Tag: RD Sharma Class 10 Maths Solutions

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Mark the correct alternative in each of the following :
Question 1.
If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
(a) 87
(b) 88
(c) 89
(d) 90
Solution:
(c) 7th term (a₇) = a + 6d = 34
13th term (a₁₃) = a + 12d = 64
Subtracting, 6d = 30 => d = 5
and a + 12 x 5 = 64 => a + 60 = 64 => a = 64 – 60 = 4
18th term (a₁₈) = a + 17d = 4 + 17 x 5 = 4 + 85 = 89

Question 2.
If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of (p + q) terms will be
(a) 0
(b) p – q
(c) p + q
(d) – (p + q)
Solution:
(d)

Question 3.
If the sum of n terms of an A.P. be 3n² + n and its common difference is 6, then its first term is
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(d)

Question 4.
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
(b) First term of an A.P. (a) = 1
Last term (l) = 11
and sum of its terms = 36
Let n be the number of terms and d be the common difference, then

Question 5.
If the sum of n terms of an A.P. is 3n² + 5n then which of its terms is 164 ?
(a) 26th
(b) 27th
(c) 28th
(d) none of these
Solution:
(b)

Question 6.
If the sum of it terms of an A.P. is 2n² + 5n, then its nth term is
(a) 4n – 3
(b) 3n – 4
(c) 4n + 3
(d) 3n + 4
Solution:
(c)

Question 7.
If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is :
(a) 13
(b) 9
(c) 21
(d) 17
Solution:
(c)

Question 8.
If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are
(a) 5, 10, 15, 20
(b) 4, 10, 16, 22
(c) 3, 7, 11, 15
(d) None of these
Solution:
(a)
4 numbers are in A.P.
Let the numbers be
a – 3d, a – d, a + d, a + 3d
Where a is the first term and 2d is the common difference
Now their sum = 50
a – 3d + a – d + a + d + a + 3d = 50
and greatest number is 4 times the least number
a + 3d = 4 (a – 3d)
a + 3d = 4a – 12d
4a – a = 3d + 12d
=> 3a = 15d

Question 9.
Let S denotes the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = S_n – k S_n-1 + S_n-2 then k =
(a) 1
(b) 2
(c) 3
(d) None of these
Solution:
(b)

Question 10.
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by

(a) S
(b) 2S
(c) 3S
(d) None of these
Solution:
(b)

Question 11.
If the sum of first n even natural number is equal to k times the sum of first n odd natural numbers, then k =

Solution:
(d)

Question 12.
If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is

Solution:
(c)

Question 13.
If S₁ is the sum of an arithmetic progression of ‘n’ odd number of terms and S₂ is the sum of the terms of the

Solution:
(a)

Question 14.
If in an A.P., S_n = n²p and S_m = m²p, where S denotes the sum of r terms of the A.P., then S_p is equal to
(a) \(\frac { 1 }{ 2 }\) p³
(b) mnp
(c) p³
(d) (m + n) p²
Solution:
(c)

Question 15.
If Sn denote the sum of the first n terms of an A.P. If S_2n = 3S_n , then S_3n : S_n is equal to
(a) 4
(b) 6
(c) 8
(d) 10
Solution:
(b)

Question 16.
In an AP, S_p = q, S_q = p and S denotes the sum of first r terms. Then, S_p+q is equal to
(a) 0
(b) – (p + q)
(c) p + q
(d) pq
Solution:
(c) In an A.P. S_p = q, S_q = p
S_p+q = Sum of (p + q) terms = Sum of p term + Sum of q terms = q + p

Question 17.
If S_n denotes the sum of the first r terms of an A.P. Then, S_3n : (S_2n – S_n) is
(a) n
(b) 3n
(b) 3
(d) None of these
Solution:
(c)

Question 18.
If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 term is
(a) 3200
(b) 1600
(c) 200
(d) 2800
Solution:
(a)

Question 19.
The number of terms of the A.P. 3, 7,11, 15, … to be taken so that the sum is 406 is
(a) 5
(b) 10
(c) 12
(d) 14
Solution:
(d)

Question 20.
Sum of n terms of the series

Solution:
(c)

Question 21.
The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is
(a) 50th
(b) 502th
(c) 508th
(d) None of these
Solution:
(d)

Question 22.

Solution:
(c)

Question 23.

Solution:
(b) S_n is the sum of first n terms
Last term nth term = S_n – S_n-1

Question 24.
The common difference of an A.P., the sum of whose n terms is S_n, is

Solution:
(a)

Question 25.

Solution:
(b)
In first A.P. let its first term be a₁ and common difference d₁
and in second A.P., first term be a₂ and common difference d₂, then

Question 26.

Solution:
(b)

Question 27.
If the first term of an A.P. is a and nth term is b, then its common difference is

Solution:
(b)

Question 28.
The sum of first n odd natural numbers is
(a) 2n – 1
(b) 2n + 1
(c) n²
(d) n² – 1
Solution:
(c)

Question 29.
Two A.P.’s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th terms is
(a) 11
(b) 3
(c) 8
(d) 5
Solution:
(d) In two A.P.’s common-difference is same
Let A and a are two A.P. ’s
First term of A is 8 and first term of a is 3
A₃₀ – a₃₀ = 8 + (30 – 1) d – 3 – (30 – 1) d
= 5 + 29d – 29d = 5

Question 30.
If 18, a, b – 3 are in A.P., the a + b =
(a) 19
(b) 7
(c) 11
(d) 15
Solution:
(d) 18, a, b – 3 are in A.P., then a – 18 = -3 – b
=> a + b = -3 + 18 = 15

Question 31.
The sum of n terms of two A.P.’s are in the ratio 5n + 4 : 9n + 6. Then, the ratio of their 18th term is

Solution:
(a)

Question 32.

Solution:
(b)

Question 33.
The sum of n terms of an A.P. is 3n² + 5n, then 164 is its
(a) 24th term
(b) 27th term
(c) 26th term
(d) 25th term
Solution:
(b)

Question 34.
If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
(a) n (n – 2)
(b) n (n + 2)
(c) n (n + 1)
(d) n (n – 1)
Solution:
(b)

Question 35.
If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
(a) 3 : 2
(b) 3 : 1
(c) 1 : 3
(d) 2 : 3
Solution:
(b)

Question 36.
The sum of first 20 odd natural numbers is
(a) 100
(b) 210
(c) 400
(d) 420 [CBSE 2012]
Solution:
(c)

Question 37.

Solution:
(a)

Question 38.

Solution:
(c)

Question 39.
The common difference of the A.P. \(\frac { 1 }{ 2b }\) ,

Solution:
(d)

Question 40.
If k, 2k – 1 and 2k + 1 are three consecutive terms of an AP, the value of k is
(a) -2
(b) 3
(c) -3
(d) 6 [CBSE 2014]
Solution:
(b) (2k – 1) – k = (2k + 1) – (2k- 1)
2k – 1 – k = 2
=> k = 3

Question 41.
The next term of the A.P. , √7 , √28, √63, …………
(a) √70
(b) √84
(c) √97
(d) √112 [CBSE 2014]
Solution:
(d)

= √(l6 x 7)= √112

Question 42.
The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals
(a) -3
(b) 4
(c) 5
(d) 2 [CBSE 2014]
Solution:
(c) 2 (3y + 5) = 3y – 1 + 5y + 1
(If a, b, c are in A.P., b – a = c – b=> 2b = a + c)
=> 6y + 10 = 8y
=> 10 = 2y
=> y = 5

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions 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 5 Arithmetic Progressions VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
Define an arithmetic progression.
Solution:
A sequence a₁, a₂, a₃, …, an is called an arithmetic progression of then exists a constant d
Such that a₂ – a₁ = d, a₃ – a₂ = d, ………… a_n – a_n-1 = d
and so on and d is called common difference

Question 2.
Write the common difference of an A.P. whose nth term is a_n = 3n + 7.
Solution:
a_n = 3n + 7
a₁ = 3 x 1 + 7 = 3 + 7 = 10
a₂ = 3 x 2 + 7 = 6 + 7 = 13
a₃ = 3 x 3 + 7 = 9 + 7 = 16
d = a₃ – a₂ or a₂ – a₁ = 16 – 13 = 3 or 13 – 10 = 3

Question 3.
Which term of the sequence 114, 109, 104, … is the first negative term ?
Solution:
Sequence is 114, 109, 104, …..
Let a_n term be negative

Question 4.
Write the value of a₃₀ – a₁₀ for the A.P. 4, 9, 14, 19, …………
Solution:

Question 5.
Write 5th term from the end of the A.P. 3, 5, 7, 9,…, 201.
Solution:

= 3 + 190 = 193
5th term from the end = 193

Question 6.
Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.
Solution:

Question 7.
Write the nth term of an A.P. the sum of whose n terms is S_n.
Solution:
Sum of n terms = S_n
Let a be the first term and d be the common difference a_n =S_n – S_n-1

Question 8.
Write the sum of first n odd natural numbers.
Solution:

Question 9.
Write the sum of first n even natural numbers.
Solution:
First n even natural numbers are
2, 4, 6, 8, ……….
Here a = 2, d = 2

Question 10.
If the sum of n terms of an A.P. is S_n = 3n² + 5n. Write its common difference.
Solution:

Question 11.
Write the expression for the common difference of an A.P. Whose first term is a and nth term is b.
Solution:
First term of an A.P. = a
and a_n = a + (n – 1) d = b .
Subtracting, b – a = (n – 1) d
d = \(\frac { b – a }{ n – 1 }\)

Question 12.
The first term of an A.P. is p and its common difference is q. Find its 10th term. [CBSE 2008]
Solution:
First term of an A.P. (a) = p
and common difference (d) = q
a₁₀ = a + (n – 1) d
= p + (10 – 1) q = p + 9q

Question 13.
For what value of p are 2p + 1, 13, 5p – 3 are three consecutive terms of an A.P.? [CBSE 2009]
Solution:

Question 14.
If \(\frac { 4 }{ 5 }\), a, 2 are three consecutive terms of an A.P., then find the value of a.
Solution:

Question 15.
If the sum of first p term of an A.P. is ap² + bp, find its common difference.
Solution:

Question 16.
Find the 9th term from the end of the A.P. 5, 9, 13, …, 185. [CBSE 2016]
Solution:
Here first term, a = 5
Common difference, d = 9 – 5 = 4
Last term, l = 185
nth term from the end = l – (n – 1) d
9th term from the end = 185 – (9 – 1) 4 = 185 – 8 x 4 = 185 – 32 = 153

