of ∆ABC will be
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 Q36.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS

More Exercises

Question 1.
In the given figure, O is the centre of the circle. If ∠ABC = 20°, then ∠AOC is equal to
(a) 20°
(b) 40°
(c) 60°
(d) 10°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q1.1
Solution:
In the given figure,
Arc AC subtends ∠AOC at the centre
and ∠ABC at the remaining part of the circle
∠AOC = 2∠ABC = 2 × 20° = 40° (b)

Question 2.
In the given figure, AB is a diameter of the circle. If AC = BC, then ∠CAB is. equal to
(a) 30°
(b) 60°
(c) 90°
(d) 45°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q2.1
Solution:
In the given figure,
AB is the diameter of the circle and AC = BC
∠ACB = 90° (angle in a semi-circle)
AC = BC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q2.2

Question 3.
In the given figure, if ∠DAB = 60° and ∠ABD = 50° then ∠ACB is equal to
(a) 60°
(b) 50°
(c) 70°
(d) 80°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q3.1
Solution:
In the given figure,
∠DAB = 60°, ∠ABD = 50°
In ∆ADB, ∆ADB = 180° – (60° + 50°)
= 180° – 110° = 70°
∠ACB = ∠ADB
(angles in the same segment) = 70° (c)

Question 4.
In the given figure, O is the centre of the circle. If ∠OAB = 40°, then ∠ACB is equal to
(a) 50°
(b) 40°
(c) 60°
(d) 70°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q4.1
Solution:
In the given figure, O is the centre of the circle.
In ∆OAB,
∠OAB = 40°
But ∠OBA = ∠OAB = 40°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q4.2

Question 5.
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to
(a) 80°
(b) 50°
(c) 40°
(d) 30°
Solution:
ABCD is a cyclic quadrilateral,
AB is the diameter of the circle circumscribing it
∠ADC = 140°, ∠BAC = Join AC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q5.1

Question 6.
In the given figure, O is the centre of the circle. If ∠BAO = 60°, then ∠ADC is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 120°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q6.1
Solution:
In the given figure, O is the centre of the circle ∠BAO = 60°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q6.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q6.3

Question 7.
In the given figure, O is the centre of the circle. If ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to
(a) 30°
(b) 45°
(c) 90°
(d) 60°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q7.1
Solution:
In the given figure, O is the centre of the circle
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q7.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q7.3
∠CAO = 105° – 45° = 60° (d)

Question 8.
In the given figure, O is the centre of a circle. If the length of chord PQ is equal to the radius of the circle, then ∠PRQ is
(a) 60°
(b) 45°
(c) 30°
(d) 15°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q8.1
Solution:
In the given figure, O is the centre of the circle
Chord PQ = radius of the circle
∆OPQ is an equilateral triangle
∴∠POQ = 60°
Arc PQ subtends ∠POQ at the centre and
∴∠PRQ at the remaining part of the circle
∴∠PRQ = \(\\ \frac { 1 }{ 2 } \) ∠POQ = \(\\ \frac { 1 }{ 2 } \) x 60° = 30° (c)

Question 9.
In the given figure, if O is the centre of the circle then the value of x is
(a) 18°
(b) 20°
(c) 24°
(d) 36°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q9.1
Solution:
In the given figure, O is the centre of the circle.
Join OA.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q9.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q9.3

Question 10.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
From Q, length of tangent PQ to the circle = 24 cm
and QO = 25 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q10.1

Question 11.
From a point which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Solution:
Let point P is 13 cm from O, the centre of the circle
Radius of the circle (OQ) = 5 cm
PQ and PR are tangents from P to the circle
Join OQ and OR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q11.2

Question 12.
If angle between two radii of a circle is 130°, the angle between the tangents at the ends of the radii is
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Solution:
Angles between two radii OA and OB = 130°
From A and B, tangents are drawn which meet at P
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q12.1

Question 13.
In the given figure, PQ and PR are tangents from P to a circle with centre O. If ∠POR = 55°, then ∠QPR is
(a) 35°
(b) 55°
(c) 70°
(d) 80°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q13.1
Solution:
In the given figure,
PQ and PR are the tangents to the circle from a point P outside it
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q13.2

Question 14.
If tangents PA and PB from an exterior point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 100°
Solution:
Length of tangents PA and PB to the circle from a point P
outside the circle with centre O, and inclined an angle of 80°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q14.1

Question 15.
In the given figure, PA and PB are tangents from point P to a circle with centre O. If the radius of the circle is 5 cm and PA ⊥ PB, then the length OP is equal to
(a) 5 cm
(b) 10 cm
(c) 7.5 cm
(d) 5√2 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q15.1
Solution:
In the given figure,
PA and PB are tangents to the circle with centre O.
Radius of the circle is 5 cm, PA ⊥ PB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q15.2

Question 16.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Solution:
AB is the diameter of a circle with radius 5 cm
At A, XAY is a tangent to the circle
CD || XAY at a distance of 8 cm from A
Join OC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q16.1

Question 17.
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other is
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Solution:
Radii of two concentric circles are 4 cm and 5 cm
AB is a chord of the bigger circle
which is tangent to the smaller circle at C.
Join OA, OC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q17.1

Question 18.
In the given figure, AB is a chord of the circle such that ∠ACB = 50°. If AT is tangent to the circle at the point A, then ∠BAT is equal to
(a) 65°
(b) 60°
(c) 50°
(d) 40°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q18.1
Solution:
In the given figure, AB is a chord of the circle
such that ∠ACB = 50°
AT is tangent to the circle at A
AT is tangent and AB is a chord
∠ACB = ∠BAT = 50°
(Angles in the alternate segments) (c)

Question 19.
In the given figure, O is the centre of a circle and PQ is a chord. If the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is
(a) 100°
(b) 80°
(c) 90°
(d) 75°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q19.1
Solution:
In the given figure, O is the centre of the circle.
PR is tangent and PQ is chord ∠RPQ = 50°
OP is radius and PR is tangent to the circle
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q19.2

Question 20.
In the given figure, PA and PB are tangents to a circle with centre O. If ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q20.1
Solution:
In the given figure,
PA and PB are tangents to the circle with centre O.
∠APB = 50°
But ∠AOB + ∠APB = 180°
∠AOB + 50° = 180°
⇒ ∠AOB = 180° – 50° = 130°
In ∆OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA
But ∠OAB + ∠OBA = 180° – ∠AOB
= 180° – 130° = 50°
∠OAB = \(\frac { { 50 }^{ 0 } }{ 2 } \) = 25° (a)

Question 21.
In the given figure, sides BC, CA and AB of ∆ABC touch a circle at point D, E and F respectively. If BD = 4 cm, DC = 3 cm and CA = 8 cm, then the length of side AB is
(a) 12 cm
(b) 11 cm
(c) 10 cm
(d) 9 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q21.1
Solution:
In the given figure,
sides BC, CA and AB of ∆ABC touch a circle at D, E and F respectively.
BD = 4 cm, DC = 3 cm and CA = 8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q21.2

Question 22.
In the given figure, sides BC, CA and AB of ∆ABC touch a circle at the points P, Q and R respectively. If PC = 5 cm, AR = 4 cm and RB = 6 cm, then the perimeter of ∆ABC is
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q22.1
Solution:
In the given figure, sides BC, CA and AB of ∆ABC
touch a circle at P, Q and R respectively
PC = 5 cm, AR = 4 cm, RB = 6 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q22.2

Question 23.
PQ is a tangent to a circle at point P. Centre of circle is O. If ∆OPQ is an isosceles triangle, then ∠QOP is equal to
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Solution:
PQ is tangent to the circle at point P centre of the circle is O.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q23.1

Question 24.
In the given figure, PT is a tangent at T to the circle with centre O. If ∠TPO = 25°, then the value of x is
(a) 25°
(b) 65°
(c) 115°
(d) 90°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q24.1
Solution:
In the given figure, PT is the tangent at T to the circle with centre O.
∠TPO = 25°
OT is the radius and TP is the tangent
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q24.2

Question 25.
In the given figure, PA and PB are tangents at ponits A and B respectively to a circle with centre O. If C is a point on the circle and ∠APB = 40°, then ∠ACB is equal to
(a) 80°
(b) 70°
(c) 90°
(d) 140°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q25.1
Solution:
In the given figure,
PA and PB are tangents to the circle at A and B respectively
C is a point on the circle and ∠APB = 40°
But ∠APB + ∠AOB = 180°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q25.2

Question 26.
In the given figure, two circles touch each other at A. BC and AP are common tangents to these circles. If BP = 3.8 cm, then the length of BC is equal to
(a) 7.6 cm
(b) 1.9 cm
(c) 11.4 cm
(d) 5.7 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q26.1
Solution:
In the given figure, two circles touch each other at A.
BC and AP are common tangents to these circles
BP = 3.8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q26.2

Question 27.
In the given figure, if sides PQ, QR, RS and SP of a quadrilateral PQRS touch a circle at points A, B, C and D respectively, then PD + BQ is equal to
(a) PQ
(b) QR
(c) PS
(d) SR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q27.1
Solution:
In the given figure,
sides PQ, QR, RS and SP of a quadrilateral PQRS
touch a circle at the points A, B, C and D respectively
PD and PA are the tangents to the circle
∴ PA = PD …(i)
Similarly, QA and QB are the tangents
∴ QA = QB …(ii)
Now PD + BQ = PA + QA = PQ (a)
[From (i) and (ii)]

Question 28.
In the given figure, PQR is a tangent at Q to a circle. If AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(a) 20°
(b) 40°
(b) 35°
(d) 45°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q28.1
Solution:
In the given figure, PQR is a tangent at Q to a circle.
Chord AB || PR and ∠BQR = 70°
BQ is chord and PQR is a tangent
∠BQR = ∠A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q28.2

Question 29.
Two chords AB and CD of a circle intersect externally at a point P. If PC = 15 cm, CD = 7 cm and AP = 12 cm, then AB is
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) none of these
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q29.1
Solution:
In the given figure,
two chords AB and CD of a circle intersect externally at P.
PC = 15 cm, CD = 7 cm, AP = 12 cm
Join AC and BD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q29.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q29.3

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

More Exercises

Question 1.
(a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q1.1
Solution:
(a) ∆ABC is an equilateral triangle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q1.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q1.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q1.4

Question 2.
(a) In the figure given below, AB is a diameter of the circle. If AE = BE and ∠ADC = 118°, find
(i) ∠BDC (ii) ∠CAE.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q2.1
(b) In the figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q2.2
Solution:
(a) Join DB, CA and CB.
∠ADC = 118° (given)
and ∠ADB = 90°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q2.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q2.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q2.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q2.6

Question 3.
(a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC).
(b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q3.1
Solution:
(a) Given: O is the centre of the circle.
To Prove : ∠AOC = 2 (∠ACB + ∠BAC).
Proof: In ∆ABC,
∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q3.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q3.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q3.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q3.5

Question 4.
(a) In the figure (i) given below, AB is the diameter of a circle. If DC is parallel to AB and ∠CAB = 25°, find :
(i)∠ADC (ii) ∠DAC.
(b) In the figure (ii) given below, the centre O of the smaller circle lies on the circumference of the bigger circle. If ∠APB = 70° and ∠BCD = 60°, find :
(i) ∠AOB (ii) ∠ACB
(iii) ∠ABD (iv) ∠ADB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q4.1
Solution:
(a) AB is diameter and DC || AB,
∠CAB = 25°, join AD,BD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q4.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q4.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q4.4

Question 5.
(a) In the figure (i) given below, ABCD is a cyclic quadrilateral. If AB = CD, Prove that AD = BC.
(b) In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ∠ABC = 50°, find ∠BDC and ∠BEC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q5.1
Solution:
(a) Given : ABDC is a cyclic quadrilateral AB = CD.
To Prove: AD = BC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q5.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q5.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q5.4

Question 6.
A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.
Solution:
Join OT, OP = 13 cm and TP = 12 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q6.2

Question 7.
Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution:
Given: Two circles with centre O and O’
touch each other internally at P.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q7.2

