Category: CBSE Class 10

RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.1
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS

Question 1.
How many balls, each of radius 1 cm, can be made from a solid sphere of lead of radius 8 cm ?
Solution:

Question 2.
How many spherical bullets each of 5 cm in diameter can be cast from a rectangular block of metal 11dm x 1m x 5 dm?
Solution:
Volume of reactangular block of metal
(V₁) = 11 dm x 1 m x 5 dm
= 11 dm x 10 dm x 5 dm = 550 dm³
Diameter of spherical bullet = 5 cm

Question 3.
A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of the two balls are 1.5 cm and 2 cm respectively. Determine the diameter of the third ball.
Solution:

Question 4.
2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.
Solution:

Question 5.
What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm ?
Solution:
Length of hollow cylinder = 16 cm
External diameter = 20 cm
and thickness = 2.5 mm = 0.25 cm
∴External radius (R) = \(\frac { 20 }{ 2 }\) = 10 cm
and internal radius (r) = 10 – 0.25 = 9.75 cm
∴ Volume of the material used in hollow cylinder
= πh (R² – r²) = π x 16 (10² – 9.75²) cm²
= 16π (100.00 – 95.0625) = 16TC (4.9375) cm³
∴ Volume of solid cylinder = 16 x 4.93 75πcm3 Diameter = 2 cm
∴Radius (r) = \(\frac { 2 }{ 2 }\) = 1 cm
Let h be the height of the cylinder, then
πr²h = 16 x 4.9375π
⇒ π (1)² h = 16 x 4.93 75π
h = 16 x 4.9375 = 79
Hence height = 79 cm

Question 6.
A cylindrical vessel having diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and jteight 21 cm which are filled completely. Find the diameter of the cylindrical vessel.
Solution:
Diameter of each small cylindrical vessels = 42 cm
∴Radius of each vessel (r) = \(\frac { 42 }{ 2 }\) = 21 cm
Height (h) = 21 cm
∴ Volume of each cylindrical vessal = πr²h = π (21 )2 (21) = 9261K cm³
and volume of both vessels = 2 x 9261π = 18522π cm³
Now volume of larger cylindrical vessel = 185 22π cm³
Let R be the radius of the vessel, then Height (H) = diameter = 2R,
∴Volume πR²H = πR² x 2R = 2πR²
∴ 2πR³ = 18522π
⇒ R³ = \(\frac { 18522 }{ 2 }\) = 9261
⇒ R³ = 9261 = (21 )³
∴ R = 21
∴Diameter = 2R = 2×21=42 cm

Question 7.
50 circular plates each of diameter 14 cm and thickness 0.5 cm are placed one above the other to form a right circular cylinder. Find its total surface area.
Solution:
Diameter of each circular plate = 14 cm
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm .
Thickness (h) = 0.5 cm
∴Height of 50 plates placing one above the other
= 0.5 x 50 = 25 cm
∴ Curved surface area of the cylinder so formed = 2πrh
= 2 x \(\frac { 2 }{ 7 }\) x 7 x 25 = 1100 cm2
and total surface area = curved surface area + 2 x base area
= 1100 + 2 x πr² =1100+ 2 x \(\frac { 22 }{ 2 }\) x 7 x 7
= 1100 + 308 = 1408 cm²

Question 8.
25 circular plates, each of radius 10.5 cm and thickness 1.6 cm, are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.
Solution:

Question 9.
Find the number of metallic circular discs with 1.5 cm base diameter and of height 0. 2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm. [NCERT Exemplar]
Solution:
Given that, lots of metallic circular disc to be melted to form a right circular cylinder. Here, a circular disc work as a circular cylinder.
Base diameter of metallic circular disc = 1.5 cm
∴Radius of metallic circular disc = \(\frac { 1.5 }{ 2 }\) cm [∵ diameter = 2 x radius]
and height of metallic circular disc i.e., = 0.2 cm
∴ Volume of a circular disc = π x (Radius)² x Height

Question 10.
How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm x 42 cm x 21 cm. [NCERT Exemplar]
Solution:

Question 11.
How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm. [NCERT Exemplar]
Solution:

Hence, the required number of spherical lead shots is 2541.

Question 12.
Three cubes of a metal whose edges are in the ratio 3:4:5 are melted and converted into a single cube whose diagonal is 12 \(\sqrt { 3 } \) cm. Find the edges of the three cubes. [NCERT Exemplar]
Solution:
Let the edges of three cubes (in cm) be 3x, 4x and 5x, respectively.
Volume of the cubes after melting is = (3x)³ + (4x)³ + (5x)³ = 216×3 cm³
Let a be the side of new cube so formed after melting. Therefore, a³ = 216x³
So, a = 6x,

Question 13.
A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed. [NCERT Exemplar]
Solution:
The volume of the solid metallic sphere  = \(\frac { 4 }{ 3 }\) π(10.5)³ cm³
Volume of a cone of radius 3.5 cm and height 3 cm = \(\frac { 1 }{ 3 }\) π (3.5)² x 3 cm³
Number of cones so formed

Question 14.
The diameter of a metallic sphere is equal to 9 cm. It is melted and drawn into a long wire of diameter 2 mm having uniform cross-section. Find the length of the wire.
Solution:

Question 15.
An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is  \(\frac { 1 }{ 4 }\) of the radius of the original ball, how many such balls are made ? Compare the surface area, of all the smaller balls combined together with that of the original ball.
Solution:
Let R be the radius of the original ball, then

Question 16.
A copper sphere of radius 3 cm is melted and recast into a right circular cone of height 3 cm. Find the radius of the base of the cone.
Solution:

Question 17.
A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.
Solution:

Question 18.
The diameters of internal and external surfaces of a hollow spherical shell are 10 cm and 6 cm respectively. If it is melted and recast into a solid cylinder of length of 2 \(\frac { 2 }{ 3 }\) cm, find the diameter of the cylinder.
Solution:

Question 19.
How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm x 10 cm x 7 cm ?
Solution:

Question 20.
The surface area of a solid metallic sphere is 616 cm². It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed. (Use π = 22/7). [CBSE 2010]
Solution:

Question 21.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied out on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap. [CBSE 2012]
Solution:

Question 22.
A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. Find the number of such cones formed. [CBSE 2014]
Solution:

Question 23.
A solid cuboid of iron with dimensions 53 cm x 40 cm x 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe. [CBSE 2015]
Solution:
Length of cuboid (l) = 53 cm
Breadth (b) = 40 cm
Height (h) = 15 cm

Volume =lbh= 53 x 40 x 15 cm³
= 31800 cm³
∴ Volume of cylindrical pipe = 31800 cm³
Inner diameter of pipe = 7 cm
and outer diameter = 8 cm
∴ Outer radius (R) = \(\frac { 8 }{ 2 }\) = 4 cm

Question 24.
The diameters of the internal and external surfaces of a holjow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder. [CBSE 2001C]
Solution:
Outer diameter of hollow spherical shell = 10 cm
and inner diameter = 6 cm
Outer radius (R) = \(\frac { 10 }{ 2 }\) = 5 cm 6
and inner radius (r) = \(\frac { 6 }{ 2 }\) = 3 cm

Question 25.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Calculate the height of the cone.
Solution:

Question 26.
A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone. 
Solution:
Interval radius of hollow sphere  (r) = 2m
and external radius (R) = 4m

Question 27.
A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball.
Solution:
Radius of big spherical ball (R) = 3 cm
∴ Volume = \(\frac { 4 }{ 3 }\) πR³ = \(\frac { 4 }{ 3 }\) n x (3)³ cm³
= \(\frac { 4 }{ 3 }\) π x 27 = 36π cm³
Similarly volume of ball of radius (r₁) = 1.5 cm
∴ Volume = \(\frac { 4 }{ 3 }\) π( 1.5)³

Question 28.
A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic metres of gravel are required to grave the path to a depth of 20 cm ?
Solution:
Diamter of pond = 40 m
∴ Radius (r) = \(\frac { 40 }{ 2 }\) =20 m
Width of path = 2 m

Question 29.
A 16 m deep well with diameter 3.5 m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7 m. Find the height of the platform.
Solution:

Volume of platform = Ibh = 27.5 x7 x h

Question 30.
A well of diameter 2 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to form an embankment of height 40 cm. Find the width of the embankment.
Solution:
Diameter of well = 2 m

Question 31.
A well with inner radius 4 m is dug 14 m deep. Earth taken out of it has been spread evenly all around a width of 3 m it to form an embankment. Find the height of the embankment.
Solution:

Question 32.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has>been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.
Solution:

Question 33.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm.
Solution:

Question 34.
A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:

Question 35.
Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen ? [Use π = 22/7]
Solution:

Question 36.
The rain water from a roof of dimensions 22 m x 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fills the cylindrical vessel, then find the rain in cm. [NCERT Exemplar]
Solution:

Question 37.
A conical flask is full of water. The flask has base-radius r and height h. The water is poured into a cylindrical flafisk of base- radius mr. Find the height of water in the cylindrical flask.
Solution:

Question 38.
A rectangular tank 15 m long and 11 m broad is required to receive entire liquid contents from a full cylindrical tank of internal diameter 21m and length 5 m. Find the least height of the tank that will serve the purpose.
Solution:

Question 39.
A hemispherical bowl of internal radius 9 cm is full of liquid. This liquid is to be fdled into cylindrical shaped small bottles each of diameter 3 cm and height 4 cm. How many bottles are necessary to empty the bowl ?
Solution:

Question 40.
A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball is dropped into the tub and the level of the water is raised by 6.75 cm. Find the radius of the ball.
Solution:

Question 41.
500 persons have to dip in a rectangular tank which is 80 m long and 50 m broad. What is the rise in the level of water inthe tank, if the average displacement of water by a person is 0.04 m3 ?
Solution:

Question 42.
A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres ?
Solution:

Question 43.
A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball of radius 9 cm is dropped into the tub and thus the level of water is raised by h cm. What is the value of h ?
Solution:

Question 44.
Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimension 16 cm x 8 cm x 8 cm when 16 spheres are packed the box is fdled with preservative liquid. Find the volume of this liquid. [Use π = 669/213]
Solution:

Question 45.
A vessel in the shape of a cuboid contains some water. If three identical spheres are immersed in the water, the level of water is increased by 2 cm. If the area of the base of the cuboid is 160 cm² and its height 12 cm, determine the radius of any of the spheres.
Solution:

Question 46.
150 spherical marbles, each of diameter 1.4 cm are dropped in a cylindrical vessel of diameter 7 cm contianing some water, which are completely immersed in water. Find the rise in the level of water in the vessel. [CBSE2014]
Solution:

Question 47.
Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which \((\frac { 2 }{ 5 } )\)² of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by Sushant? [CBSE 2014]
Solution:
Radius of vessel (R) = 2.5 cm
and height (h) = 11 cm

Question 48.
16 glass spheres each of radius 2 cm are packed into 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 water fdled in the box. |NCERT Exemplar]
Solution:
Given, dimensions of the cuboidal
= 16 cm x 8 cm x 8 cm
∴ Volume of the cuboidal =16x8x8
= 1024 cm³
Also, given radius of one glass sphere = 2cm
∴ Volume of one glass sphere = \((\frac { 4 }{ 3 } )\)πr³
= \((\frac { 4 }{ 3 } )\) x \((\frac { 22 }{ 7 } )\) x (2)³
= \((\frac { 704 }{ 21 } )\) = 33.523 cm³
Now, volume of 16 glass spheres = 16 x 33.523 = 536.37 cm³
∴ Required volume of water = Volume of cuboidal – Volume of 16 glass spheres
= 1024 – 536.37 = 487.6 cm³

Question 49.
Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour? [NCERT Exemplar]
Solution:
Given, radius of tank, r₁ = 40 cm
Let height of water level in tank in half an hour = h₁
Also, given internal radius of cylindrical pipe, r₁ = 1 cm
and speed of water = 80 cm/s /., in 1 water flow = 80 cm
∴In 30 (min) water flow = 80 x 60 x 30 = 144000 cm
According to the question,
Volume of water in cylindrical tank = Volume of water flow the circular pipe in half an hour
⇒ πr₁²h₁ = πr₂²h₂
⇒ 40 x 40 x h₁ = 1 x 1 x 14400
∴ h₁ = \((\frac {144000 }{ 40 x 40 } )\) = 90 cm
Hence, the level of water in cylindrical tank rises 90 cm in half an hour.

Question 50.
Water in a canal 1.5 m wide and 6 m deep is flowing with a speed of 10km/hr. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired ?
Solution:

Question 51.
A famer runs a pipe of internal diameter 20 cm from the canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled ?
Solution:

Question 52.
A cylindrical tank full of water is emptied by a pipe at the rate of 225 litres per minute. How much time will it take to empty half the tank, if the diameter of its base is 3 m and its height is 3.5 m? [Use π = 22/7] [ [CBSE 2014]
Solution:

Question 53.
A cylindrical tank full of water is emptied by a pipe at the rate of 225 litres per minute. How much time will it take to empty half the tank, if the diameter of its base is 3 m and its height is 3.5 m? [Use π  = 22/7] [CBSE 2014]
Solution:

Question 54.
Water flows at the rate of 15 km/hr through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?
[NCERT Exemplar]
Solution:
Given, length of the pond = 50 m and width of the pond = 44 m

Question 55.
A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing with a speed of 20 km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired? [NCERT Exemplar]
Solution:
Volume of water flows in the canal in one hour = width of the canal x depth of the canal x speed of the canal water
= 3 x 1.2 x 20 x 1000 m³ = 72000 m³
In 20 minutes the volume of water
= \(\frac { 72000\times 20 }{ 60 }\) m³ = 24000 m³
Area irrigated in 20 minutes, if 8 cm,
i. e., 0.08 m standing water is required
= \((\frac { 2 }{ 5 } )\)
= \((\frac { 2 }{ 5 } )\) m²
= 300000 m²= 30 hectares.

Question 56.
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm2, find the volume of cylinder. (Use re = 22/7) [CBSE 2016]
Solution:
Let the radius and height of cylinder be r and h respectively.
r + h = 37 cm …(i) [given]
Total surface area of cylinder = 1628 cm²
2πr(r + h) = 1628
⇒ 2πr(37) = 1628

Question 57.
A tent of height 77 dm is in the form a right circular cylinder of diameter 36 m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at ₹3.50 per m². [Use π = 22/7]
Solution:
Total height of the tent = 77 dm
Height of the cylindrical part (h₁) = 44 dm = 4.4 m
Height of conical part (h₂) = 77 – 44 = 33 dm = 3.3 m
Diameter of the base of the tent = 36 m

Question 58.
The largest sphere is to be curved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.
Solution:
Largest sphere to be curved must have its diameter = diameter of cylinder = height of cylinder

Question 59.
A right angled triangle whose sides are 3 cm, 4 cm and 5 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two cones so formed. Also, find their curved surfaces.
Solution:
In ∆ABC, ∠B = 90°
and sides are 3 cm, 4 cm and 5 cm
Here diagonal is 5 cm

(i) When it is revolved along BC i.e., 4 cm side, then
Radius of the base of the cone (r₁) = 3 cm
and height (h₁) = 4 cm and slant height l = 5 cm
∴ Volume = \((\frac { 2 }{ 5 } )\) πr2h = \((\frac { 1 }{ 3 } )\) π x 3 x 3 x 4 cm³
= 12π cm³
and surface area = πrl = π x 3 x 5 cm2 = 157π cm2
(ii) When it is revolved along 3 cm side, then
r = 4 cm and h = 3 cm and l = 5 cm
∴ Volume = \((\frac { 1 }{ 3 } )\) πr2h =-\((\frac { 1 }{ 3 } )\) π x 4 x 4 x 3 cm3
= 16πcm³
and curved surface area = πrl
= π x 4 x 5 = 20n cm²
∴ Difference of their volumes = I6π- 12π = 4πcm³

Question 60.
A 5 m wide cloth is used to make a conical tent of base diameter 14 ir. and height 24 m. Find the cost of cloth used at the rate of ₹25 per metre. [Use π = 22/7] [CBSE 2014]
Solution:
Diameter of base of conical tent = 14 m
Radius (r) = \((\frac { 14 }{ 3 } )\)= 7 cm
Height (h) = 24 m

Question 61.
The volume of a hemi-sphere is 2425 \((\frac { 1 }{ 2 } )\) cm³, find its curved surface area. [2012]
Solution:

Question 62.
The difference between the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm². If the volume of nretal used in making the cylinder is 176 cm³, find the outer and inner diameters of the cylinder. [CBSE 2010]
Solution:
Height of hollow right cylinder = 14 cm
Difference between outer and inner curved surface area = 88 cm²
and volume of metal used =176 cm³

Question 63.
The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 cm² of the surface is 10 paise. Find the total cost to paint the vessel ail over.
Solution:
Outer diameter of the hemispherical vessel = 25.2 cm
and inner diameter = 21 cm

Question 64.
Prove that the surface area of a sphere is equal to the curved surface area of the circumscribed cylinder.
Solution:
Let r be the radius of the sphere, then surface area = 4πr² ….(i)
Then radius of the circumscribed cylinder = radius of the sphere = r

Question 65.
If the total surface area of a solid hemisphere is 462 cm², find its volume (Take π = 22/7).
Solution:

Question 66.
Water flows at the rate of 10 m / minutes through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm? [NCERT Exemplar]
Solution:
Given, speed of water flow = 10 m min^-1 = 1000 cm/min

Question 67.
A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone. [NCERT Exemplar]
Solution:
(i) Whenever we placed a solid right circular cone in a right circular cylinder with full of water, then volume of a solid right circular cone is equal to the volume of water failed from the cylinder.
(ii) Total volume of water in a cylinder is equal to the volume of the cylinder.
(iii) Volume of water left in the cylinder = Volume of the right circular cylinder – Volume of a right circular cone.

