## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C. https://ncertmcq.com/selina-concise-mathematics-class-10-icse-solutions/

Other Exercises

Question 1.
Find the sum of the first 22 terms of the A.P. : 8, 3, – 2,…..
Solution:
A.P. = 8, 3, – 2,…..
Here, a = 8, d = 3 – 8 = – 5, n = 22

Question 2.
How many terms of the A.P. : 24, 21, 18, must be taken so that their sum is 78 ?
Solution:
Let n term of the given A.P. be taken
and A.P. = 24, 21, 18……
Let n be the number of terms.
Here, a = 24, d = 21 – 24 = – 3, Sn = 78

Question 3.
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Solution:
nth term (Tn) = 8n – 5
T1 = 8 – 5 = 3
T2 = 8 x 2 – 5 = 16 – 5 = 11
T3 = 8 x 3 – 5 = 24 – 5 = 19
A.P. is 3, 11, 19,
Here, a = 3,d = 11 – 3 = 8 and n = 28

Question 4.
Find the sum of :
(i) all odd natural numbers less than 50.
(ii) first 12 natural numbers each of which is a multiple of 7.
Solution:
(i) Sum of all odd natural numbers less then 50
1 + 3 + 5 + 7 +…….+ 49
Here a = 1, d = 3 – 1 = 2
Let n be the number of term, then
49 = a + (n – 1)d
=> 49 = 1 + (n – 1) x 2
=> 49 – 1 = (n – 1) x 2

Question 5.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Solution:
Sum of first 51 terms of an A.P. in which T2 = 14 and T3 = 18
d = T3 – T2 = 18 – 14 = 4
and T2 = a + d
=> 14 = a + 4
=> a = 14 – 4 = 10
A.P. = 10, 14, 18, 22,….

Question 6.
The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first tt terms.
Solution:
S7 = 49,
S17 = 289

Question 7.
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Solution:
First term of an AP (a) = 5
Last term = 45
and Sn = 1000

Question 8.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Solution:
All natural numbers between 250 and 1000 which are divisible by 9 are
252, 261, 270……999

Question 9.
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
Solution:
In an A.P.
T1 = a = 34, l = 700, d = 18
Let n be the number of terms, then
Tn = a + (n – 1 )d
=> 700 = 34 + (n – 1) x 18
=> 700 – 34 = 18(n – 1)
=> 180 – 0 = 666

Question 10.
In an A.P. the first term is 25, nth term is – 17 and the sum of n terms is 132. Find n and the common difference.
Solution:
In an A.P.
First term (a) = 25
nth term = – 17
and Sn = 132
Let d be the common difference

Question 11.
If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of this A.P.
Solution:
In an A.P.
8th term = 37
15th term = 12th term + 15
Let a be the first term and d be the common difference, then

Question 12.
Find the sum of all multiples of 7 lying between 300 and 700.
Solution:
Multiples of 7 lying between 300 and 700 are 301, 308, 315, 322,…., 693

Question 13.
The sum of n natural numbers is 5n² + 4n. Find its 8th term.
Solution:
Sum of n natural number = 5n² + 4n
Sn = 5n² + 4n
S1(a) = 5 x (1)² + 4 x 1
= 5 + 4 = 9
S2 = 5(2)² + 4 x 2 = 20 + 8 = 28
S2 – S1 = T2 = 28 – 9 = 19
=> a + d = 19 => 9 + d = 19
d = 19 – 9 = 10
a = 9, d = 10
T8 = a + (n – 1 )d = 9 + (8 – 1) x 10
= 9 + 7 x 10 = 9 + 70 = 79

Question 14.
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
Solution:
In an A.P.
T= 11
T8 = 2T4 + 5
Now, a + 3d = 11
a + 7d = 2 x 11 + 5 = 22 + 5 = 27
Subtracting, 4d = 16
=> d = $$\\ \frac { 16 }{ 4 }$$ = 4

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C 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 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.

Solution:
Diameter of the biggest hole = 30 cm.
Radius (r) = $$\\ \frac { 30 }{ 2 }$$ = 15 cm
and height (h) = 70 cm.

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.

Solution:
Edge of cubical part = 28 cm
and radius of cylindrical part (r) = $$\\ \frac { 28 }{ 2 }$$ = 14cm

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 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 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 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.

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 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 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.

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

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 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 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

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

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.

Solution:
In the given figure,
The total height of the toy rocket = 26 cm
Diameter of cylindrical portion = 3 cm

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

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

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

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

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

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

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

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.

Solution:
(i) In the given figure,
Height of cylindrical portion (h) = 8 cm
Radius (r) = 3 cm
Scale = 1 : 200

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4, drop a comment below and we will get back to you at the earliest.

## 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)
$$\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) or$$

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

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

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)

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

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
$$\frac { AP }{ AB } =\frac { 3 }{ 4 }$$

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)

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.

(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)

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)

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

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 = $$\sqrt { { \left( 3+2 \right) }^{ 2 }+{ \left( -7-5 \right) }^{ 2 } }$$

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.

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

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

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).
$$5=\frac { { m }_{ 1 }\times 7+{ m }_{ 2 }\times 2 }{ { m }_{ 1 }+{ m }_{ 2 } }$$

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.

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)

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

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,

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.

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,

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

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

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

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.

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).

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)

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

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

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
$$\left( \frac { 10+6 }{ 2 } ,\frac { 5-3 }{ 2 } \right) or\left( \frac { 16 }{ 2 } ,\frac { 2 }{ 2 } \right) or(8,1)$$

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 $$\frac { AP }{ PB } =\frac { AQ }{ QC } =\frac { 1 }{ 2 }$$

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.

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

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

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).

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

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.

## Selina Concise Mathematics Class 10 ICSE Solutions 2020-21 Edition

Learninsta.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Books of Concise Mathematics for Class 10 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines. By studying these Selina ICSE Solutions for Class 10 Maths you can easily get good marks in ICSE Class 10 Board Examinations.

## Understanding ICSE Mathematics Class 10 ML Aggarwal Solved Solutions

Get Latest Edition of ML Aggarwal Class 10 Solutions PDF Download on LearnInsta.com. It provides step by step solutions for ML Aggarwal Maths for Class 10 ICSE Solutions Pdf Download. You can download the Understanding ICSE Mathematics Class 10 ML Aggarwal Solved Solutions with Free PDF download option, which contains chapter wise solutions. APC Maths Class 10 Solutions ICSE all questions are solved and explained by expert Mathematic teachers as per ICSE board guidelines. By studying these ML Aggarwal Class 10 ICSE Solutions you can easily get good marks in ICSE Class 10 Board Examinations. You also refer Selina Concise Mathematics Class 10 Solutions for more practice.