Question 8.
From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that
(i) ∠AOP = ∠BOP.
(ii) OP is the perpendicular bisector of the chord AB.
Solution:
Given: From a point P, outside the circle with centre O.
PA and PB are the tangents to the circle,
OA, OB and OP are joined.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q8.2

Question 9.
(a) The figure given below shows two circles with centres A, B and a transverse common tangent to these circles meet the straight line AB in C. Prove that:
AP : BQ = PC : CQ.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q9.1
(b) In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q9.2
Solution:
(a) Given: Two circles with centres A and B
and a transverse common tangent to these circles meets AB at C.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q9.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q9.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q9.5

Question 10.
In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q10.1
Solution:
In the given figure, two chords with centre A and B touch externally.
PM is a tangent to the circle with centre A
and QN is tangent to the circle with centre B.
PM = 15 cm, QN = 12 cm, PA = 17 cm, QB = 13 cm.
We have to find AB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q10.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q10.3

Question 11.
Two chords AB, CD of a circle intersect externally at a point P. If PB = 7 cm, AB = 9 cm and PD = 6 cm, find CD.
Solution:
∵ AB and CD are two chords of a circle
which intersect each other at P, outside the circle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q11.2

Question 12.
(a) In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q12.1
(b) In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also, find the length of the tangent drawn from P to the circle. .
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q12.2
Solution:
Given : (a) AB is a chord of a circle with centre O
and PT is tangent and CD is the diameter of the circle
which meet at P.
AP = 16 cm, AB = 12 cm, OP = 2 cm
∴PB = PA – AB = 16 – 12 = 4 cm
∵ABP is a secant and PT is tangent.
∴PT² = PA × PB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q12.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q12.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q12.5

Question 13.
In the figure given below, chord AB and diameter PQ of a circle with centre O meet at X. If BX = 5 cm, OX = 10 cm and.the radius of the circle is 6 cm, compute the length of AB. Also, find the length of tangent drawn from X to the circle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q13.1
Solution:
Chord AB and diameter PQ meet at X
on producing outside the circle
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q13.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q13.3

Question 14.
(a) In the figure (i) given below, ∠CBP = 40°, ∠CPB = q° and ∠DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that ∠AQD = 2 q° and the points C, P, B and Q are concyclic, find the values of p and q.
(b) In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, ∠AOB = 130° and ∠ACE = 20°, find:
(i)∠BEC (ii) ∠ACB
(iii) ∠BCD (iv) ∠CED.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q14.1
Solution:
(a) (i) Given : ABCD is a cyclic quadrilateral.
Ext. ∠PBC = ∠ADC
⇒ 40° = ∠ADC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q14.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q14.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q14.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q14.5

Question 15.
(a) In the figure (i) given below, APC, AQB and BPD are straight lines.
(i) Prove that ∠ADB + ∠ACB = 180°.
(ii) If a circle can be drawn through A, B, C and D, Prove that it has AB as a diameter
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q15.1
(b) In the figure (ii) given below, AQB is a straight line. Sides AC and BC of ∆ABC cut the circles at E and D respectively. Prove that the points C, E, P and D are concyclic.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q15.2
Solution:
(a) Given: In the figure, APC,
AQB and BPD are straight lines.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q15.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q15.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q15.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q15.6

Question 16.
(a) In the figure (i) given below, chords AB, BC and CD of a circle with centre O are equal. If ∠BCD = 120°, find
(i) ∠BDC (ii) ∠BEC
(iii) ∠AEC (iv) ∠AOB.
Hence Prove that AOAB is equilateral.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. The chord BC of the circle is parallel to the radius OD and the lines OC and BD meet at E. Prove that
(i) ∠CED = 3 ∠CBD (ii) CD = DA.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q16.1
Solution:
(a) In ∆BCD, BC = CD
∠CBD = ∠CDB
But ∠BCD + ∠CBD + ∠CDB = 180°
(∵ Angles of a triangle)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q16.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q16.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q16.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q16.5

Question 17.
(a) In the adjoining figure, (i) given below AB and XY are diameters of a circle with centre O. If ∠APX = 30°, find
(i) ∠AOX (ii) ∠APY (iii) ∠BPY (iv) ∠OAX.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q17.1
(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If ∠CBP = 25° and ∠CAP = 40°, find :
(i) ∠ADB (ii) ∠AOB (iii) ∠ACB (iv) ∠APB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q17.2
Solution:
(a) AB and XY are diameters of a circle with centre O.
∠APX = 30°.
To find :
(i) ∠AOX (ii) ∠APY
(iii) ∠BPY (iv) ∠OAX
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q17.3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q17.4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q17.5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test Q17.6

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2

More Exercises

Question 1.
Shweta deposits Rs. 350 per month in a recurring deposit account for one year at the rate of 8% p.a. Find the amount she will receive at the time of maturity.
Solution:
Deposit per month = Rs 350,
Rate of interest = 8% p.a.
Period (x) = 1 year
= 12 months
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q1.1

Question 2.
Salom deposited Rs 150 per month in a bank for 8 months under the Recurring Deposit Scheme. ‘What will be the maturity value of his deposit if the rate of interest is 8% per annum ?
Solution:
Deposit per month = Rs. 150
Rate of interest = 8% per
Period (x) = 8 month
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q2.1

Question 3.
Mrs. Goswami deposits Rs. 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value. (2009)
Solution:
Deposit per month (P) = Rs. 1000
Period = 3 years = 36 months
Rate = 8%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q3.1

Question 4.
Kiran deposited Rs. 200 per month for 36 months in a bank’s recurring deposit account. If the banks pays interest at the rate of 11% per annum, find the amount she gets on maturity ?
Solution:
Amount deposited month (P) = Rs. 200
Period (n) = 36 months,
Rate (R) = 11% p.a.
Now amount deposited in 36 months = Rs. 200 x 36 = Rs 7200
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q4.1

Question 5.
Haneef has a cumulative bank account and deposits Rs. 600 per month for a period of 4 years. If he gets Rs. 5880 as interest at the time of maturity, find the rate of interest.
Solution:
Interest = Rs. 58800
Monthly deposit (P) = Rs. 600
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q5.1

Question 6.
David opened a Recurring Deposit Account in a bank and deposited Rs. 300 per month for two years. If he received Rs. 7725 at the time of maturity, find the rate of interest per annum. (2008)
Solution:
Deposit during one month (P) = Rs. 300
Period = 2 years = 24 months.
Maturity value = Rs. 7725
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q6.1

Question 7.
Mr. Gupta-opened a recurring deposit account in a bank. He deposited Rs. 2500 per month for two years. At the time of maturity he got Rs. 67500. Find :
(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.
Solution:
Deposit per month = Rs. 2500
Period = 2 years = 24 months
Maturity value = Rs. 67500
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q7.1

Question 8.
Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs 800 per month for \(1 \frac { 1 }{ 2 } \) years. If he received Rs 15084 at the time of maturity, find the rate of interest per annum.
Solution:
Money deposited by Shahrukh per month (P)= Rs 800
r = ?
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q8.1

Question 9.
Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs 1200 as interest at the time of maturity, find:
(i) the monthly instalment
(ii) the amount of maturity. (2016)
Solution:
Interest = Rs 1200
Period (n) = 2 years = 24 months
Rate (r) = 6% p.a.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q9.1

Question 10.
Mr. R.K. Nair gets Rs 6,455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly instalment.
Solution:
Let monthly instalment is Rs P
here n = 1 year = 12 months
n = 12
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q10.1

Question 11.
Samita has a recurring deposit account in a bank of Rs 2000 per month at the rate of 10% p.a. If she gets Rs 83100 at the time of maturity. Find the total time for which the account was held.
Solution:
Deposit per month = Rs 2000,
Rate of interest = 10%, Let period = n months
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Ex 2 Q11.2

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test

More Exercises

Question 1.
1. A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being ₹ 18000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax is 8%, find:
(i) the VAT paid by the shopkeeper. .
(ii) the total amount that the consumer pays for the washing machine.
Solution:
(i) S.P. of washing machine
= \(\left( 1-\frac { 10 }{ 100 } \right) \) x ₹18000
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q1.2

Question 2.
A manufacturing company sold an article to its distributor for ₹22000 including VAT. The distributor sold the article to a dealer for ₹22000 excluding tax and the dealer sold it to a consumer for ₹25000 plus tax (under VAT). If the rate of sales tax (under VAT) at each stage is 10%, find :
(i) the sale price of the article for the manufacturing company.
(ii) the amount of VAT paid by the dealer.
Solution:
S.P. of an article for a manufacturer = ₹22000 including VAT
C.P. for the distributor = ₹22000
Rate of VAT = 10%
S.P. for the distributor of ₹22000 excluding VAT
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q2.1

Question 3.
The marked price of an article is ₹7500. A shopkeeper sells the article to a consumer at the marked prices and charges sales tax at . the rate of 7%. If the shopkeeper pays a VAT of ₹105, find the price inclusive of sales tax of the article which the shopkeeper paid to the wholesaler.
Solution:
Marked price of an article = ₹7500
Rate of S.T. = 7%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q3.1

Question 4.
A shopkeeper buys an article at a discount of 30% and pays sales tax at the rate of 6%. The shopkeeper sells the article to a consumer at 10% discount on the list price and charges sales tax at the’ same rate. If the list price of the article is ₹3000, find the price inclusive of sales tax paid by the shopkeeper.
Solution:
List price of an article = ₹3000
Rate of discount = 30%
and rate of S.T. = 6%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q4.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q4.2

Question 5.
Mukerjee purchased a movie camera for ₹27468. which includes 10% rebate on the list price and then 9% sales tax (under VAT) on the remaining price. Find the list price of the movie camera.
Solution:
Let list price of the movie camera = x
Rebate = 10%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q5.1

Question 6.
A retailer buys an article at a discount of 15% on the printed price from a wholesaler. He marks up the price by 10%. Due to competition in the market, he allows a discount of 5% to a buyer. If the buyer pays ₹451.44 for the article inclusive of sales tax (under VAT) at 8%, find :
(i) the printed price of the article
(ii) the profit percentage of the retailer.
Solution:
(i) Let the printed price of the article = ₹100
Then, retailer’s cost price
= ₹100-₹15 = ₹85
Now, marked price for the retailer
= ₹100 + ₹10 = ₹110
Rate of discount allowed = 5%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test Q6.2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test are helpful to complete your math homework.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax MCQS

More Exercises

A retailer purchases a fan for ₹ 1200 from a wholesaler and sells it to a consumer at 15% profit. If the rate of sales tax (under VAT) at every stage is 8%, then choose the correct answer from the given four options for questions 1 to 5 :

Question 1.
The selling price of the fan by the retailer (excluding tax) is
(a) ₹ 1200
(b) ₹ 1380
(c) ₹ 1490.40
(d) ₹ 11296
Solution:
Cost price of a fan = ₹ 1200
Profit = 15%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax MCQS Q1.1

Question 2.
VAT paid by the wholesaler is
(a) ₹ 96
(b) ₹ 14.40
(c) ₹ 110.40
(d) ₹ 180
Solution:
Rate of VAT =8%
.’. VAT paid by wholesaler = ₹ 1200 x \(\\ \frac { 8 }{ 100 } \)
= ₹ 96 (a)

Question 3.
VAT paid by the retailer
(a) ₹ 180
(b) ₹ 110.40
(c) ₹ 96
(d) ₹ 14.40
Solution:
VAT deducted by the retailer = ₹ \(\\ \frac { 1380\times 8 }{ 100 } \)
VAT paid by wholesalers = ₹ 96 .
Net VAT paid by his = ₹ 110.40 – 96.00
= ₹14.40 (d)

Question 4.
VAT collected by the Government on the sale of fan is
(a) ₹14.40
(b) ₹96
(c) ₹110.40
(d) ₹180
Solution:
VAT collected by the govt, on the sale of fan
= ₹ \(\\ \frac { 11040 }{ 100 } \)
= ₹110.40 (c)