Question 17.
For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k – 1 form on A.P.? [CBSE 2016]
Solution:
(3k + 3) – (2k + 1) = (5k – 1) – (3k + 3)
3k + 3 – 2k – 1 = 5k – 1 – 3k – 3
k + 2 = 2k – 4
2k – k = 2 + 4
k = 6

Question 18.
Write the nth term of the A.P.
\(\frac { 1 }{ m }\) , \(\frac { 1 + m }{ m }\) , \(\frac { 1 + 2m }{ m }\) , ……… [CBSE 2017]
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions 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 14 Surface Areas and Volumes VSAQS

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 VSAQS

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.
The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. What is the ratio of their volumes ?
Solution:
Radii of the bases of a cylinder and a cone = 3:4
and ratio in their heights = 2:3
Let r_1, r₂ be the radii and h₁ and h₂ be their heights
heights of the cylinder and cone respectively,

Question 2.
If the heights of two right circular cones are in the ratio 1 : 2 and the perimeters of their bases are in the ratio 3 : 4. What is the ratio of their volumes ?
Solution:
Ratio in the heights of two cones =1:2 and ratio in the perimeter of their bases = 3:4
Let r₁, r₂ be the radii of two cones and ht and h2 be their heights

Question 3.
If a cone and sphere have equal radii and equal volumes what is the ratio of the diameter of the sphere to the height of the cone ?
Solution:
Let r be the radius of a cone, then
radius of sphere = r
Let h be the height of cone
Now volume of cone = volume of sphere

Question 4.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. What is the ratio of their volumes?
Solution:
Let r and h be the radius and heights of a cone, a hemisphere and a cylinder
∴ Volume of cone =  \((\frac { 1 }{ 3 } )\) πr²h
Volume of hemisphere = \((\frac { 2 }{ 3 } )\) πr³

Question 5.
The radii of two cylinders are in the ratio 3 : 5 and their heights are in the ratio 2 : 3. What is the ratio of their curved surface areas ?
Solution:
Radii of two cylinders are in the ratio = 3:5
and ratio in their heights = 2:3
Let r₁, r₂ be the radii and h₁, h₂ be the heights of the two cylinders respectively, then

Question 6.
Two cubes have their volumes in the ratio 1 : 27. What is the ratio of their surface areas ?
Solution:
Ratio in the volumes of two cubes = 1 : 27
Let a₁ and a₂ be the sides of the two cubes respectively then volume of the first area = a₁³
and volume of second cube = a_2³

Question 7.
Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii ?
Solution:
Ratio the heights of two right circular cylinders = 1:2
Let r₁,r₂ be their radii and h₁, h₂ be their

Question 8.
If the volumes of two cones are in the ratio 1 : 4, and their diameters are in the ratio 4 : 5, then write the ratio of their weights.
Solution:
Volumes of two cones are in the ratio =1:4 and their diameter are in the ratio = 4:5
Let r₁ and r₂ be the radii and h₁ ,h₂ be their

Question 9.
A sphere and a cube have equal surface areas. What is the ratio of the volume of the sphere to that of the cube ?
Solution:
Surface areas of a sphere and a cube are equal
Let r be the radius of sphere and a be the edge of cube,

Question 10.
What is the ratio of the volume of a cube to that of a sphere which will fit inside it?
Solution:
A sphere is fit inside the cube
Side of a cube = diameter of sphere
Let a be the side of cube and r be the radius of the sphere, then

Question 11.
What is the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height ?
Solution:
Diameters (or radii), and heights of a cylinder a cone and a sphere are equal,
Let r and h be the radius and height be the cone cylinder, cone and sphere respectively, thus their volumes will be

Question 12.
A sphere of maximum volume is cut-out from a solid hemisphere of radius r. What is the ratio of the volume of the hemisphere to that of the cut-out sphere?
Solution:
r is the radius of a hemisphere, then
the diameter of the sphere which is cut out of the hemisphere will be r

Question 13.
A metallic hemisphere is melted and recast in the shape of a cone with the same base radius R as that of the hemisphere. If H is the height of the cone, then write the value of \((\frac { H }{ R } )\).
Solution:
R is the radius of a hemisphere 2

Question 14.
A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and height are in the ratio 5 : 12, write the ratio of the total surface area of the cylinder to that of the cone.
Solution:
Radius and height of a cone and a cylinder be r and h respectively

Question 15.
A cylinder, a cone and a hemisphere are of equal base and have the same height. What is the ratio of their volumes ?
Solution:
Let r and h be the radii and heights of the cylinder cone and hemisphere respectively, then
Volume of cylinder = πr²h
Volume of cone = \((\frac { 1 }{ 3 } )\) πr²h
Volume of hemisphere = \((\frac { 2 }{ 3 } )\) πr³

Question 16.
The radii of two cones are in the ratio 2 : 1 and their volumes are equal. What is the ratio of their heights ?
Solution:
Radii of two cones are in the ratio = 2:1
Let r₁, r₂ be the radii of two cones and h₁, h₂ be their heights respectively,

Question 17.
Two cones have their heights in the ratio 1 : 3 and radii 3:1. What is the ratio of their volumes ?
Solution:
Ratio in heights of two cones = 1:3
and ratio in their ratio = 3:1
Let r₁, r₂ be their radii and h₁, h₂ be their
heights, then

Question 18.
A hemisphere and a cone have equal bases. If their heights are also equal, then what is the ratio of their curved surfaces ?
Solution:
Bases of a hemisphere and a cone are equal
and their heights are also equal
Let r and h be their radii and heights
respectively
∴ r = h₁

Question 19.
If r₁ and r₂ denote the radii of the circular bases of the frustum of a cone such that r₁ > r₂ then write the ratio of the height of the cone of which the frustum is a part to the height of the frustum.
Solution:
r₁ , r₂ are the radii of the bases of a frustum and r₁ > r₂
Let h₁ be the height of cone and h₂ be the height of smaller cone
∴ Height of frustum = h₁ – h₂

Question 20.
If the slant height of the frustum of a cone is 6 cm and the perimeters of its circular bases are 24 cm and 12 cm respectively. What is the curved surface area of the frustum ?
Solution:
Slant height of a frustum (l) = 6 cm
Perimeter of upper base (P₁) = 24 cm
and perimeter of lower base (P₂) = 12 cm

Question 21.
If the areas of circular bases of a frustum of a cone are 4 cm² and 9 cm² respectively and the height of the frustum is 12 cm. What is the volume of the frustum ?
Solution:
In a frustum,
Area of upper base (A₁) = 4 cm²
and area of lower base (A₂) = 9 cm²

Question 22.
The surface area of a sphere is 616 cm². Find its radius.
Solution:
Surface area of a sphere = 616 cm²
Let r be the radius, then

Question 23.
A cylinder and a cone are of the same base radius and of same height. Find the ratio of the value of the cylinder to that of the cone. [CBSE 2009]
Solution:
Let r be the radius of the base of the cylinder
small as of cone
and let height of the cylinder = h
Then height of cone = h
∴ Volume of cylinder =  πr²h
and volume of cone = \((\frac { 1 }{ 3 } )\)  πr²h

Question 24.
The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two circular ends is 4 cm, write the height of the frustum. [CBSE 2010]
Solution:
Slant height of frustum (l) = 5 cm
Difference between the upper and lower radii = 4 cm
Let h be height and upper radius r₁ and lower radius = r₂

Question 25.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?
Solution:
Volume of hemisphere = Surface area of hemisphere (given)

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes 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 14 Surface Areas and Volumes Revision Exercise

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 Revision Exercise

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.
A metallic sphere 1 dm in diameter is beaten into a circular sheet of uniform thickness equal to 1 mm. Find the radius of the sheet.
Solution:
Diameter of a sphere = 1 dm = 10 cm
∴  Radius (r) = \((\frac { 10 }{ 2 } )\) = 5 cm
Volume of metal used in the sphere = \((\frac { 4 }{ 3 } )\) πr³
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise

Question 2.
Three solid spheres of radii 3,4 and 5 cm respectively are melted and converted into a single solid sphere. Find the radius of this sphere.
Solution:
Radius of first sphere (r₁) = 3 cm

Question 3.
A spherical shell of lead, whose external diameter is 18 cm, is melted and recast into a right circular cylinder, whose height is 8 cm and diameter 12 cm. Determine the internal diameter of the shell.
Solution:
Diameter of the cylinder = 12 cm
∴  Radius (r₁) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height (h) = 8 cm
∴ Volume = πr₁²h = π(6)² x 8 cm³
= π x 36 x 8 = 288π cm³
Now volume of metal used in spherical shell = 288π cm
External diameter = 18 cm 18
∴  External radius (R) = \((\frac { 18 }{ 2 } )\) = 9 cm
Let r be the internal radius, then
Volume of the metal = \((\frac { 4 }{ 3 } )\) π (R³ – r³ )

Question 4.
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of two well = 10 m
∴ Radius (r) = \((\frac { 10 }{ 2 } )\) = 5 m
Depth (h) = 8.4 m
∴ Volume of earth dug out = πr²h

Question 5.
 In the middle of a rectangular field measuring 30 m x 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.
Solution:
Diameter of well = 7 m
∴ Radius (r) = \((\frac { 7 }{ 2 } )\) m
Depth (h) = 10 m
∴ Volume of earth dug out = πr²h

Question 6.
The inner and outer radii of a hollow cylinder are 15 cm and 20 cm, respectively. The cylinder is melted and recast into a solid cylinder of the same height. Find the radius of the base of new cylinder.
Solution:
Inner radius of hollow cylinder (r) = 15 cm
Outer radius (R) = 20 cm
Let h be the height of the hollow cylinder,
Then volume of metal used = πR (R² – r²)
= πh (20²- 15²) cm³
=  πh (400 – 225) cm³
= 175 πh cm³
Volume of the new cylinder = 175nh cm³
Height = h
Let R be the radius of new cylinder,
then πR²h = 175 πh
⇒ R²= 175
⇒ R = \(\sqrt { 175 } \)
= 13.2
∴ Radius = 13.2 cm

Question 7.
Two cylindrical vessels are filled with oil. Their radii are 15 cm, 12 cm and heights 20 cm, 16 cm respectively. Find the radius of a cylindrical vessel 21 cm in height, which will just contain the oil of the two given vessels.
Solution:
Radius of first cylinder (r₁) = 15 cm
and radius of second cylinder (r₂) = 12 cm
Height of the first cylinder (h₁) = 20 cm
and height of second cylinder (h₂) = 16 cm
∴ Volume of both of cylinders