APC Understanding ICSE Mathematics Class 10 ML Aggarwal Solutions 2018 Edition for 2019 Examinations

ML Aggarwal Class 10 Maths Chapter 1 Value Added Tax

ML Aggarwal Class 10 Maths Chapter 2 Banking

ML Aggarwal Class 10 Maths Chapter 3 Shares and Dividends

ML Aggarwal Class 10 Maths Chapter 4 Linear Inequations

ML Aggarwal Class 10 Maths Chapter 5 Quadratic Equations in One Variable

ML Aggarwal Class 10 Maths Chapter 6 Factorization

ML Aggarwal Class 10 Maths Chapter 7 Ratio and Proportion

ML Aggarwal Class 10 Maths Chapter 8 Matrices

ML Aggarwal Class 10 Maths Chapter 9 Arithmetic and Geometric Progressions

ML Aggarwal Class 10 Maths Chapter 10 Reflection

ML Aggarwal Class 10 Maths Chapter 11 Section Formula

ML Aggarwal Class 10 Maths Chapter 12 Equation of a Straight Line

ML Aggarwal Class 10 Maths Chapter 13 Similarity

ML Aggarwal Class 10 Maths Chapter 14 Locus

ML Aggarwal Class 10 Maths Chapter 15 Circles

ML Aggarwal Class 10 Maths Chapter 16 Constructions

ML Aggarwal Class 10 Maths Chapter 17 Mensuration

ML Aggarwal Class 10 Maths Chapter 18 Trigonometric Identities

ML Aggarwal Class 10 Maths Chapter 19 Trigonometric Tables

ML Aggarwal Class 10 Maths Chapter 20 Heights and Distances

ML Aggarwal Class 10 Maths Chapter 21 Measures of Central Tendency

ML Aggarwal Class 10 Maths Chapter 22 Probability

### FAQs on ML Aggarwal Class 10 Solutions

1.  How do I download the PDF of ML Aggarwal Solutions in Class 10?

All you have to do is tap on the direct links available on our LearnInsta.com page to access the Class 10 ML Aggarwal Solutions in PDF format. You can download them easily from here for free of cost.

2. Where can I find the solutions for ML Aggarwal Maths Solutions for Class 10?

You can find the Solutions for ML Aggarwal Maths for Class 10 from our page. View or download them as per your convenience and aid your preparation to score well.

3. What are the best sources for Class 10 Board Exam Preparation?

Aspirants preparing for their Class 10 Board Exams can make use of the quick and direct links available on our website LearnInsta.com regarding ML Aggarwal Solutions.

4. Is solving ML Aggarwal Solutions Chapterwise benefit you during your board exams?

Yes, it can be of huge benefit during board exams as you will have indepth knowledge of all the topics by solving Chapterwsie ML Aggarwal Solutions.

5. Where to download Class 10 Maths ML Aggarwal Solutions PDF?

Candidates can download the Class 10 Maths ML Aggarwal Solutions PDF from the direct links available on our page. We don’t charge any amount from you and they are absolutely free of cost.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A.

Other Exercises

Question 1.
How much money will be required to buy 400, ₹ 12.50 shares at a premium of ₹ 1?
Solution:
Number of shares purchased = 400
Rate of each share = ₹ 12.50
M.V. = ₹ 1 premium = ₹ 12.50 + ₹ 1 = ₹ 13.50
Amount of in vestment = ₹ 400 x ₹ 13.50 = ₹ 5400

Question 2.
How much money will be required to buy 250, ₹ 15 shares at a discount of ₹ 1.50?
Solution:
Number of shares = 250
M.V. = at ₹ 15 at a discount of ₹ 1.50 = ₹ 15 – ₹ 1.50 = ₹ 13.50
Amount of investment = ₹ 13.50 x 250 = ₹ 3375

Question 3.
A person buys 120 shares at a nominal value of ₹ 40 each, which he sells at ₹ 42.50 each. Find his profit and profit percent.
Solution:
No. of shares = 120
Nominal value of each share = ₹ 40.00
Profit at each share = ₹ 42.50 – ₹ 40.00 = ₹ 2.50
Total profit = 2.50 x 120 = ₹ 300
Cost price of 120 shares = ₹ 40 x 120 = ₹ 4,800

Question 4.
Find the cost of 85 shares of Rs. 60 each when quoted at ₹ 63.25
Solution:
No. of shares = 85
Market value of cach share = ₹ 63.25
Total cost = ₹ 63.25 x 85 = ₹ 5,376.25

Question 5.
A man invests ₹ 800 in buying 75 shares and when they are selling at a premium of ₹ 1.15, he sells all the shares. Find his profit and profit percent.
Solution:
Investment = ₹ 800
In first case face value of each share = ₹ 5
and market value of each share = ₹ 5.00 + ₹ 1.15 = ₹ 6.15
Gain on each share of ₹ 5 = ₹ 1.15

Question 6.
Find the annual income derived from 125, ₹ 120 shares paying 5% dividend.
Solution:
Amount of investment = ?
Number of shares purchased = 125 at ₹ 120, 5% dividend
Amount of investment = ₹ 125 x 120 = ₹ 15000
His annual income = 15000 x $$\frac { 5 }{ 100 }$$ = ₹ 750

Question 7.
A man invests ₹ 3,072 in a company paying 5% per annum when its ₹ 10 share can be bought for ₹ 16 each. Find:
(i) his annual income;
(ii) his percentage income on his investment.
Solution:
Total investment = ₹ 3,072
Market value of each shares = ₹ 16

Question 8.
A man invests ₹ 7,770 in a company paying 5 percent dividend when a share of nominal value of ₹ 100 sells at a premium of ₹ 5. Find :
(i) the number of shares bought;
(ii) annual income ;
(iii) percentage income ;
Solution:
Investment = ₹ 7770
Nominal value of each share = 100
Market value = 100 + 5 = 105

Question 9.
A man buys ₹ 50 shares of a company paying 12 percent dividend, at a premium of ₹ 10. Find :
(i) the market value of 320 shares ;
(ii) his annual income ;
(iii) his profit percent.
Solution:
(i) Market value of each share = ₹ 50 + ₹ 10 = ₹ 60
Market value of 320 shares = ₹ 60 x 320 = ₹ 19,200
(ii) Rate of dividend = 12%
Face value of 320 shares = Rs. 50 x 320 = Rs. 16,000

Question 10.
A man buys of Rs. 75 shares at a discount of Rs. 15 of a company paying 20% dividend. Find :
(i) the market value of 120 shares ;
(ii) his annual income ;
(iii) his profit percent.
Solution:
(i) Market value of one share = Rs. 75 – 15 = Rs. 60
Market value of 120 shares = Rs. 60 x 120 = Rs. 7,200
(ii) Rate of dividend = 20%
Face value of 120 shares = Rs. 75 x 120 = Rs. 9,000

Question 11.
A man has 300, ₹ 50 shares of a company paying 20% dividend. Find his net income after paying 3% income tax.
Solution:
No. of shares = 300
Face value of 50 shares = Rs. 50 x 300 = Rs. 15,000
Rate of dividend = 20%

Question 12.
A company pays dividend of 15 % on its ten-rupee shares from which it deducts income tax at the rate of 22%. Find the annual income of a man who owns one thousand shares of this company.
Solution:
No. of shares = 1,000
Face Value of each share = Rs. 10
Rate of dividend = 15%
Rate of income tax = 22%
Face value of 1,000 shares = 1,000 x 10 = Rs. 10,000
Total dividend = Rs. 10,000 x $$\frac { 15 }{ 100 }$$ = Rs. 1,500
Income tax deducted = Rs. 1500 x $$\frac { 22 }{ 100 }$$ = Rs. 330
Net income = Rs.1500 – Rs. 330 = Rs. 1170