Now, given that
Height of a right circular cone = 120 cm
Radius of a right circular cone = 60 cm
∴Volume of a right circular cone = \((\frac { 1 }{ 3 } )\)πr² x h
= \((\frac { 1 }{ 3 } )\) x \((\frac { 22 }{ 7 } )\) x 60 x 60 x 120
=\((\frac { 22 }{ 7 } )\) x 20 x 60 x 120
= 144000π cm³
∴Volume of a right circular cone = Volume of water failed from the cylinder = 1440007c cm3 [from point (i)]
Given that, height of a right circular cylinder = 180 cm
and radius of a right circular cylinder = Radius of a right circular cone = 60 cm
∴Volume of a right circular cylinder = πr² x h
= π x 60 x 60 x 180 = 64800071 cm3 So, volume of a right circular cylinder = Total volume of water in a cylinder = 64800071 cm3 [from point (ii)]
From point (iii),
Volume of water left in the cylinder = Total volume of water in a cylinder – Volume of water failed from the cylinder when solid cone is placed in it
= 648000π- 144000π
= 504000π = 504000 x \((\frac { 22 }{ 7 } )\) = 1584000 cm³ 1584000
= \((\frac { 1584000 }{ (10)6 } )\) m³ = 1.584 m³
Hence, the required volume of water left in the cylinder is 1.584 m³.

Question 68.
A heap of rice in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of rice. How much canvas cloth is required to cover the heap? [NCERT Exemplar]
Solution:
Given that, a heap of rice is in the form of a cone.
Height of a heap of rice i.e., cone (h) = 3.5m
and diameter of a heap of rice i. e., cone = 9 m
Radius of a heap of rice i.e., cone (r) = \((\frac { 9 }{ 2 } )\) m

Question 69.
A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the concial heap is 24 cm, find the radius and slant height of the’ heap. [NCERT Exemplar]
Solution:
Given, radius of the base of the bucket = 18 cm
Height of the bucket = 32 cm
So, volume of the sand in cylindrical bucket
= πr²h = π (18)²x 32= 10368π
Also, given height of the conical heap (h) = 24 cm
Let radius of heap be r cm
Then, volume of the sand in the heap = \((\frac { 1 }{ 3 } )\) πr²h

Question 70.
A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl? [NCERT Exemplar]
Solution:
Given, radius of hemispherical bowl, r = 9cm
and radius of cylindrical bottles, R = 1.5 cm
and height, h = 4 cm
∴Number of required cylindrical bottles

Question 71.
A factory manufactures 120,000 pencils daily. The pencils are cylindrical in shape each of length 25 cm and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at ₹0.05 per dm². [NCERT Exemplar]
Solution:
Given, pencils are cylindrical in shape.
Length of one pencil = 25 cm
and circumference of base = 1.5 cm

Now, cost of colouring 1 dm2 curved surface of the pencils manufactured in one day = ₹0.05
∴ Cost of colouring 45000 dm2 curved surface = ₹2250

Question 72.
The\((\frac { 3 }{ 4 } )\)th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.
Solution:
Let height of water in cylindrical vessel = h cm
Volume of water (in cylinder) = \((\frac { 3 }{ 4 } )\) Volume of water (in come)
∵ Volume of cylinder = nr²h & Volume of cone = \((\frac { 1 }{ 3 } )\) πr²h

Hence, height of water in cylindrical vessel (h) = 1.5 cm

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.1 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
Write the discriminant of the following quadratic equations :
(i) 2x² – 5x + 3 = 0
(ii) x² + 2x + 4 = 0
(iii) (x – 1) (2x – 1) = 0
(iv) x² – 2x + k = 0, k ∈ R
(v) √3 x² + 2√2 x – 2√3 = 0
(vi) x² – x + 1 = 0
Solution:

Question 2.
In the following, determine whether the given quadratic equations have real roots and if so, And the roots :
(i) 16x² = 24x + 1
(ii) x² + x + 2 = 0
(iii) √3 x² + 10x – 8√3 = 0
(iv) 3x² – 2x + 2 = 0
(v) 2x² – 2√6 x + 3 = 0
(vi) 3a²x² + 8abx + 4b² = 0, a ≠ 0
(vii) 3x² + 2√5 x – 5 = 0
(viii) x² – 2x + 1 = 0
(ix) 2x² + 5√3 x + 6 = 0
(x) √2 x² + 7x + 5√2= 0 [NCERT]
(xi) 2x² – 2√2 x + 1 = 0 [NCERT]
(xii) 3x² – 5x + 2 = 0 [NCERT]
Solution:

Question 3.
Solve for x :

Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5 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.

RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.1
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.2
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.3
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.4
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex VSAQS
• RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

Mark the correct alternative in each of the following:
Question 1.
Which of the following is not a measure of central tendency :
(a) Mean
(b) Median
(c) Mode
(d) Standard deviation
Solution:
Standard deviation is not a measure of central tendency. Only mean, median and mode are measures. (d)

Question 2.
The algebraic sum of the deviations of a frequency distribution from its mean is
(a) always positive
(b) always negative
(c) 0
(d) a non-zero number
Solution:
The algebraic sum of the deviations of a frequency distribution from its mean is zero
Let x₁, x₂, x₃, …… x_n are observations and \(\overline { X }\) is the mean

Question 3.
The arithmetic mean of 1, 2, 3, ….. , n is

Solution:
Arithmetic mean of 1, 2, 3, …… n is

Question 4.
For a frequency distribution, mean, median and mode are connected by the relation
(a) Mode = 3 Mean – 2 Median
(b) Mode = 2 Median – 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean
Solution:
The relation between mean, median and mode is: Mode = 3 Median – 2 Mean (c)

Question 5.
Which of the following cannot be determined graphically ?
(a) Mean
(b) Median
(c) Mode
(d) None of these
Solution:
Mean cannot be determind graphically, (a)

Question 6.
The median of a given frequency distribution is found graphically with the help of
(a) Histogram
(b) Frequency curve
(c) Frequency polygon
(d) Ogive
Solution:
Median of a given frequency can be found graphically by an ogive, (d)

Question 7.
The mode of a frequency distribution can be determined graphically from
(a) Histogram
(b) Frequency polygon
(c) Ogive
(d) Frequency curve
Solution:
Mode of frequency can be found graphically by an ogive, (c)

Question 8.
Mode is
(a) least frequent value
(b) middle most value
(c) most frequent value
(d) None of these
Solution:
Mode is the most frequency value of observation or a class, (c)

Question 9.
The mean of n observations is \(\overline { X }\) . If the first item is increased by 1, second by 2 and so on,
then the new mean is

Solution:
Mean of n observations = \(\overline { X }\)
By adding 1 to the first item, 2 to second item and so on, the new mean will be
Let x₁, x₂, x_3,…..  x_n are the items whose mean is \(\overline { X }\) , then mean of
(x₁+ 1) + (x₂ + 2) + (x₃ + 3) + …… (x_n + n)

Question 10.
One of the methods of determining mode is
(a) Mode = 2 Median – 3 Mean
(b) Mode = 2 Median + 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean
Solution:
Mode = 3 Median – 2 Mean (c)

Question 11.
If the mean of the following distribution is 2.6, then the value of y is

(a) 3
(b) 8
(c) 13
(d) 24
Solution:

Question 12.
The relationship between mean, median and mode for a moderately skewed distribution is
(a) Mode = 2 Median – 3 Mean
(b) Mode = Median – 2 Mean
(c) Mode = 2 Median – Mean
(d) Mode = 3 Median – 2 Mean
Solution:
The relationship between mean, median and mode is Mode = 3 Median – 2 Mean, (d)

Question 13.
The mean of a discrete frequency distribution x_i /f_i ; i= 1, 2, …… n is given by

Solution:

Question 14.
If the arithmetic mean of x, x + 3, x + 6, x + 9, and x + 12 is 10, then x =
(a) 1
(b) 2
(c) 6
(d) 4
Solution:

Question 15.
If the median of the data : 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5, then x =
(a) 27
(b) 25
(c) 28
(d) 30
Solution:

Question 16.
If the median of the data : 6, 7, x – 2, x, 17, 20, written in ascending order, is 16. Then x =
(a) 15
(b) 16
(c) 17
(d) 18
Solution:

Question 17.
The median of first 10 prime numbers is
(a) 11
(b) 12
(c) 13
(d) 14
Solution:

Question 18.
If the mode of the data : 64,60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =
(a) 44
(b) 45
(c) 46
(d) 48
Solution:
Mode of 64, 60, 48, x, 43, 48, 43, 34 is 43
∵ By definition mode is a number which has maximum frequency which is 43
∴ x = 43
∴ x + 3 = 43 + 3 = 46 (c)

Question 19.
If the mode of the data : 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
(a) 15
(b) 16
(c) 17
(d) 19
Solution:
Mode of 16, 15, 17, 16, 15, x, 19, 17, 14 is 15
∵By definition mode of a number which has maximum frequency which is 15
∴ x = 15 (a)

Question 20.
The mean of 1, 3, 4, 5, 7, 4 is m. The number 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then p + q =
(a) 4
(b) 5
(c) 6
(d) 7
Solution:

Question 21.
If the mean of a frequency distribution is 8.1 and Σf_ix_i = 132 + 5k, Σf_i = 20, then k =
(a) 3
(b) 4
(c) 5
(d) 6
Solution:

Question 22.
If the mean of 6, 7, x, 8, y, 14 is 9, then
(a)x+y = 21
(b)x+y = 19
(c) x -y = 19
(d) v -y = 21
Solution:

Question 23.
The mean of n observations is \(\overline { x }\) If the first observation is increased by 1, the second by 2, the third by 3, and so on, then the new mean is

Solution:

Question 24.
If the mean of first n natural numbers is \(\frac { 5n }{ 9 }\) then n =
(a) 5
(b) 4
(c) 9
(d) 10
Solution:

Question 25.
The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is
(a) 25
(b) 18
(c) 20
(d) 22
Solution:
Arithmetic mean = 24
Mode = 12
∴ But mode = 3 median – 2 mean
⇒ 12 = 3 median – 2 x 24
⇒ 12 = 3 median =-48
⇒ 12 + 48 = 3 median
⇒ 3 median = 60
Median = \(\frac { 60 }{ 3 }\) = 20 (c) 

Question 26.
The mean of first n odd natural number is

Solution:

Question 27.
The mean of first n odd natural numbers is \(\frac { n2 }{ 81 }\) , then n = 81
(a) 9
(b) 81
(c) 27
(d) 18
Solution:

Question 28.
If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36
Solution:
Difference of mode and median = 24
Mode = 3 median – 2 mean
⇒ Mode – median = 2 median – 2 mean
⇒ 24 = 2 (median – mean)
⇒ Median – mean = \(\frac { 24 }{ 2 }\) = 12 (a)

Question 29.
If the arithmetic mean of 7, 8, x, 11, 14 is x, then x =
(a) 9
(b) 9.5
(c) 10
(d) 10.5
Solution:

Question 30.
If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a) 4
(b) 8
(c) 6
(d) 10
Solution:
Mode of a series = Its mean + 12
Mean = mode – 12
Mode = 3 median – 2 mean
Mode = 3 median – 2 (mode -12)
⇒ Mode = 3 median – 2 mode + 24
⇒ Mode + 2 mode – 3 median = 24
⇒ 3 mode – 3 median = 24
⇒ 3 (mode – median) = 24
⇒ Mode – medain = \(\frac { 24 }{ 3 }\) = 8 (b)

Question 31.
If the mean of first n natural number is 15, then n =
(a) 15
(b) 30
(c) 14
(d) 29
Solution:

Question 32.
If the mean of observations x₁, x₂, …, x_n is \(\overline { x }\) , then the mean of x₁ + a, x₂ + a,…, x_n + a is

Solution:

Question 33.
Mean of a certain number of observations is \(\overline { x }\) If each observation is divided by m (m ≠ 0) and increased by n, then the mean of new observation is

Solution:
Mean of some observations = \(\overline { x }\)
If each observation is divided by m and increased by n
Then mean will be = \(\frac { \overline { x } }{ m }\) +n

Question 34.
If u_i= \(\frac { xi-25\quad }{ 10 }\) Σf_iu_i = 20, Σf_i = 100, then \(\overline { x }\)
(a) 23
(b) 24
(c) 27
(d) 25
Solution:

Question 35.
If 35 is removed from the data : 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by
(a) 2
(b) 1.5
(c) 1
(d) 0.5
Solution:

Question 36.
While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes.
(b) centred at the class marks of the classes.
(c) centred at the upper limit of the classes.
(d) centred at the lower limit of the classes.
Solution:
In computing the mean of grouped data, the frequencies are centred at the class marks of the classes. (b)

Question 37.

Solution:

Question 38.
For the following distribution:

the sum of the lower limits of the median and modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Solution:

Now, \(\frac { N }{ 2 }\) = \(\frac { 66 }{ 2 }\) = 33, which lies in the interval 10-15.
Therefore, lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 15-20.
Therefore, lower limit of modal class is 15.
Hence, required sum is 10 + 15 = 25. (b)

Question 39.
For the following distribution:

the modal class is
(a) 10-20
(b) 20-30
(c) 30-40
(d) 50-60
Solution:

Here, we see that the highest frequency is 30, which lies in the interval 30-40. (c)

Question 40.
Consider the following frequency distribution:

The difference of the upper limit of the median class and the lower limit of the modal class is
(a) 0
(b) 19
(c) 20
(d) 38
Solution:

Here, \(\frac { N }{ 2 }\) = \(\frac { 67 }{ 2 }\) = 33.5 which lies in the interval 125-145.
Hence, upper limit of median class is 145.
Here, we see that the highest frequency is 20 which lies in 125-145.
Hence, the lower limit of modal class is 125.
∴ Required difference = Upper limit of median class – Lower limit of modal class
= 145 – 125 = 2 (C)

Question 41.
In the formula \(\overline { X }\) = a + \(\frac { \Sigma fidi }{ \Sigma fi }\) for finding the mean of grouped data di’s are deviations from a of
(a) lower limits of classes
(b) upper limits of classes
(c) mid-points of classes
(d) frequency of the class marks
Solution:
We know that, d_i = x_i – a
i .e , d_i‘s are the deviation from a mid-points of the classes. (c)

Question 42.
The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data gives its
(a) mean
(b) median
(c) mode
(d) all the three above
Solution:
Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa. (b)

Question 43.
Consider the following frequency distribution:

The upper limit of the median class is
(a) 17
(b) 17.5
(c) 18
(d) 18.5
Solution:
Given, classes are not continuous, so we make continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.

Here, \(\frac { N }{ 2 }\) = \(\frac { 57 }{ 2 }\) = 28.5, which lies in the interval 11.5-17.5.
Hence, the upper limit is 17.5.

Hope given RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 15 Statistics Ex VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 15 Statistics VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.1
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.2
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.3
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.4
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex VSAQS
• RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

Question 1.
Define Mean.
Solution:
The mean of a set of observations is equal to their sum divided by the total number of observations. Mean is also called an average.

Question 2.
What is the algebraic sum of deviations of a frequency distribution about its mean ?
Solution:
The algebraic sum of deviation of a frequency distribution about its mean is zero.

Question 3.
Which measure of central tendency is given by the x-coordinates of the point of intersection of the ‘more than’ ogive and ‘less than’ ogive ? (C.B.S.E. 2008)
Solution:
Median is given by the x-coordinate of the point of intersection of the more than ogive and less than ogive.

Question 4.
What is the value of the median of the data using the graph in the following figure of less than ogive and more than ogive ?

Solution:
Median = 4, because the coordinates of the point of intersection of two ogives at x-axis is 4.

Question 5.
Write the empirical relation between mean, mode and median.
Solution:
The empirical relation is Mode = 3
Median – 2 Mean

Question 6.
Which measure of central tendency can be determined graphically ?
Solution:
Median can be determinded graphically.

Question 7.
Write the modal class for the following frequency distribution:

Solution:
The modal class is 20-25 as it has the maximum frequency of 75 in the given distribution.

Question 8.
A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the students of the class.

Solution:
Median marks
Here N = 40, then \(\frac { N }{ 2 }\) = \(\frac { 40 }{ 2 }\) = 20
From 20 on y-axis, draw a line parallel to the x-axis meeting the curve at P and from P, draw a perpendicular on x-axis meeting it at M. Then M is the median which is 50.

Question 9.
Write the median class for the following frequency distribution:

Solution:

Here N = 100, then \(\frac { N }{ 2 }\) = 50
Which lies in the class 40-50 (∵32 < 50 < 60)
∴ Required class interval is 40-50

Question 10.
In the graphical representation of a frequency distribution, if the distance between mode and mean isk times the distance between median and mean, then write the value of k.
Solution:
We know that
Mode = 3 median – 2 mean ….(i)
Now mode – mean = k (median – mean) , ….(ii)
But mode – mean = 3 median – 2 mean [from (i)]
⇒ Mode – mean = 3 (median – mean) ….(iii)
Comparing (ii) and (iii)
k = 3

Question 11.
Find the class marks of classes 10-25 and 35-55. (C.B.S.E. 2008)
Solution:

Question 12.
Write the median class of the following distribution :

Solution:


Here n = 50
∴ Median = \(\frac { n + 1 }{ 2 }\) = \(\frac { 5 + 1 }{ 2 }\) = 25.5 which lies in the class 30-40
Hence median class = 30-40.

Hope given RD Sharma Class 10 Solutions Chapter 15 Statistics VSAQS 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.

RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Solve the following quadratic equations by factorization.
Question 1.
(x – 4) (x + 2) = 0
Solution:
(x – 4) (x + 2) = 0
Either x – 4 = 0, then x = 4
or x + 2, = 0, then x = -2
Roots are x = 4, -2

Question 2.
(2x + 3) (3x – 7) = 0
Solution:

Question 3.
3x² – 14x – 5 = 0 (C.B.S.E. 1999C)
Solution:

Roots are x = 5, \(\frac { -1 }{ 3 }\)

Question 4.
9x² – 3x – 2 = 0
Solution:

Question 5.

Solution:

Question 6.
6x² + 11x + 3 = 0
Solution:

Question 7.
5x² – 3x – 2 = 0
Solution:

Question 8.
48x² – 13x – 1 =0
Solution:

Roots are x = \(\frac { 1 }{ 3 }\) , \(\frac { -1 }{ 16 }\)

Question 9.
3x² = – 11x – 10
Solution:

Question 10.
25x (x + 1) = -4
Solution:

Question 11.
16x – \(\frac { 10 }{ x }\) = 27 [CBSE 2014]
Solution:

Question 12.