Question 8.
A cylindrical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.
Solution:
Diameter of cylindrical bucket = 28 cm
∴ Radius (r) = \((\frac { 28 }{ 2 } )\) = 14 cm
Height (h) = 72 cm
∴ Volume of water filled in it = πr²h
= \((\frac { 22 }{ 7 } )\) x 14 x 14 x 72 cm³ = 44352 cm³
∴ Volume of water in rectangular tank = 44352 cm³
Length of tank (l) = 66 cm
and breadth (b) = 28 cm
Let h₁ be its height
∴  Ibh₁ = 44352
⇒  66 x 28 h₁= 44352
⇒ h₁ = \((\frac { 44352 }{ 66×28 } )\) = 24
∴Height of water in the tank = 24 cm

Question 9.
A cubic cm of gold is drawn into a wire 0.1 mm in diameter, find the length of the wire.
Solution:
Volume of solid gold = 1 cm³
Diameter of cylinderical wire = 0.1 mm

Question 10.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 3 m
Radius (r) = \((\frac { 3 }{ 2 } )\) m
Depth (h) = 14 m

Question 11.
A conical vessel whose internal radius is 10 cm and height 48 cm is full of water. Find the volume of water. If this water is poured into a cylindrical vessel with internal radius 20 cm, find the height to which the water level rises in it.
Solution:
Internal radius of the conical vessel (r₁) = 10 cm
Height (h₁) = 48 cm

Question 12.
The vertical height of a conical tent is 42 dm and the diameter of its base is 5.4 m. Find the number of persons it can accommodate if each person is to be allowed 29.16 cubic dm.
Solution:
Vertical height of conical tent (h) = 42 dm
and diameter base (b) = 5.4 dm
∴ Radius (r) = \((\frac { 5.4 }{ 2 } )\) = 2.7 dm

Question 13.
A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them.
Solution:
Let r and h be the radius and height of a circular cylinder and also of a cone, then curved surface area of the cylinder = 2πrh
and curved surface area of cone

Question 14.
A sphere of diameter 5 cm is dropped into a cylindrical vessel partly filled with water. The diameter of the base of the vessel is 10 cm. If the sphere is completely submerged, by how much will the level of water rise ?
Solution:
Diameter of sphere = 5 cm

Question 15.
A spherical ball of iron has been melted – and made into smaller balls. If the radius of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made ?
Solution:
Let the radius of larger ball = r

Question 16.
Find the depth of a cylindrical tank of radius 28 m, if its capacity is equal to that of a rectangular tank of size 28 m x 16 m x 11 m.
Solution:
Dimensions of a rectangular tank = 28m x 16m x 11m
∴ Volume = 28 x 16 x 11 m³ = 4928 m³
∴  Volume of cylindrical tank = 4928 m³
Radius of the cylindrical tank = 28 m
Let h be depth of the tank, then

Question 17.
A hemispherical bowl of internal radius IS cm contains a liquid. The liquid is to be filled into cylindrical-shaped bottles of diameter S cm and height 6 cm. How many bottles are necessary to empty the bowl? (C.B.S.E. 2001C)
Solution:
Internal radius of hemispherical bowl (r) = 15 cm

Question 18.
In a cylindrical vessel of diameter 24 cm, filled up with sufficient quantity of water, a solid spherical ball of radius 6 cm is completely immersed. Find the increase in height of water level.
Solution:
Diameter of the cylindrical vessel = 24 cm

Question 19.
 A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere of lead (r₁) = 7 cm

Question 20.
A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4\((\frac { 2 }{ 3 } )\)cm and height 3 cm. Find the number of cones so formed. (C.B.S.E. 2004)
Solution:

Question 21.
The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross section. If the length of the wire is 108 m, find its diameter. (C.B.S.E. 1994)
Solution:
Diameter of copper sphere – 18 cm 18
Radius (R) = \((\frac { 18 }{ 2 } )\) = 9 cm 4
Volume = \((\frac { 4 }{ 3 } )\) π (R³)

Question 22.
A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere (R) = 7 cm

Question 23.
A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained. (C.B.S.E. 2004)
Solution:
Radius of sphere (R) = 10.5 cm
∴ Volume of sphere =\((\frac { 4 }{ 3 } )\) πR³

Question 24.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1:2:3.
Solution:
Let radius of a cone, a hemisphere and a cylinder be r
and height in each case = h
∴ h = r

Question 25.
A hollow sphere of internal and external diameters 4 and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Solution:
Outer diameter of a hollow sphere = 8 cm
∴ Outer radius (R) = \((\frac { 8 }{ 2 } )\) = 4 cm
and inner diameter = 4 cm
∴ Inner radius (r)=\((\frac { 4 }{ 2 } )\) =2 cm

Question 26.
The largest sphere is carved out of a cube of the side 10.5 cm. Find the volume of the sphere.
Solution:
Side of a cube = 10.5 cm
∵ The largest sphere is carved out of the cube,
∴ Diameter of the cube = side of the cube = 10.5 cm

Question 27.
Find the weight of a hollow sphere of metal having internal and external diameters as 20 cm and 22 cm, respectively if 1 cm³ of metal weighs 21 g.
Solution:
Internal diameter of a hollow sphere = 20 cm
and external diameter = 22 cm
∴ Outer radius (R) = \((\frac { 22 }{ 2 } )\) = 11 cm 20
and inner radius = \((\frac { 20 }{ 2 } )\) = 10 cm

Question 28.
A solid sphere of radius ‘r’ is melted and recast into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm, its height 24 cm and thinkness 2 cm, find the value of ‘r’.
Solution:
Radius of solid sphere = r

Question 29.
Lead spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18 cm and water rises by 40 cm find the number of lead spheres dropped in the water.
Solution:
Diameter of cylindrical diameter = 18 cm

Question 30.
The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm and height 4 cm arc cut off. Find the volume of the remaining solid.
Solution:
Diameter of right solid cylinder = 7 cm

Question 31.
A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. (Use TC = 3.1416)
Solution:
Total height of the solid =108 cm
Each diameter of base of hemispherical part = 36 cm

Question 32.
The surface area of a sphere is the same as the curved surface area of a cone having the radius of the bases as 120 cm and height 160 cm. Find the radius of the sphere.
Solution:

Question 33.
 A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them.
Solution:

Question 34.
A rectangular vessel of dimensions 20 cm x 16 cm x 11 cm is full of water. This water is poured into a conical vessel. The top of the conical vessel has its radius 10 cm. If the conical vessel is filled completely, determine its height. (Use π = 22/7)
Solution:
Dimension of rectangular vessel are 20 cm x 16 cm x 11 cm
Volume of vessel = 20 x 16 x 11 cm³= 3520 cm³
∴ Volume of water in conical vessel = 3520 cm³
Radius of the top of vessel = 10 cm
Let h be its height, then

Question 35.
If r₁ and r₂ be the radii of two solid metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is (r₁³ + r₁³ )1/3.
Solution:

Question 36.
A solid metal sphere of 6 cm diameter is melted and a circular sheet of thickness 1 cm is prepared. Determine the diameter of the sheet.
Solution:
Diameter of solid sphere = 6 cm 6

Question 37.
A hemispherical tank full of water is \((\frac { 25 }{ 7 } )\) emptied by a pipe at the rate of  litres per second. How much time will it take to half-empty the tank, if the tank is 3 metres in diameter ?
Solution:

Question 38.
Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:

Question 39.
The radius of the base of a right circular cone of semi-vertical angle a is r. Show that its volume is \((\frac { 1 }{ 3 } )\) πr³ cot a and curved surface area is πr² cosec α.
Solution:
Radius of circular cone = r
and semi vertical angle = α
Let AO = h and slant height AC = l
In ΔAOC, AO ⊥ BC

Question 40.
An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cubic cm of iron weighs 7.5 gm.
Solution:
Diameter of cylindrical portion = 20 cm
∴ Radius (r) = \((\frac { 20 }{ 2 } )\)  = 10 cm
Height of (h₁) = 2.8 m = 280 cm
and height of cone (h₂) = 42 cm

Question 41.
A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.
Solution:
Diameter of the tent = 105 m
∴ Radius (r) = \((\frac { 105 }{ 2 } )\)  m
Height of cylindrical part (h₁) = 3m
Slant height of conical part (h₂) – 53 m

∴ Total surface area of the tent = curved surface area of the conical part + curved surface area of the cylindrical area

Question 42.
Height of a solid cylinder is 10 cm and diameter 8 cm. Two equal conical hole have been made from its both ends. If the diameter of the holes is 6 cm and height 4 cm, find (i) volume of the cylinder, (ii) volume of one conical hole, (iii) volume of the remaining solid.
Solution:
Height of the solid cylinder (h₁) = 10 cm
Diameter = 8 cm
∴Radius (r₁) = \((\frac { 8 }{ 2 } )\) = 4 cm

Question 43.
The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm, and height 4 cm are cut off. Find the volume of the remaining solid.
Solution:
Diameter of the base of a cylinder = 7 cm
∴ Radius (r₁) = \((\frac { 7 }{ 2 } )\) cm
Height of cylinder (h₁) = 15 cm

Question 44.
A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. (Useπ = 3.1416)
Solution:
Total height of the solid =108 cm
Diameter of base of each hemisphere = 36 cm

Question 45.
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:
Radius of cylinder (r) = 7 cm
and height (h) = 14 cm

The diameter of the largest sphere curved out of the given cylinder = diameter of the cylinder
= 2 x 7 = 14 cm

Question 46.
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the base of the cylinder or the cone is 24 m. The height of the cylinder is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas required for making the tent. (Use π = 22/7)
Solution:
Diameter of the base of the cone = 24 m
∴  Radius (r) = \((\frac { 24 }{ 2 } )\) = 12 m
Height of the cylindrical part (h₁) = 11 m
Total height of the tent = 16 m
Height of the conical part (h₂)
= 16- 11 = 5 m

Question 47.
Area of the canvas required for the tent = 1320 m2 47. A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm find the total surface area and volume of the toy. (C.B.S.E. 2000, 2002)
Solution:
Radius of the toy (r) = 3.5 cm
Total height of the toy = 15.5 cm
∴ Height of the conical part = 15.5 – 3.5 = 12 cm
Slant height of the conical part (l)

