Ncert Solutions For Class 7 Maths Ex 9.2

NCERT Solutions For Class 7 Maths Ex 9.2 PDF Free Download

NCERT Solutions for Class 7 Maths Exercise 9.2 Chapter 9 Rational Numbers in simple PDF are given here. Operations on rational numbers in that addition, subtraction, multiplication and division of rational numbers are the topics covered in this exercise of NCERT Solutions for Class 7 Chapter 9. The NCERT Solutions for Class 7 Maths is the best study material for students who aspire to score good marks in maths.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 9 Rational Numbers – Exercise 9.2

 

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ncert sol class 7 math ch 9 ex 2
ncert sol class 7 math ch 9 ex 2
ncert sol class 7 math ch 9 ex 2
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ncert sol class 7 math ch 9 ex 2
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ncert sol class 7 math ch 9 ex 2
ncert sol class 7 math ch 9 ex 2

 

Access another exercise of NCERT Solutions For Class 7 Chapter 9 – Rational Numbers

Exercise 9.1 Solutions

Access answers to Maths NCERT Solutions for Class 7 Chapter 9 – Rational Numbers Exercise 9.2

1. Find the sum:

(i) (5/4) + (-11/4)

Solution:-

We have:

= (5/4) – (11/4)

= [(5 – 11)/4] … [∵ denominator is same in both the rational numbers]

= (-6/4)

= -3/2 … [∵ Divide both numerator and denominator by 3]

(ii) (5/3) + (3/5)

Solution:-

Take the LCM of the denominators of the given rational numbers.

LCM of 3 and 5 is 15

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(5/3)= [(5×5)/ (3×5)] = (25/15)

(3/5)= [(3×3)/ (5×3)] = (9/15)

Then,

= (25/15) + (9/15) … [∵ denominator is same in both the rational numbers]

= (25 + 9)/15

= 34/15

(iii) (-9/10) + (22/15)

Solution:-

Take the LCM of the denominators of the given rational numbers.

LCM of 10 and 15 is 30

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(-9/10)= [(-9×3)/ (10×3)] = (-27/30)

(22/15)= [(22×2)/ (15×2)] = (44/30)

Then,

= (-27/30) + (44/30) … [∵ denominator is same in both the rational numbers]

= (-27 + 44)/30

= (17/30)

(iv) (-3/-11) + (5/9)

Solution:-

We have,

= 3/11 + 5/9

Take the LCM of the denominators of the given rational numbers.

LCM of 11 and 9 is 99

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(3/11)= [(3×9)/ (11×9)] = (27/99)

(5/9)= [(5×11)/ (9×11)] = (55/99)

Then,

= (27/99) + (55/99) … [∵ denominator is same in both the rational numbers]

= (27 + 55)/99

= (82/99)

(v) (-8/19) + (-2/57)

Solution:-

We have

= -8/19 – 2/57

Take the LCM of the denominators of the given rational numbers.

LCM of 19 and 57 is 57

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(-8/19)= [(-8×3)/ (19×3)] = (-24/57)

(-2/57)= [(-2×1)/ (57×1)] = (-2/57)

Then,

= (-24/57) – (2/57) … [∵ denominator is same in both the rational numbers]

= (-24 – 2)/57

= (-26/57)

(vi) -2/3 + 0

Solution:-

We know that any number or fraction is added to zero the answer will be the same number or fraction.

Hence,

= -2/3 + 0

= -2/3

(vii) NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Image 14 + NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Image 15

Solution:-

First we have to convert mixed fraction into improper fraction.

=
NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Image 16= -7/3

=
NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Image 17= 23/5

We have, -7/3 + 23/5

Take the LCM of the denominators of the given rational numbers.

LCM of 3 and 5 is 15

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(-7/3)= [(-7×5)/ (3×5)] = (-35/15)

(23/5)= [(23×3)/ (15×3)] = (69/15)

Then,

= (-35/15) + (69/15) … [∵ denominator is same in both the rational numbers]

= (-35 + 69)/15

= (34/15)

2. Find

(i) 7/24 – 17/36

Solution:-

Take the LCM of the denominators of the given rational numbers.