Solution:

Question 13.
x – \(\frac { 1 }{ x }\) = 3, x ≠ 0 [NCERT, CBSE 2010]
Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
a²x² – 3abx + 2b² = 0
Solution:

Question 17.
9x² – 6b²x – (a⁴ – b⁴) = 0 [CBSE 2015]
Solution:

Question 18.
4x² + 4bx – (a² – b²) = 0
Solution:

Question 19.
ax² + (4a² – 3b)x- 12ab = 0
Solution:

Question 20.
2x² + ax – a² = 0
Solution:

Question 21.

Solution:

x² = 16
x = ±4

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Roots are 4, \(\frac { -2 }{ 9 }\)

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.

Solution:

Question 34.

Solution:

Question 35.

Solution:

Question 36.

Solution:

Question 37.

Solution:

Question 38.

Solution:

Question 39.

Solution:

Question 40.

Solution:

Question 41.
x² – (√2 + 1) x + √2 = 0
Solution:

Question 42.
3x² – 2√6x + 2 = 0 [NCERT, CBSE 2010]
Solution:

Question 43.
√2 x² + 7x + 5√2 = 0
Solution:

Question 44.
\(\frac { m }{ n }\) x² + \(\frac { n }{ m }\) = 1 – 2x
Solution:

Question 45.

Solution:

Question 46.

Solution:

Question 47.

Solution:

Question 48.

Solution:

Question 49.

Solution:

Question 50.
x² + 2ab = (2a + b) x
Solution:

Question 51.
(a + b)² x² – 4abx – (a – b)2 = 0
Solution:

Question 52.
a (x² + 1) – x (a² + 1) = 0
Solution:

Question 53.
x² – x – a (a + 1) = 0
Solution:

Question 54.
x² + (a + \(\frac { 1 }{ a }\)) x + 1 = 0
Solution:

Question 55.
abx² + (b² – ac) x – bc = 0 (C.B.S.E. 2005)
Solution:

Question 56.
a²b²x² + b²x – a²x – 1 = 0 (C.B.S.E. 2005)
Solution:

Question 57.

Solution:

Question 58.

Solution:

Question 59.

Solution:


⇒ x = 0
x = 0, -7

Question 60.

Solution:

Question 61.

Solution:

Question 62.

Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3 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.

RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6

Other Exercises

• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.1
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.2
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.3
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.4
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex VSAQS
• RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

Question 1.
Draw an ogive by less than method for the following data :

Solution:

Take numbers of rooms along the x-axis and c.f along the y-axis
Plot the points (1, 4), (2, 13), (3, 35), (4, 63), (5, 87), (6, 99), (7, 107), (8, 113), (9, 118) and (10, 120) on the graph and join them and with free hand to get an ogive as shown. This is the less than ogive.

Question 2.
The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

Prepare a cumulative frequency table by less than method and draw on ogive.
Solution:

Take marks along ,t-axis and no. of students (c.f) along 3 -axis. Now plot the points (640, 16), (680, 61), (720, 217), (760, 501), (800, 673), (840, 732) and (880, 750) on the graph and join them with free hand. This is the less than ogive.

Question 3.
Draw an ogive to represent the following frequency distribution:

Solution:
Representing the classes in exclusive form :


Represent class intervals along x-axis and c.f. along y-axis. Now plot the points (4.5, 2), (9.5, 8), (14.5, 18), (19.5, 23) and (24.5, 26) on the graph and join then in free hand to get an ogive as shown.

Question 4.
The monthly profits (in Rs.) of 100 shops are distributed as follows:

Draw the frequency polygon for it.
Solution:


Represent profits per shop along x-axis and no. of shop (c.f.) along y-axis.
Plot the points (50, 12), (100, 30), (150, 57), (200, 77), (250, 94) and (300, 100) on the graph and join them with ruler. This is the cumulative polygon as shown.

Question 5.
The following distribution gives the daily income of 50 workers of a factory:

Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive.
Solution:


Now plot the points (120, 12), (140, 26), (160, 34), (180, 40) and (200,50) on the graph and join them with free hand to get an ogive which is less than.

Question 6.
The following table gives production yield per hectare of wheat of 100 farms of a village:

Draw ‘less than’ ogive and ‘more than’ ogive.
Solution:
(i) Less than

Now plot the points (55, 2), (60, 10), (65, 22), (70, 46), (75, 84) and (80, 100) on the graph and join them in free hand to get a less than ogive.
(ii) More than

Now plot the points (50, 100), (55, 84), (60, 46), (65, 22), (70, 10), (75, 2) and (80, 0) on the graph and join than in free hand to get a more than ogive as shown below :

Question 7.
During the medical check-up of 35 students of a class, their weights were recorded as follows :

Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula. (C.B.S.E. 2009)
Solution:

Plot the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35) on the graph and join them in free hand to get an ogive as shown.
Here N = 35 which is odd
∴ \(\frac { N }{ 2 }\) = \(\frac { 25 }{ 2 }\) = 17.5
From 17.5 on y-axis draw a line parallel to x-axis meeting the curve at P. From P, draw PM ⊥ x-axis
∴ Median which is 46.5 (approx)
Now N = 17.5 lies in the class 46 – 48 (as 14 < 17.5 < 28)
∴ 46-48 is the median class
Here l= 46, h = 2,f= 14, F= 14

Question 8.
The annual rainfall record of a city for 66 days is given in the following tab

Calculate the median rainfall using ogives of more than type and less than type. [NCERT Exemplar]
Solution:
We observe that, the annual rainfall record of a city less than 0 is 0. Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0-10. So, the total annual rainfall record of a city for less than 10 cm is 0 + 22 = 22 days. Continuing in this manner, we will get remaining less than 20, 30, 40, 50 and 60.
Also, we observe that annual rainfall record of a city for 66 days is more than or equal to 0 cm. Since, 22 days lies in the interval 0-10. So, annual rainfall record for 66 – 22 = 44 days is more than or equal to 10 cm. Continuing in this manner we will get remaining more than or equal to 20, 30 , 40, 50, and 60.
Now, we construct a table for less than and more than type.

To draw less than type ogive we plot the points (0,0), (10,22), (20,32), (30, 40), (40, 55), (50, 60), (60, 66) on the paper and join them by free hand.
To draw the more than type ogive we plot the points (0, 66), (10, 44), (20, 34), (30, 26), (40, 11), (50, 6) and (60, 0) on the graph paper and join them by free hand.

∵ Total number of days (n) = 66
Now, \(\frac { n }{ 2 }\) = 33
Firstly, we plot a line parallel to X-axis at intersection point of both ogives, which further intersect at (0, 33) on Y- axis. Now, we draw a line perpendicular to X-axis at intersection point of both ogives, which further intersect at (21.25, 0) on X-axis. Which is the required median using ogives.
Hence, median rainfall = 21.25 cm.

Question 9.
The following table gives the height of trees:

Draw ‘less than’ ogive and ‘more than’ ogive.
Solution:
(i) First we prepare less than frequency table as given below:

Now we plot the points (7, 26), (14, 57), (21, 92), (28, 134), (35, 216), (42, 287), (49, 341), (56, 360) on the graph and join then in a frequency curve which is ‘less than ogive’
(ii) More than ogive:
First we prepare ‘more than’ frequency table as shown given below:

Now we plot the points (0, 360), (7, 341), (14, 287), (21, 216), (28, 134), (35, 92), (42, 57), (49, 26), (56, 0) on the graph and join them in free hand curve to get more than ogive.

Question 10.
The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:

Draw both ogives for the above data and hence obtain the median.
Solution:

Now plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) on the graph and join them to get a more than curve.
Less than curve:

Now plot the points (10, 3), (15, 7), (20, 10), (25, 14), (30, 16), (35, 28) and (40, 30) on the graph and join them to get a less them ogive. The two curved intersect at P. From P, draw PM 1 x-axis, M is the median which is 22.5
∴ Median = Rs. 22.5 lakh

Hope given RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer. [NCERT]
Solution:
Let the first number = x
Then second number = x + 1
Their product = 306
x (x + 1) = 306
=> x² + x – 306 = 0
Required quadratic equation will be x² + x – 306 = 0

Question 2.
John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John had x marbles.
Solution:
No. of marbles John and Jivanti have = 45
Let number of marbles John has = x
Then number of marbles Jivanti has = 45 -x
Every one lost 5 marbles, then John’s marbles = x – 5
and Jivanti’s marbles = 45 – x – 5 = 40 – x
According to the condition,
(x – 5) (40 – x) = 128
=> 40x – x² – 200 + 5x – 128 = 0
=> -x² + 45x – 328 = 0
=> x² – 45x + 328 = 0

Question 3.
A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation of find x.
Solution:
Let the number of toys in a day = x
Cost of each toy = x – 55
on a particular cost of production = Rs. 750
x (x – 55) = 750
=> x² – 55x – 750 = 0
Hence required quadratic equation will be x² – 55x – 750 = 0

Question 4.
The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.
Solution:
Let the base of a right triangle = x
Its height = x – 7
and hypotenuse = 13 cm
=> By Pythagoras Theorem
(Hypotenuse)² = (Base)² + (Height)²
(13 )² = x² + (x – 7)²
=> 169 = x² + x² – 14x + 49
=> 2x² – 14x + 49 – 169 = 0
=> 2x² – 14x – 120 = 0
=> x² – 7x – 60 = 0 (Dividing by 2)
Hence required quadratic equation will be x² – 7x – 60 = 0

Question 5.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.
Solution:
Distance between Mysore and Bangalore = 132 km
Let average speed of passenger train=x km/ hr
Then average speed of express train = (x + 11) km/hr

Question 6.
A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.
Solution:
Total distance = 360 km
Let the uniform speed of the train = x km/hr
Time taken = \(\frac { 360 }{ x }\)

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.2
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.3
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.4
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.5
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.6
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.7
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.8
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.9
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.11
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.12
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations VSAQS
• RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations MCQS

Question 1.
Which of the following are quadratic equations ?