Question 48.
A cylindrical container is filled with ice-cream, whose diameter is 12 cm and height is 15 cm. The whole ice-cream is distributed to 10 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice-cream.
Solution:
Diameter of the cylindrical container = 12 cm
Radius (r₁) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height (h₁) = 15 cm

Question 49.
Find the volume of a solid in the form of a right circular cylinder with hemi-spherical ends whose total length is 2.7 m and the diameter of each hemispherical end is 0.7 m.
Solution:
Total length of solid = 2.7 m
Diameter of each hemisphere at the ends = 0.7 cm

Question 50.
A tent of height 8.25 m is in the form of a right circular cylinder with diameter of base 30 m and height 5.5 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs. 45 per m².
Solution:
Total height of the tent = 8.25 m
Height of cylindrical part (h₁) = 5.5 m
∴ Height of conical part (h₂) = 8.25 – 5.5 = 2.75m

Question 51.
An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm³ of iron has 8 gram mass approximately. (Use π = 355/115)
Solution:
Diameter of the base of the cylindrical pole = 12 cm
∴ Radius (r) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height of cylindrical portion (h₁) = 110 cm
and height of conical portion (h₂) = 9 cm

Question 52.
The interior of a building is in the form of a cylinder of base radius 12 m and height 3.5 m, surmounted by a cone of equal base and slant height 12.5 m. Find the internal curved surface area and the capacity of the building.
Solution:
Radius of the building (r) = 12m
Height of the cylindrical portion (h₁) = 3.5 m and
slant height of conical portion (l) = 12.5 m

Question 53.
A right angled triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume of the double cone thus generated.
Solution:
In right angled ΔABC, ∠B = 90°
AB = 3 cm and BC = 4 cm

Now revolving the triangle along CA,

Question 54.
A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. Determine the surface area of the toy (Use π = 3.14).
Solution:
Diameter of the base of the toy = 6 cm
∴ Radius (r) = \((\frac { 6 }{ 2 } )\) = 3 cm
Height of conical portion (h) = 4 cm

Question 55.
Find the mass of a 3.5 m long lead pipe, if the external diameter of the pipe is 2.4 cm, thickness of the metal is 2 mm and the mass of 1 cm3 of lead is 11.4 grams.
Solution:
External diameter of a cylindrical pipe = 2.4 cm
Radius (R) = \((\frac { 2.4 }{ 2 } )\) = 1.2 cm
Thickness of the pipe = 2 mm =\((\frac { 2 }{ 10 } )\) = 0.2 cm
∴ Inner radius (r) = 1.2 – 0.2 = 1.0 cm
Height (length) of the pipe (h) = 3.5 m
= 350 cm
Volume of the mass of the pipe = πh (R² – r²)

Question 56.
A solid is in the form of a cylinder with hemispherical ends. Total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the solid.
Solution:
Total height of the solid = 19 cm
Diameter of the cylinder = 7 cm
Radius (r) = \((\frac { 7 }{ 2 } )\) cm
Height of the cylinder = 19 – 2 x \((\frac { 7 }{ 2 } )\) cm
= 19-7 =12cm

Question 57.
A golf ball has diameter equal to 4.2 cm. Its surface has 200 dimples each of radius 2 mm. Calculate the total surface area which is exposed to the surroundings assuming that the dimples are hemi-spherical.
Solution:
Diameter of the golf ball = 4.2 cm
∴  Radius (R) = \((\frac { 4.2 }{ 2 } )\) =2.1 cm
Radius of each hemispherical dimples (r)  = 2 mm = \((\frac { 2 }{ 10 } )\) = \((\frac { 2 }{ 5 } )\) cm
Curved surface area of one dimple = 2πr²

Question 58.
The radii of the ends of a bucket of height 24 cm are 15 cm and 5 cm. Find its capacity. (Take π = 22/7).
Solution:
Height of the bucket (frustum) (h) = 24 cm
Upper radius (r₁) = 15 cm
and lower radius (r₂) = 5 cm

Question 59.
The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity in litres and the amount of sheet required to make this bucket.
Solution:
Height of the bucket (frustum) (h) = 30 cm
Upper radius (r₁) = 21 cm
and lower radius (r₂) = 7 cm

Question 60.
The radii of the ends of a frustum of a right circular cone are 5 metres and 8 metres and its lateral height is 5 metres. Find the lateral surface and volume of the frustum.
Solution:
Upper radius of a frustum (r₁) = 8 m
and lower radius (r₂) = 5 m
Lateral height (l) = 5m

Question 61.
A frustum of a cone is 9 cm thick and the diameters of its circular ends are 28 cm and 4 cm. Find the volume and lateral surface area of the frustum. (Take π = 22/7)
Solution:
Upper diameter = 28 cm
and lower diameter = 4 cm
Height (h) = 9 cm

Question 62.
A bucket is in the form of a frustum of a cone and holds 15.25 litres of water. The diameters of the top and bottom are 25 cm and 20 cm respectively. Find its height and area of tin used in its construction.
Solution:
Water in a bucket (frustum) = 15.25l
Upper diameter = 25 cm
and lower diameter = 20 cm
∴ Upper radius (r₁) = \((\frac { 25 }{ 2 } )\) cm
and lower radius (r₂) = \((\frac { 20 }{ 2 } )\) cm =10 cm
Volume = 15.25 /= 1525 x 10 cm³ = 15250 cm³
Let h be its height

Question 63.
If a cone of radius 10 cm is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts. (C.B.S.E. 2000C)
Solution:
Radius of the cone (r₁) = 10 cm
Cone is divided into 2 parts Such that PQ || AB

Question 64.
A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 pnetres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres
Solution:
Radius of the cylinder (r) = 14 m
and total height of the tent = 13.5 m
Height of the cylindrical part (h₁) = 3 m
Height of conical part (h₂) = 13.5-3.0 = 10.5m

Question 65.
An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height be 22 cm, the diameter of the cylindrical portion 8 cm and the diameter of the top of the funnel 18 cm, find the area of the tin required. (Use : π = 22/7).
Solution:
Upper diameter of the frustum = 18 cm
and  lower diameter = 8m

Question 66.
A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys with the shape of a right circular cone mounted on a hemisphere of radius 3 cm. If the height of the toy is 12 cm, find the number of toys so formed. (C.B.S.E. 2006C)
Solution:
Diameter of solid cylinder = 12 cm
and height (h₁) = 15 cm

Question 67.
A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of ₹21 per litre. (Use π = 22/7)
Solution:
Upper radius (R) = 20 cm
Lower radius (r) = 8 cm
Height (h) = 24 cm

Question 68.
 A cone of maximum size is carved out from a cube of edge 14 cm. Find the . surface area of the cone and of the remaining solid left out after the cone carved out. [NCERT Exemplar]
Solution:
The cone of maximum size that is carved out from a cube of edge 14 cm will be of base radius 7 cm and the height 14 cm.

Question 69.
A cone of radius 4 cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base. Compare the volumes of two parts. [NCERT Exemplar]
Solution:
Let h be the height of the given cone. One dividing the cone through the mid-point of its axis and parallel to its base into two parts, we obtain the following figure:

Question 70.
A wall 24 m, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm x 16 cm x 10 cm. If the mortar occupies \((\frac { 1 }{ 10 } )\) th of the volume of the wall, then find the number of bricks used in constructing the wall. [NCERT Exemplar]
Solution:
Given that, a wall is constructed with the help of bricks and mortar.
∴ Number of bricks

Question 71.
A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket. [NCERT Exemplar] 
Solution:

Question 72.
Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm. [NCERT Exemplar]
Solution:

Question 73.
Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape formed. [NCERT Exemplar]
Solution:
If two cones with same base and height are joined together along their bases, then the shape so formed is look like as figure shown.

Question 74.
From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid. [NCERT Exemplar]
Solution:
Given that, side of a solid cube (a) = 1 cm
Height of conical cavity i.e., cone, h = 7 cm

Since, the height of conical cavity and the side of cube is equal that means the conical cavity fit vertically in the cube.
Radius of conical cavity i. e., cone, r = 3 cm
⇒ Diameter = 2 x r = 2 x 3 = 6 cm
Since, the diameter is less than the side of a cube that means the base of a conical cavity is not fit inhorizontal face of cube.
Now, volume of cube = (side)³ = a³ = (7)³ = 34³ cm³
and volume of conical cavity i.e., cone

Question 75.
Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacitites are 2 : 1. Find the heights and capacities of the cones. Also, find the volume of the remaining portion of the cylinder. [NCERT Exemplar]

Solution:

Question 76.
An icecream cone full of icecream having radius 5 cm and height 10 cm as shown’in the figure. Calculate the volume of icecream, provided that its 1/6 parts is left unfilled with icecream.

Solution:

 

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RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Question 1.
Show that the sequence defined by a_n = 5n – 7 is an A.P., find its common difference.
Solution:

Question 2.
Show that the sequence defined by a_n = 3n² – 5 is not an A.P.
Solution:

Question 3.
The general term of a sequence is given by a_n = -4n + 15. Is the sequence an A.P.? If so, find its 15th term and the common difference.
Solution:
General term of a sequence
a_n = -4n + 15

Question 4.
Write the sequence with nth term :
(i) a_n = 3 + 4n
(ii) a_n = 5 + 2n
(iii) a_n = 6 – n
(iv) a_n = 9 – 5n
Show that all of the above sequences form A.P.
Solution:

Question 5.
The nth term of an A.P. is 6n + 2. Find the common difference. [CBSE 2008]
Solution:

Question 6.
Justify whether it is true to say that the sequence, having following nth term is an A.P.
(i) a_n = 2n – 1
(ii) a_n = 3n² + 5
(iii) a_n = 1 + n + n²
Solution:

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RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Question 1.
Write the first five terms of each of the following sequences whose nth terms are:

Solution:

Question 2.
Find the indicated terms in each of the following sequences whose nth terms are:

Solution:

Question 3.
Find the next five terms of each of the following sequences given by :

Solution:

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RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3

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 14.3

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.
A bucket has top and bottom diameters of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm. Also, find the cost of tin sheet used for making the bucket at the rate of ?1.20 per dm2. (Use % = 3.14)
Solution:
Upper diameter = 40 cm
and lower diameter = 20 cm
∴  Upper radius (r₁) = \((\frac { 40 }{ 2 } )\) = 20 cm
and lower radius (r₂) = \((\frac { 20 }{ 2 } )\) = 10 cm
Depth or height (h) = 12 cm
Volume of the bucket