Question 13.
A man invests Rs. 8,800 in buying shares of a company of face value of rupees hundred each at a premium of 10%. If he earns Rs. 1,200 at the end of the year as dividend find:
(i) the number of shares he has in the company;
(ii) the dividend percent per share. [2001]
Solution:
Investment = Rs. 8,800
Face value of each share = Rs. 100
Market value of each share = Rs. 100 + 10 = Rs. 110

Question 14.
A man invests Rs. 1,680 in buying shares of nominal value Rs. 24 and selling at 12% premium. The dividend on the shares is 15% per annum. Calculate :
(i) The number of shares he buys ;
(ii) The dividend he receives annually. [1999]
Solution:
Investment = Rs. 1680
Nominal value of each share = Rs. 24
Market value of each share = Rs. 24 + 12% of 24
= Rs. 24 + 2.88 = Rs. 26.88
Rate of dividend = 15%
(i) No. of shares = $$\frac { 1680 }{ 26.88 }$$ = 62.5
(ii) Face value of 62.5 shares = 62.5 x 24 = Rs. 1500
Amount of dividend = 1500 x $$\frac { 15 }{ 100 }$$ = Rs. 225

Question 15.
By investing Rs. 7,500 in a company paying 10 percent dividend, an annual income of Rs. 500 is received. What price is paid for each of Rs. 100 share? [1990]
Solution:
Investment = Rs. 7,500
Rate of dividend = 10%
Total income = Rs. 500

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3A 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.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking (Recurring Deposit Accounts) Ex 2B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B.

Other Exercises

Question 1.
Pramod deposits ₹ 600 per month in a Recurring Deposit Account for 4 years. If the rate of interest is 8% per year; calculate the maturity value of his account.
Solution:
Deposit per month (P) = ₹ 600
Rate of interest (r) = 8%
Period (n) = 4 years = 48 months.
According to formula,

Maturity value = ₹ 600 x 48 + ₹ 4,704 = ₹ 28,800 + ₹ 4,704 = ₹ 33504

Question 2.
Ritu has a Recurring Deposit Account in a bank and deposits ₹ 80 per month for 18 months. Find the rate of interest paid by the bank if the maturity value of this account is ₹ 1,554.
Solution:
Let rate of interest = r%,
n = 18,
P = ₹ 80
and A is maturity value.
Using formula

Question 3.
The maturity value of a R.D. Account is ₹ 16,176. If the monthly installment is ₹ 400 and the rate of interest is 8%; find the time (period) of this R.D. Account.
Solution:
Here maturity value (A) = ₹ 16,176
Rate = 8%,
P = ₹ 400
Let period = n (No. of months)
Using formula :
I = A – P x n = 16,176 – 400 x n = 16,716 – 400n.

⇒ 48,528 – 1,200n = 4n² + 4n
⇒ 4n² + 4n + 1200n – 48,528 = 0
⇒ 4n² + 1,204n – 48,528 = 0
⇒ n² + 301n — 12,132 = 0 (dividing by 4)
⇒ n² – 36n + 337n – 12,132 = 0
⇒ n (n – 36) + 337 (n – 36) = 0
⇒ (n – 36) (n + 337) = 0
Either n = 36 months or n = -337, which is not possible.
Time = 36 months = 3 years

Question 4.
Mr. Bajaj needs ₹ 30,000 after 2 years. What least money (in multiple of ₹ 5) must he deposit every month in a recurring deposit account to get required money at the end of 2 years, the rate of interest being 8% p.a. ?
Solution:
Amount of maturity = ₹ 30000
Period (n) = 2 years = 24 months
Rate = 8% p.a.
Let x be the monthly deposit

Amount of monthly deposit in the multiple of ₹ 5 = ₹ 1155

Question 5.
Rishabh has a recurring deposit account in a post office for 3 years at 8% p.a. simple interest. If he gets ₹ 9,990 as interest at the time of maturity, find :
(i) the monthly installment.
(ii) the amount of maturity.
Solution:
Total interest = ₹ 9990
Period (n) = 3 years = 36 months
Rate of interest (r) = 8%
(i) Let monthly installment = x

Monthly installment = ₹ 2250
(ii) Amount of maturity = Principal + Interest
= 36 x 2250 + 9990
= ₹ 81000 + 9990 = ₹ 90990

Question 6.
Gopal has a cumulative deposit account and deposits ₹ 900 per month for a period of 4 years. If he gets ₹ 52,020 at the time of maturity, find the rate of interest.
Solution:
Maturity value = ₹ 52,020
Monthly installment (P) = ₹ 900
Total principal = ₹ 900 x 48 = ₹ 43200
Amount of interest = ₹ 52020 – ₹ 43200 = ₹ 8820
Let rate of interest = r%

Question 7.
Deepa has a 4 year recurring deposit account in a bank and deposits ₹ 1,800 per month. If she gets ₹ 1,800 per month. If she gets ₹ 1,08,450 at the time of maturity, find the rate of interest.
Solution:
Deposit per month = ₹ 1800
Period = 4 years = 48 months
Maturity value = ₹ 108450
Total principal = ₹ 1800 x 48 = ₹ 86400
Amount of interest = ₹ 108450 – 86400 = ₹ 22050
Let r be the rate of interest

Question 8.
Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets ₹ 8,088 from the bank after 3 years, find the value of his monthly installment. (2013)
Solution:
Let monthly installment = ₹ x
Period (n) = 3 x 12 months = 36 months

Question 9.
Sharukh opened a Recurring Deposit Account in a bank and deposited ₹ 800 per month for 1$$\frac { 1 }{ 2 }$$ years. If he received ₹ 15,084 at the time of maturity, find the rate of interest per annum. (2014)
Solution:
Money deposited per month (P) = ₹ 800
r = ?

Question 10.
Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly installment is ₹ 1,000, find the :
(i) interest earned in 2 years
(ii) maturity value. (2015)
Solution:
Period (n) = 2 years = 2 x 12 = 24 months
Rate of interest (r) = 6%
Monthly installment (P) = ₹ 1000

Question 11.
Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1200 as interest at the time of maturity, find :
(i) the monthly installment
(ii) the amount of maturity
Solution:
(i) Interest = ₹ 1200,
n = 2 x 12 = 24 months,
r = 6%

⇒ P = ₹ 800
So the monthly installment is ₹ 800
(ii) Total sum deposited = P x n = ₹ 800 x 24 = ₹ 19200
The amount that Mohan will get at the time of maturity = Total sum deposited + Interest Received
= ₹ 19200 + ₹ 1200 = ₹ 20400
Hence, the amount of maturity is ₹ 20400

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2B 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.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking (Recurring Deposit Accounts) Ex 2A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A.