Solution:

Question 2.
In each of the following, determine whether the given values are solutions of the given equation or not :

Solution:

Question 3.
In each of the following, find the value of k for which the given value is a solution of the given equation.

Solution:

Question 4.
Determine, if 3 is a root of the equation given below :

Solution:

Question 5.
If x = \(\frac { 2 }{ 3 }\) and x = -3 are the roots of the equation ax² + 7x + b = 0, find the values of a and b.
Solution:

 

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.1 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5

Other Exercises

• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.1
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.2
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.3
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.4
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.6
• RD Sharma Class 10 Solutions Chapter 15 Statistics Ex VSAQS
• RD Sharma Class 10 Solutions Chapter 15 Statistics MCQS

Question 1.
Find the mode of the following data :
(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
Solution:

We see that 5 occurs in maximum times which is 5
∴ Mode = 5

we see that 3 occurs in maximum times i.e. 5
∴ Mode = 3


Here we see that 15 occurs in maximum times i.e. 54
∴ Mode = 15

Question 2.
The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows :

Find the model shirt size worn by the group.
Solution:

We see that frequency of 40 is maximum which is 41
∴Mode = 40

Question 3.
Find the mode of the following distribution.

Solution:

Question 4.
Compare the modal ages of two groups of students appearing for an entrance test :

Solution:
(i) For group A

We see that class 18-20 has the maximum frequency
∴ It is a modal class

(ii)  For group B

We see that class 18-20 has the maximum frequency
∴ It is a modal class

Question 5.
The marks in science of 80 students of class X are given below: Find the mode of the marks obtained by the students in science.

Solution:

We see that class 50-60 has the maximum frequency 20
∴ It is a modal class

Question 6.
The following is the distribution of height of students of a certain class in a certain city :

Find the average height of maximum number of students.
Solution:
Writing the classes in exclusive form,

Here model class is 165.5 – 168.5
and l = 165.5, h = 3, f= 142, f₁= 118, f₂= 127

Question 7.
The following table shows the ages of the patients admitted in a hospital during a year :

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:

(i) We see that class 35-45 has the maximum frequency 23
∴ It is a modal class

= 35.375 = 35.37 years
We see that mean is less than its mode

Question 8.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.
Solution:


We see that class 60-80 has the maximum frequency 61
∴ It is the modal class
Here l = 60, f= 6, f₁ = 52, f₂ = 38,h = 20
∴ Modal of life time (in hrs.)

Question 9.
The following table gives the daily income of 50 workers of a factory :

Find the mean, mode and median of the above data. (C.B.S.E. 2009)
Solution:

Here i = 20 and AM = 150

(ii) Max frequency  = 14
∴ Model class = 120 – 140

Question 10.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:

Solution:
Let assumed mean (A) = 32.5

(i)  We see that the class 30-35 has the maximum frequency
∴ It is the modal class
Here l = 30,f = 10, f₁ = 9, f₂ = 3, h = 5

Question 11.
Find the mean, median and mode of the following data:

Solution:
Let assumed mean (A) =175

(i) Here N = 25, \(\frac { 5 }{ 3 }\) = \(\frac { 25 }{ 2 }\) = 12.5 or 13 which lies in the class 150-200
l= 150, F= 10, f= 6, h = 50

Question 12.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

Solution:

We see that class 40-50 has the maximum frequency
∴ It is a modal class

Question 13.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them:

Solution:
Let assumed mean (A) = 135

Here N = 68 , \(\frac { N }{ 2 }\) = \(\frac { 68 }{ 2 }\) = 34

Question 14.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, And the modal size of the surnames.
Solution:
Let Assumed mean (A) = 8.5

(i) Here N = 100, \(\frac { N }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50 which lies in the class 7-10
Here l = 7, F = 36, f= 40, h = 3

Question 15.
Find the mean, median and mode of the following data (C.B.S.E. 2008)

Solution:
Let assumed mean A = 70

(i) Here N = 50, then \(\frac { N }{ 2 }\) = \(\frac { 50 }{ 2 }\) = 25 which lies in the class 60-80
∴ l= 60, F= 24, f = 12 , h = 20

Question 16.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. .Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

(i) We see that the c ass 1500-2000 has maximum frequency 40
∴ It is a modal class
Here l = 1500, f = 40 , f₁ = 24 , f₂ = 33 , h =500

Question 17.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.
Solution:

We see that class 4000-5000 has the maximum frequency 18
∴It is a modal class

Question 18.
The frequency distribution table of agriculture holdings in a village is given below:

Find the modal agriculture holdings of the village.
Solution:
Here the maximum class frequency is 80,
and the class corresponding to this frequency is 5-7.
So, the modal class is 5-7.
l (lower limit of modal class) = 5
f₁ (frequency of the modal class) = 80
f₀ (frequency of the class preceding the modal class) = 45
f₂(frequency of the class succeeding the modal class) = 55
h (class size) = 2

Hence, the modal agricultural holdings of the village is 6.2 hectares.

Question 19.
The monthly income of 100 families are given as below:

Solution:
In a given data, the highest frequency is 41, which lies in the interval 10000-15000.
Here, l = 10000,f₁ = 41, f₀ = 26,f₂ = 16 and h = 5000

Hope given RD Sharma Class 10 Solutions Chapter 15 Statistics Ex 15.5 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.10
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

Mark the correct alternative in each of the following.
Question 1.
The value of k for which the system of equations. kx – y = 2, 6x – 2y = 3 has a unique solution is
(a) = 3
(b) ≠ 3
(c) ≠ 0
(d) = 0
Solution:
(b)

Question 2.
The value of k for which the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely number of solutions, is
(a) 1
(b) 3
(c) 6
(d) 0
Solution:
(c)

Question 3.
The value of k for which the system of equations x + 2y – 3 = 0 and 5x + ky + 1 = 0 has no solution, is
(a) 10
(b) 6
(c) 3
(d) 1
Solution:
(a)

Question 4.
The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0, has a non-zero solution, is
(a) 0
(b) 2
(c) 6
(d) 8
Solution:
(c)

Question 5.
If the system of equations
2x + 3y = 7
(a + b) x + (2a – b) y = 21
has infinitely many solutions, then
(a) a = 1, b = 5
(b) a = 5, b = 1
(c) a = -1, b = 5
(d) a = 5, b = -1
Solution:
(b)

Question 6.
If the system of equations 3x + y = 1 , (2k – 1) x + (k – 1) y = 2k + 1 is inconsistant, then k =
(a) 1
(b) 0
(c) -1
(d) 2
Solution:
(d)

Question 7.
If am ≠ bl, then the system of equations
ax + by = c
lx + my = n
(a) has a unique solution
(b) has no solution
(c) has infinitely many solutions
(d) may or may not have a solution.
Solution:
(a)

The given system as a unique solution

Question 8.
If the system of equations
2x + 3y = 7
2ax + (a + b) y = 28
has infinitely many solutions, then
(a) a = 2b
(b) b = 2a
(c) a + 2b = 0
(d) 2a + b = 0
Solution:
(b)

Question 9.
The value of k for which the system of equations
x + 2y = 5
3x + ky + 15 = 0 has no solution is
(a) 6
(b) – 6
(c) \(\frac { 3 }{ 2 }\)
(d) None of these
Solution:
(a)

k = 6

Question 10.
If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a + b represent coincident lines, then a and b satisfy the equation
(a) a + 5b = 0
(b) 5a + b = 0
(c) a – 56 = 0
(d) 5a – b – 0
Solution:
(c)

Question 11.
If a pair of linear equations in two variables is consistent, then the lines represented by two equations are
(a) intersecting
(b) parallel
(c) always coincident
(d) intersecting or coincident
Solution:
(d) The system of equation is coincident
The line of there equations are intersecting or coincident

Question 12.
The area of the triangle formed by the line \(\frac { x }{ a } +\frac { y }{ b } =1\) with the coordinate axes a is
(a) ab
(b) 2ab
(c) \(\frac { 1 }{ 2 }\) ab
(d) \(\frac { 1 }{ 4 }\) ab
Solution:
(c)
The triangle is formed by the line \(\frac { x }{ a } +\frac { y }{ b } =1\) with co-ordinates
It will intersect x-axis at a and y-axis at y
Area of the triangle so formed = \(\frac { 1 }{ 2 }\) x a x b = \(\frac { 1 }{ 2 }\) ab