Question 2.
A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.
Solution:
Base diameter of frustum = 20 cm 20
∴  Radius (r₁) = \((\frac { 20 }{ 2 } )\) = 10 cm
and diameter of top = 12 cm

Question 3.
The slant height of the frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface of the frustum.
Solution:
Perimeter of the top of frustum = 18 cm

Question 4.
The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.
Solution:
Perimeter of the top of frustum = 44 cm

Question 5.
If the radii of the circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm, find the capacity of the bucket. (Use π = 22/7). (C.B.S.E. 2000)
Solution:

Question 6.
The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If its volume be \((\frac { 1 }{ 125 } )\)  of the volume of the original cone, determine at what height above the base the section is made.
Solution:
Total height of the cone (h₁) = 20 cm
Let a cone whose height is h₂ is cut off Then height of the remaining portion (frustum)

Question 7.
If the radii of the circular ends of a bucket 24 cm high are 5 cm and 15 cm respectively, find the surface area of the bucket.
Solution:
Height of the bucket (frustum) (A) = 24 cm
Radius of the top (r₁) = 15 cm 1
and radius of the bottom (r₂) = 5 cm

Question 8.
The radii of the circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 12 cm. Find the total surface area and the volume of the frustum.
Solution:
Height of the frustum (A) = 12 cm
Radius of the top (r₁) = 12 cm
and radius of the bottom (r₂) = 3 cm

Question 9.
A tent consists of a frustum of a cone capped by a cone. If the radii of the ends of the frustum be 13 m and 7 m, the height of the frutum be 8 m and the slant height of the conical cap be 12 m, find the canvas required for the tent. (Take : π = 22/7)
Solution:
Radius of the bottom of the tent (r₁) = 13 m
and radius of the top (r₂) = 7 m
Height of frustum portion (h₁) = 8 m
Slant height of the conical cap (l₂) = 12 m
Let l₁ be the slant height of the frustum portion, then

Question 10.
A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ₹44 per litre which the container can hold. [NCERT Exemplar]
Solution:
Given that, height of milk container (h) = 16 cm
Radius of lower end of milk container (r) = 8 cm
and radius of upper end of milk container (R) = 20 cm

∴ Volume of the milk container made of metal sheet in the form of a frustum of a cone

Question 11.
A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container, at the rate of ₹25 per litre. (Use π = 3.14) [NCERT Exemplar]
Solution:

Question 12.
A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm³ of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making. (Use π = 3.14). [CBSE 2006C]
Solution:
Volume of frustum (bucket) = 12308.8 cm³
Upper radius (r₁) = 20 cm
and lower radius (r₂) = 12 cm

Question 13.
A bucket made of aluminium sheet is of height 20 cm and its upper and lower ends are of radius 25 cm and 10 cm respectively. Find the cost of making the bucket if the aluminium sheet costs Rs. 70 per 100 cm². (Use π = 3.14) (C.B.S.E. 2006C)
Solution:
Height of bucket (frustum) (h) = 20 cm
Upper radius (r₁) = 25 cm
and lower radius (r₂) = 10

Question 14.
The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area. (C.B.S.E. 2005)
Solution:
Upper radius of frustum (r₁) = 3.3 cm
and lower radius (r₂) = 27 cm
Slant height (l) = 10 cm

Question 15.
A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the volume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used is Rs. 20 per 100 cm². (Use π = 3.14) (CBSE 2008)
Solution:
Lower radius of bucket (r) = \((\frac { 16 }{ 2 } )\) = 8 cm
and upper radius (R) = \((\frac { 40 }{ 2 } )\) = 20 cm
Height (h) = 16

Question 16.
A solid is in the shape of a frustum of a cone. The diameter of the two circular ends are 60 cm and 36 cm and the leight is 9 ³cm. Find the area of its whole surface and the volume. [CBSE 2010]
Solution:
In a solid frustum upper diameter = 60 cm
∴ Radius (r₁) = \((\frac { 60 }{ 2 } )\) = 30 cm
Lower diameter = 36 cm 36
∴ Radius (r₂) = \((\frac { 36 }{ 2 } )\) = 18 cm
Height (h) = 9 cm

Question 17.
A milk container is made of metal sheet in the shape of frustum of a cone whose volucrn is 10459\((\frac { 3 }{ 7 } )\) cm³. The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm². (Use π = 22.7) [CBSE 2010]
Solution:
Volume of frustum = 10459\((\frac { 3 }{ 7 } )\) cm3 73216
= \((\frac { 73216 }{ 7 } )\) cm³
Lower radius (r₂) = 8 cm
and upper radius (r₁) = 20 cm

Question 18.
A solid cone of base radius 10 cm is cut into two parts through the mid-pint of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone. [CBSE 2013]
Solution:
Radius of solid cone (r) = 10 cm
Let total height = h
In ΔAOB,
C is mid point of AO and CD || OB

Question 19.
A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of ₹10 per 100 cm². (Use π = 3.14). [CBSE 2013]
Solution:

Question 20.
In the given figure, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use π = 22/7 and \(\sqrt { 5 } \) = 2.236).  [ CBSE 2015]

Solution:
Total height of cone = 12 cm
Radius of its base = 6 cm
A cone of height 4 cm is cut out
Height of the so formed frustum = 12 – 4 = 8 cm
Let r be the radius of the cone cut out

Question 21.
The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.
Solution:
Let the height and radius of the given cone be H and R respectively.
The cone is divided into two parts by drawing a plane through the mid point of its axis and parallel to the base.
Upper part is a smaller cone and the bottom part is the frustum of the cone.

Question 22.
A bucket, made of metal sheet, is in the form of a cone whose height is 35 cm and radii of circular ends are 30 cm and 12 cm. How many litres of milk it contains if it is full to the brim? If the milk is sold at ₹40 per litre, find the amount received by the person. [CBSE 2017]
Solution:
Radii of the bucket in the form of frustum of cone = 30 cm
and 12 cm Depth of the bucket = 35 cm

Question 23.
A reservoir in the form of the frustum of a right circular cone contains 44 x 10⁷ litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir. (Take : π = 22/7)
Solution:
A reservoir is a frustum in shape and its upper radius (r₁) = 100 m
Lower radius (r₂) = 50 m
and capacity of water in it = 44 x 10⁷ litres

P.Q. A metallic right circular cone 20 cm high and whose vertical angle is 90° is cut into two parts at the middle point of its axis by a plane parallel to the base. If the frustum so obtained be drawn into a wire  \((\frac { 1 }{ 16 } )\)cm, find the length of the wire.

Solution: In the cone ABC, ∠A = 90°, AL ⊥ BC and = 20 cm
It is cut into two parts at the middle point M on the axis AL
AL bisects ∠A also

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2

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 14.2

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.
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent.
Solution:
Diameter of the base of the tent = 24 m
∴ Radius (r)= \((\frac { 24 }{ 2 } )\)  = 12m
Total height of the tent = 16 m
Height of the cylindrical portion (h₁) = 11 m
Height of the conical portion (h₂) =16-11 = 5 m

Question 2.
A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area and the volume of the rocket.
Solution:
Radius of the base of the rocket (r) = 2.5 m
Height of cylindrical portion (h₁) = 21 m
Slant height of the conical portion (l) = 8 m
Let height of conical portion = h₂

Question 3.
A tent of height 77 dm is in the form of 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 Rs. 3.50 per m² . (Use π = 22/7).
Solution:
Total height of the tent = 77 dm
Height of cylindrical part (h₁) = 44 dm
= 4.4 m
Height of conical part (h₂) = 7.7 – 4.4 = 3.3 m
Diameter of the base of the tent = 36 m

Question 4.
A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Determine the surface area of the toy. (Use π = 3.14).
Solution:
Diameter of the base of the toy = 6 cm
∴ Radius (r) = \((\frac { 6 }{ 2 } )\)  = 3 cm
Height (h) = 4 cm

Total surface area of the toy = curved surface area of the conical part + surface area of the hemispherical part
= πrl + 2πr² = πr (l + 2r)
= 3.14 x 3 (5 + 6) = 3.14 x 3 x 11 cm²
= 3.14 x 33 = 103.62 cm²

Question 5.
A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the ~ heights of the cylindrical and conical portions are 10 cm and 6 cm, respectively. Find the total surface area of the solid. (Use π = 22/7)
Solution:
Radius of the common base (r) = 3.5 m
Height of cylindrical part (h₁) = 10 cm
Height of conical part (h₂) = 6 cm

Now total surface area of the solid = curved surface of conical part + curved surface of cylindrical part + curved surface of hemispherical part

Question 6.
A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm. (C.B.S.E. 2002)
Solution:
Radius of the base of the cylindrical part (r) = 5 cm

Height of cylindrical part (h₁) = 13 cm
Height of the conical part (h₂) = 30 – (13 + 5) = 30- 18 = 12 cm

Question 7.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub. (Take π = 22/7) (C.B.S.E. 2000C)
Solution:
Radius of the cylindrical tub (R) = 5 cm
and height (h₁) = 9.8 cm
Radius of the solid (r) = 3.5 cm
and height of cone (h₁) = 5 cm

= 38.5 x 4 = 154 cm3
∴ Water flowed out of the tub = 154 cm³
Remaining water in the tub = 770 – 154
= 616 cm³

Question 8.
A circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20 m. The heights of the cylindrical and conical portions are 4.2 m and 2.1 m respectively. Find the volume of the tent.
Solution:
Radius of the tent (r) = 20 m
Height of the conical part (h₁) = 2.1 m
and height of the cylindrical part (h₂) = 4.2 m

Question 9.
A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank.
Solution:
Diameter of cylindrical part = 21 cm 21
∴ Radius (r) = \((\frac { 21 }{ 2 } )\) cm
Height of cylindrical part (h₁) = 18 cm
and height of each conical part (h₂) = 9 cm

Question 10.
A conical hole is drilled in a circular cylinder of height 12 cm and base radius 5 cm. The height and the base radius of the cone are also the same. Find the whole surface and volume of the remaining cylinder.
Solution:
Base radius of the cylinder (r) = 5 cm
and height (h) = 12 cm
∴ Volume = πr²h = n (5)2 x 12 cm3 = 300π cm³
∵ The base and height of the cone drilled are the same as those of the cylinder
∴ Volume of cone = \((\frac { 1 }{ 3 } )\)πr²h