Other Exercises

Question 1.
Manish opens a Recurring Deposit Account with the Bank of Rajasthan and deposits ₹ 600 per month for 20 months. Calculate the maturity value of this account, if the bank pays interest at the rate of 10% per annum.
Solution:
Recurring Deposit per month = ₹ 600
Period (n) = 20 months
Rate of interest (r) = 10% p.a.
Total principal for 1 month

Maturity value = ₹ 600 x 20 + ₹ 1,050 = ₹ 12,000 + ₹ 1,050 = ₹ 13,050

Question 2.
Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited ₹ 640 per month for 4$$\frac { 1 }{ 2 }$$ years. Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.
Solution:
Recurring deposit per month = ₹ 640
Period (n) = 4$$\frac { 1 }{ 2 }$$ years = 54 months
Rate of interest (r) = 12%
Total principal for 1 month

Maturity value = ₹ 640 x 54 + ₹ 9,504 = ₹ 34,560 + ₹ 9,504 = ₹ 44,064

Question 3.
Each of A and B opened a recurring deposit accounts in a bank. If A deposited ₹ 1,200 per month for 3 years and B deposited ₹ 1,500 per month for 2$$\frac { 1 }{ 2 }$$ years; find, on maturity, who will it get more amount and by how much ? The rate of interest paid by the bank is 10% per annum.
Solution:
A’s deposit per month (P) = ₹ 1200
Period = 3 years = 36 months
Total principal for one month

and maturity value = ₹ 1500 x 30 + Interest
= ₹ 45000 + 5812.50
= ₹ 50812.50
It is clear that B’s maturity value is greater Difference = ₹ 50812.50 – ₹ 49860 = ₹ 952.50

Question 4.
Ashish deposits a certain sum of money every month in a Recurring Deposit Account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets ₹ 12,715 as the maturity value of this account, what sum of money did he pay every month ?
Solution:
Let Recurring deposit per month = ₹ x
Period (n) = 12 months
Rate of interest (r) = 11%
Maturity value = ₹ 12,715 ………. (i)
Total principal for one month

Recurring deposit per month ₹ 1000 p.m.

Question 5.
A man has a Recurring Deposit Account in a bank for 3$$\frac { 1 }{ 2 }$$ years. If the rate of interest is 12% per annum and the man gets ₹ 10206 on maturity, find the value of monthly installments.
Solution:
Let Recurring deposit per month = ₹ x
Period (n) = 3$$\frac { 1 }{ 2 }$$ years = 42 months
Rate of interest (r) = 12% p.a.
Amount of maturity = ₹ 10206 ……… (i)

Amount of each installment = ₹ 200

Question 6.
(i) Puneet has a Recurring Deposit Account in the Bank of Baroda and deposits ₹ 140 per month for 4 years If he gets ₹ 8,092 on maturity, find the rate of interest given by the bank.
(ii) David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years. If he received ₹ 7725 at the time of maturity, find the rate of interest per annum. (2008)
Solution:
(i) Recurring deposit per month = ₹ 140
Period (n) = 4 years = 48 months
Let Rate of interest (r) = r % p.a.
Amount of maturity = ₹ 8,092
Total principal for one month

Question 7.
Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposits Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and interest is calculated at the end of every month ? [I.C.S.E. 2001]
Solution:
Amount of Recurring deposit = ₹ 150
Period (n) = 8 months
Rate of interest (r) = 8% p.a.
Total principal for one month

Amount of maturity value = ₹ 150 x 8 + ₹ 36 = ₹ 1,200 + ₹ 36 = ₹ 1,236

Question 8.
Mrs. Geeta deposited ₹ 350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is ₹ 5,565; find the rate of interest per annum.
Solution:
Amount of recurring deposit per month = ₹ 350
Period (n) = 1 year 3 months = 15 months
Let rate of interest = r % p.a.
Amount of maturity = ₹ 5565
Amount of interest = ₹ 5,565 – ₹ 350 x 15 = ₹ 5,565 – 5,250 = ₹ 315 ….(i)
Now, total principal for one month

Question 9.
A recurring deposit account of ₹ 1,200 per month has a maturity value of ₹ 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.
Solution:
Amount of recurring deposit per month = ₹ 1,200
Rate of interest (r) = 8% p.a.
Let period = n months
Amount of maturity = ₹ 12,440
Amount of interest = ₹ 12440 – ₹ 1200 x n ….(i)
Total principal for one month

from (i) and (ii), we get,
4n (n + 1) = 12440 – 1200n
⇒ 4n² + 4n = 12440 – 1200n
⇒ 4n² + 1204n – 12440 = 0
Dividing by 4,
⇒ n² + 301n – 3110 = 0
⇒ n² + 311n – 10n – 3110 = 0
⇒ n (n + 311) – 10 (n + 311) = 0
⇒ (n + 311) (n – 10) = 0
Given n + 311 = 0, then n = – 311 Which is not possible,
or n – 10 = 0, then n = 10
Period = 10 months.

Question 10.
Mr. Gulati has a Recurring Deposit Account of ₹ 300 per month. If the rate of interest is 12% and the maturity value of this account is ₹ 8,100; find the time (in years) of this Recurring Deposit Account.
Solution:
Amount of recurring deposit per month = ₹ 300
Let Period = n months
Rate of interest (r) = 12% p.a.
Amount of maturity = ₹ 8,100
Interest = 8,100 – 300 x n ……. (i)
Total principal for 1 month

⇒ 3n (n + 1) = 16,200 – 600 n
⇒ 3n² + 3n = 16,200 – 600 n
⇒ 3n² + 603n – 16,200 = 0
Dividing by 3, we get,
⇒ n² + 201n – 5,400 = 0
⇒ n² + 225n – 24n – 5,400 = 0
⇒ n(n + 225) – 24 (n + 225) = 0
⇒ (n + 225) (n – 24) = 0
Either n + 225 = 0, then n = – 225 Which is not possible
or n – 24 = 0, then n = 24
Period = 24 months or 2 years.

Question 11.
Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2,500 per month for two years. At the time of maturity he got ₹ 67,500. Find :
(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum. (2010)
Solution:
(i) Total amount deposited by Mr. Gupta in 24 months = ₹ 2500 x 24 = ₹ 60,000
Maturity amount = ₹ 67,500
Total interest earned by Mr. Gupta = ₹ 67,500 – ₹ 60,000 = ₹ 7,500
(ii) Total principal for 1 month

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking Ex 2A 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.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C.

Other Exercises

Question 1.
By investing 745,000 in 10% 7100 shares, Sharad gets 73,000 as divided. Find the market value of each share.
Solution:
Total investment = ₹ 45000 at 10% of ₹ 100 shares
and amount of dividend = ₹ 3000

Question 2.
Mrs. Kulkarni invests ₹ 1,31,040 in buying ₹ 100 shares at a discount of 9%. She sells shares worth ₹ 72,000 at a premium of 10% and the rest at a discount of 5%. Find her total gain or loss on the whole.
Solution:
Total investment = ₹ 1,31,040 in ₹ 100 share at discount of 9%
Market value of each share = ₹ 100 – ₹ 9 = ₹ 91

Question 3.
A man invests a certain sum in buying 15% ₹ 100 shares at 20% premium. Find:
(i) his income from one share.
(ii) the number of shares bought to have an income, from the dividend, ₹ 6,480.
(iii) sum invested.
Solution:
Face value of each share = ₹ 100
Market value of each share = ₹ 100 + ₹ 20 = ₹ 120
Rate of dividend = 15%
(i) Income from one share = ₹ 15
(ii) and number of shares when amount of dividend
= $$\frac { 6480 }{ 15 }$$ = 432
(iii) and sum invested = ₹ 432 x 120 = ₹ 51,840

