# RD SHARMA Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

## Chapter 4 - Quadratic Equations Exercise 4.82

If the equation x^{2} + 4x + k = 0 has real and distinct root, then

(a) k < 4

(b) k > 4

(c) k ≥ 4

(d) k ≤ 4

We know for the quadratic equation

ax^{2} + bx + c = 0

condition for roots to be real and distinct is

D = b^{2} - 4ac > 0 ..........(1)

for the given question

x^{2 }+ 4x + k = 0

a = 1, b = 4, c = k

from (1)

16 - 4k > 0

k < 4

So, the correct option is (a).

## Chapter 4 - Quadratic Equations Exercise 4.83

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

Any quadratic equation having roots 0 or 1 are only possible quadratic equation because on squaring 0 or 1, it remains same.

Hence, 2 solutions are possible, one having roots 1 and 1, while the other having roots 0 and 1.

So, the correct option is (c).

If the equation x^{2} - ax + 1 = 0 has two distinct roots, then

(a) |a| = 2

(b) |a| < 2

(c) |a| > 2

(d) None of these

For the equation x^{2} - ax + 1 = 0 has two distinct roots, condition is

(-a)^{2 }- 4 (1) (1) > 0

a^{2 }- 4 > 0

a^{2 }> 4

|a| > 2

So, the correct option is (c).

If the equation ax^{2} + 2x + a = 0 has two distinct roots, if

(a) a = ± 1

(b) a = 0

(c) a = 0, 1

(d) a = -1, 0

For any quadratic equation

ax^{2 } + bx + c = 0

having two distinct roots, condition is

b^{2} - 4ac > 0

For the equation ax^{2} + 2x + a = 0 to have two distinct roots,

(2)^{2} - 4 (a) (a) > 0

4 - 4a^{2} > 0

4(1 - a^{2}) > 0

1 - a^{2} > 0 since 4 > 0

that is, a^{2} - 1 < 0

Hence -1 < a < 1, only integral solution possible is a = 0

So, the correct option is (b).

The positive value of k for which the equation x^{2} + kx + 64 = 0 and x^{2} - 8x + k = 0 will both have real roots, is

(a) 4

(b) 8

(c) 12

(d) 16

For any quadratic equation

ax^{2} + bx + c = 0

having real roots, condition is

b^{2} - 4ac ≥ 0 .......(1)

According to question

x^{2 }+ kx + 64 = 0 have real root if

k^{2 }- 4 × 64 ≥ 0

k^{2} ≥ 256

|k|≥ 16 ........(2)

Also, x^{2} - 8x + k = 0 has real roots if

64 - 4k ≥ 0

k ≤ 16 .........(3)

from (2), (3) the only positive solution for k is

k = 16

So, the correct option is (d).

If 2 is a root of the equation x^{2 }+ bx + 12 = 0 and the equation x^{2} + bx + q = 0 has equal roots, then q =

(a) 8

(b) -8

(c) 16

(d) -16

It is given that 2 is a root of equation x^{2 } + bx + 12 = 0

Hence

(2)^{2 }+ b(2) + 12 = 0

4 + 2b + 12 = 0

2b + 16 = 0

b = -8 .........(1)

It is also given that x^{2} + bx + q = 0 has equal root

so, b^{2} - 4(q) = 0 ........(2)

from (1) & (2)

(-8)^{2 }- 4q =0

q = 16

So, the correct option is (c).

For any quadratic equation

ax^{2} + bx + c = 0

Having equal roots, the condition is

b^{2 } - 4ac = 0

For the equation

(a^{2} + b^{2}) x^{2} - 2 (ac + bd)x + c^{2} + d^{2 }= 0

to have equal roots, we have

(-2(ac + bd))^{2 }- 4 (c^{2 }+ d^{2}) (a^{2 }+ b^{2}) = 0

4 (ac + bd)^{2 }- 4 (a^{2}c^{2 }+ b^{2}c^{2} + d^{2}a^{2 }+ b^{2}d^{2}) = 0

(a^{2}c^{2 }+ b^{2}d^{2} + 2abcd) - (a^{2}c^{2 }+ b^{2}c^{2} + a^{2}d^{2}+ b^{2}d^{2}) = 0

2abcd - b^{2}c^{2} - a^{2}d^{2} = 0

b^{2}c^{2} - a^{2}d^{2 }- 2abcd = 0

(bc - ad)^{2 }= 0

bc = ad

So, the correct option is (b).

