﻿ NCERT Solutions For Class 9 & 10 of Maths & Science Syllabus

# NCERT Solutions For For Class 9 & 10

## Maths & Science NCERT Solutions Class 9 & 10

The question paper framed by the Central Board of Secondary Education (CBSE) is mainly based on the content published in NCERT text books. Byju’s provides the NCERT solutions, exercise solved problems and solved examples for class 9, 10, 11 & 12 for Physics, Chemistry, and Math. Being the first source of priority, NCERT books provide the selective and precise content, compared to other publishers.

NCERT Solutions for CBSE class 9 Mathematics

Chapter 4 – Linear Equations in Two Variables

1. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) 2x + 3y = 9.35                  (ii) -2x + 3y = 6                      (iii) x – y/5 – 10 = 0       (iv) x = 3y

(v) 2x = -5y                            (vii) 3x + 2 = 0                       (vi) y – 2 = 0                           (viii) 5 = 2x

(i) 2x + 3y = 9.35

⇒ 2x + 3y – 9.35 = 0

Compare this equation with ax + by + c = 0, you get

a = 2x, b = 3 and c = -9.35

(ii) -2x + 3y = 6

⇒ -2x + 3y – 6 = 0

Compare this equation with ax + by + c = 0, you get

a = -2, b = 3 and c = -6

(iii) x – y/5 – 10 = 0

Compare this equation with ax + by + c = 0, you get

a = 1, b = -1/5 and c = -10

(iv) x = 3y

⇒ x – 3y = 0

Compare this equation with ax + by + c = 0, you get

a = 1, b = -3 and c = 0

(v) 2x = -5y

⇒ 2x + 5y = 0

Compare this equation with ax + by + c = 0, you get

a = 2, b = 5 and c = 0

(vi) y – 2 = 0

⇒ 0x + y – 2 = 0

Compare this equation with ax + by + c = 0, you get

a = 0, b = 1 and c = -2

(vii) 3x + 2 = 0

⇒ 3x + 0y + 2 = 0

Compare this equation with ax + by + c = 0, you get

a = 3, b = 0 and c = 2

(viii) 5 = 2x

⇒ -2x + 0y + 5 = 0

Compare this equation with ax + by + c = 0, you get

a = -2, b = 0 and c = 5

2.The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

(Take the cost of a notebook to be x and that of a pen to be y).

Let the cost of pen be y and the cost of notebook be x.

A/q,

Cost  of a notebook = twice the pen = 2y.

∴2y = x

⇒ x – 2y = 0

This is a linear equation in two variables to represent this statement.

NCERT Solutions for CBSE class 10 Mathematics

Chapter 1 – Real Numbers

1. Prove that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is an integer.

Let take as any positive integer and b = 6.

Then applying Euclid’s algorithm, we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6.

So possible forms are 6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5

6q + 0

6 is divisible by 2, so it is an even number.

6q + 1

6 is divisible by 2 but 1 is not divisible by 2, so it is an odd number.

6q + 2

6 is divisible by 2 and 2 is also divisible by 2, so it is an even number.

6q  +3

6 is divisible by 2 but 3 is not divisible by 2 so it is an odd number.

6q + 4

6 is divisible by 2 and 4 is also divisible by 2, so it is an even number.

6q + 5

6 is divisible by 2 but 5 is not divisible by 2, so it is an odd number.

So odd numbers will be in the form of 6q + 1, or 6q + 3, or 6q + 5.

1. Apply Euclid’s division algorithm to find the HCF of:

(i) 135 and 225

(ii) 867 and 255

(iii) 196 and 38220

(i) 225 > 135, we always divide larger number with smaller one.

Divide 225 by 135, we get 1 quotient and 90 as remainder so that

225= 135 × 1 + 90

Divide 135 by 90, we get 1 quotient and 45 as remainder so that

135= 90 × 1 + 45

Divide 90 by 45, we get 2 quotient and no remainder, so we can write it as

90 = 2 × 45 + 0

As there is no remainder, divisor 45 is the HCF.

(ii) 867 > 255, we always divide bigger number with smaller one.

Divide 867 by 255, then we get quotient 3 and remainder is 102, so we can write it as

867 = 255 × 3 + 102

Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as

255 = 102 × 2 + 51

Divide 102 by 51, we get quotient 2 and no remainder, so we can write it as

102 = 51 × 2 + 0

As there is no remainder, so divisor 51 is the HCF.

(iii) 38220 > 196, always divide bigger number with smaller one.

Divide 38220 by 196, we get quotient 195 and no remainder. So we can write it as

38220 = 196 × 195 + 0

Since there is no remainder, the divisor 196 is the HCF.

1. Apply Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

[Hint: Let x be any positive integer, then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

a2 = (3q)2 or (3q + 1)2 or (3q + 2)2

a2 = (9q)2 or 9q2 + 6q + 1 or 9q2 + 12q + 4

= 3 × (3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1

= 3k1 or 3k2 + 1 or 3k3 + 1

Where k1, k2, and k3 are some positive integers.

Hence, the square of any positive integer is either of the form 3m or 3m + 1.

1. Apply Euclid’s division lemma to prove that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

a = 3q or 3q + 1 or 3q + 2

Therefore, every number can be represented in these three forms. We have three cases.

Case 1: When a = 3q,

a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,

Where m is an integer such that m = 3q3

Case 2: When a = 3q + 1,

a3 = (3q +1)3

a3= 27q3 + 27q2 + 9q + 1

a3 = 9(3q3 + 3q2 + q) + 1

a3 = 9m + 1

Where m is an integer such that m = (3q3 + 3q2 + q)

Case 3: When a = 3q + 2,

a3 = (3q +2)3

a3= 27q3 + 54q2 + 36q + 8

a3 = 9(3q3 + 6q2 + 4q) + 8

a3 = 9m + 8

Where m is an integer such that m = (3q3 + 6q2 + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1,

or 9m + 8.

1. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?