Question 13.
The area of the triangle formed by the lines y = x, x = 6 and y = 0 is
(a) 36 sq. units
(b) 18 sq. units
(c) 9 sq. units
(d) 72 sq. units
Solution:
(b)
The triangle formed by the lines y = x, x = 6 and y = 0 will be an isosceles right triangle whose sides will be 6 units
Area = \(\frac { 1 }{ 2 }\) x 6 x 6 = 18 sq. units

Question 14.
If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =
(a) 1
(b) \(\frac { 1 }{ 2 }\)
(c) 3
(d) 6
Solution:
(d)

Question 15.
If the system of equations kx – 5y = 2, 6x + 2y = 7 has no solution, then k =
(a) -10
(b) -5
(c) -6
(d) -15
Solution:
(d)

Question 16.
The area of the triangle formed by the lines x = 3, y = 4 and x = y is
(a) \(\frac { 1 }{ 2 }\) sq. unit
(b) 1 sq. unit
(c) 2 sq. unit
(d) None of these
Solution:
(a)
The triangle is formed by three lines x = 3, y = 4 and x = y
Its sides containing the right angle will be 1 and 1 units
Area of triangle so formed = \(\frac { 1 }{ 2 }\) x 1 x 1 sq. units = \(\frac { 1 }{ 2 }\) sq. units

Question 17.
The area of the triangle formed by the lines 2x + 3y = 12, x – y – 1 = 0 and x = 0 is
(a) 7 sq. units
(b) 7.5 sq. units
(c) 6.5 sq. units
(d) 6 sq. units

Solution:
(b) In the given graph as shown
The triangle formed by the lines
2x + 3y = 12, x – y – 1 =0 and x = 0
Its base BD = 4 + 1 = 5 units
and perpendicular from P on BD = 6 units
Area = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 5 x 3 = \(\frac { 15 }{ 2 }\) sq. units
= 7.5 square units

Question 18.
The sum of the digits of a two digit number is 9. if 27 is added to it, the digits of the number get reversed. The number is
(a) 25
(b) 72
(c) 63
(d) 36
Solution:
(d) Since the sum of the digits of a two-digit number is 9, therefore
x + y = 9 …(i)
It says if the digits are reversed, the new number is 27 less than the original.
Since we are looking at the number like xy, to separate them, it is actually 10x + y for x is a tens digit.
10y+ x = 10x + y + 27
Simplify it, we get 9y = 9x + 27
y = x + 3 …(ii)
Substitute (ii) into (i), we will have
x + (x + 3) = 9
=> 2x + 3 = 9
=> 2x = 6
=> x = 3
Put back into equation (i),
=> 3 + y = 9 => y = 6
The original number is 36.

Question 19.
If x = a, y = b is the solution of the systems of equations x – y = 2 and x + y = 4, then the values of a and b are, respectively
(a) 3 and 1
(b) 3 and 5
(c) 5 and 3
(d) -1 and -3
Solution:
(a) Since, x = a and y = b is the solution of the equations x – y = 2 and x + y = 4, then these values will satisfy that equations.
a – b = 2 …(i)
and a + b = 4 …(ii)
By adding (i) and (ii), we get
2a = 6
Therefore, a = 3
By putting a = 3 in (i), we get
3 – b = 2 Therefore, b = 1
Thus, a = 3 ; b = 1

Question 20.
For what value k, do the equations 3x – y + 8 = 0 and 6x – ky + 16 = 0 represent coincident lines?
(a) \(\frac { 1 }{ 2 }\)
(b) – \(\frac { 1 }{ 2 }\)
(c) 2
(d) -2
Solution:
(c)
Let 3x – y + 8 = 0 …(i)
and 6x – – ky + 16 = 0 …(ii)
Here, a₁ = 3, b₁ = -1, c₁ = 8

Question 21.
Aruna has only ₹ 1 and ₹ 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹ 75, then the number of ₹ 1 and ₹ 2 coins are, respectively
(a) 35 and 15
(b) 35 and 20
(c) 15 and 35
(d) 25 and 25
Solution:
(d) Let number of ₹ 1 coins = x
and number of ₹ 2 coins = y
Now, by given condition x + y = 50 …(i)
Also, x x 1 + y x 2 = 15
=>x + 2y = 75 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
(x + 2y) – (x + y) = 75 – 50
=> y = 25
When y = 25, then x = 25

Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables 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.

RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11

Other Exercises

• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.10
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

Question 1.
If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increased by 33 square units. Find the area of the rectangle.
Solution:
Let the length of rectangle = x units
and breadth = y units
Area = Length x breadth = x x y = xy sq. units
According to the condition given,
(x + 2) (y – 2) = xy – 28
=> xy – 2x + 2y – 4 = xy – 28
=> -2x + 2y = -28 + 4 = -24
=> x – y = 12 ….(i)
(Dividing by -2)
and (x – 1) (y + 2) = xy + 33
=> xy + 2x – y – 2 = xy + 33
=> 2x – y = 33 + 2
=> 2x – y = 35 ….(ii)
Subtracting (i), from (ii)
x = 23
Substituting the value of x in (i)
23 – y = 12
=> -y = 12 – 23 = -11
=> y = 11
Area of the rectangle = xy = 23 x 11 = 253 sq. units

Question 2.
The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and breadth is increased by 5 metres. Find the dimensions of the rectangle.
Solution:
Let the length of a rectangle = x m
and breadth = y m
Area = Length x breadth = xy sq. m
According to the condition given,
(x + 7)(y – 3) = xy
=> xy – 3x + 7y – 21 = xy
-3x + 7y = 21 ….(i)
and (x – 7) (y + 5) = xy
=>xy + 5x – 7y = xy
5x – 7y = 35 …(ii)
(i) from (ii)
2x = 56
=> x = 28
Substituting the value of x in (i)
-3 x 28 + 7y = 21
– 84 +7y = 21
=> 7y = 21 + 84 = 105
y = 15
Length of the rectangle = 28 m and breadth = 15 m

Question 3.
In a rectangle, if the length is increased by 3 metres and breadth is decreased by 4 metres, the area of the rectangle is reduced by 67 square metres. If length reduced by 1 metre and breath increased by 4 metres, the area is increased by is 89 sq. metres. Find the dimensions of the rectangle.
Solution:
Let length of rectangle = x m
and breadth = y m
Area = Length x breadth = xy m²
According to the given condition,
(x + 3) (y – 4) = xy – 67
=> xy – 4x + 3y – 12 = xy – 67
=>-4x + 3y = -67+ 12 = -55
=> 4x – 3y = 55 ….(i)
and (x -1) (y + 4) = xy + 89
=> xy + 4x – y – 4 = xy + 89
=> 4x – y = 89 + 4 = 93 ….(ii)
Subtracting (i) from (ii)
2y = 38
=> y = 19
Substituting the value of y in (i)
4x – 3 x 19 = 55
=> 4x – 57 = 55
=> 4x = 55 + 57 = 112
=> x = 28
Length of the rectangle = 28 m and breadth = 19 m

Question 4.
The incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves ₹ 1250, find their incomes.
Solution:
Incomes of X and Y are in the ratio 8 : 7
and their expenditures = 19 : 16
Let income of X = ₹ 8x
and income of Y = ₹ 7x
and let expenditures of X = 19y
and expenditure of Y = 16y
Saving in each case is save i.e. ₹ 1250
8x – 19y = 1250 ….(i)
and 7x – 16y = 1250 ….(ii)
Multiplying (i) by 7 and (ii) by 8, we get

Question 5.
A and B each has some money. If A gives ₹ 30 to B, then B will have twice the money left with A, but if B gives ₹ 10 to A, then A will have thrice as much as is left with B. How much money does each have?
Solution:
Let A’s money = 7 x
and B’s money = 7 y
According to the given conditions,
2(x – 30) = 7 + 30
=> 2x – 60 = y + 30
=> 2x – y = 30 + 60 = 90 ….(i)
and x + 10 = 3 (y – 10)
=> x + 10 = 3y – 30
=> x – 3y = -30 – 10 = -40 ………(ii)
Multiplying (i) by 3 and (ii) by 1,

Question 6.
ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y – 5)°, ∠C = (+4x)° and ∠D = (7x + 5)°. Find the four angles. [NCERT]
Solution:

Question 7.
2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it ?
Solution:
Let one man can do a work is = x days

Question 8.
In a ∆ABC, ∠A = x°, ∠B = (3x – 2)°, ∠C = y°. Also ∠C – ∠B = 9°. Find the three angles.
Solution:
We know that sum of three angles of a triangle = 180°
∠A + ∠B + ∠C = 180°
=> x° + (3x – 2)° + y° = 180°
=> x + 3x – 2 + y = 180
=> 4x + y = 180 + 2 = 182 …(i)
∠C – ∠B = 9
y – (3x – 2) = 9
=> y – 3x + 2 = 9
=> y – 3x = 9 – 2
=> -3x + y = 7 ….(ii)
Subtracting (ii) from (i)
7x = 175 => x = 25
Substituting the value of x in (ii)
-3 x 25 + y = 7
=> -75 + y = 7
=> y = 7 + 75 = 82
∠A = x° = 25°
∠B = 3x – 2 = 3 x 25 – 2 = 73°
∠C = 7 = 82°