Question 5.
The cost of the fan to the consumer inclusive of tax is
(a) ₹1296
(b) ₹1380
(c) ₹1310.40
(d) ₹1490.40
Solution:
Cost of fan to the consumer inclusive tax
= ₹1380 + 110.40
= ₹ 1490.40 (d)

A shopkeeper bought a TVfrom a distributor at a discount of 25% of the listed price of ₹ 32000. The shopkeeper sells the TV to a consumer at the listed price. If the sales tax (under VAT) is 6% at every stage, then choose the correct answer from the given four options for questions 6 to 8 :

Question 6.
VAT paid by the distributor is
(a) ₹1920
(b) ₹1400
(c) ₹480
(d) ₹8000
Solution:
List price of T.V. set = ₹32000
Discount = 25%
Rate of VAT = 6%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax MCQS Q6.1
= ₹1440 (b)

Question 7.
VAT paid by the shopkeeper is
(a) ₹480
(b) ₹1440
(c) ₹1920
(d) ₹8000
Solution:
Total VAT charged by the shopkeeper
= ₹32000 x \(\\ \frac { 6 }{ 100 } \)
= ₹1920
VAT already paid by distributor = ₹ 1440
Net VAT paid by shopkeeper
= ₹1920 – ₹1440
= ₹480 (a)

Question 8.
The cost of the TV to the consumer inclusive of tax is
(a) ₹8000
(b) ₹32000
(c) ₹33920
(d) none of these
Solution:
Cost of T.V. to the consumer inclusive of VAT = ₹32000 + 1920
= ₹33920 (c)

A wholesaler buys a computer from a manufacturer for ₹ 40000. He marks the price of the computer 20% above his cost price and sells it to a retailer at a discount of 10% on the marked price. The retailer sells the computer to a consumer at the marked price. If the rate of sales tax (under VAT) is 10% at every stage, then choose the correct answer from the given four options for questions 9 to 15 :

Question 9.
The marked price of the computer is
(a) ₹40000
(b) ₹48000
(c) ₹50000
(d) none of these
Solution:
C.R of computer for manufacturer = ₹40000
After marking 20% above the C.R, the price
= ₹40000 x \(\\ \frac { 100+20 }{ 100 } \)
= ₹40000 x \(\\ \frac { 120 }{ 100 } \)
= ₹48000 (b)

Question 10.
Cost of the computer to the retailer (excluding tax) is
(a) ₹36000
(b) ₹40000
(c) ₹43200
(d) ₹47520
Solution:
Rate of discount = 10%
.’. Sales price after discount
= ₹48000 x \(\\ \frac { 100-10 }{ 100 } \)
= ₹48000 x \(\\ \frac { 90 }{ 100 } \)
= ₹43200 (c)

Question 11.
Cost of the computer to the retailer inclusive of tax is
(a) ₹47520
(b) 43200
(c) 44000
(d) none of these
Solution:
Rate of sales tax (VAT) = 10%
Sales tax charged = ₹43200 x \(\\ \frac { 10 }{ 100 } \)
= ₹4320
Cost price of T.V. including S.T.
= ₹43200 + ₹4320
= ₹47520 (a)

Question 12.
VAT paid by the manufacturer is
(a) ₹4000
(b) ₹4320
(c) ₹320
(d) none of these
Solution:
VAT paid by the manufacturer
= ₹40000 x \(\\ \frac { 10 }{ 100 } \)
= ₹4000 (a)

Question 13.
VAT paid by the wholesaler is
(a) ₹4000
(b) ₹4320
(c) ₹320
(d) ₹480
Solution:
VAT paid by the wholesaler = ₹43200 x \(\\ \frac { 10 }{ 100 } \)
= ₹4320
VAT already paid = ₹4000
Net VAT paid by = ₹4320 – ₹4000
= ₹320 (c)

Question 14.
VAT paid by the retailer is
(a) ₹4000
(b) ₹4320
(c) ₹320
(d) ₹480
Solution:
VAT paid by retailer = 48000 x \(\\ \frac { 10 }{ 100 } \) = ₹4800
VAT already paid = ₹4320
Net VAT to be paid = ₹4800 – 4320
= ₹480 (d)

Question 15.
Consumer’s cost price inclusive of VAT is
(a) ₹47520
(b) ₹48000
(c) ₹52800
(d) ₹44000
Solution:
Sol. Sale price to consumer = ₹4800
VAT paid by the consumer = ₹48000 x \(\\ \frac { 10 }{ 100 } \)
= ₹4800
Consumers cost price inclusive of VAT = ₹48000 + ₹4800
= ₹52800 (c)

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1

More Exercises

Question 1.
A manufacturing company sells a T.V. to a trader A for ₹ 18000. Trader A sells it to a trader B at a point of ₹ 750 and trader B sells it to a consumer at a profit of ₹ 900. If the rate of sales tax (under VAT) is 10%, find
(i) the amount of tax received by the Government.
(ii) the amount paid by the consumer for the T.V.
Solution:
Sale price of a T.V. to trader A = ₹ 18000
Rate of VAT tax = 10%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q1.1
(i) Total tax paid to Govt. = ₹ (I800 + 75 + 90) = ₹ 1965
(ii) Amount paid by the consumer to trader B = ₹ 18000 + 750 + 900 + Tax 1965 = ₹ 21615

Question 2.
A manufacturer sells a washing machine to a wholesaler for ₹ 15000. The wholesaler sells it to a trader at a profit of ₹ 1200 and the trader sells it to a consumer at a profit of ₹ 1800. If the rate of VAT is 8%, find :
(i) The amount of VAT received by the State Government on the sale of this machine from the manufacturer and the wholesaler.
(ii) The amount that the consumer pays for the machine.
Solution:
Total amount under VAT = ₹ 15000 + ₹ 1200 + ₹ 1800 = ₹ 18000
(i) VAT = 8% of ₹ 18000
= \(\frac { 8 }{ 100 }\) x 18000 = ₹ 1440
(ii) Consumer pays for the machine = ₹ 18000 + ₹ 1440 = ₹ 19440

Question 3.
A manufacturer buys raw material for ₹ 40000 and pays sales tax at the rate of 4%. He sells the ready stock for ₹ 78000 and charges sales tax at the rate of 7.5%. Find the VAT paid by the manufacturer.
Solution:
Cost price of raw material = ₹ 40000
Rate of sales tax = 4%
Total tax = ₹ \(\frac { 40000 x 4 }{ 100 }\) = ₹ 1600
Selling price = ₹ 78000
Rate of sales tax = 7.5%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q3.1
VAT paid by the manufacturer = ₹ 5850 – ₹ 1600 = ₹ 4250

Question 4.
A shopkeeper buys a camera at a discount of 20% from the wholesaler, the printed price of the camera being ₹ 1600 and the rate of sales tax is 6%. The shopkeeper sells it to the buyer at the printed price and charges sales tax at the same rate. Find
(i) the price at which the camera can be bought.
(ii) the VAT (Value Added Tax) paid by the shopkeeper.
Solution:
Printed price of the camera (MP) = ₹ 1600
Rate of discount = 20%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q4.1
(i) Price of camera = ₹ 1600 + ₹ 96 = ₹ 1696
(ii) VAT paid by the shopkeeper= ₹ 96 – ₹ 76.80 = ₹ 19.20

Question 5.
The printed price of an article is ₹ 60000. The wholesaler allows a discount of 20% to the shopkeeper. The shopkeeper sells the article to the customer at the printed price. Sales tax (under VAT) is charged at the rate of 6% at every stage. Find :
(i) the cost to the shopkeeper inclusive of tax.
(ii) VAT paid by the shopkeeper to the Government.
(iii) the cost to the customer inclusive of tax.
Solution:
Printed price of an article = ₹ 60000
Rate of discount allowed = 20%
Total discount = ₹ 60000 x \(\frac { 20 }{ 100 }\) = ₹ 12000
S.P. after discount = ₹ 60000 – ₹ 12000 = ₹ 48000
Rate of VAT = 6%
(i) Amount paid by the shopkeeper
= ₹ 48000 + ₹ 48000 x \(\frac { 6 }{ 100 }\)
= ₹ 48000 + ₹ 2880 = ₹ 50880
(ii) The price at which the shopkeeper sold to the customer = ₹ 60000
Profit = ₹ 60000 – ₹ 48000 = ₹ 12000
VAT paid by the customer to the Govt.
= ₹ 12000 x \(\frac { 6 }{ 100 }\) = ₹ 720
(iii) Total cost to the customer = ₹ 60000 + VAT inclusive of tax
= ₹ 60000 + \(\frac { 60000 x 6 }{ 100 }\)
= ₹ 60000 + ₹ 3600 = ₹ 63600

Question 6.
A shopkeeper bought a TV at a discount of 30% of the listed price of ₹ 24000. The shopkeeper offers a discount of 10% of the listed price to his customer. If the VAT (Value Added Tax) is 10%, find : the amount paid by the customer, the VAT to be paid by the shopkeeper.
Solution:
List price = ₹ 24000
Discount = 30%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q6.1
(i) Amount paid by customer = ₹ 21600 + ₹ 2160 = ₹ 23760
(ii) Total VAT to be paid by shopkeeper = ₹ 2160 – ₹ 1680 = ₹ 480

Question 7.
A shopkeeper sells an article at the listed price of ₹ 1500 and the rate of VAT is 12% at each stage of sale. If the shopkeeper pays a VAT of ₹ 36 to the Government, what was the amount inclusive of tax at which the shopkeeper purchased the articles from the wholesaler?
Solution:
List price (M.P.) of an article = ₹ 1500
Rate of VAT = 12%
Total VAT = ₹ \(\frac { 1500 x 12 }{ 100 }\) = ₹ 180
But VAT paid by the shopkeeper = ₹ 36
Total VAT paid by wholeseller = ₹ 180 – ₹ 36 = ₹ 144
Rate of VAT = 12%
S.P. of the whole seller = \(\frac { 144 x 100 }{ 12 }\) = ₹ 1200
Total amount paid by the wholeseller including VAT = ₹ 1200 + ₹ 144 = ₹ 1344

Question 8.
A shopkeeper buys an article whose list price is ₹ 800 at some rate of discount from a wholesaler. He sells the article to a consumer at the list price and charges sales tax at the prescribed rate of 7.5%. If the shopkeeper has to pay a VAT of ₹ 6, find the rate of discount at which he bought the article from the wholesaler.
Solution:
List price (MP) of an article = ₹ 800
S.P. of the shopkeeper = ₹ 800
Rate of VAT = 7.5%
Total VAT = ₹ \(\frac { 800 x 7.5 }{ 100 }\) = ₹ 60
VAT paid by the shopkeeper = ₹ 6
VAT paid by the wholeseller = ₹ 60 – ₹ 6 = ₹ 54
Rate of VAT = 7.5%
S.P. of wholeseller
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q8.1

Question 9.
A manufacturing company ‘P’ sells a Desert cooler to a dealer A for ₹ 8100 including sales tax (under VAT). The dealer A sells it to a dealer B for ₹ 8500 plus sales tax and the dealer B sells it to a consumer at a profit of ₹ 600. If the rate of sales tax (under VAT) is 8%, find
(i) the cost price of the cooler for the dealer A.
(ii) the amount of tax received by the Government.
(iii) the amount which the consumer pays for the cooler.
Solution:
Manufactures ‘P’ selling price for Desert cooler including sales tax (VAT) = ₹ 8100
Rate of sales tax (VAT) = 8%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q9.1
Cost price of dealer A = ₹ 7500
and sale price of dealer A = ₹ 8500
Gain = ₹ 8500 – ₹ 7500 = ₹ 1000
or cost price of dealer B = ₹ 8500
Gain = ₹ 600
S.P. of dealer B = ₹ 8500 + ₹ 600 = ₹ 9100
Consumers cost price = ₹ 8500 + ₹ 600 = ₹ 9100
(ii) Tax paid to the Govt.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q9.2
= ₹ 600 + ₹ 80 + ₹ 48 = ₹ 728
The amount which the consumer pays = ₹ 7500 + ₹ 1000 + ₹ 600 + ₹ 728 = ₹ 9828