Question 11.
A tent is in the form of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of equal base and height 7.5 m. Find the capacity of the tent and the cost of the canvas at Rs. 100 per square metre.
Solution:
Diameter of the base of the tent = 20 m
∴ Radius (r) =\((\frac { 20 }{ 2 } )\) = 10 m
Height of cylindrical part (h₁) = 2.5 m
and height of conical part (h₂) = 7.5 m
Slant height of the conical part (l)

Question 12.
A boiler is in the form of a cylinder 2 m long with hemispherical ends each of 2 metre diameter. Find the volume of the boiler.
Solution:
Diameter of the cylinder = 2 m
∴ Radius (r) = \((\frac { 1 }{ 3 } )\) = 1 m
Height (length) of cylindrical part (h) = 2 m

Question 13.
A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylinder is \((\frac { 14 }{ 3 } )\) m and the diameter of hemisphere is 3.5 m. Calculate the volume and the internal surface area of the solid.
Solution:
Diameter of the cylindrical part = 3.5 m
∴Radius (r) = \((\frac { 3.5 }{ 2 } )\) = 1.75 = \((\frac { 7 }{ 4 } )\) m
and height (h) = \((\frac { 14 }{ 3 } )\) m

Question 14.
A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm, find the cost of polishing its surface at the rate of Rs. 10 per dm² . (C.B.S.E. 2006C)
Solution:
Total height = 104 cm
Radius of hemispherical part (r) = 7 cm
Height of cylinder (h) = 104 cm – 2 x 7 cm = 104- 14 = 90 cm

Total outer surface area = curved surface area of the cylindrical part + 2 x curved surface area of each hemispherical part

Question 15.
A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimeters of cork dust will be required ?
Solution:
Diameter of inner cylinder = 14 cm
∴Radius (r) = \((\frac { 14 }{ 2 } )\) = 7 cm
Diameter of outer cylinder = 16 cm
∴ Radius (R) = \((\frac { 16 }{ 2 } )\) = 8 cm
Height (h) = 42cm

Question 16.
A cylindrical road roller made of iron is 1 m long. Its internal diameter is 54 cm and the thickness of the iron sheet used in making the roller is 9 cm. Find the mass of the roller, if 1 cm³ of iron has 7.8 gm mass. (Use π = 3.14)
Solution:
Length of roller (h) = 1 m = 100 cm
Inner diameter = 54 cm
Thickness of iron sheet = 9 cm
∴ Inner radius (r) = \((\frac { 52 }{ 2 } )\) = 27 cm
and outer radus (R) = 27 + 9 = 36 cm
∴ Volume of the mass = πR²h – πr²h
= πh (R²- r²)
= 3.14 x 100 (36² – 27²) cm³
= 314 x (36 + 27) (36- 27) cm³
= 314 x 63 x 9 cm^c
= 178038 cm³
Weight of 1 cm³ of iron = 7.8 gm
∴ Total weight = 178038 x 7.8 gm
= 1388696.4 gm
= 1388.6964 kg
= 1388.7 kg

Question 17.
A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
Diameter of hemisphere = 14 cm
Total height =13 cm
Radius of hemisphere = \((\frac { 14 }{ 2 } )\) = 7 cm
∴ Height of cylindrical part =13-7 = 6 cm

∴  Inner surface area of the vessel = inner surface area of cylindrical part + inner surface area of hemispherical part

Question 18.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius of cone (r) – 3.5 cm
Total height of the toy = 15.5 cm

Height of the conical part (h) = 15.5 – 3.5 = 12 cm
∴ Slant height of the cone (l)

Question 19.
The difference between outside and inside surface areas of cylindrical metallic pipe 14 cm long is 44 m². If the pipe is made of 99 cm³ of metal, find the outer and inner radii of the pipe.
Solution:
In cylindrical metallic pipe,
length of pipe = 14 cm

Difference between outside and inside
surface area = 44 m²
Volume of pipe material = 99 cm³
Let R and r be the outer and inner radii of the pipe respectively, then Outer surface area – inner surface area = 44 cm²

Question 20.
A right circular cylinder having diameter 12 cm and height 15 cm is full ice-cream. The ice-cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Solution:
Height of cylinder (H) = 15 cm
and diameter =12 cm

Question 21.
A solid iron pole having cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that the mass of 1 cm3 of iron is 8 gm.
Solution:
Diamter of the base = 12 cm
∴ Radius (r) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height of the cylindrical portion (h₁)= 110 cm
and height of conical portion (h₂) = 9 cm

Question 22.
A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover.
Solution:
Height of conical part (h) = 2 cm
Diameter of base = 4 cm

Now volume of the cylinder which circum scribes the toy = πr²h
= π (2)² x 4 = 16π cm³
∴  Difference of their volumes = 16π – 8π = 8πcm³

Question 23.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottoms. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of conical part = 60 cm
and height (h) = 120 cm

Question 24.
A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the value of water (i) displaced out of the cylinder. (ii) left in the cylinder. (C.B.S.E. 2009)
Solution:
Internal diameters of cylindrical vessel = 10 cm
∴ Radius (r) = \((\frac { 10 }{ 2 } )\) =5 cm
and height (h) = 10.5 cm
∴ Volume of water filled in it

Question 25.
A hemispherical depression is cut out from one face of a cubical wooden block of edge 21 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Determine the volume and total surface area of the remaining block. [CBSE 2010]
Solution:
Edge of cube = 21 cm
∴ Diameter of the hemisphere curved out of

Question 26.
A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is 2/3 of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy. (Use π = 22/7).
Solution:
Radius of base of the conical part (r) = 21 cm

Question 27.
A solid is in the shape of a cone surmounted on a hemi-sphere, the radius of each of them is being 3.5 cm and the total height of solid is 9.5 cm. Find the volume of the solid. [CBSE 2012]
Solution:

Question 28.
An wooden toy is made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius
3.5 cm, find the volume of wood in the toy. (Use π = 22/7). [CBSE 2013]
Solution:
Height of cylindrical part (h) = 10 cm
Radius of the base (r) = 3.5 cm

Question 29.
The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left. (Use π = 22/7). [CBSE 2014]
Solution:

Question 30.
From a solid cylinder of height 2.8 cm and diameter 4.2 cm, conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid, (take π = 22/7). [CBSE 2014]
Solution:
Diameter of solid cylinder= 4.2 cm
∴ Radius (r) =\((\frac { 4.2 }{ 2 } )\) = 2.1 cm
Height (h) = 2.8 cm

Question 31.
The largest cone is curved out from one face of solid cube of side 21 cm. Find the volume of the remaining solid. [CBSE 2015]
Solution:
Side of a solid cube (a) = 21 cm
∴ Volume = a³, = (21 )³ cm³
= 9261 cm³
Diameter of the base of cone = 21 cm
Now radius of cone curved from it (r) =\((\frac { 21 }{ 2 } )\) cm
and height (h) = 21 cm

Question 32.
A solid wooden toy is in the form of a hemisphere surmounted by a Cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is 166 \((\frac { 5 }{ 6 } )\) cm³ . Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of ₹10 per cm². (Take π  = 22/7). [CBSE 2015]
Solution:

Question 33.
In the given figure, from a cuboidal solid metalic block, of dimensions 15 cm x 10 cm x 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. (Take  π = 22/7) [CBSE 2015]
Solution:

Radius of hole = \((\frac { 7 }{ 2 } )\)cm and height = 5 cm
Length of block (l) = 15 cm
Breadth (b) = 10 cm and height = 5 cm
∴ Surface area = 2(lb + bh + hl)
= 2(15 x 10 + 10 x 5 + 5 x 15) cm²
= 2(150 + 50 + 75) = 2 x 275 = 550 cm²
Area of circular holes of both sides = 2 x πr²

Question 34.
 A building is in the form of a cylinder surmounted by a hemi-spherical vaulted 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? [NCERT Exemplar]
Solution:
Let total height of the building = Internal diameter of the dome = 2rm

Radius of building (or dome) = \((\frac { 2r }{ 2 } )\) = r m
Height of cylinder = 2r-r = rm

Question 35.
A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm x 5 cm x 4 cm. The radius of each of the conical depression is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand. [NCERT Exemplar]
Solution:
Given that, length of cuboid pen stand (l) = 10 cm
Breadth of cuboid pen stand (b) = 5 cm
and height of cuboid pen stand (h) = 4 cm

∴ Volume of cuboid pend stand = l x b x h= 10 x 5 x 4 = 200 cm³
Also, radius of conical depression (r) = 0.5 cm
and height (depth) of a conical depression (h₁) = 2.1 cm
∴ Volume of a conical depression = πrh₁

Question 36.
 A building is in the form of a cylinder surmounted by a hemispherical dome. The base diameter of the dome is equal to \((\frac { 2 }{ 3 } )\) of the total height of the building. Find the height of the building, if it contains 67 \((\frac { 1 }{ 21 } )\) m³ of air.
Solution:
Let the radius of the hemispherical dome be r metres and the total height of the building be h metres.
Since the base diameter of the dome is equal to \((\frac { 2 }{ 3 } )\) of the total height, therefore
2 r = \((\frac { 2 }{ 3 } )\)h. This implies r = \((\frac { h }{ 3 } )\). Let H metres be the height of the cylindrical portion.
Therefore, H = h – \((\frac { h }{ 3 } )\) = \((\frac { 2 }{ 3 } )\)h metres.
Volume of the air inside the building = Volume of air inside the dome + Volume of the air inside the
cylinder = \((\frac { 2 }{ 3 } )\) πr³ + πr²H, where H is the height of the cylindrical portion

Question 37.
A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy. [NCERT Exemplar]
Solution:
Let r be the radius of the hemisphere and the cone and h be the height of the cone.
Volume of the toy=Volume of the hemisphere + Volume of the cone

= \((\frac { 1408 }{ 7 } )\) cm³
A cube circumsrcibes the given solid. Therefore, edge of the cube should be 8 cm. Volume of the cube = 83 cm3 = 512 cm3 Difference in the volume of the cube and

Question 38.
A circus tent is in the shape of a cylinder surmounted by a conical top of same diameter. If their common diameter is 56 m, the height of the cylindrical part is 6 m and the total height of the tent above the ground is 27 m, find the area of the canvas used in making the tent. [ICBSE 2017]
Solution:
We have, diameter of base of cylinder = d = 56 m
Radius of base of cylinder = r=  \((\frac { d }{ 2 } )\)=  \((\frac { 52 }{ 2 } )\)= 28 m
Height of tent = 27 m
Height of cylinder = 6 m