Question 4.
Gagan invested 80% of his savings in 10% ₹ 100 shares at 20% premium and the rest of his savings in 20% ₹ 50 shares at 20% discount. If his incomes from these shares is ₹ 5,600, calculate:
(i) his investment in shares on the whole.
(ii) the number of shares of first kind that he bought
(iii) percentage return, on the shares bought, on the whole.
Solution:
(i) Total income = ₹ 5600
Let total investment = ₹ x

Question 5.
Aishwarya bought 496, ₹ 100 shares at ₹ 132 each. Find:
(i) investment made by her.
(ii) income of Aishwarya from these shares, if the rate of dividend is 7.5%.
(iii) how much extra must Aishwarya invest in order to increase her income by ₹ 7,200?
Solution:
Number of shares = 496
Market value of each share = ₹ 132
(i) Total investment = 496 x 132 = ₹ 65472
(ii) Rate of dividend = 7.5%
Income = 496 x 7.5 = ₹ 3720
(iii) New income (increase in income) = ₹ 7200
Market value of share = ₹ 132
Rate of income = 7.5%
Exit investment

Question 6.
Gopal has some ₹ 100 shares of company A, paying 10% dividend. He sells a certain number of these shares at a discount of 20% and invests the proceeds in ₹ 100 shares at ₹ 60 of company B paying 20% dividend. If his income, from the shares sold, increases by ₹ 18,000, find the number of shares sold by Gopal.
Solution:
Let number of share purchased = x
Face value of these shares = ₹ 100 x x = 100x
dividend = 10%

Question 7.
A man invests a certain sum of money in 6% hundred rupee shares at ₹ 12 premium. When the shares fell to ₹ 96, he sold out all the shares bought and invested the proceed in 10%, ten rupee shares at ₹ 8. If the change in his income is ₹ 540, find the sum invested originally.
Solution:
Let investment = ₹ x
Dividend at the rate of 6% at 12% premium

Question 8.
Mr. Gupta has a choice to invest in ten rupee shares of two firms at ₹ 13 or at 716. If the first firm pays 5% dividend and the second firm pays 6% dividend per annum, find:
(i) which firm is paying better ?
(ii) If Mr. Gupta invests equally in both the firms and the difference between the returns from them is ₹ 30, Find how much in all does he invest ?
Solution:
Face value of each share = ₹ 10
Market value of first firm’s share = ₹ 13
and market value of second firm’s share = ₹ 16
Dividend from first firm = 5%
and dividend from second firm = 6%
(i) Let investment in each firm = ₹ 13 x 16
Income from first firm’s shares

It is clear from the above that first firm’s shares are better.
(ii) Difference in income = ₹ 8.00 – ₹ 7.80 = ₹ 0.20
If difference is ₹ 0.20 then investment in each firm = ₹ 13 x 16
and if difference is ₹ 30, then investment

Total investment in both firms = ₹ 31200 x 2 = ₹ 62,400

Question 9.
Ashok invested ₹ 26,400 in 12%, ₹ 25 shares of a company. If he receives a dividend of ₹ 2,475, find the :
(i) number of shares he bought.
(ii) market value of each share.
Solution:
(i) Given,
Investment = ₹ 26400
Rate of dividend = 12%
Dividend earned = ₹ 2475
Face value of one share = ₹ 25
Total dividend earned = No. of shares x Rate of dividend x Face value of one share

Question 10.
A man invested ₹ 45,000 in 15% ₹ 100 shares quoted at ₹ 125. When the market value of these shares rose to Rs. 140, he sold some shares, just enough to raise ₹ 8,400. calculate :
(i) the number of shares he still holds;
(ii) the dividend due to him on these remaining shares. [2004]
Solution:

Question 11.
Mr. Tiwari invested ₹ 29,040 in 15% ₹100 shares quoted at a premium of 20%. Calculate :
(i) the number of shares bought by Mr. Tiwari.
(ii) Mr. Tiwari’s income from the investment.
(iii) the percentage return on his investment.
Solution:
Mr. Tiwari’s investment = ₹ 29040
Face value of each share = ₹ 100
Market value of each share = ₹ 100 + ₹ 20 = ₹ 120
Rate of income = 15%
(i) Number of shares purchased

Question 12.
A dividend of 12% was declared on ₹ 150 shares selling at a certain price. If the rate of return is 10%, calculate :
(i) the market value of the shares.
(ii) the amount to be invested to obtain an annual dividend of ₹ 1,350.
Solution:
Let market value of each share = x
Rate of return on investment = 10%
Face value of each share = ₹ 150
Dividend rate = 12%
(i) Now, rate of return x market value = Rate of dividend x Face value
⇒ 10 x x = 12 x 150

Amount of investment in ₹ 5 shares = ₹ 5 x ₹ 180 = ₹ 13500

Question 13.
Divide ₹ 50,760 into two parts such that if one part is invested in 8% ₹ 100 shares at 8% discount and the other in 9% ₹ 100 shares at 8% premium, the annual incomes from both the investments are equal.
Solution:
Total investment = ₹ 50,760
Let first part of investment = x
Then second part = ₹ 50,760 – x
Rate of dividend in first part = 8% ₹100 at discount = 8%
M.V. of each share = ₹ 100 – 8 = ₹ 92
Rate of dividend second part = 9% ₹ 100 at premium = 8%
M.V. of each share = 100 + 8 = ₹ 108
But annual income from both part is same

Question 14.
Mr. Shameem invested 33$$\frac { 1 }{ 3 }$$ % of his savings in 20% ₹ 50 shares quoted at ₹ 60 and the remainder of the savings in 10% ₹ 100 shares quoted at ₹ 110. If his total income from these investments is ₹ 9,200 ; find :
(i) his total savings
(ii) the number of ₹ 50 shares.
(iii) the number of ₹ 100 shares.
Solution:
Let total investment = x

Question 15.
Vivek invests ₹ 4500 in 8% ₹ 10 shares at ₹ 15. He sells the shares when the price rises to ₹ 30, and invests the proceeds in 12% ₹ 100 shares at ₹ 125. Calculate,
(i) the sale proceeds
(ii) the number of ₹ 125 shares he buys.
(iii) the change in his annual income from dividend.
Solution:
(i) By investing ₹ 15, share bought = ₹ 10
By investing ₹ 4500, share bought = $$\frac { 10 }{ 15 }$$ x 4500 = ₹ 3000
Total face value of ₹ 10 shares = ₹ 3000, Income = 8%
= $$\frac { 8 }{ 100 }$$ x 3000 = ₹ 240
By selling Rs. 10 share money received = ₹ 30
By selling Rs. 3000 shares money = $$\frac { 30 }{ 10 }$$ x 3000 = ₹ 9000
(ii) By investing ₹ 125, no. of share of ₹ 100 bought = 1
By investing ₹ 9000, no. of share of ₹ 100 bought = $$\frac { 1 }{ 125 }$$ x 9000 = 72
No. of ₹ 125 shares bought = 72
(iii) By investing ₹ 125 in Rs. 100 share, income = ₹ 12
By investing ₹ 9000 in ₹ 100 share, income = $$\frac { 12 }{ 125 }$$ x 9000 = ₹ 864
Increase in income = ₹ 864 – ₹ 240 = ₹ 624