Question 10.
A manufacturer marks an article for ₹ 5000. He sells it to a wholesaler at a discount of 25% on the marked price and the wholseller sells it to a retailer at a discount of 15% on the marked price. The retailer sells it to a consumer at the marked price and at each stage the VAT is 8%.
Calculate the amount of VAT received by the Government from :
(i) the wholesaler.
(ii) the retailer.
Solution:
Marked price (M.P.) of an article = ₹ 5000
Discount given to the wholesaler = 25%
Cost price of wholesaler or S.P. of the manufacturer
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q10.2
(i) VAT received from the wholesaler = ₹ 340 – ₹ 300 = ₹ 40
(ii) and VAT received by the retailer = ₹ 400 – ₹ 340 = ₹ 60

Question 11.
A manufacturer listed the price of his goods at ₹ 160 per article. He allowed a discount of 25% to a wholesaler who in his turn allowed a discount of 20% on the listed price to a retailer. The rate of sales tax on the goods is 10%. If the retailer sells one article to a consumer at a discount of 5% on the listed price, then find
(i) the VAT paid by the wholesaler.
(ii) the VAT paid by the retailer.
(iii) the VAT received by the Government.
Solution:
List price (MP) of the goods = ₹ 160 per article
Rate of discount = 25%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q11.2
(i) VAT paid by the wholesaler = ₹ 12.80 – ₹ 12 = ₹ 0.80
(ii) VAT paid by the retailer = 15.20 – 12.80 = ₹ 2.40
(iii) Total VAT paid to the Govt. = ₹ 15.20

Question 12.
Kiran purchases an article for ₹ 5, 400 which includes 10% rebate on the marked price and 20% sales tax (under VAT) on the remaining price. Find the marked price of the article.
Solution:
Let market price of the article be ₹ x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q12.1
Hence, market price of the article is ₹ 5000

Question 13.
A shopkeeper buys an article for ₹ 12000 and marks up its price by 25%. The shopkeeper gives a discount of 10% on the marked up price. He gives a further off-season discount of 5% on the balance. But the sales tax (under VAT) is charged at 8% on the remaining price. Find :
(i) the amount of VAT which a customer has to pay.
(ii) the final price he has to pay for the article.
Solution:
Cost price of an article = ₹ 12000
Rate of mark up in price = 25%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q13.1
(i) Amount of sales tax = ₹ 1026
(ii) Price to be paid = ₹ 12825 + ₹ 1026 = ₹ 13851

Question 14.
In a particular tax period, Mr. Sunder Dass, a shopkeeper pruchased goods worth ₹ 960000 and paid a total tax of ₹ 62750 (under VAT). During this period, his sales consisted of taxable turnover of ₹ 400000 of goods taxable at 6% and ₹ 480000 for goods taxable at 12.5%. He also sold tax exempted goods worth ₹ 95640 in the same period. Calculate his tax liability (under VAT) for this period.
Solution:
Cost price of good purchased by Sunder Dass = ₹ 960000
Tax paid (VAT) = ₹ 62750
Sale of goods worth = ₹ 400000
Rate of VAT = 6%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q14.1
Tax paid to Govt. (VAT) = ₹ 62750
Tax liability = ₹ 84000 – ₹ 62750 = ₹ 21250

Question 15.
In the tax period ended March 2015, M/S Hari Singh & Sons purchased floor tiles worth ₹ 800000 taxable at 7.5% and sanitary fittings worth ₹ 750000 taxable at 10%. During this period, the sales turnover for floor tiles and sanitary fittings are worth ₹ 840000 and ₹ 920000 respectively. However, the floor tiles worth ₹ 60000 were returned by the firm during the same period. Calculate the tax liability (under VAT) of the firm for this tax period.
Solution:
Cost of floor tiles = ₹ 800000
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q15.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Ex 1 Q15.2
Liability of tax of the firm = 155000 – (135000 + 4500) = ₹ 155000 – ₹ 139500 = ₹ 15500

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax EX 1 are helpful to complete your math homework.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test

More Exercises

Question 1.
A game consists of spinning an arrow which comes to rest at one of the regions 1, 2 or 3 (shown in the given figure). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test Q1.1
Solution:
In a game,
No, the outcomes are not equally likely.
Outcome 3 is more likely to occur than the outcomes of 1 and 2.

Question 2.
In a single throw of a die, find the probability of getting
(i) a number greater than 5
(ii) an odd prime number
(iii) a number which is multiple of 3 or 4.
Solution:
In a single throw of a die
Number of total outcomes = 6 (1, 2, 3, 4, 5, 6)
(i) Numbers greater than 5 = 6 i.e., one number
Probability = \(\\ \frac { 1 }{ 6 } \)
(ii) An odd prime number 2 i.e., one number
Probability = \(\\ \frac { 1 }{ 6 } \)
(iii) A number which is a multiple of 3 or 4 which are 3, 6, 4 = 3 numbers
Probability = \(\\ \frac { 3 }{ 6 } \) = \(\\ \frac { 1 }{ 2 } \)

Question 3.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Rohana will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that :
(i) She will buy it?
(ii) She will not buy it?
Solution:
In a lot, there are 144 ball pens in which defective ball pens are = 20
and good ball pens are = 144 – 20 = 124
Rohana buys a pen which is good only.
(i) Now the number of possible outcomes = 144
and the number of favourable outcomes = 124
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test Q3.1

Question 4.
A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
(ii) acceptable to the trader?
Solution:
Number of total mobiles = 48
Number of good mobiles = 42
Number having minor defect = 3
Number having major defect = 3
(i) Acceptable to Varnika = 42
Probability = \(\\ \frac { 42 }{ 48 } \) = \(\\ \frac { 7 }{ 8 } \)
(ii) Acceptable to trader = 42 + 3 = 45
Probability = \(\\ \frac { 45 }{ 48 } \) = \(\\ \frac { 15 }{ 16 } \)

Question 5.
A bag contains 6 red, 5 black and 4 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is
(i) white
(ii) red
(iii) not black
(iv) red or white.
Solution:
Total number of balls = 6 + 5 + 4 = 15
Number of red balls = 6
Number of black balls = 5
Number of white balls = 4
(i) Probability of a white ball will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 4 }{ 15 } \)
(ii) Probability of red ball will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 6 }{ 15 } \) = \(\\ \frac { 2 }{ 5 } \)
(iii) Probability of not black ball will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 15-5 }{ 15 } \)
= \(\\ \frac { 10 }{ 15 } \)
= \(\\ \frac { 2 }{ 3 } \)
(iv) Probability of red or white ball will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6+4 }{ 15 } \)
= \(\\ \frac { 10 }{ 15 } \)
= \(\\ \frac { 2 }{ 3 } \)

Question 6.
A bag contains 5 red, 8 white and 7 black balls. A ball is drawn from the bag at random. Find the probability that the drawn ball is:
(i) red or white
(ii) not black
(iii) neither white nor black
Solution:
Total number of balls in a bag = 5 + 8 + 7 = 20
(i) Number of red or white balls = 5 + 8 = 13
Probability of red or white ball will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 13 }{ 20 } \)
(ii) Number of ball which are not black = 20 – 7 = 13
Probability of not black ball will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 13 }{ 20 } \)
(iii) Number of ball which are neither white nor black
= Number of ball which are only red = 5
Probability of neither white nor black ball will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 20 } \)
= \(\\ \frac { 1 }{ 4 } \)

Question 7.
A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is :
(i) white or blue
(ii) red or black
(iii) not white
(iv) neither white nor black ?
Solution:
Number of total balls = 5 + 7 + 4 + 2 = 18
Number of white balls = 5
number of red balls = 7
number of black balls = 4
and number of blue balls = 2.
(i) Number of white and blue balls = 5 + 2 = 7
Probability of white or blue balls will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 7 }{ 18 } \)
(ii) Number of red and black balls = 7 + 4 = 11
Probability of red or black balls will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 11 }{ 18 } \)
(iii) Number of ball which are not white = 7 + 4 + 2 = 13
Probability of not white balls will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 13 }{ 18 } \)
(iv) Number of balls which are neither white nor black = 18 – (5 + 4) = 18 – 9 = 9
Probability of ball which is neither white nor black will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 9 }{ 18 } \) = \(\\ \frac { 1 }{ 2 } \)

Question 8.
A box contains 20 balls bearing numbers 1, 2, 3, 4,……, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) prime number
(iv) not divisible by 10?
Solution:
In a box, there are 20 balls containing 1 to 20 number
Number of possible outcomes = 20
(i) Numbers which are odd will be,
1, 3, 5, 7, 9, 11, 13, 15, 17, 19 = 10 balls.
Probability of odd ball will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 10 }{ 20 } \) = \(\\ \frac { 1 }{ 2 } \)
(ii) Numbers which are divisible by 2 or 3 will be
2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 = 13 balls
Probability of ball which is divisible by 2 or 3 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 13 }{ 20 } \)
(iii) Prime numbers will be 2, 3, 5, 7, 11, 13, 17, 19 = 8
Probability of prime number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 8 }{ 20 } \) = \(\\ \frac { 2 }{ 5 } \)
(iv) Numbers not divisible by 10 will be
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19 = 18
Probability of prime number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 18 }{ 20 } \) = \(\\ \frac { 9 }{ 10 } \)

Question 9.
Find the probability that a number selected at random from the numbers 1, 2, 3,……35 is a
(i) prime number
(ii) multiple of 7
(iii) multiple of 3 or 5.
Solution:
Numbers are 1, 2, 3, 4, 5,…..30, 31, 32, 33, 34, 35
Total = 35
(i) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
which are 11
Probability of prime number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 11 }{ 35 } \)
(ii) Multiple of 7 are 7, 14, 21, 28, 35 which are 5
Probability of multiple of 7 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 5 }{ 35 } \) = \(\\ \frac { 1 }{ 7 } \)
(iii) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12 ,15, 18, 20, 21, 24, 25, 27, 30, 33, 35.
Which are 16 in numbers
Probability of multiple of 3 or 5 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 16 }{ 35 } \)

Question 10.
Cards marked with numbers 13, 14, 15,…..60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is
(i) divisible by 5
(ii) a number which is a perfect square.
Solution:
Number of cards which are marked with numbers
13, 14, 15, 16, 17,….to 59, 60 are = 48
(i) Numbers which are divisible by 5 will be
15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10
Probability of number divisible by 5 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 10 }{ 48 } \) = \(\\ \frac { 5 }{ 24 } \)
(ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers.
Probability of number which is a perfect square will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 4 }{ 48 } \) = \(\\ \frac { 1 }{ 12 } \)

Question 11.
The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has
(i) an odd number
(ii) a perfect square number.
Solution:
Cards in a box are from 14 to 99 = 86
No. of total cards = 86
One card is drawn at random
Cards bearing odd numbers are 15, 17, 19, 21, …, 97, 99
Which are 43
(i) P(E) = \(\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events } \)
= \(\\ \frac { 43 }{ 86 } \)
= \(\\ \frac { 1 }{ 2 } \)
(ii) Cards bearing number which are a perfect square
= 16, 25, 36, 49, 64, 81
Which are 6
P(E) = \(\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events } \)
= \(\\ \frac { 6 }{ 86 } \)
= \(\\ \frac { 3 }{ 43 } \)

Question 12.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, find the number of balls in the bags.
Solution:
Number of red balls = 5
and let number of blue balls = x
Total balls in the bag = 5 + x
and that of red balls = \(\\ \frac { 5 }{ 5+x } \)
According to the condition,
\(\frac { x }{ 5+x } =4\times \frac { 5 }{ 5+x } =>\frac { x }{ 5+x } =\frac { 20 }{ 5+x } \)
x ≠ – 5
x = 20
Hence, number of blue balls = 20
and number of balls in the bag = 20 + 5 = 25

Question 13.
A bag contains 18 balls out of which x balls are white.
(i) If one ball is drawn at random from the bag, what is the probability that it is white ball?
(ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be \(\\ \frac { 9 }{ 8 } \) times that of probability of white ball coming in part (i). Find the value of x.
Solution:
Total numbers of balls in a bag = 18
No. of white balls = x
(i) One ball is drawn a random
Probability of being a white ball = \(\\ \frac { x }{ 18 } \)
(ii) If 2 more white balls an put, then number of white balls = x + 2
and probability is \(\\ \frac { 9 }{ 8 } \) times
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test Q13.1