For any quadratic equation

ax^{2 }+ bx + c = 0

having equal roots, condition is

b^{2} - 4ac = 0

According to question, quadratic equation is

(a^{2} + b^{2})x^{2} - 2b(a + c)x + b^{2} + c^{2 }= 0

having equal roots, so

(2b(a + c))^{2} - 4(a^{2} + b^{2}) (b^{2} + c^{2}) = 0

4b^{2} (a + c)^{2} - 4(a^{2}b^{2} + a^{2}c^{2 }+ b^{4} + b^{2}c^{2}) = 0

b^{2}(a^{2} + c^{2} + 2ac) - (a^{2}b^{2} + a^{2}c^{2 }+ b^{2}c^{2} + b^{4}) = 0

a^{2}b^{2} + b^{2}c^{2} + 2acb^{2} - a^{2}b^{2} - a^{2}c^{2 }- b^{2}c^{2} - b^{4} = 0

2acb^{2} - a^{2}c^{2} - b^{4} = 0

a^{2}c^{2} + b^{4} - 2acb^{2} = 0

(ac - b^{2})^{2} = 0

ac = b^{2}

So, the correct option is (b).

If the equation x^{2} - bx + 1 = 0 does not possess real roots, then

(a) -3 < b < 3

(b) -2 < b < 2

(c) b < 2

(d) b < -2

For any quadratic equation ax^{2} + bx + c = 0 having no real roots, condition is

b^{2} - 4ac < 0

For the equation, x^{2} - bx + 1 = 0 having no real roots

b^{2} - 4 < 0

b^{2} < 4

-2 < b < 2

So, the correct option is (b).

If x = 1 is a common root of the equations ax^{2} + ax + 3 = 0 and x^{2} + x + b = 0, then ab =

(a) 3

(b) 3.5

(c) 6

(d) -3

If p and q are the roots of the equation x^{2 }-px + q = 0, then

(a) p = 1, q = -2

(b) b = 0, q = 1

(c) p = -2, q = 0

(d) p = -2, q = 1

If p, q are the roots of equation x^{2 }- px + q = 0, then p and q satisfies the equation

Hence

(p)^{2 }- p(p) + q = 0

p^{2 }- p^{2 }+ q = 0

q = 0

and (q)^{2 }- p(q) + q = 0

q^{2 }- p(q) + q = 0

0 = 0

p can take any value

p = -2 and q = 0

So, the correct option is (c).

If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax^{2} + bx + 1 = 0 having real roots is

(a) 10

(b) 7

(c) 6

(d) 12

For the ax^{2 }+ bx + 1 = 0 having real roots condition is

b^{2 }- 4(a) (1) ≥ 0

b^{2} ≥ 4a

For a = 1

b^{2} ≥ 4

b ≥ 2

b can take value 2, 3, 4

Here, 3 possible solutions are possible .......(1)

For a = 2

b^{2} ≥ 8

Here, b can take value 3, 4 ......(2)

Here, 2 solutions are possible

For a = 3

b^{2 }≥ 12

possible value of b is 4

Hence, only 1 possible solution ......(3)

For a = 4

b^{2} ≥ 16

possible value of b is 4

Hence, only 1 possible solution .......(4)

from (1), (2), (3), (4)

Total possible solutions are 7

So, the correct option is (b).

## Chapter 4 - Quadratic Equations Exercise 4.84

If one root of the equation x^{2}^{ }+ ax + 3 = 0 is 1, then its other root is

(a) 3

(b) -3

(c) 2

(d) -2

x^{2 }+ ax + 3 = 0

product of roots = 3

One root is 1. Hence other root is 3.

So, the correct option is (a).