Question 16.
Mr. Parekh invested ₹ 52,000 on ₹ 100 shares at a discount of ₹ 20 paying 8% dividend. At the end of one year he sells the shares at a premium of ₹ 20. Find
(i) The annual dividend.
(ii) The profit earned including his dividend.
Solution:
Investment = ₹ 52000,
N.V. of 1 share = ₹ 100
M.V. of 1 share for 1 st year = ₹ 100 – 20 = ₹ 80
No. of shares = $$\frac { 52000 }{ 80 }$$ = 650
(i) Annual dividend = $$\frac { 8 }{ 100 }$$ x 650 x 100 = ₹ 5200
(ii) S.P of 1 share = ₹ 100 + 20 = ₹ 120
S.P. of 650 shares = ₹ 120 x 650 = ₹ 78000
C.P. of 650 shares = ₹ 100 x 650 = ₹ 65000
Profit = S.P. – C.P. = ₹ 78000 – ₹ 65000 = ₹ 13000
Profit including dividend = ₹ 13000 + ₹ 5200 = ₹ 18200

Question 17.
Salman buys 50 shares of face value ₹ 100 available at ₹ 132.
(i) What is his investment ?
(ii) If the dividend is 7.5%, what will be his annual income ?
(iii) If he wants to increase his annual income by ₹ 150, how many extra shares should he buy?
Solution:
F.V. = ₹ 100
(i) M. V. = ₹ 132,
no. of shares = 50
Investment = no. of shares x M.V. = 50 x 132 = ₹ 6600
(ii) Income per share = 7.5% of N.V.
= $$\frac { 75 }{ 10 x 100 }$$ x 100 = ₹ 7.5
Annual incomes = 7.5 x 50 = ₹ 375
(iii) New annual income = 375 + 150 = ₹ 525
No. of shares = $$\frac { 525 }{ 7.5 }$$ = 70
No. of extra share = 70 – 50 = 20

Question 18.
Salman invests a sum of money in ₹ 50 shares, paying 15% dividend quoted at 20% premium. If his annual dividend is ₹ 600, calculate:
(i) the number of shares he bought
(ii) his total investment
(iii) the rate of return on his investment. (2014)
Solution:
Nominal value = ₹ 50

Question 19.
Rohit invested ₹ 9,600 on ₹ 100 shares at ₹ 20 premium paying 8% dividend. Rohit sold the shares when the price rose to ₹ 160. He invested the proceeds (excluding dividend) in 10% ₹ 50 shares at ₹ 40. Find the :
(i) original number of shares.
(ii) sale proceeds.
(iii) new number of shares.
(iv) change in the two dividends. (2015)
Solution:
Investment by Rohit = ₹ 9600
Rate of dividend = 8% on 100 shares at ₹ 20 premium
Market value = ₹ 100 + ₹ 20 = ₹ 120

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3C 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.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B.

Other Exercises

Question 1.
A man buys 75, Rs. 100 shares paying 9 per cent dividend. He buys shares at such a price that he gets 12 per cent of his money. At what price did he buy the shares?
Solution:
No. of shares = 75
Value of each share = Rs. 100
Rate of dividend = 9%
Let the market value of each share = x
Thus. 12% of x = 9% of 100

Market value of each share = Rs. 75

Question 2.
By purchasing Rs. 25 gas shares for Rs. 40 each, a man gets 4 per cent profit on his investment. What rate per cent is the company paying ? What is his dividend if he buys 60 shares?
Solution:
Face value of each share = Rs. 25
Market value of each share = Rs. 40
Profit = 4% of his investment
Let the rate of dividend = x %

Question 3.
Hundred rupee shares of a company are available in the market at a premium of Rs. 20. Find the rate of dividend given by the company, when a man’s return on his investment is 15 percent.
Solution:
Face value of each share = Rs. 100
Market value of each share = Rs. 120
Total dividend = 15% on his investment
Let the rate of dividend = x %
Then x % of 100 = 15 % of 120

x = 18%
Rate of dividend = 18%

Question 4.
Rs. 50 shares of a company are quoted at a discount of 10%. Find the rate of dividend given by the company, the return on the investment on these shares being 20 percent.
Solution:
Face value of each share = Rs. 50

Rate of dividend = 18%

Question 5.
A company declares 8 per cent dividend to the share holders. If a man receives Rs. 2,840 as his dividend, find the nominal value of his shares.
Solution:
Rate of dividend = 8%
Total dividend = Rs. 2.840
Let Nominal value of shares = x
Then 8% of x = Rs. 2840

Nominal value of his shares = Rs. 35,500

Question 6.
How much should a man invest in Rs. 100 shares selling at Rs. 110 to obtain an annual income of Rs. 1,680, if the dividend declared is 12% ?
Solution:
Face value of each share = Rs. 100
Market value of each share = Rs. 110
Total annual income = Rs. 1,680.
Rate of dividend = 12%
Let total amount of shares = x
Then x x 12% = 1,680

Question 7.
A company declares a dividend of 11.2% to all its share-holders. If its Rs. 60 share is available in the market at a premium of 25%, how much should Rakesh invest in buying the shares of this company in order to have an annual income of Rs. 1,680 ?
Solution:
Face value of each share = Rs. 60
Market value = Rs. 60 x $$\frac { 125 }{ 100 }$$ = Rs. 75
Rate of dividend = 11.2%
Annual income = Rs. 1,680
Let the face value of shares = x
Dividend = x x 11.2%
x x 11.2% = 1680

Question 8.
A man buys 400, twenty rupee shares at a premium of Rs. 4 each and receives a dividend of 12%. Find :
(i) the amount invested by him.
(ii) his total income from the shares.
(iii) percentage return on his money.
Solution:
No. of shares = 400
Face value of one share = Rs. 20
Market value of one share = Rs. 20 + 4 = Rs. 24
Rate of dividend = 12%
(i) Amount invested by the man = Rs. 24 x 400 = Rs. 9600

Question 9.
A man buys 400, twenty-rupee shares at a discount of 20% and receives a return of 12% on his money. Calculate
(i) the amount invested by him.
(ii) the rate of dividend paid by the company.
Solution:
No. of shares = 400
Face value of one share = Rs. 20
Market value of one share = Rs. 20 x $$\frac { 80 }{ 100 }$$ = Rs. 16
Amount of shares = Rs. 20 x 400 = Rs. 8,000
(i) Amount invested = Rs. 16 x 400 = Rs. 6,400
(ii) Total dividend = Rs. 6,400 x $$\frac { 12 }{ 100 }$$ = Rs. 768
Rate of dividend = $$\frac { 768 x 100 }{ 8000 }$$= 9.6 %

Question 10.
A company with 10,000 shares of Rs. 100 each, declares an annual dividend of 5%.
(i) What is the total amount of dividend paid by the company ?
(ii) What should be the annual income of a man who has 72 shares, in the company ?
(iii) If he received only 4% of his investment, find the price he paid for each share.
Solution:
No. of shares = 10000
Face value of each share = 100
Rate of dividend = 5%
Amount of shares = Rs. 100 x 10,000 = Rs. 10,00,000

Question 11.
A lady holds 1800, Rs. 100 shares of a company that pays 15 % dividend annually. Calculate her annual dividend. If she had bought these shares at 40% premium, what is the return she gets as percent on her investment. Give your answer to the nearest integer.
Solution:
No. of shares = 1800
Face value of each share = Rs. 100
Rate of dividend = 15 %
Market value of each share Rs. 140
Total value of shares = Rs. 1800 x 100 = Rs. 1,80,000