Question 14.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :
(i) a red face card
(ii) neither a club nor a spade
(iii) neither an ace nor a king of red colour
(iv) neither a red card nor a queen
(v) neither a red card nor a black king.
Solution:
Number of cards in a pack of well-shuffled cards = 52
(i) Number of a red face card = 3 + 3 = 6
Probability of red face card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 6 }{ 52 } \) = \(\\ \frac { 3 }{ 26 } \)
(ii) Number of cards which is neither a club nor a spade = 52 – 26 = 26
Probability of card which’ is neither a club nor a spade will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 26 }{ 52 } \) = \(\\ \frac { 1 }{ 2 } \)
(iii) Number of cards which is neither an ace nor a king of red colour
= 52 – (4 + 2) = 52 – 6 = 46
Probability of card which is neither ace nor a king of red colour will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 46 }{ 52 } \) = \(\\ \frac { 23 }{ 26 } \)
(iv) Number of cards which are neither a red card nor a queen are
= 52 – (26 + 2) = 52 – 28 = 24
Probability of card which is neither red nor a queen will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 24 }{ 52 } \) = \(\\ \frac { 6 }{ 13 } \)
(v) Number of cards which are neither red card nor a black king
= 52 – (26 + 2) = 52 – 28 = 24
Probability of cards which is neither red nor a black king will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 24 }{ 52 } \) = \(\\ \frac { 6 }{ 13 } \)

Question 15.
From pack of 52 playing cards, blackjacks, black kings and black aces are removed and then the remaining pack is well-shuffled. A card is drawn at random from the remaining pack. Find the probability of getting
(i) a red card
(ii) a face card
(iii) a diamond or a club
(iv) a queen or a spade.
Solution:
Total number of cards = 52
Black jacks, black kings and black aces are removed
Now number of cards = 52 – (2 + 2 + 2) = 52 – 6 = 46
One card is drawn
(i) No. of red cards = 13 + 13 = 26
∴Probability = \(\\ \frac { 26 }{ 46 } \) = \(\\ \frac { 13 }{ 23 } \)
(ii) Face cards = 4 queens, 2 red jacks, 2 kings = 8
∴Probability = \(\\ \frac { 8 }{ 46 } \) = \(\\ \frac { 4 }{ 23 } \)
(iii) a diamond on a club = 13 + 10 = 23
∴Probability = \(\\ \frac { 23 }{ 46 } \) = \(\\ \frac { 1 }{ 2 } \)
(iv) A queen or a spade = 4 + 10 = 14
∴Probability = \(\\ \frac { 14 }{ 46 } \) = \(\\ \frac { 7 }{ 23 } \)

Question 16.
Two different dice are thrown simultaneously. Find the probability of getting:
(i) sum 7
(ii) sum ≤ 3
(iii) sum ≤ 10
Solution:
(i) Numbers whose sum is 7 will be (1, 6), (2, 5), (4, 3), (5, 2), (6, 1), (3, 4) = 6
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 6 }{ 36 } \) = \(\\ \frac { 1 }{ 6 } \)
(ii) Sum ≤ 3
Then numbers can be (1, 2), (2, 1), (1, 1) which are 3 in numbers
∴Probability will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 3 }{ 36 } \) = \(\\ \frac { 1 }{ 12 } \)
(iii) Sum ≤ 10
The numbers can be,
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, .6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),
(6, 1), (6, 2), (6, 3), (6, 4) = 33
Probability will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \) = \(\\ \frac { 33 }{ 36 } \) = \(\\ \frac { 11 }{ 12 } \)

Question 17.
Two dice are thrown together. Find the probability that the product of the numbers on the top of two dice is
(i) 6
(ii) 12
(iii) 7
Solution:
Two dice are thrown together
Total number of events = 6 × 6 = 36
(i) Product 6 = (1, 6), (2, 3), (3, 2). (6, 1) = 4
Probability = \(\\ \frac { 4 }{ 36 } \) = \(\\ \frac { 1 }{ 9 } \)
(ii) Product 12 = (2, 6), (3, 4), (4, 3), (6, 2) = 4
Probability = \(\\ \frac { 4 }{ 36 } \) = \(\\ \frac { 1 }{ 9 } \)
(iii) Product 7 = 0 (no outcomes)
Probability = \(\\ \frac { 0 }{ 36 } \) = 0

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability MCQS

More Exercises

Choose the correct answer from the given four options (1 to 28):

Question 1.
Which of the following cannot be the probability of an event?
(a) 0.7
(b) \(\\ \frac { 2 }{ 3 } \)
(c) – 1.5
(d) 15%
Solution:
– 1.5 (negative) can not be a probability as a probability is possible 0 to 1. (c)

Question 2.
If the probability of an event is p, then the probability of its complementary event will be
(a) p – 1
(b) p
(c) 1 – p
(d) \(1- \frac { 1 }{ p } \)
Solution:
Complementary of p is 1 – p
Probability of complementary even of p is 1 – p. (c)

Question 3.
Out of one digit prime numbers, one selecting an even number is
(a) \(\\ \frac { 1 }{ 2 } \)
(b) \(\\ \frac { 1 }{ 4 } \)
(c) \(\\ \frac { 4 }{ 9 } \)
(d) \(\\ \frac { 2 }{ 5 } \)
Solution:
One digit prime numbers are 2, 3, 5, 7 = 4
Probability of an even prime number (i.e , 2) = \(\\ \frac { 1 }{ 4 } \) (b)

Question 4.
Out of vowels, of the English alphabet, one letter is selected at random. The probability of selecting ‘e’ is
(a) \(\\ \frac { 1 }{ 26 } \)
(b) \(\\ \frac { 5 }{ 26 } \)
(c) \(\\ \frac { 1 }{ 4 } \)
(d) \(\\ \frac { 1 }{ 5 } \)
Solution:
Vowels of English alphabet are a, e, i, o, u = 4
One letter is selected at random.
The probability of selecting ’e’ = \(\\ \frac { 1 }{ 5 } \) (d)

Question 5.
When a die is thrown, the probability of getting an odd number less than 3 is
(a) \(\\ \frac { 1 }{ 6 } \)
(b) \(\\ \frac { 1 }{ 3 } \)
(c) \(\\ \frac { 1 }{ 2 } \)
(d) 0
Solution:
A die is thrown
Total number of events = 6
Odd number less than 3 is 1 = 1
Probability = \(\\ \frac { 1 }{ 6 } \) (a)

Question 6.
A fair die is thrown once. The probability of getting an even prime number is
(a) \(\\ \frac { 1 }{ 6 } \)
(b) \(\\ \frac { 2 }{ 3 } \)
(c) \(\\ \frac { 1 }{ 3 } \)
(d) \(\\ \frac { 1 }{ 2 } \)
Solution:
A fair die is thrown once
Total number of outcomes = 6
Prime numbers = 2, 3, 5 and even prime is 2
Probability of getting an even prime number = \(\\ \frac { 1 }{ 6 } \) (a)

Question 7.
A fair die is thrown once. The probability of getting a composite number is
(a) \(\\ \frac { 1 }{ 3 } \)
(b) \(\\ \frac { 1 }{ 6 } \)
(c) \(\\ \frac { 2 }{ 3 } \)
(d) 0
Solution:
A fair die is thrown once
Total number of outcomes = 6
Composite numbers are 4, 6 = 2
Probability = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \) (a)

Question 8.
If a fair dice is rolled once, then the probability of getting an even number or a number greater than 4 is
(a) \(\\ \frac { 1 }{ 2 } \)
(b) \(\\ \frac { 1 }{ 3 } \)
(c) \(\\ \frac { 5 }{ 6 } \)
(d) \(\\ \frac { 2 }{ 3 } \)
Solution:
A fair dice is thrown once.
Total number of outcomes = 6
Even numbers or a number greater than 4 = 2, 4, 5, 6 = 4
Probability = \(\\ \frac { 4 }{ 6 } \) = \(\\ \frac { 2 }{ 3 } \) (d)

Question 9.
Rashmi has a die whose six faces show the letters as given below :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability MCQS Q9.1
If she throws the die once, then the probability of getting C is
(a) \(\\ \frac { 1 }{ 3 } \)
(b) \(\\ \frac { 1 }{ 4 } \)
(c) \(\\ \frac { 1 }{ 5 } \)
(d) \(\\ \frac { 1 }{ 6 } \)
Solution:
A die having 6 faces bearing letters A, B, C, D, A, C
Total number of outcomes = 4
Probability of getting C = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \) (a)

Question 10.
If a letter is chosen at random from the letters of English alphabet, then the probability that it is a letter of the word ‘DELHI’ is
(a) \(\\ \frac { 1 }{ 5 } \)
(b) \(\\ \frac { 1 }{ 26 } \)
(c) \(\\ \frac { 5 }{ 26 } \)
(d) \(\\ \frac { 21 }{ 26 } \)
Solution:
Total number of English alphabets = 26
Letter of Delhi = D, E, L, H, I. = 5
Probability = \(\\ \frac { 5 }{ 26 } \) (c)

Question 11.
A card is drawn from a well-shuffled pack of 52 playing cards. The event E is that the card drawn is not a face card. The number of outcomes favourable to the event E is
(a) 51
(b) 40
(c) 36
(d) 12
Solution:
Number of playing cards = 52
Probability of a card which is not a face card = (52 – 12) = 40
Number of possible events = 40 (b)

Question 12.
A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is
(a) 4
(b) 13
(c) 48
(d) 51
Solution:
Total number of cards = 52
Balance 52 – 1 = 51
Number of possible events = 51 (d)

Question 13.
If one card is drawn from a well-shuffled pack of 52 cards, the probability of getting an ace is
(a) \(\\ \frac { 1 }{ 52 } \)
(b) \(\\ \frac { 4 }{ 13 } \)
(c) \(\\ \frac { 2 }{ 13 } \)
(d) \(\\ \frac { 1 }{ 13 } \)
Solution:
Total number of cards = 52
Number of aces = 4
Probability of card being an ace = \(\\ \frac { 4 }{ 52 } \) = \(\\ \frac { 1 }{ 13 } \) (d)

Question 14.
A card is selected at random from a well- shuffled deck of 52 cards. The probability of its being a face card is
(a) \(\\ \frac { 3 }{ 13 } \)
(b) \(\\ \frac { 4 }{ 13 } \)
(c) \(\\ \frac { 6 }{ 13 } \)
(d) \(\\ \frac { 9 }{ 13 } \)
Solution:
Total number of cards = 52
No. of face cards = 3 × 4 = 12
.’. Probability of face card = \(\\ \frac { 12 }{ 52 } \) = \(\\ \frac { 3 }{ 13 } \) (a)

Question 15.
A card is selected at random from a pack of 52 cards. The probability of its being a red face card is
(a) \(\\ \frac { 3 }{ 26 } \)
(b) \(\\ \frac { 3 }{ 13 } \)
(c) \(\\ \frac { 2 }{ 13 } \)
(d) \(\\ \frac { 1 }{ 2 } \)
Solution:
Total number of card = 52
No. of red face card = 3 × 2 = 6
.’. Probability = \(\\ \frac { 6 }{ 52 } \) = \(\\ \frac { 3 }{ 26 } \) (a)

Question 16.
If a card is drawn from a well-shuffled pack of 52 playing cards, then the probability of this card being a king or a jack is
(a) \(\\ \frac { 1 }{ 26 } \)
(b) \(\\ \frac { 1 }{ 13 } \)
(c) \(\\ \frac { 2 }{ 13 } \)
(d) \(\\ \frac { 4 }{ 13 } \)
Solution:
Total number of cards 52
Number of a king or a jack = 4 + 4 = 8
.’. Probability = \(\\ \frac { 8 }{ 52 } \) = \(\\ \frac { 2 }{ 13 } \) (c)