If (a^{2 }+ b^{2}) x^{2 }+ 2(ab + bd) x + c^{2} + d^{2 }= 0 has no real roots, then

(a) ad = bc

(b) ab = cd

(c) ac = bd

(d) ad ≠ bc

If any quadratic equation ax^{2 }+ bx + c has no real roots then b^{2 }- 4ac < 0 ......(1)

According to the question, the equation is

(a^{2 }+ b^{2}) x^{2 }+ 2(ac + bd) x + c^{2} + d^{2 }= 0

from (1)

4(ac + bd)^{2 }- 4(a^{2} + b^{2}) (c^{2} + d^{2}) < 0

a^{2}c^{2} + b^{2}d^{2} + 2abcd - (a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2}) < 0

a^{2}c^{2} + b^{2}d^{2} + 2abcd - a^{2}c^{2} - a^{2}d^{2} - b^{2}c^{2} - b^{2}d^{2} < 0

2abcd - a^{2}d^{2} - b^{2}c^{2} < 0

-(ad - bc)^{2 }< 0

(ad - bc)^{2 }> 0

For this condition to be true ad ≠ bc

So, the correct option is (d).

If x = 1 is a common root of ax^{2 }+ ax + 2 = 0 and x^{2 }+ x + b = 0 then, ab =

(a) 1

(b) 2

(c) 4

(d) 3

It is given that x = 1 is root of equation ax^{2} + ax + 2 = 0

Hence, a(1)^{2} + a(1) + 2 = 0

2a + 2 = 0

a = -1 ......(1)

It is given that x = 1 is also root of x^{2} + x + b = 0

Hence, (1)^{2 }+ (1) + b = 0

b = -2 ......(2)

from (1) & (2)

ab = (-1) (-2)

ab = 2

So, the correct option is (b).

If sin α and cos α are the roots of the equation ax^{2 }+ bx + c = 0, then b^{2} =

(a) a^{2 }- 2ac

(b) a^{2 }+ 2ac

(c) a^{2 }- ac

(d) a^{2 }+ ac

If 2 is a root of the equation x^{2 }+ ax + 12 = 0 and the quadratic equation x^{2 } + ax + q = 0 has equal roots, then q =

(a) 12

(b) 8

(c) 20

(d) 16

Given, 2 is a root of equation x^{2 }+ ax + 12 = 0

so (2)^{2 }+ a(2) + 12 = 0

4 + 2a + 12 = 0

a = -8 .....(1)

Given x^{2 }+ ax + q = 0 has equal roots so

a^{2 }- 4q = 0 ......(2)

from (1) & (2)

(-8)^{2 }- 4q = 0

4q = 64

q = 16

So, the correct option is (d).

If the sum of the roots of the equation x^{2} - (k + 6)x + 2(2k - 1) = 0 is equal to half of their product, then k =

(a) 6

(b) 7

(c) 1

(d) 5

If a and b are roots of the equation x^{2} + ax + b = 0, then a + b =

(a) 1

(b) 2

(c) -2

(d) -1

If a and b are roots of the equation x^{2 }+ ax + b = 0

Then, sum of roots = -a

a + b = -a

2a + b = 0 .......(1)

product of roots = b

ab = b

ab - b = 0

b(a - 1) = 0 ........(2)

from (1) and (2)

-2a(a - 1) = 0...(From (1), we have b = -2a)

a(a - 1) = 0

a = 0 or a = 1

if a = 0 b = 0

if a = 1 b = -2

Now a and b can't be zero at same time, so correct solution is

a = 1 and b = -2

a + b = -1

So, the correct option is (d).

A quadratic equation whose one root is 2 and the sum of whose roots is zero, is

(a) x^{2 }+ 4 = 0

(b) x^{2 }- 4 = 0

(c) 4x^{2 }- 1 = 0

(d) x^{2 }- 2 = 0

Given sum of roots is zero and one root is 2.

So the other root must be -2

so any quadratic equation having root 2 and -2 is

(x - 2) (x - (-2)) = 0

(x - 2) (x + 2) = 0

x^{2 }- 4 = 0

So, the correct option is (b).

If one root of the equation ax^{2 }+ bx + c = 0 is three times the other, then b^{2} : ac =

(a) 3 : 1

(b) 3 : 16

(c) 16 : 3

(d) 16 : 1

If one root of the equation 2x^{2 }+ kx + 4 = 0 is 2, then the other root is

(a) 6

(b) -6

(c) -1

(d) 1

## Chapter 4 - Quadratic Equations Exercise 4.85

If one root of the equation 4x^{2 }- 2x + (λ - 4) = 0 is a reciprocal of the other, then λ =

(a) 8

(b) -8

(c) 4

(d) -4

Any quadratic equation, ax^{2 }+ bx + c = 0 has real and equal roots if b^{2 }- 4ac = 0

For the question, equation is 16x^{2 }+ 4kx + 9 = 0,

(4k)^{2 }- 4 × 16 × 9 = 0

16k^{2 }- 36 × 16 = 0

k^{2} - 36 = 0

k^{2 }= 36

k = ± 6

So, the correct option is (c).