Question 12.
A man invests Rs. 11,200 in a company paying 6 percent per annum when its Rs. 100 shares can be bought for Rs. 140 find:
(i) his annual dividend.
(ii) his percentage return on his investment.
Solution:
Investment = Rs. 11,200
Rate of dividend = 6%
Market value of a share = Rs. 140
No. of shares = Rs. 11,200 ÷ Rs. 140 = 80
Face value of 80 shares = 80 x Rs. 100 = Rs. 8,000

Question 13.
Mr. Sharma has 60 shares of N.V. ₹ 100 and sells them when they are at a premium of 60%. He invests the proceeds in shares of nominal value ₹ 50, quoted at 4% discount, and paying 18% dividend annually. Calculate:
(i) the sale proceeds ;
(ii) the number of shares he buys and
(iii) his annual dividend from the shares.
Solution:
(i) No. of shares = 60
Face value of each share = Rs. 100
Total amount = Rs. 100 x 60 = Rs. 6,000
Market value = Rs. 160
His sale proceed = Rs. 160 x 60 = Rs. 9,600
(ii) In second case :
Nominal value of each share = Rs.50
and Market value = Rs. 50 x $$\frac { 96 }{ 100 }$$ = Rs. 48
Rate of dividend = 18%
No. of shares he purchased = $$\frac { 9600 }{ 48 }$$ = 200
(iii) Face value of 200 shares = 200 x Rs. 50 = Rs. 10,000
Dividend = Rs. 10000 x $$\frac { 18 }{ 100 }$$ = Rs. 1,800

Question 14.
A company with 10,000 shares of nominal value ₹ 100 declares an annual dividend of 8% to shareholders.
(i) Calculate the total amount of dividend paid by the company.
(ii) Ramesh had bought 90 shares of the company at ₹ 150 per share. Calculate the dividend he receives and the percentage of return on his investment.
Solution:
(i) No. of shares = 10,000
Nominal value of each share = Rs. 100
Dividend = 8%
Total face value of 10,000 shares = Rs. 100 x 10,000 = Rs. 10,00,000
and amount of dividend = Rs. $$\frac { 1000000 x 8 }{ 100 }$$ = Rs. 8000
(ii) In second case :
Ramesh bought = 90 shares
Market value of each share = Rs. 150
His investment = Rs. 150 x 90 = Rs. 13,500
Face value of 90 shares = Rs. 100 x 90 = Rs. 9,000

Question 15.
Which is the better investment:
16% Rs. 100 shares at 80 or 20% Rs. 100 shares at 120 ?
Solution:
In first case :
Income on Rs. 80 = Rs. 16

From above, it is clear that first investment is better.

Question 16.
A man has a choice to invest in hundred rupee shares of two firms at Rs. 120 or at Rs. 132. The first firm pays r. dividend of 5% per annum and the second firm pays a dividend of 6% per annum. Find :
(i) Which company is giving a better return.
(ii) If a man invests Rs. 26,400 with each firm, how much will be the difference between the annual returns from the two firms.
Solution:
In first case :
Market value of share = Rs. 120
and dividend = 5%
Income on Rs. 120 = Rs. 5

= Rs. 1,200
Difference = Rs. 1,200 – Rs. 1,100 = Rs. 100

Question 17.
A man bought 360, ten rupee shares of a company paying 12 percent per annum. He sold the shares when their price rose to Rs. 21 per share and invested the proceeds in five rupee shares paying 4.5 per cent per annum at Rs 3.50 per share. Find the annual change in his income.
Solution:
No. of shares bought = 360
Face value of each share = Rs. 10
Dividend = 12%
Cost price of 360 shares = Rs. 360 x 10 = Rs. 3,600
Market value = Rs. 21
Selling price = Rs. 21 x 360 = Rs. 7,560
In second case :
Face value of each share = Rs. 5
Market value of each share = Rs. 3.5

Difference in his income = Rs. 486 Rs. 432 = Rs. 54

Question 18.
A man sold 400 (Rs. 20) shares of a company paying 5% at Rs. 18 and invested the proceeds in (Rs. 10) shares of another company paying 7% at Rs. 12. How many (Rs. 10) shares did he buy and what was the change in his income?
Solution:
In first case :
No. of shares sold = 400
Face value of each share = Rs. 20
Market value = Rs. 18
Income = 5%
Amount of his investment = Rs. 18 x 400 = Rs. 7,200
and amount of shares = Rs. 20 x 400 = Rs. 8000
In second case :
Market value of each share = Rs. 12
Face value of each share = Rs. 10
Rate of dividend = 7%
No. of shares purchased = $$\frac { 7200 }{ 12 }$$ = 600
Face value of 600 shares = Rs. 10 x 600 = Rs. 6,000
Now, income in first case = Rs. 8000 x $$\frac { 5 }{ 100 }$$ = Rs. 400
and income in second case = Rs. 6000 x $$\frac { 7 }{ 100 }$$ = Rs. 420
Increase in income = Rs. 420 – 400 = Rs. 20

Question 19.
Two brothers A and B invest Rs. 16,000 each in buying shares of two companies. A buys 3% hundred-rupee shares at 80 and B buys ten rupee shares at par. If they both receive equal dividend at the end of the year, find the rate percent of the dividend received by B.
Solution:
A’s investment = Rs. 16,000
Face value of each share = Rs. 100
Market value of each share = Rs. 80
and rate of dividend = 3%
No. of shares purchased = $$\frac { 16000 }{ 80 }$$ = 200
Amount of shares = 200 x Rs. 100 = Rs. 20,000
and amount of dividend = 20,000 x $$\frac { 3 }{ 100 }$$ = Rs. 600
B’s investment = Rs. 16,000
Face value of each share = Rs. 10
and amount of dividend = Rs. 600
Rate of dividend = $$\frac { 600 }{ 16000 }$$ x 100 = 3.75 %

Question 20.
A man invests Rs. 20,020 in buying shares of nominal value Rs. 26 at 10% premium. The dividend on the shares is 15% per annum Calculate :
(i) The number of shares he buys.
(ii) The dividend he receives annually.
(iii) The rate of interest he gets on his money [2003]
Solution:
Total investment = Rs. 20,020
Nominal value of each share = Rs. 26.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 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.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C.