Question 17.
The probability that a non-leap year selected at random has 53 Sundays is.
(a) \(\\ \frac { 1 }{ 365 } \)
(b) \(\\ \frac { 2 }{ 365 } \)
(c) \(\\ \frac { 2 }{ 7 } \)
(d) \(\\ \frac { 1 }{ 7 } \)
Solution:
Number of a non-leap year 365
Number of Sundays = 53
In a leap year, there are 52 weeks or 364 days
One days is left
Now we have to find the probability of a Sunday out of remaining 1 day
∴ Probability = \(\\ \frac { 1 }{ 7 } \) (d)

Question 18.
A bag contains 3 red balk, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is
(a) \(\\ \frac { 1 }{ 5 } \)
(b) \(\\ \frac { 1 }{ 3 } \)
(c) \(\\ \frac { 7 }{ 15 } \)
(d) \(\\ \frac { 8 }{ 1 } \)
Solution:
In a bag, there are
3 red balls + 5 white balls + 7 black balls
Total number of balls = 15
One ball is drawn at random which is neither
red not black
Number of outcomes = 5
Probability = \(\\ \frac { 5 }{ 15 } \) = \(\\ \frac { 1 }{ 3 } \) (b)

Question 19.
A bag contains 4 red balls and 5 green balls. One ball is drawn at random from the bag. The probability of getting either a red ball or a green ball is
(a) \(\\ \frac { 4 }{ 9 } \)
(b) \(\\ \frac { 5 }{ 9 } \)
(c) 0
(d) 1
Solution:
In a bag, there are
4 red balls + 5 green balls
Total 4 + 5 = 9
One ball is drawn at random
Probability of either a red or a green ball = \(\\ \frac { 9 }{ 9 } \) = 1 (d)

Question 20.
A bag contains 5 red, 4 white and 3 black balls. If a. ball is drawn from the bag at random, then the probability of the ball being not black is
(a) \(\\ \frac { 5 }{ 12 } \)
(b) \(\\ \frac { 1 }{ 3 } \)
(c) \(\\ \frac { 3 }{ 4 } \)
(d) \(\\ \frac { 1 }{ 4 } \)
Solution:
In a bag, there are
5 red + 4 white + 3 black balls = 12
One ball is drawn at random
Probability of a ball not black = \(\\ \frac { 5+4 }{ 12 } \) = \(\\ \frac { 9 }{ 12 } \) = \(\\ \frac { 3 }{ 4 } \) (c)

Question 21.
One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is
(a) \(\\ \frac { 1 }{ 5 } \)
(b) \(\\ \frac { 3 }{ 5 } \)
(c) \(\\ \frac { 4 }{ 5 } \)
(d) \(\\ \frac { 1 }{ 3 } \)
Solution:
There are t to 40 = 40 tickets in a bag
No. of tickets which is multiple of 5 = 8
(5, 10, 15, 20, 25, 30, 35, 40)
Probability = \(\\ \frac { 8 }{ 40 } \) = \(\\ \frac { 1 }{ 5 } \) (a)

Question 22.
If a number is randomly chosen from the numbers 1,2,3,4, …, 25, then the probability of the number to be prime is
(a) \(\\ \frac { 7 }{ 25 } \)
(b) \(\\ \frac { 9 }{ 25 } \)
(c) \(\\ \frac { 11 }{ 25 } \)
(d) \(\\ \frac { 13 }{ 25 } \)
Solution:
There are 25 number bearing numbers 1, 2, 3,…,25
Prime numbers are 2, 3, 5, 7, 11, 13, 17 19, 23 = 9
Probability being a prime number = \(\\ \frac { 9 }{ 25 } \) (b)

Question 23.
A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is
(a) \(\\ \frac { 1 }{ 10 } \)
(b) \(\\ \frac { 9 }{ 100 } \)
(c) \(\\ \frac { 1 }{ 9 } \)
(d) \(\\ \frac { 1 }{ 100 } \)
Solution:
In a box, there are
90 cards bearing numbers 1 to 90
Perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9
Probability of being a perfect square = \(\\ \frac { 9 }{ 90 } \) = \(\\ \frac { 1 }{ 10 } \) (a)

Question 24.
If a (fair) coin is tossed twice, then the probability of getting two heads is
(a) \(\\ \frac { 1 }{ 4 } \)
(b) \(\\ \frac { 1 }{ 2 } \)
(c) \(\\ \frac { 3 }{ 4 } \)
(d) 0
Solution:
A coin is tossed twice
Number of outcomes = 2 x 2 = 4
Probability of getting two heads (HH = 1) = \(\\ \frac { 1 }{ 4 } \) (a)

Question 25.
If two coins are tossed simultaneously, then the probability of getting atleast one head is
(a) \(\\ \frac { 1 }{ 4 } \)
(b) \(\\ \frac { 1 }{ 2 } \)
(c) \(\\ \frac { 3 }{ 4 } \)
(d) 1
Solution:
Two coins are tossed
Total outcomes = 2 × 2 = 4
Probability of getting atleast one head (HT,TH,H,H) = \(\\ \frac { 3 }{ 4 } \) (c)

Question 26.
Lakshmi tosses two coins simultaneously. The probability that she gets almost one head
(a) 1
(b) \(\\ \frac { 3 }{ 4 } \)
(c) \(\\ \frac { 1 }{ 2 } \)
(d) \(\\ \frac { 1 }{ 7 } \)
Solution:
Two coins are tossed
Total number of outcomes = 2 × 2 = 4
Probability of getting atleast one head = (HT, TH, RH = 3) = \(\\ \frac { 3 }{ 4 } \) (b)

Question 27.
The probability of getting a bad egg in a lot of 400 eggs is 0.035. The number of bad eggs in the lot is
(a) 7
(b) 14
(c) 21
(d) 28
Solution:
Total number of eggs 400
Probability of getting a bad egg = 0.035
Number of bad eggs = 0.035 of 400 = \(400 \times \frac { 35 }{ 1000 } \) = 14 (b)

Question 28.
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets she has bought?
(a) 40
(b) 240
(c) 480
(d) 750
Solution:
For a girl,
Winning a first prize = 0.08
Number of total tickets = 6000
Number of tickets she bought = 0.08 of 6000 = \(6000 \times \frac { 8 }{ 100 } \) = 480 (c)

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability MCQS help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability MCQS, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22

More Exercises

Question 1.
A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Anjali takes out a ball from the bag without looking into it. What is the probability that she takes out
(i) yellow ball ?
(ii) red ball ?
(iii) blue ball ?
Solution:
Number of balls in the bag = 3.
(i) Probability of yellow ball = \(\\ \frac { 1 }{ 3 } \)
(ii) Probability of red ball = \(\\ \frac { 1 }{ 3 } \)
(iii) Probability of blue ball = \(\\ \frac { 1 }{ 3 } \)

Question 2.
A box contains 600 screws, one-tenth are rusted. One screw is taken out at random from this box. Find the probability that it is a good screw.
Solution:
Number of total screws = 600
Rusted screws = \(\\ \frac { 1 }{ 10 } \) of 600 = 60
∴ Good screws = 600 – 60 = 540
Probability of a good screw
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 540 }{ 600 } \)
= \(\\ \frac { 9 }{ 10 } \)

Question 3.
In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.
Solution:
Number of prized tickets = 5
Number of blank tickets = 995
Total number of tickets = 5 + 995 = 1000
Probability of prized ticket
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 1000 } \)
= \(\\ \frac { 1 }{ 200 } \)

Question 4.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens = 12
Number of good pens = 132
Total number of pens =12 + 132 = 144
Probability of good pen
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 132 }{ 144 } \)
= \(\\ \frac { 11 }{ 12 } \)

Question 5.
If the probability of winning a game is \(\\ \frac { 5 }{ 11 } \), what is the probability of losing ?
Solution:
Probability of winning game = \(\\ \frac { 5 }{ 11 } \)
⇒ P(E) = \(\\ \frac { 5 }{ 11 } \)
We know that P (E) + P (\(\overline { E } \)) = 1
where P (E) is the probability of losing the game.
\(\\ \frac { 5 }{ 11 } \) + P (\(\overline { E } \)) = 1
⇒ P (\(\overline { E } \)) = \(1- \frac { 5 }{ 11 } \)
= \(\\ \frac { 11-5 }{ 11 } \)
= \(\\ \frac { 6 }{ 11 } \)

Question 6.
Two players, Sania and Sonali play a tennis match. It is known that the probability of Sania winning the match is 0.69. What is the probability of Sonali winning ?
Solution:
Probability of Sania’s winning the game = 0.69
Let P (E) be the probability of Sania’s winning the game
and P (\(\overline { E } \)) be the probability of Sania’s losing
the game or probability of Sonali, winning the game
P (E) + P (\(\overline { E } \)) = 1
⇒ 0.69 + P (\(\overline { E } \)) = 1
⇒ P(\(\overline { E } \)) = 1 – 0.69 = 0.31
Hence probability of Sonali’s winning the game = 0.31

Question 7.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random’ from in bag. What is the probability that the ball drawn is .
(i) red ?
(ii) not red ?
Solution:
Number of red balls = 3
Number of black balls = 5
Total balls = 3 + 5 = 8
Let P (E) be the probability of red balls,
then P (\(\overline { E } \)) will be the probability of not red balls.
P (E) + P (\(\overline { E } \)) = 1
(i) But P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 8 } \)
(ii) P (\(\overline { E } \)) = 1 – P(E)
= \(1- \frac { 6 }{ 11 } \)
= \(\\ \frac { 8-3 }{ 8 } \)
= \(\\ \frac { 5 }{ 8 } \)

Question 8.
There are 40 students in Class X of a school of which 25 are girls and the.others are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of
(i) a girl ?
(ii) a boy ?
Solution:
Number of total students = 40
Number of girls = 25
Number of boys = 40 – 25 = 15
(i) Probability of a girl
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 25 }{ 40 } \)
= \(\\ \frac { 5 }{ 8 } \)
(ii) Probability of a boy
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 15 }{ 40 } \)
= \(\\ \frac { 3 }{ 8 } \)

Question 9.
A letter is chosen from the word ‘TRIANGLE’. What is the probability that it is a vowel ?
Solution:
There are three vowels: I, A, E
.’. The number of letters in the word ‘TRIANGLE’ = 8.
Probability of vowel
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 8 } \)

Question 10.
A letter of English alphabet is chosen at random. Determine the probability that the letter is a consonant.
Solution:
No. of English alphabet = 26
No. of vowel = 5
No. of constant = 25 – 5 = 21
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 21 }{ 26 } \)

Question 11.
A bag contains 5 black, 7 red and 3 white balls. A ball is drawn at random from the bag, find the probability that the ball drawn is:
(i) red
(ii) black or white
(iii) not black.
Solution:
In a bag,
Number of black balls = 5
Number of red balls = 7
and number of white balls = 3
Total number of balls in the bag
= 5 + 7 + 3 = 15
(i) Probability of red balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7 }{ 15 } \)
(ii) Probability of black or white balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5+3 }{ 15 } \)
= \(\\ \frac { 8 }{ 15 } \)
(iii) Probability of not black balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7+3 }{ 15 } \)
= \(\\ \frac { 10 }{ 15 } \)
= \(\\ \frac { 2 }{ 3 } \)

Question 12.
A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) black?
(ii) blue or black?
(iii) not black?
(iv) green?
Solution:
Total number of marbles in the box
= 7 + 8 + 5 = 20
Since, a marble is drawn at random from the box
(i) Probability (of a black Marble)
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 20 } \)
= \(\\ \frac { 1 }{ 4 } \)
(ii) Probability (of a blue or black marble)
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7+5 }{ 20 } \)
= \(\\ \frac { 12 }{ 20 } \)
= \(\\ \frac { 3 }{ 5 } \)
(iii) Probability (of not black marble)
= 1 – P (of black 1)
= \(1- \frac { 1 }{ 4 } \)
= \(\\ \frac { 4-1 }{ 4 } \)
= \(\\ \frac { 3 }{ 4 } \)
(iv) P (of a green marble) = 0
(∴ Since, a box does not contain a green marble,
so the probability of green marble will be zero)

