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**NCERT Solutions for CBSE class 9 Mathematics**

**Chapter 4** **– Linear Equations in Two Variables **

**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**

**Answer**

(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).

Answer

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 **

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

**Answer**

Let take as any positive integer and *b* = 6.

Then applying Euclid’s algorithm, we get a = 6*q* + *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 6*q *+ 0, 6*q *+ 1, 6*q *+ 2, 6*q *+ 3, 6*q* + 4, 6*q* + 5

6*q* + 0

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

6*q* + 1

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

6*q* + 2

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

6*q* +3

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

6*q* + 4

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

6*q* + 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.

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

(i) 135 and 225

(ii) 867 and 255

(iii) 196 and 38220

**Answer**

(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.

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

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

**Answer**

Let a be any positive integer and *b* = 3.

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

And *r* = 0, 1, 2 because 0 ≤ *r* < 3

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

Or,

*a*^{2} = (3*q*)^{2} or (3*q* + 1)^{2} or (3*q* + 2)^{2}

*a*^{2} = (9*q*)^{2} or 9*q*^{2} + 6*q* + 1 or 9*q*^{2} + 12*q* + 4

= 3 × (3*q*^{2}) or 3(3*q*^{2} + 2*q*) + 1 or 3(3*q*^{2} + 4*q* + 1) + 1

= 3*k*_{1} or 3*k*_{2} + 1 or 3*k*_{3} + 1

Where *k*_{1}, *k*_{2}, and *k*_{3} are some positive integers.

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

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

**Answer**

Let a be any positive integer and b = 3

*a* = 3*q* + *r*, where *q* ≥ 0 and 0 ≤ *r* < 3

∴ *a* = 3q or 3*q* + 1 or 3*q* + 2

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

Case 1: When *a* = 3*q*,

*a*^{3} = (3*q*)^{3} = 27*q*^{3} = 9(3*q*)^{3} = 9*m*,

Where *m* is an integer such that *m* = 3*q*^{3}

Case 2: When *a *= 3q + 1,

*a*^{3} = (3*q* +1)^{3}

*a*^{3}= 27*q*^{3} + 27*q*^{2} + 9*q* + 1

*a*^{3} = 9(3*q*^{3} + 3*q*^{2} + *q*) + 1

*a*^{3} = 9*m* + 1

Where *m* is an integer such that *m* = (3*q*^{3} + 3*q*^{2} + *q*)

Case 3: When *a* = 3*q* + 2,

*a*^{3} = (3*q* +2)^{3}

*a*^{3}= 27*q*^{3} + 54*q*^{2} + 36*q* + 8

*a*^{3} = 9(3*q*^{3} + 6*q*^{2} + 4q) + 8

*a*^{3} = 9*m* + 8

Where *m* is an integer such that *m* = (3*q*^{3} + 6*q*^{2} + 4*q*)

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

or 9*m* + 8.

- 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?

**Answer**

HCF (616, 32) gives the maximum number of columns in which they can march.

We can apply Euclid’s algorithm to find the HCF.

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF of (616, 32) is 8.

Therefore, they can march in 8 columns each.

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