Height of conical portion = 27 – 6 = 21 m
Radius of conical portion, r = 28 m

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 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 MCQS

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

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

Mark the correct alternative in each of the following :
Question 1.
If the equation x² + 4x + k = 0 has real and distinct roots, then
(a) k < 4
(b) k > 4
(c) k ≥ 4
(d) k ≤ 4
Solution:
(a) In the equation x² + 4x + k = 0
a = 1, b = 4, c = k
D = b² – 4ac = (4)² – 4 x 1 x k = 16 – 4k
Roots are real and distinct
D > 0
=> 16 – 4k > 0
=> 16 > 4k
=> 4 > k
=> k < 4

Question 2.
If the equation x² – ax + 1 = 0 has two distinct roots, then
(a) |a| = 2
(b) |a| < 2
(c) |a| > 2
(d) None of these
Solution:
(c) In the equation x² – ax + 1 = 0
a = 1, b = – a, c = 1
D = b² – 4ac = (-a)² – 4 x 1 x 1 = a² – 4
Roots are distinct
D > 0
=> a² – 4 > 0
=> a² > 4
=> a² > (2)²
=> |a| > 2

Question 3.
If the equation 9x² + 6kx + 4 = 0, has equal roots, then the roots are both equal to
(a) ± \(\frac { 2 }{ 3 }\)
(b) ± \(\frac { 3 }{ 2 }\)
(c) 0
(d) ± 3
Solution:
(a)

Question 4.
If ax² + bx + c = 0 has equal roots, then c =

Solution:
(d) In the equation ax² + bx + c = 0
D = b² – 4ac
Roots are equal
D = 0 => b² – 4ac = 0
=> 4ac = b²
=> c = \(\frac { { b }^{ 2 } }{ 4a }\)

Question 5.
If the equation ax² + 2x + a = 0 has two distinct roots, if
(a) a = ±1
(b) a = 0
(c) a = 0, 1
(d) a = -1, 0
Solution:
(a) In the equation ax² + 2x + a = 0
D = b² – 4ac = (2)² – 4 x a x a = 4 – 4a²
Roots are real and equal
D = 0
=> 4 – 4a² = 0
=> 4 = 4a²
=> 1 = a²
=> a² = 1
=> a² = (±1)²
=> a = ±1

Question 6.
The positive value of k for which the equation x² + kx + 64 = 0 and x² – 8x + k = 0 will both have real roots, is
(a) 4
(b) 8
(c) 12
(d) 16
Solution:
(d) In the equation x² + kx + 64 = 0

Question 7.

Solution:
(b)

Which is not possible
x = 3 is correct

Question 8.
If 2 is a root of the equation x² + bx + 12 = 0 and the equation x² + bx + q = 0 has equal roots, then q =
(a) 8
(b) – 8
(c) 16
(d) -16
Solution:
(c)

Question 9.
If the equation (a² + b²) x² – 2 (ac + bd) x + c² + d² = 0 has equal roots, then
(a) ab = cd
(b) ad = bc
(c) ad = √bc
(d) ab = √cd
Solution:
(b)

Question 10.
If the roots of the equation (a² + b²) x² – 2b (a + c) x + (b² + c²) = 0 are equal, then ;
(a) 2b = a + c
(b) b² = ac
(c) b = \(\frac { 2ac }{ a + c }\)
(d) b = ac
Solution:
(b)

Question 11.
If the equation x² – bx + 1 = 0 does not possess real roots, then
(a) -3 < b < 3
(b) -2 < b < 2
(c) b > 2
(d) b < -2
Solution:
(b)

Question 12.
If x = 1 is a common root of the equations ax² + ax + 3 = 0 and x² + x + b = 0, then ab =
(a) 3
(b) 3.5
(c) 6
(d) -3
Solution:
(a) In the equation
ax² + ax + 3 = 0 and x² + x + b = 0
Substituting the value of x = 1, then in ax² + ax + 3 = 0

Question 13.
If p and q are the roots of the equation x² – px + q + 0, then
(a) p = 1, q = -2
(b) p = 0, q = 1
(c) p = -2, q = 0
(d) p = -2, q = 1
Solution:
(a)

Question 14.
If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax² + bx + 1 = 0 having real roots is
(a) 10
(b) 7
(c) 6
(d) 12
Solution:
(b)
ax² + bx + 1 = 0
D = b² – 4a = b² – 4a
Roots are real
D ≥ 0
=> b² – 4a ≥ 0
=> b² ≥ 4a
Here value of b can be 2, 3 or 4
If b = 2, then a can be 1,
If b = 3, then a can be 1, 2
If b = 4, then a can be 1, 2, 3, 4
No. of equation can be 7

Question 15.
The number of quadratic equations having real roots and which do not change by squaring their roots is
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c) There can be two such quad, equations whose roots can be 1 and 0
The square of 1 and 0 remains same
No. of quad equation are 2

Question 16.
If (a² + b²) x² + 2(ab + bd) x + c² + d² = 0 has no real roots, then
(a) ad = bc
(b) ab = cd
(c) ac = bd
(d) ad ≠ bc
Solution:
(d)

Question 17.
If the sum of the roots of the equation x² – x = λ (2x – 1) is zero, then λ =
(a) -2
(b) 2
(c) – \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ 2 }\)
Solution:
(c)

Question 18.
If x = 1 is a common root of ax² + ax + 2 = 0 and x² + x + b = 0 then, ab =
(a) 1
(b) 2
(c) 4
(d) 3
Solution:
(b)

Question 19.
The value of c for which the equation ax² + 2bx + c = 0 has equal roots is

Solution:
(a)

Question 20.
If x² + k (4x + k – 1) + 2 = 0 has equal roots, then k =

Solution:
(b)

Question 21.
If the sum and product of the roots of the equation kx² + 6x + 4k = 0 are equal, then k =

Solution:
(b)

Question 22.
If sin α and cos α are the roots of the equations ax² + bx + c = 0, then b² =
(a) a² – 2ac
(b) a² + 2ac
(b) a² – ac
(d) a² + ac
Solution:
(b)

Question 23.
If 2 is a root of the equation x² + ax + 12 = 0 and the quadratic equation x² + ax + q = 0 has equal roots, then q =
(a) 12
(b) 8
(c) 20
(d) 16
Solution:
(d)

Question 24.
If the sum of the roots of the equation x² – (k + 6) x + 2 (2k – 1) = 0 is equal to half of their product, then k =
(a) 6
(b) 7
(c) 1
(d) 5
Solution:
(b) In the quadratic equation
x² – (k + 6) x + 2 (2k – 1) = 0
Here a = 1, b = – (k + 6), c = 2 (2k – 1)

Question 25.
If a and b are roots of the equation x² + ax + b = 0, then a + b =
(a) 1
(b) 2
(c) -2
(d) -1
Solution:
(d) a and b are the roots of the equation x² + ax + b = 0
Sum of roots = – a and product of roots = b
Now a + b = – a
and ab = b => a = 1 ….(i)
2a + b = 0
=> 2 x 1 + b = 0
=> b = -2
Now a + b = 1 – 2 = -1

Question 26.
A quadratic equation whose one root is 2 and the sum of whose roots is zero, is
(a) x² + 4 = 0
(b) x² – 4 = 0
(c) 4x² – 1 = 0
(d) x² – 2 = 0
Solution:
(b) Sum of roots of a quad, equation = 0
One root = 2
Second root = 0 – 2 = – 2
and product of roots = 2 x (-2) = – 4
Equation will be
x² + (sum of roots) x + product of roots = 0
x² + 0x + (-4) = 0
=> x² – 4 = 0

Question 27.
If one root of the equation ax² + bx + c = 0 is three times the other, then b² : ac =
(a) 3 : 1
(b) 3 : 16
(c) 16 : 3
(d) 16 : 1
Solution:
(c)

Question 28.
If one root of the equation 2x² + kx + 4 = 0 is 2, then the other root is
(a) 6
(b) -6
(c) -1
(d) 1
Solution:
(d) The given quadratic equation 2x² + kx + 4 = 0
One root is 2
Product of roots = \(\frac { c }{ a }\) = \(\frac { 4 }{ 2 }\) = 2
Second root = \(\frac { 2 }{ 2 }\) = 1

Question 29.
If one root of the equation x² + ax + 3 = 0 is 1, then its other root is
(a) 3
(b) -3
(c) 2
(d) -2
Solution:
(a) The quad, equation is x² + ax + 3 = 0
One root =1
and product of roots = \(\frac { c }{ a }\) = \(\frac { 3 }{ 1 }\) = 3
Second root = \(\frac { 3 }{ 1 }\) = 3

Question 30.
If one root of the equation 4x² – 2x + (λ – 4) = 0 be the reciprocal of the other, then λ =
(a) 8
(b) -8
(c) 4
(d) -4
Solution:
(a)

Question 31.
If y = 1 is a common root of the equations ay² + ay + 3 = 0 and y² + y + b = 0, then ab equals
(a) 3
(b) – \(\frac { 1 }{ 2 }\)
(c) 6
(d) -3 [CBSE 2012]
Solution:
(a)

Question 32.
The values of k for which the quadratic equation 16x² + 4kx + 9 = 0 has real and equal roots are
(a) 6, – \(\frac { 1 }{ 6 }\)
(b) 36, -36
(c) 6, -6
(d) \(\frac { 3 }{ 4 }\) , – \(\frac { 3 }{ 4 }\) [CBSE 2014]
Solution:
(c) 16x² + 4kx + 9 = 0
Here a = 16, b = 4k, c = 9
Now D = b² – 4ac = (4k)² – 4 x 16 x 9 = 16k² – 576
Roots are real and equal
D = 0 or b² – 4ac = 0
=> 16k² – 576 = 0
=> k² – 36 = 0
=> k² = 36 = (± 6)²
k = ± 6
k = 6, -6

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations 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 4 Quadratic Equations VSAQS

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

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

Answer each of the following questions either in one word or one sentence or as per requirement of the question :
Question 1.
Write the value of k for which the quadratic equation x² – kx + 4 = 0 has equal roots.
Solution:
x² – kx + 4 = 0
Here a = 1, b = – k, c = 4
Discriminant (D) = b² – 4ac
= (-k)² – 4 x 1 x 4 = k² – 16
The roots are equal
D = 0 => k² – 16 = 0
=> (k + 4) (k – 4) = 0.
Either k + 4 = 0, then k = – 4
or k – 4 = 0, then k = 4
k = 4, -4

Question 2.
What is the nature of roots of the quadratic equation 4x² – 12x – 9 = 0 ?
Solution:
4x² – 12x – 9 = 0
Here a = 4, b = -12, c = – 9
Discriminant (D) = b² – 4ac = (-12)² – 4 x 4 x (-9)
= 144 + 144 = 288
D > 0
Roots are real and distinct

Question 3.
If 1 + √2 is a root of a quadratic equation with rational co-efficients, write its other root.
Solution:
The roots of the quadratic equation with rational co-efficients are conjugate
The other root will be 1 – √2

Question 4.
Write the number of real roots of the equation x² + 3 |x| + 2 = 0.
Solution:

Question 5.
Write the sum of the real roots of the equation x² + |x| – 6 = 0.
Solution:

Question 6.
Write the set of values of ‘a’ for which the equation x² + ax – 1 = 0, has real roots.
Solution:
x² + ax – 1=0
Here a = 1, b = a, c = -1
D = b² – 4ac = (a)² – 4 x 1 x (-1) = a² + 4
Roots are real
D ≥ 0 => a² + 4 ≥ 0
For all real values of a, the equation has real roots.