Question 13.
A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :
(i) white
(ii) red or black
(iii) not green
(iv) neither white nor black.
Solution:
In a bag,
Number of red balls = 6
Number of white balls = 8
Number of green balls = 5
and number of black balls = 3
Total number of balls in the bag
= 6 + 8 + 5 + 3 = 22
(i) Probability of white balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 22 } \)
= \(\\ \frac { 4 }{ 11 } \)
(ii) Probability of red or black balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6+3 }{ 22 } \)
= \(\\ \frac { 9 }{ 22 } \)
(iii) Probability of not green balls i.e. having red, white and black balls.
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6+8+3 }{ 22 } \)
= \(\\ \frac { 17 }{ 22 } \)
(iv) Probability of neither white nor black balls i.e. red and green balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6+5 }{ 22 } \)
= \(\\ \frac { 11 }{ 22 } \)
= \(\\ \frac { 1 }{ 2 } \)

Question 14.
A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. It is equally likely that one of the coins will fall down when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be Rs 5 coin?
Solution:
In a piggy bank, there are
100, 50 p coin
50, Rs 1 coin
20, Rs 2 coin
10, Rs 5 coin
Total coins = 100 + 50 + 20 + 10 = 180
One coin is drawn at random Probability of
(i) 50 p coins = \(\\ \frac { 100 }{ 180 } \)
= \(\\ \frac { 5 }{ 9 } \)
(ii) Will not be Rs 5 coins
= 100 + 50 + 20 = 170
Probability = \(\\ \frac { 170 }{ 180 } \) = \(\\ \frac { 17 }{ 18 } \)

Question 15.
A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Peter, a trader, will only accept the shirts which are good, but Salim, another trader, will only reject the shirts which have major defects. One shirts is drawn at random from the carton. What is the probability that
(i) it is acceptable to Peter ?
(ii) it is acceptable to Salim ?
Solution:
In a carton, there the 100 shirts.
Among these number of shirts which are good = 88
number of shirts which have minor defect = 8
number of shirt which have major defect = 4
Total number of shirts = 88 + 8 + 4 = 100
Peter accepts only good shirts i.e. 88
Salim rejects only shirts which have major defect i.e. 4
(i) Probability of good shirts which are acceptable to Peter
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 88 }{ 100 } \)
= \(\\ \frac { 22 }{ 25 } \)
(ii) Probability of shirts acceptable to Salim
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 88+8 }{ 100 } \)
= \(\\ \frac { 96 }{ 100 } \)
= \(\\ \frac { 24 }{ 25 } \)

Question 16.
A die is thrown once. What is the probability that the
(i) number is even
(ii) number is greater than 2 ?
Solution:
Dice is thrown once
Sample space = {1, 2, 3, 4, 5, 6}
(i) No. of ways in favour = 3
(∵ Even numbers are 2, 4, 6)
Total ways = 6
Probability = \(\\ \frac { 3 }{ 6 } \) = \(\\ \frac { 1 }{ 2 } \)
(ii) No. of ways in favour = 4
(Numbers greater than 2 are 3, 4, 5, 6)
Total ways = 6
Probability = \(\\ \frac { 4 }{ 6 } \) = \(\\ \frac { 2 }{ 2 } \)

Question 17.
In a single throw of a die, find the probability of getting:
(i) an odd number
(ii) a number less than 5
(iii) a number greater than 5
(iv) a prime number
(v) a number less than 8
(vi) a number divisible by 3
(vii) a number between 3 and 6
(viii) a number divisible by 2 or 3.
Solution:
A die is thrown and on its faces, numbers 1 to 6 are written.
Total numbers of possible outcomes = 6
(i) Probability of an odd number,
odd number are 1, 3 and 5
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 6 } \)
= \(\\ \frac { 1 }{ 2 } \)
(ii) A number less them 5 are 1, 2, 3, 4
Probability of a number less than 5 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 4 }{ 6 } \)
= \(\\ \frac { 2 }{ 3 } \)
(iii) A number greater than 5 is 6
Probability of a number greater than 5 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 6 } \)
(iv) Prime number is 2, 3, 5
Probability of a prime number is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 6 } \)
= \(\\ \frac { 1 }{ 2 } \)
(v) Number less than 8 is nil
P (E) = 0
(vi) A number divisible by 3 is 3, 6
Probability of a number divisible by 3 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 2 }{ 6 } \)
= \(\\ \frac { 1 }{ 3 } \)
(vii) Numbers between 3 and 6 is 4, 5
Probability of a number between 3 and 6 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 2 }{ 6 } \)
= \(\\ \frac { 1 }{ 3 } \)
(viii) Numbers divisible by 2 or 3 are 2, 4 or 3,
Probability of a number between 2 or 3 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 2 }{ 6 } \)
= \(\\ \frac { 1 }{ 3 } \)

Question 18.
A die has 6 faces marked by the given numbers as shown below:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22 Q18.1
The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than – 3.
(iii) the smallest integer ?
Solution:
Total outcomes n(S)= 6
(i) a positive integer = (1, 2, 3)
No. of favourables n(E) = 3
Probability = \(\\ \frac { n(E) }{ n(S) } \)
= \(\\ \frac { 3 }{ 6 } \)
= \(\\ \frac { 1 }{ 2 } \)
(ii) Integer greater than -3
= (1, 2, 3, -1, -2)
No. of favourables n(E) = 5
Probability = \(\\ \frac { n(E) }{ n(S) } \)
= \(\\ \frac { 5 }{ 6 } \)
(iii) Smallest integer = -3
No. of favourables n(E) = 1
Probability = \(\\ \frac { n(E) }{ n(S) } \)
= \(\\ \frac { 1 }{ 6 } \)

Question 19.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number ?
(iii) a number greater than 2?
(iv) a number less than 9?
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22 Q19.1
Solution:
On the face of a game, numbers 1 to 8 is shown.
Possible outcomes = 8
(i) Probability of number 8 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 8 } \)
(ii) Odd number are 1, 3, 5, 7
Probability of a number which is an odd will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 4 }{ 8 } \)
= \(\\ \frac { 1 }{ 2 } \)
(iii) A number greater than 2 are 3, 4, 5, 6, 7, 8 which are 6
Probability of number greater than 2 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6 }{ 8 } \)
= \(\\ \frac { 3 }{ 4 } \)
(iv) A number less than 9 is 8.
Probability of a number less than 9 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 8 } \)

Question 20.
Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year.
Solution:
In January, there are 31 days and in an ordinary year,
there are 365 days but in a leap year, there are 366 days.
(i) In January of an ordinary year, there are 31 days i.e. 4 weeks and 3 days.
Probability of Monday will be = \(\\ \frac { 3 }{ 7 } \)
(ii) In January of a leap year, there are 31 days i.e. 4 weeks and 3 days
Probability of Monday will be = \(\\ \frac { 3 }{ 7 } \)

Question 21.
Find the probability that the month of February may have 5 Wednesdays in
(i) a leap year
(ii) a non-leap year.
Solution:
In the month of February, there are 29 days in a leap year
while 28 days in a non-leap year,
(i) In a leap year, there are 4 complete weeks and 1 day
Probability of Wednesday = P (E) = \(\\ \frac { 1 }{ 7 } \)
(ii) and in a non leap year, there are 4 complete weeks and 0 days
Probability of Wednesday P (E) = \(\\ \frac { 0 }{ 7 } \) = 0

Question 22.
Sixteen cards are labelled as a, b, c,…, m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median.
Solution:
Here, sample space (S) = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p)
∴n(S) = 16
(i) Vowels (V) = {a, e, i, o}
∴n(V) = 4
∴P(a vowel) = \(\\ \frac { n(V) }{ n(S) } \) = \(\\ \frac { 4 }{ 16 } \) = \(\\ \frac { 1 }{ 4 } \)
(ii) Consonants (C) = {b, c, d, f, g, h, j, k, l, m, n, p}
∴n(C) = 12
∴P (a consonant) = \(\\ \frac { n(C) }{ n(S) } \) = \(\\ \frac { 12 }{ 16 } \) = \(\\ \frac { 3 }{ 4 } \)
(iii) None of the letters of the word MEDIAN (N) = {b, c, f, g, h, j, k, l, o, p)
∴n(N) = 10
∴P (N) = \(\\ \frac { n(N) }{ n(S) } \) = \(\\ \frac { 10 }{ 16 } \) = \(\\ \frac { 5 }{ 8 } \)

Question 23.
An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?
Solution:
Integers between 0 and 100 = 99
(i) Number divisible by 7 are
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98 = 14
Probability = \(\\ \frac { 14 }{ 99 } \)
(ii) Not divisible by 7 are 99 – 14 = 85
Probability = \(\\ \frac { 85 }{ 99 } \)

Question 24.
Cards marked with numbers 1, 2, 3, 4, 20 are well shuffled and a card is drawn at random.
What is the probability that the number on the card is
(i) a prime number
(ii) divisible by 3
(iii) a perfect square ? (2010)
Solution:
Number cards is drawn from 1 to 20 = 20
One card is drawn at random
No. of total (possible) events = 20
(i) The card has a prime number
The prime number from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19
Actual No. of events = 8
P(E) = \(\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events } \)
= \(\\ \frac { 8 }{ 20 } \)
= \(\\ \frac { 2 }{ 5 } \)
(ii) Numbers divisible by 3 are 3, 6, 9, 12, 15, 18
No. of actual events = 6
P(E) = \(\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events } \)
= \(\\ \frac { 6 }{ 20 } \)
= \(\\ \frac { 3 }{ 10 } \)
(iii) Numbers which are perfect squares = 1, 4, 9, 16 = 4
P(E) = \(\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events } \)
= \(\\ \frac { 4 }{ 20 } \)
= \(\\ \frac { 1 }{ 5 } \)

Question 25.
A box contains 25 cards numbered 1 to 25. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) even
(ii) prime
(iii) multiple of 6
Solution:
Number of card in a box = 25 numbered 1 to 25
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
i.e. number of favourable outcomes = 12
Probability of an even number will be
P(E) = \(\\ \frac { 12 }{ 25 } \)
(ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23
i.e. number of primes = 9
Probability of primes will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 25 } \)
(iii) Multiples of 6 are 6, 12, 18, 24
Number of multiples = 4
Probability of multiples of 6 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 4 }{ 25 } \)

Question 26.
A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) Odd
(ii) prime
(iii) divisible by 3
(iv) divisible by 3 and 2 both
(v) divisible by 3 or 2
(vi) a perfect square number.
Solution:
Number of cards in a box =15 numbered 1 to 15
(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15
Number of odd numbers = 8
Probability of odd numbers will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 15 } \)
(ii) Prime number are 2, 3, 5, 7, 11, 13
Number of primes is 6
Probability of prime number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6 }{ 15 } \)
= \(\\ \frac { 2 }{ 5 } \)
(iii) Numbers divisible by 3 are 3, 6, 9, 12, 15
which are 5 in numbers
Probability of number divisible by 3 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 15 } \)
= \(\\ \frac { 1 }{ 3 } \)
(iv) Divisible by 3 and 2 both are 6, 12
which are 2 in numbers.
Probability of number divisible by 3 and 2
Both will be = \(\\ \frac { 2 }{ 15 } \)
(v) Numbers divisible by 3 or 2 are
2, 3, 4, 6, 8, 9, 10, 12, 14, 15 which are 10 in numbers
Probability of number divisible by 3 or 2 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 10 }{ 15 } \)
= \(\\ \frac { 2 }{ 3 } \)
(v) Perfect squares number are 1, 4, 9 i.e., 3 number
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 15 } \)
= \(\\ \frac { 1 }{ 5 } \)