## Chapter 4 - Quadratic Equations Exercise Ex. 4.1

solutions of the equation 2x^{2} - x + 9 = x^{2} + 4x + 3 , x = 2 and x = 3

= 2x^{2 }- x + 9 – x^{2 }+ 4x + 3

= x^{2} - 5x + 6 = 0

Here, LHS = x^{2} - 5x + 6 and RHS = 0

Substituting x = 2 and x = 3

= x^{2} - 5x + 6

= (2)^{2} – 5(2) + 6

=10-10

=0

= RHS

= x^{2} - 5x + 6

= (3)2 – 5(3) + 6

= 9 - 15 + 6

=15 - 15

=0

= RHS

x = 2 and x = 3 both are the solutions of the given quadratic equation.

## Chapter 4 - Quadratic Equations Exercise Ex. 4.10

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 possible to do so? If yes, at what distances from the two gates should the pole be erected?

## Chapter 4 - Quadratic Equations Exercise Ex. 4.11

Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm^{2}. Find the sides of the squares.

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m^{2}? If so, find its length and breadth.

Sum of the areas of two squares is 640 m^{2}. If the difference of their perimeter is 64 m, find the sides of the two squares.

Sum of the areas of two squares is 400 cm^{2}. If the difference of their perimeters is 16 cm, find the sides of two squares.

The area of a rectangular plot is 528 m^{2}. The length of the plot (in meters) is one meter more than twice its breadth. Find the length and the breadth of the plot.

In the centre of a rectangular lawn of dimension 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m^{2}. Find the length and breadth of the pond.

## Chapter 4 - Quadratic Equations Exercise Ex. 4.12

To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

Let us assume that the larger pipe takes 'x' hours to fill the pool.

So, as per the question, the smaller pipe takes 'x + 10' hours to fill the same pool.

## Chapter 4 - Quadratic Equations Exercise Ex. 4.13

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 his marks, would have been 180. Find his marks in the two subjects.

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, find the number of articles produced and the cost of each article.

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 minutes. Find t.

## Chapter 4 - Quadratic Equations Exercise Ex. 4.2

The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.

John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John had x marbles.

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.

The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.

## Chapter 4 - Quadratic Equations Exercise Ex. 4.3

Solve the following quadratic equation by factorisation:

Solve the following quadratic equation by factorisation:

9x^{2 }- 6b^{2}x - (a^{4} - b^{4}) = 0

Solve the following quadratic equation by factorisation:

2x^{2 }+ ax - a^{2} = 0

Solve the following quadratic equation by factorisation:

Solve the following quadratic equations by factorization:

Solve the following quadratic equation by factorisation:

Solve the following quadratic equation by factorisation:

Solve the following quadratic equation by factorisation:

Solve the following quadratic equation by factorisation:

Solve the following quadratic equation by factorisation:

Solve the following quadratic equation by factorisation:

Solve the following quadratic equations by factorization:

Solve the following quadratic equations by factorization:

Solve the following quadratic equations by factorization:

Solve the following quadratic equation by factorisation:

Solve the following quadratic equation by factorisation:

Solve the following quadratic equation by factorisation:

## Chapter 4 - Quadratic Equations Exercise Ex. 4.4

## Chapter 4 - Quadratic Equations Exercise Ex. 4.5

3x^{2} - 5x + 2 = 0

Solve for x:

Solve for x:

Solve for x:

## Chapter 4 - Quadratic Equations Exercise Ex. 4.6

Determine the nature of the roots of the following quadratic equations:

2x^{2} - 3x + 5 = 0

Determine the nature of the roots of the following quadratic equations:

2x^{2} - 6x + 3 = 0

Find the values of k for which the roots are real and equal in each of the following equations:

4x^{2}
- 2 (k + 1)x + (k + 1) = 0

4x^{2}
- 2 (k + 1)x + (k + 1) = 0
Comparing with ax^{2} + bx + c = 0, we get
a = 4, b = -2(k + 1), c = k + 1
According to the question, roots are real and equal.
Hence, b^{2} - 4ac = 0

In the following, determine the set of values of k for which the given quadratic equation has real roots:

2x^{2} + x + k = 0

Find the values of k for which the following equations have real and equal roots:

x^{2}
+ k(2x + k - 1) + 2 = 0

x^{2} + k(2x + k - 1) + 2 = 0
x^{2} + 2kx + k(k - 1) + 2 = 0
Comparing with ax^{2} + bx + c = 0, we get
a = 1, b = 2k, c = k(k - 1) + 2
According to the question, roots are real and equal.
Hence, b^{2} - 4ac = 0

Find the values of k for which the roots are real and equal in each of the following equations:

2x^{2} + kx + 3 = 0

Find the values of k for which the roots are real and equal in each of the following equations:

kx (x - 2) + 6 = 0

Find the values of k for which the roots are real and equal in each of the following equations:

x^{2} - 4kx + k = 0

Find the value of k for which the roots are real and equal in the following equation:

Find the value of p for which the roots are real and equal in the following equation:

px(x - 3) + 9 = 0

Find the values of k for which the following equations have real roots.

4x^{2}
+ kx + 3 = 0

4x^{2}
+ kx + 3 = 0
Comparing with ax^{2} + bx + c = 0, we get
a = 4, b = k, c = 3
According to the question, roots are real and equal.
Hence, b^{2} - 4ac = 0

Find the values of k for which the quadratic equation (3k + 1)x^{2} + 2(k + 1)x + 1 = 0 has equal roots. Also, find the roots.

Find the values of p for which the quadratic equation (2p + 1)x^{2} - (7p + 2)x + (7p - 3) = 0 has equal roots. Also, find these roots.

If -5 is a root of the quadratic equation, 2x^{2} + px - 15 = 0, and the quadratic equation p(x^{2} + x) + k = 0 has equal roots, find the value of k.

If 2 is a root of the quadratic equation 3x^{2} + px - 8 = 0 and the quadratic equation 4x^{2} - 2px + k = 0 has equal roots, find the value of k.

If 1 is root of the quadratic equation 3x^{2} + ax - 2 = 0 and the quadratic equation a(x^{2} + 6x) - b = 0 has equal roots, find the value of b.

Find the value of p for which the quadratic equation (p + 1)x^{2} - 6(p + 1)x + 3(p + 9) = 0, p ≠ -1 has equal roots. Hence, find the roots of equation.

## Chapter 4 - Quadratic Equations Exercise Ex. 4.7

The sum of a number and its square is 63/4. Find the numbers.

The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers.

The sum of two numbers is 9. The sum of their reciprocals is 1/2. Find the numbers.

Find two consecutives odd positive integers, sum of whose squares is 970.

The difference of two natural numbers is 3 and the difference of their reciprocal is . Find the numbers.

The sum of the squares of two consecutive odd numbers is 394. Find the numbers.

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.

The sum of the squares of two consecutive even numbers is 340. Find the numbers.

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is. Find the original fraction.

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

A natural number when increased by 12 equals 160 times its reciprocal. Find the number.

## Chapter 4 - Quadratic Equations Exercise Ex. 4.8

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed, If it takes 3 hours to complete total journey, what is its original average speed?

**Concept Insight:** Use the relation s =d/t to crack this question and remember here distance is constant so speed and time will vary inversely.

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.

A motor boat whose speed in still water is 18 km/hr takes 1 hour more to go 24 km up stream than to return downstream to the same spot. Find the speed of the stream.

A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.

Let the speed of a car be x km/hr. According to the question, time is hr. Distance = Speed × Time 2592 = x = 72 km/hr Hence, the time taken by a car to cover a distance of 2592 km is 36 hrs.

## Chapter 4 - Quadratic Equations Exercise Ex. 4.9

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now?

At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

## Chapter 8 - Quadratic Equations Exercise Ex. 8.6

Find the value of p for which the roots are real and equal in the following equation:

4x^{2} + px + 3 = 0

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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