Question 7.
In there any real value of ‘a’ for which the equation x² + 2x + (a² + 1) = 0 has real roots ?
Solution:
x² + 2x + (a² + 1) = 0
D = (-b)² – 4ac = (2)² – 4 x 1 (a² + 1) = 4 – 4a² – 4 = – 4a²
For real value of x, D ≥ 0
But – 4a² ≤ 0
So it is not possible
There is no real value of a

Question 8.
Write the value of λ, for which x² + 4x + λ is a perfect square.
Solution:
In x² + 4x + λ
a = 1, b = 4, c = λ
x² + 4x + λ will be a perfect square if x² + 4x + λ = 0 has equal roots
D = b² – 4ac = (4)² – 4 x 1 x λ = 16 – 4λ
D = 0
=> 16 – 4λ = 0
=> 16 = 4A
=> λ = 4
Hence λ = 4

Question 9.
Write the condition to be satisfied for which equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 have equal roots.
Solution:
In ax² + 2bx + c = 0

Question 10.
Write the set of values of k for which the quadratic equation has 2x² + kx – 8 = 0 has real roots.
Solution:
In 2x² + kx – 8 = 0
D = b²- 4ac = (k)² – 4 x 2 x (-8) = k² + 64
The roots are real
D ≥ 0
k² + 64 ≥ 0
For all real values of k, the equation has real roots.

Question 11.
Write a quadratic polynomial, sum of whose zeros is 2√3 and their product is 2.
Solution:
Sum of zeros = 2√3
and product of zeros = 2
The required polynomial will be

Question 12.
Show that x = – 3 is a solution of x² + 6x + 9 = 0 (C.B.S.E. 2008)
Solution:
The given equation is x² + 6x + 9 = 0
If x = -3 is its solution then it will satisfy it
L.H.S. = (-3)² + 6 (-3) + 9 = 9 – 18 + 9 = 18 – 18 = 0 = R.H.S.
Hence x = – 3 is its one root (solution)

Question 13.
Show that x = – 2 is a solution of 3x² + 13x + 14 = 0. (C.B.S.E. 2008)
Solution:
The given equation is 3x² + 13x + 14 = 0
If x = – 2 is its solution, then it will satisfy it
L.H.S. = 3(-2)² + 13 (- 2) + 14 =3 x 4 – 26 + 14
= 12 – 26 + 14 = 26 – 26 = 0 = R.H.S.
Hence x = – 2 is its solution

Question 14.
Find the discriminant of the quadratic equation 3√3 x² + 10x + √3 =0. (C.B.S.E. 2009)
Solution:

Question 15.
If x = \(\frac { -1 }{ 2 }\), is a solution of the quadratic equation 3x² + 2kx – 3 = 0, find the value of k. [CBSE 2015]
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations 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 4.13

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13

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.
A piece of cloth costs Rs. 35. If the piece were 4 m longer and each metre costs Re. one less, the cost would remain unchanged. How long is the piece ?
Solution:
Let the length of piece of cloth = x m
Total cost = Rs. 35
Cost of 1 m cloth = Rs. \(\frac { 35 }{ x }\)
According to the condition,

=> x (x + 14) – 10 (x + 14) = 0
=> (x + 14) (x – 10) = 0
Either x + 14 = 0, then x = – 14 which is not possible being negative
or x – 10 = 0, then x = 10
Length of piece of cloth = 10 m

Question 2.
Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic ?
Solution:
Let the number of students = x
and total budget = Rs. 480
Share of each students = Rs. \(\frac { 480 }{ x }\)
According to the condition,

=> x (x + 16) – 24 (x + 16) = 0
=> (x + 16) (x – 24) = 0
Either x + 16 = 0, then x = -16 which is not possible being negative
or x – 24 = 0, then x = 24
Number of students = 24
and number of students who attended the picnic = 24 – 8 = 16

Question 3.
A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
Solution:
Let cost price ofjfie article = Rs. x
Selling price = Rs. 24
Gain = x %
According to the condition,

Question 4.
Out of a group of swans, \(\frac { 7 }{ 2 }\) times the square root of the total number are playing on the share of a pond. The two remaining ones are swinging in water. Find the total number of swans.
Solution:
Let the total number of swans = x
According to the condition,

Number of total swans = 16

Question 5.
If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys mope for Rs. 360. Find the original price of the toy. (C.B.S.E. 2002C)
Solution:
List price of the toy = Rs. x
Total amount = Rs. 360
Reduced price of each toy = (x – 2)
According to the condition,

=> x (x – 20) + 18 (x – 20) = 0
=> (x – 20) (x + 18) = 0
Either x – 20 = 0, then x = 20
or x + 18 = 0, then x = -18 which is not possible being negative
Price of each toy = Rs. 20

Question 6.
Rs. 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons.
Solution:
Total amount = Rs. 9000
Let number of persons = x
Then each share = Rs. \(\frac { 9000 }{ x }\)
Increased persons = (x + 20)
According to the condition,

Either x + 45 = 0, then x = -45 which is not possible being negative
or x – 25 = 0, then x = 25
Number of persons = 25

Question 7.
Some students planned a picnic. The budget for food was Rs. 500. But 5 of them failed to go and thus the cost of food for each number increased by Rs. 5. How many students attended the picnic? (C.B.S.E. 1999)
Solution:
Let number of students = x
Total budget = Rs. 500
Share of each student = Rs. \(\frac { 500 }{ x }\)
No. of students failed to go = 5
According to the given condition,

Question 8.
A pole has to be erected at a point on the boundary of a. circular park of diameter 13 metres in such a way that the difference, of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it the possible to do so ? If yes, at what distances from the two gates Should the pole be erected ?
Solution:
In a circle, AB is the diameters and AB = 13 m

Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 5 = 0, then x = 5
P is at a distance of 5 m from B and 5 + 7 = 12 m from A

Question 9.
In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of their marks, would have been 180. Find his marks in the two subjects. (C.B.S.E. 2008)
Solution:
Sum of marks in Mathematics and Science = 28
Let marks in Math = x
Then marks in Science = 28 – x
According to the condition,
(x + 3) (28 – x – 4) = 180
=> (x + 3) (24 – x) = 180
=> 24x – x² + 72 – 3x = 180
=> 21x – x² + 72 – 180 = 0
=> – x² + 21x – 108 = 0
=> x² – 21x + 108 = 0
=> x² – 9x – 12x + 108 = 0
=> x (x – 9) – 12 (x – 9) – 0
=> (x – 9)(x – 12) = 0
Either x – 9 = 0, then x = 9
or x – 12 = 0, then x = 12
(i) If x = 9, then Marks in Maths = 9 and marks in Science = 28 – 9 = 19
(ii) If x = 12, then Marks in Maths = 12 and marks in Science = 28 – 12 = 16

Question 10.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in two subjects. [NCERT]
Solution:
Sum of marks in Mathematics and English = 30
Let marks obtained in Mathematics = x
Then in English = 30 – x
According to the condition,
(x + 2) (30 – x – 3) = 210
=> (x + 2) (27 – x) = 210
=> 27x – x² + 54 – 2x – 210 = 0
=> – x² + 25x – 156 = 0
=> x² – 25x + 156 = 0
=> x² – 12x – 13x +156 = 0
=> x (x – 12) – 13 (x – 12) = 0
=> (x – 12) (x – 13) = 0
Either x – 12 = 0, then x = 12
or x – 13 = 0, then x = 13
(i) If x = 12, then
Marks in Maths =12 and in English = 30 – 12 = 18
(ii) If x = 13, then
Marks in Maths = 13 and in English = 30 – 13 = 17

Question 11.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, And the number of articles produced and the cost of each article. [NCERT]
Solution:
Total cost = Rs. 90
Let number of articles = x
Then price of each articles = 2x + 3
x (2x + 3) = 90
=> 2x² + 3x – 90 = 0
=> 2x² – 12x + 15x – 90 = 0
=> 2x (x – 6) + 15 (x – 6) = 0
=> (x – 6) (2x + 15) = 0
Either x – 6 = 0, then x = 6
or 2x + 15 = 0 then 2x = -15 => x = \(\frac { -15 }{ 2 }\) which is not possible being negative
x = 6
Number of articles = 6
and price of each article = 2x + 3 = 2 x 6 + 3 = 12 + 3 = 15

Question 12.
At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than \(\frac { { t }^{ 2 } }{ 4 }\) minutes. Find t.
Solution:
We know that, the time between 2 pm to 3 pm = 1 h = 60 minutes
Given that, at t minutes past 2 pm, the time needed by the min. hand of a clock to show 3 pm was found to be 3 min. less than \(\frac { { t }^{ 2 } }{ 4 }\) min.
i.e., t = (\(\frac { { t }^{ 2 } }{ 4 }\) – 3) = 60
=> 4t + t² – 12 = 240
=> t² + 4t – 252 = 0
=> t² + 18t – 14t – 252 = 0 [by splitting the middle term]
=> t (t + 18) – 14 (t + 18) = 0
=> (t + 18) (t – 14) = 0 [since, time cannot be negative, so t ≠ -18]
t = 14 min.
Hence, the required value of t is 14 minutes

Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.13 are helpful to complete your math homework.

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