Question 27.
A box contains 19 balls bearing numbers 1, 2, 3,…., 19. A ball is drawn at random
from the box. Find the probability that the number on the ball is :
(i) a prime number
(ii) divisible by 3 or 5
(iii) neither divisible by 5 nor by 10
(iv) an even number.
Solution:
In a box, number of balls = 19 with number 1 to 19.
A ball is drawn
Number of possible outcomes = 19
(i) Prime number = 2, 3, 5, 7, 11, 13, 17, 19
which are 8 in number
Probability of prime number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 19 } \)
(ii) Divisible by 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18
which are 8 in number
Probability of number divisible by 3 or 5 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 19 } \)
(iii) Numbers which are neither divisible by 5 nor by 10 are
1, 2, 3, 4, 6, 7, 8, 9, 11, 12,
13, 14, 16, 17, 18, 19
which are 16 in numbers
Probability of there number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 16 }{ 19 } \)
(iv) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18
which are 9 in numbers.
Probability of there number will be
Number of favourable outcome
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 19 } \)

Question 28.
Cards marked with numbers 13, 14, 15, …, 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is
(i) divisible by 5

(ii) a perfect square number.
Solution:
Number of card bearing numbers 13,14,15, … 60 = 48
One card is drawn at random.
(i) Card divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10
Probability = \(\\ \frac { 10 }{ 48 } \)
= \(\\ \frac { 5 }{ 24 } \)
(ii) A perfect square = 16, 25, 36, 49 = 4
Probability = \(\\ \frac { 4 }{ 48 } \)
= \(\\ \frac { 1 }{ 12 } \)

Question 29.
Tickets numbered 3, 5, 7, 9,…., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.
Solution:
In a box there are 14 tickets with number
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
Number of possible outcomes = 14
(i) Prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29
which are 9 in number
Probability of prime will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 14 } \)
(ii) Number less than 16 are 3, 5, 7, 9, 11, 13, 15
which are 7 in numbers,
Probability of number less than 16 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7 }{ 14 } \)
= \(\\ \frac { 1 }{ 2 } \)
(iii) Numbers divisible by 3 are 3, 9, 15, 21, 27
which are 5 in number
Probability of number divisible by 3 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 14 } \)

Question 30.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
There are 90 discs in a box containing numbered from 1 to 90.
Number of possible outcomes = 90
(i) Two digit numbers are 10 to 90 which are 81 in numbers.
Probability of two digit number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 81 }{ 90 } \)
= \(\\ \frac { 9 }{ 10 } \)
(ii) Perfect squares are 1, 4, 9, 16, 25, 36,49, 64, 81
which are 9 in numbers.
Probability of square will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 90 } \)
= \(\\ \frac { 1 }{ 10 } \)
(iii) Number divisible by 5 are
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
which are 18 in numbers.
Probability of number divisible by 5 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 18 }{ 90 } \)
= \(\\ \frac { 1 }{ 5 } \)

Question 31.
Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is
(i) an even number
(ii) a number less than 14
(iii) a number which is a perfect square
(iv) a prime number less than 30.
Solution:
Number of cards with numbered from 2 to 101 are placed in a box
Number of possible outcomes = 100 one card is drawn
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16,….., 96, 98, 100
which are 50 in numbers.
Probability of even number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 50 }{ 100 } \)
= \(\\ \frac { 1 }{ 2 } \)
(ii) Numbers less than 14 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
which are 12 in numbers
Probability of number less than 14 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 12 }{ 100 } \)
= \(\\ \frac { 3 }{ 25 } \)
(iii) Perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100 which are 9 in numbers
Probability of perfect square number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 100 } \)
(iv) Prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
which are 10 in numbers Probability of prime numbers, less than 30 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 10 }{ 100 } \)
= \(\\ \frac { 1 }{ 10 } \)

Question 32.
A bag contains 15 balls of which some are white and others are red. If the probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag.
Solution:
In a bag, there are 15 balls.
Some are white and others are red.
Probability of red ball = 2 probability of white ball
Let number of white balls = x
Then, number of red balls = 15 – x
\(2\times \frac { 15-x }{ 15 } =\frac { x }{ 15 } \)
⇒ 2(15 – x) = x
⇒ 30 – 2x = x
⇒ 30 = x + 2x
⇒ x = \(\\ \frac { 30 }{ 3 } \) = 10
Number of red balls = 10
and Number of white balls = 15 – 10 = 5

Question 33.
A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is twice that of a red ball, find the number of balls in the bag.
Solution:
In a bag, there are 6 red balls, and some blue balls
Probability of blue ball = 2 × probability of red ball
Let number of blue balls = x
and number of red balls = 6
Total balls = x + 6
Probability of a blue ball = 2
⇒ \(\frac { x }{ x+6 } =2\times \frac { 6 }{ x+6 } \)
⇒ \(\frac { x }{ x+6 } =\frac { 12 }{ x+6 } \)
⇒ x = 12
Number of balls = x + 6 = 12 + 6 = 18

Question 34.
A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A blue is selected at random. Find the probability that it is
(i) white
(ii) not red.
Solution:
In a bag, there are 24 balls
Since, there are x balls red, 2 × balls white and 3 × balls blue
x + 2x + 3x = 24
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22 Q34.1

Question 35.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(i) ‘2’ of spades
(ii) a jack .
(iii) a king of red colour
(iv) a card of diamond
(v) a king or a queen
(vi) a non-face card
(vii) a black face card
(viii) a black card
(ix) a non-ace
(x) non-face card of black colour
(xi) neither a spade nor a jack
(xii) neither a heart nor a red king
Solution:
In a playing card, there are 52 cards
Number of possible outcome = 52
(i) Probability of‘2’ of spade will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 52 } \)
(ii) There are 4 jack card Probability of jack will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 4 }{ 52 } \)
= \(\\ \frac { 1 }{ 13 } \)
(iii) King of red colour are 2 in number
Probability of red colour king will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 2 }{ 52 } \)
= \(\\ \frac { 1 }{ 26 } \)
(iv) Cards of diamonds are 13 in number
Probability of diamonds card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 13 }{ 52 } \)
= \(\\ \frac { 1 }{ 4 } \)
(v) Number of kings and queens = 4 + 4 = 8
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 52 } \)
= \(\\ \frac { 2 }{ 13 } \)
(vi) Non-face cards are = 52 – 3 × 4 = 52 – 12 = 40
Probability of non-face card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 40 }{ 52 } \)
= \(\\ \frac { 10 }{ 13 } \)
(vii) Black face cards are = 2 × 3 = 6
Probability of black face card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6 }{ 52 } \)
= \(\\ \frac { 3 }{ 26 } \)
(viii) No. of black cards = 13 x 2 = 26
Probability of black card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 26 }{ 52 } \)
= \(\\ \frac { 1 }{ 2 } \)
(ix) Non-ace cards are 12 × 4 = 48
Probability of non-ace card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 48 }{ 52 } \)
= \(\\ \frac { 12 }{ 13 } \)
(x) Non-face card of black colours are 10 × 2 = 20
Probability of non-face card of black colour will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 20 }{ 52 } \)
= \(\\ \frac { 5 }{ 13 } \)
(xi) Number of card which are neither a spade nor a jack
= 13 × 3 – 3 = 39 – 3 = 36
Probability of card which is neither a spade nor a jack will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 36 }{ 52 } \)
= \(\\ \frac { 9 }{ 13 } \)
(xii) Number of cards which are neither a heart nor a red king
= 3 × 13 = 39 – 1 = 38
Probability of card which is neither a heart nor a red king will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 38 }{ 52 } \)
= \(\\ \frac { 19 }{ 26 } \)

Question 36.
All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting
(i) a black face card
(ii) a queen
(iii) a black card
(iv) a heart
(v) a spade
(vi) ‘9’ of black colour
Solution:
In a pack of 52 cards
All the three face cards of spade are = 3
Number of remaining cards = 52 – 3 = 49
One card is drawn at random
(i) Probability of a black face card which are = 6 – 3 = 3
Probability = \(\\ \frac { 3 }{ 49 } \)
(ii) Probability of being a queen which are 4 – 1 = 3
Probability = \(\\ \frac { 3 }{ 49 } \)
(iii) Probability of being a black card = (26 – 3 = 23)
Probability = \(\\ \frac { 23 }{ 49 } \)
(iv) Probability of being a heart = \(\\ \frac { 13 }{ 49 } \)
(v) Probability of being a spade = (13 – 3 = 10)
Probability = \(\\ \frac { 10 }{ 49 } \)
(vi) Probability of being 9 of black colour (which are 2) = \(\\ \frac { 2 }{ 49 } \)

Question 37.
From a pack of 52 cards, a blackjack, a red queen and two black kings fell down. A card was then drawn from the remaining pack at random. Find the probability that the card drawn is
(i) a black card
(ii) a king
(iii) a red queen.
Solution:
In a pack of 52 cards, a blackjack, a red queen, two black being felt down.
Then number of total out comes = 52 – (1 + 1 + 2) = 48
(i) Probability of a black card (which are 26 – 3 = 23) = \(\\ \frac { 23 }{ 48 } \)
(ii) Probability of a being (4 – 2 = 2) = \(\\ \frac { 2 }{ 48 } \) = \(\\ \frac { 1 }{ 24 } \)
(iii) Probability of a red queen = (2 – 1 = 1) = \(\\ \frac { 1 }{ 48 } \)

Question 38.
Two coins are tossed once. Find the probability of getting:
(i) 2 heads
(ii) at least one tail.
Solution:
Total possible outcomes are . HH, HT, TT, TH, i.e., 4
(i) Favourable outcomes are HH, i.e., 1
So, P(2 heads)
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 4 } \)
(ii) Favourable outcomes are HT, TT, TH, i.e., 3
So, P (at least one tail) = \(\\ \frac { 3 }{ 4 } \)

Question 39.
Two different coins are tossed simultaneously. Find the probability of getting :
(i) two tails
(ii) one tail
(iii) no tail
(iv) atmost one tail.
Solution:
Two different coins are tossed simultaneously
Number of possible outcomes = (2)² = 4
Number of event having two tails = 1 i.e. (T, T)
(i) Probability of two tails will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 4 } \)
(ii) Number of events having one tail = 2 i.e. (TH) and (HT)
Probability of one tail will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 4 } \)
(iii) Number of events having no tail = 1 i.e. (HH)
Probability of having no tail will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 4 } \)
(iv) Atmost one tail
Number Of events having at the most one tail = 3 i.e. (TH), (HT, (TT)
Probability of at the most one tail will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 4 } \)

Question 40.
Two different dice are thrown simultaneously. Find the probability of getting:
(i) a number greater than 3 on each dice
(ii) an odd number on both dice.
Solution:
When two different dice are thrown simultaneously,
then the sample space S of the random experiment =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) .
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
It consists of 36 equally likely outcomes.
(i) Let E be the event of ‘a number greater than 3 on each dice’.
E = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
No. of favourable outcomes (E) = 9
P (number greater than 3 on each dice) = \(\\ \frac { 9 }{ 36 } \) = \(\\ \frac { 1 }{ 4 } \)
(ii) Let E be the event of ‘an odd number on both dice’.
E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
No. of favourable outcomes (E) = 9
∴ P (Odd on both dices) = \(\\ \frac { 9 }{ 36 } \) = \(\\ \frac { 1 }{ 4 } \)

Question 41.
Two different dice are thrown at the same time. Find the probability of getting :
(i) a doublet
(ii) a sum of 8
(iii) sum divisble by 5
(iv) sum of atleast 11.
Solution:
Two different dice are thrown at the same time
Possible outcomes will be (6)² i.e. 36
(i) Number of events which doublet = 6
i.e. (1, 1), (2, 2) (3, 3), (4, 4), (5, 5) and (6, 6)
.’. Probability of doublets will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6 }{ 36 } \)
= \(\\ \frac { 1 }{ 6 } \)
(ii) Number of event in which the sum is 8 are
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5
Probability of a sum of 8 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 36 } \)
(iii) Number of event when sum is divisible by
5 are (1, 4), (4, 1), (2, 3), (3, 2), (4, 6),
(5, 5) = 7 in numbers
Probability of sum divisible by 5 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7 }{ 36 } \)
(iv) Sum of atleast 11, will be in following events
(5, 6), (6, 5), (6, 6)
Probability of sum of atleast 11 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 36 } \)
= \(\\ \frac { 1 }{ 12 } \)

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