Other Exercises

[In this exercise, all the prices are excluding tax/VAT unless specified].
Question 1.
Madan purchases a compact computer system for ₹ 47,700 which includes 10% rebate on the marked price and then 6% Sales Tax on the remaining price. Find the marked price of the computer.
Solution:
Sales price of the computer = ₹ 47700
Rate of Sales Tax = 6%

Hence marked price = ₹ 50,000

Question 2.
An article is marked at ₹ 500. The wholesaler sells it to a retailer at 20% discount and charges sales-tax on the remaining price at 12.5%. The retailer, in turn, sells the article to a customer at its marked price and charges sales-tax at the same rate. Calculate :
(i) the price paid by the customer.
(ii) the amount of VAT paid by the retailer.
Solution:
Marked price of an article = ₹ 500
Rate of discount = 20%

(i) Price paid by-the customer = ₹ 500 + ₹ 62.50 = ₹ 562.50
(ii) VAt paid by the retailer = ₹ 62.50 – ₹ 50.00 = ₹ 12.50

Question 3.
An article is marked at ₹ 4,500 and the rate of sales-tax on it is 6%. A trader buys this article at some discount and sells it to a customer at the marked price. If the trader pays ₹ 81 as VAT; find :
(i) how much per cent discount does the trader get ?
(ii) the total money paid by the trader, including tax, to buy the article.
Solution:
Marked price = ₹ 4500
Rate of S.T. = 6%
VAT paid by the trader = ₹ 81
C.P. for the customer = ₹ 4500

Question 4.
A retailer sells an article for ₹ 5,350 including 7% Sales Tax on the listed price. If he had bought it at a discount and has made a profit of 25% on the whole, And the rate of discount he gets.
Solution:
S.P. of an article of the retailer (including S.T.) = ₹ 5350
Rate of S.T. = 7%

Question 5.
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 tax at the same rate. Find :
(i) The price at which the camera can be bought from the shopkeeper.
(ii) The VAT (Value Added Tax) paid by the shopkeeper. (2008)
Solution:
Printed price (M.P) = ₹ 1600
Rate of discount = 20%

(i) The buyers bought the camera = ₹ 1600 + ₹ 96 = ₹ 1696
(ii) VAT paid by the shopkeeper = ₹ 96 – ₹ 76.80 = ₹ 19.20

Question 6.
Tarun bought an article for ₹ 8000 and spent ₹ 1000 for transportation. He marked the article at ₹ 11,700 and sold it to a customer. If the customer had to pay 10% sales tax, find:
(i) The customer’s price
(ii) Tarun’s profit percent.
Solution:
Tarun’s cost price = ? 8000
Transportation charges = ₹ 1000
Tarun’s total cost = (₹ 8000 + ₹ 1000) = ₹ 9000
Tarun’s selling price = ₹ 11,700
Rate of sales tax = 10%

Question 7.
A shopkeeper sells an article at the listed price of ₹ 1,500 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 price, inclusive of Tax, at which the shopkeeper purchased the articles from the wholesaler?
Solution:
Tax charged by shopkeeper

Question 8.
A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being ₹ 18,000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax under is 8%, find:
(i) the VAT paid by the shopkeeper.
(ii) the total amount that the consumer pays for the washing machine. (2014)
Solution:
(i) S.P. of washing machine

VAT paid by shopkeeper = Tax charged – Tax paid = ₹ 1296 – ₹ 1152 = ₹144
(ii) Price paid by customer = ₹ 16200 + ₹ 1296 = ₹ 17496

Question 9.
Mohit, a dealer in electronic goods, buys a high class TV set for ₹ 61,200. He sells this TV set to Geeta, Geeta to Rohan and Rohan sells it to Manoj. If the profit at each stage is ₹ 2,000 and the rate of VAT at each stage is 12.5%, find :
(i) total amount of tax (under VAT) paid to the Government.
(ii) Money paid by Manoj to buy the TV set.
Solution:
For Mohit,
C.P. of electronic goods = ₹ 61200
Amount of profit in each case = ₹ 2000

Net VAT = ₹ 8400 – ₹ 8150 = ₹ 250
(i) Total amount of VAT paid to govt. = ₹ 7650 + ₹ 250 + ₹ 250 + ₹ 250 = ₹ 8400
(ii) Money paid by Manoj to buy the T.V. set = ₹ 67200 + ₹ 8400 = ₹ 75600

Question 10.
A shopkeeper buys an article at a discount of 30% of the list price which is ₹ 48,000. In turn, the shopkeeper sells the article at 10% discount. If the rate of VAT is 10%, find the VAT to be paid by the shopkeeper.
Solution:
Market value (M.P.) of an article = ₹ 48000
Rate of discount = 30%
Amount of discount = ₹ 48000 x $$\frac { 30 }{ 100 }$$ = ₹ 14400
(i) Cost price of shopkeeper = ₹ 48000 – ₹ 14400 = ₹ 33600
(ii) C.P. for shopkeeper = ₹ 33600
Rate of VAT = 10%
Amount of VAT paid by shopkeeper

Net VAT paid by the shopkeeper = VAT recovered from customer – VAT paid to dealer
= ₹ 4320 – ₹ 3360 = ₹ 960

Question 11.
A company sells an article to a dealer for ₹ 40,500 including VAT (sales-tax). The dealer sells it to some other dealer for ₹ 42,500 plus tax. The second dealer sells it to a customer at a profit of ₹ 3,000. If the rate of sales-tax under VAT is 8%, find :
(i) the cost of the article (excluding tax) to the first dealer.
(ii) the total tax (under VAT) received by the Government.
(iii) the amount that a customer pays for the article.
Solution:
For a company,
S.P. of an article including VAT = ₹ 40500
Rate of VAT = 8%
(i) Amount of article excluding VAT

= ₹ 425 x 108 = ₹ 45900
Total tax (VAT) = ₹ 45900 – ₹ 42500 = ₹ 3000
S.P. for the second dealer = ₹ 42500 + ₹ 3000 = ₹ 45500
(ii) VAT = ₹ $$\frac { 45500 x 8 }{ 100 }$$ = ₹ 3640
(iii) Total amount paid by customer = ₹ 45500 + ₹ 3640 = ₹ 49140

Question 12.
A wholesaler buys a TV from the manufacturer for ₹ 25,000. He marks the price of the TV 20% above his cost price and sells it to a retailer at 10% discount on the marked price. If the rate of VAT is 8%, find the :
(i) marked price.
(ii) retailer’s cost price inclusive of tax.
(iii) VAT paid by the wholesaler. (2015)
Solution:
S.P. for the manufactures = ₹ 25000
or C.P. for a whole seller a T.V. = ₹ 25000

C.P. for the retailer = ₹ 27000 + ₹ 2160 = ₹ 29160
(iii) VAT paid by the whole seller = ₹ 2160 – ₹ 2000 = ₹ 160

Question 13.
A dealer buys an article at a discount of 30 % from the wholesaler, the marked price being ₹ 6,000. The dealer sells it to a shopkeeper at a discount of 10% on the marked price. If the rate of VAT is 6%, find :
(i) the price paid by the shopkeeper including the tax.
(ii) the VAT paid by the dealer. (2016)
Solution:
Market price of the article = ₹ 6000
A dealer buys an article at a discount of 30% from the wholesaler.
Price of the article which the dealer paid to the wholesaler = 6000 – 30% of 6000

Amount of the article inclusive of sales tax at which the dealer bought it
= ₹ 4200 + ₹ 252 = ₹ 4452
Dealer sells the article at a discount of 10% to the shopkeeper.
Price of the article which the shopkeeper paid to the dealer = ₹ 6000 – 10% of 6000

Amount of the article inclusive of sales tax at which the shopkeeper bought it
= ₹ 5400 + ₹ 324 = ₹ 5724
The value added by dealer = ₹ 5400 – ₹ 4200 = ₹ 1200
Amount of VAT paid by dealer = 6% of 1200
= $$\frac { 6 }{ 100 }$$ x 1200 = ₹ 72
Price paid by shopkeeper including tax is ₹ 5724
VAT paid by dealer is ₹ 72

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 1 Value Added Tax Ex 1C 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.