Question 9.
In a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x – 5)°. Find the four angles.
Solution:
ABCD is a cyclic quadrilateral
and ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x – 5)°
The sum of opposite angles = 180°
∠A + ∠C=180° and ∠B + ∠D = 180°
=> 2x + 4 + 2y + 10 = 180°
=> 2x + 2y + 14 = 180°
=> 2x + 2y = 180 – 14 = 166
=> x + y = 83 ….(i)
(Dividing by 2)
and y + 3 + 4x – 5 = 180°
=> 4x + 7 – 2 = 180°
4x + 7 = 180°+ 2= 182° ….(ii)
Subtracting (i) from (ii)
3x = 99
=> x = 33
Substituting the value of x in (i)
33 + y = 83
=> y = 83 – 33 = 50
∠A = 2x + 4 = 2 x 33 + 4 = 66 + 4 = 70°
∠B = y + 3 = 50 + 3 = 53°
∠C = 2y + 10 = 2 x 50 + 10 = 100 + 10 = 110°
∠D = 4x – 5 = 4 x 33 – 5 = 132 – 5 = 127°

Question 10.
Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test ?
Solution:
Let number of right answers questions = x
and number of wrong answers questions = y
Now according to the given conditions,
3x – y = 40 ….(i)
and 4x – 2y = 50
=> 2x – y = 25 ….(ii)
(Dividing by 2)
Subtracting (ii) from (i), x = 15
Substituting the value of x in (i)
3 x 15 – y = 40
=> 45 – y = 40
=> y = 45 – 40 = 5
x = 15 and y = 5
Total number of questions = 15 + 5 = 20

Question 11.
In a ∆ABC, ∠A = x°, ∠B = 3x° and ∠C = 7°. If 3y – 5x = 30, prove that the triangle is right angled.
Solution:
In a ∆ABC ,
∠A = x°, ∠B = 3x° and ∠C = 7°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
=> x + 3x + y = 180
=>4x + y = 180 ………(i)
and 3y – 5x = 30 ….(ii)
from (i) y = 180 – 4x
Substituting the value of y in (ii)
3 (180 – 4x) – 5x = 30
540 – 12x – 5x = 30
=> -17x = -540 + 30 = -510
=> 17x = 510
=> x = 30
y = 180 – 4x = 180 – 4 x 30 = 180 – 120 = 60
∠A = x = 30°
∠B = 3x = 3 x 30° = 90°
∠C = y = 60°
∠B of ∆ABC = 90°
∆ABC is a right angled triangle.

Question 12.
The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is ₹ 89 and for a journey of 20 km, the charge paid is ₹ 145. What will a person have to pay for travelling a distance of 30 km?
Solution:
Let the fixed charges = ₹ x
and charges per km = ₹ y
According to the given conditions

Question 13.
A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charged and the cost of food per day. [CBSE 2000]
Solution:
Let the fixed charge of the college hostel = ₹ x
and daily charges = ₹ y
According to the given conditions,
x + 20y = 1000 …(i)
x + 26y =1180 …(ii)
Subtracting (i) from (ii), we get
6y = 180
=> y = 30
Substituting the value of y in (i)
x + 20 x 30 = 1000
=> x + 600 = 1000
=> x = 1000 – 600 = 400
Fixed charges = ₹ 400 and daily charges = ₹ 30

Question 14.
Half the perimeter of a garden whose length is 4 m more them its width is 36 m. Find the dimensions of the garden.
Solution:
Let the length of the garden = x m
and width = y m
x – y = 4 …(i)
and x + y = 36 ….(ii)
Adding we get,
2x = 40
=> x = 20
and subtracting,
2y = 32
y = 16
Length of the garden = 20 m and width = 16 m

Question 15.
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them
Solution:
In two supplementary angles,
Let larger angle = x
and smaller angle = y
x – y = 18°
But x + y = 180°
Adding we get,
2x = 198°
=> x = 9°
Subtracting the get,
2y = 162°
=> y = 81°
Angles are 99° and 81°

Question 16.
2 women and 5 men can together finish a piece of embroidery in 4 days while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery and that taken by 1 man alone.
Solution:
Let one woman can do the work in = x days
and one man can do the same work = y days

Question 17.
Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes ₹ 50 and ₹ 100 she received.
Solution:
Total amount with drawn = ₹ 2000
Let number of ₹ 50 notes = x
and of ₹ 100 = y
According to the conditions
x + y = 25 …(i)
and 50x + 100y = 2000
=> x + 2y = 40 ….(ii)
(Dividing by 50)
Subtracting (i) from (ii), y = 15
Substituting the value of y in (i)
x + y = 25
x + 15 = 25
=> x = 25 – 15 = 10
Number of 50 rupees notes = 10
and number of 100 rupee notes = 15

Question 18.
There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room.
Solution:
Let A examination room has students = x
and B room has students = y
According to the given conditions,
x – 10 = y + 10
=> x – y = 10 + 10
=> x – y = 20 ….(i)
and x + 20 = 2 (y – 20)
=> x + 20 = 2y – 40
=> x – 2y = – 40 – 20 = – 60 ….(ii)
Subtracting (ii) from (i), y = 80
Substituting the value of y in (i)
x – 80 = 20
=> x = 20 + 80 = 100
In examination room A, the students are = 100
and in B room = 80

Question 19.
A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs ₹ 216 and one full and one half reserved first class tickets cost ₹ 327. What is the basic first class full fare and what is the reservation charge?
Solution:
Let the rate of fare of full ticket = ₹ x
and rate of reservation = ₹ y
According to the given conditions,
x + y = 216 …(i)

Question 20.
A wizard having powers of mystic in candations and magical medicines seeing a cock, fight going on, spoke privately to both the owners of cocks. To one he said; if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that. Going to the other, he promised in the same way to give three fourths. From both of them his gain would be only 12 gold coins. Find the stake of money each of the cock- owners have.
Solution:
Let the first owner of cock has state money = x gold coins
and second owner has = y gold coins
According to the given conditions,

Question 21.
The students of a class are made to stand in rows. If 3 student are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the number of students in the class.
Solution:
Let number of rows = x
and number of students in each row = y
Now according to the conditions given
Number of total students = xy
(y + 3)(x – 1) = xy
=> xy – y + 3x – 3 = xy
=>3x – y = 3 …(i)
and (y – 3) (x + 2) = xy
=> xy + 2y – 3x – 6 = xy
=> 2y – 3x = 6 …(ii)
Adding (i) and (ii), y = 9
Substituting the value of y in (i)
3x – 9 = 3
=> 3x = 3 + 9 = 12
=> x = 4
Number of total students = xy = 4 x 9 = 36

Question 22.
One says, “give me hundred, friend ! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their respective capital ?
Solution:
Let first person has amount of money = Rs. x
and second person has = Rs. y
Now according to the conditions given,
x + 100 = 2(y – 100)
=> x + 100 = 2y – 200
=> x – 2y = – 200 – 100
=> x – 2y = -300 ………(i)
and 6(x – 10) = (y + 10)
=> 6x – 60 = y + 10
=> 6x – y = 10 + 60
=> 6x – y = 70 ……….(ii)
From (i), x = 2y – 300
Substituting the value of x in (ii)
6(2y – 300) – y = 70
=> 12y – 1800 – y = 70
=> 11y = 70 + 1800 = 1870
y = 170
x = 2y – 300 = 2 x 170 -300 = 340 – 300 = 40
First person has money = ₹ 40
and second person has = ₹ 170

Question 23.
A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of ₹ 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got ₹ 1028. Find the cost price of the saree and the list price (price before discount) of the sweater. [NCERT Exemplar]
Solution:
Let the cost price of the saree and the list price of the sweater be ₹ x and ₹ y, respectively.

Question 24.
In a competitive examination, one mark is aw awarded for each correct answer while \(\frac { 1 }{ 2 }\) mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly. [NCERT Exemplar]
Solution:
Let x be the number of correct answer of the questions in a competitive examination,
then (120 – x) be the number of wrong answers of the questions.
Then, by given condition,

Hence, Jayanti answered correctly 100 questions.

Question 25.
A shopkeeper gives books pn rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid ₹ 22 for a book kept for 6 days, while Anand paid ₹ 16 for the book kept for four days. Find the fixed charges and charge for each extra day. [NCERT Exemplar]
Solution:
Let Latika takes a fixed charge for the first two day is ₹ x and additional charge for each day thereafter is ₹ y.
Now, by first condition,
Latika paid ₹ 22 for a book kept for six days
i.e., x + 4y = 22 …(i)
and by second condition,
Anand paid ₹ 16 for a book kept for four days
i.e., x + 2y = 16 …(ii)
Now, subtracting Eq. (ii) from Eq. (i), we get
2y = 6 => y = 3
On putting the value of y in Eq. (ii), we get
x + 2 x 3 = 16
x = 16 – 6 = 10
Hence, the fixed charge = ₹ 10
and the charge for each extra day = ₹ 3

Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11 are helpful to complete your math homework.

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