LCM of 24 and 36 is 72

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(7/24)= [(7×3)/ (24×3)] = (21/72)

(17/36)= [(17×2)/ (36×2)] = (34/72)

Then,

= (21/72) – (34/72) … [∵ denominator is same in both the rational numbers]

= (21 – 34)/72

= (-13/72)

(ii) 5/63 – (-6/21)

Solution:-

We can also write -6/21 = -2/7

= 5/63 – (-2/7)

We have,

= 5/63 + 2/7

Take the LCM of the denominators of the given rational numbers.

LCM of 63 and 7 is 63

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(5/63)= [(5×1)/ (63×1)] = (5/63)

(2/7)= [(2×9)/ (7×9)] = (18/63)

Then,

= (5/63) + (18/63) … [∵ denominator is same in both the rational numbers]

= (5 + 18)/63

= 23/63

(iii) -6/13 – (-7/15)

Solution:-

We have,

= -6/13 + 7/15

LCM of 13 and 15 is 195

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(-6/13)= [(-6×15)/ (13×15)] = (-90/195)

(7/15)= [(7×13)/ (15×13)] = (91/195)

Then,

= (-90/195) + (91/195) … [∵ denominator is same in both the rational numbers]

= (-90 + 91)/195

= (1/195)

(iv) -3/8 – 7/11

Solution:-

Take the LCM of the denominators of the given rational numbers.

LCM of 8 and 11 is 88

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(-3/8)= [(-3×11)/ (8×11)] = (-33/88)

(7/11)= [(7×8)/ (11×8)] = (56/88)

Then,

= (-33/88) – (56/88) … [∵ denominator is same in both the rational numbers]

= (-33 – 56)/88

= (-89/88)

(v) –NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Image 18 – 6

Solution:-

First we have to convert the mixed fraction into improper fraction,


NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Image 19= -19/9

We have, -19/9 – 6

Take the LCM of the denominators of the given rational numbers.

LCM of 9 and 1 is 9

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(-19/9)= [(-19×1)/ (9×1)] = (-19/9)

(6/1)= [(6×9)/ (1×9)] = (54/9)

Then,

= (-19/9) – (54/9) … [∵ denominator is same in both the rational numbers]

= (-19 – 54)/9

= (-73/9)

3. Find the product:

(i) (9/2) × (-7/4)

Solution:-

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

The above question can be written as (9/2) × (-7/4)

We have,

= (9×-7)/ (2×4)

= -63/8

(ii) (3/10) × (-9)

Solution:-

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

The above question can be written as (3/10) × (-9/1)

We have,

= (3×-9)/ (10×1)

= -27/10

(iii) (-6/5) × (9/11)

Solution:-

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (-6×9)/ (5×11)

= -54/55

(iv) (3/7) × (-2/5)

Solution:-

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (3×-2)/ (7×5)

= -6/35

(v) (3/11) × (2/5)

Solution:-

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (3×2)/ (11×5)

= 6/55

(vi) (3/-5) × (-5/3)

Solution:-

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (3×-5)/ (-5×3)

On simplifying,

= (1×-1)/ (-1×1)

= -1/-1

= 1

4. Find the value of:

(i) (-4) ÷ (2/3)

Solution:-

We have,

= (-4/1) × (3/2) … [∵ reciprocal of (2/3) is (3/2)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-4×3) / (1×2)

= (-2×3) / (1×1)

= -6

(ii) (-3/5) ÷ 2

Solution:-

We have,

= (-3/5) × (1/2) … [∵ reciprocal of (2/1) is (1/2)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-3×1) / (5×2)

= -3/10

(iii) (-4/5) ÷ (-3)

Solution:-

We have,

= (-4/5) × (1/-3) … [∵ reciprocal of (-3) is (1/-3)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-4× (1)) / (5× (-3))

= -4/-15

= 4/15

(iv) (-1/8) ÷ 3/4

Solution:-

We have,

= (-1/8) × (4/3) … [∵ reciprocal of (3/4) is (4/3)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-1×4) / (8×3)

= (-1×1) / (2×3)

= -1/6

(v) (-2/13) ÷ 1/7

Solution:-

We have,

= (-2/13) × (7/1) … [∵ reciprocal of (1/7) is (7/1)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-2×7) / (13×1)

= -14/13

(vi) (-7/12) ÷ (-2/13)

Solution:-

We have,

= (-7/12) × (13/-2) … [∵ reciprocal of (-2/13) is (13/-2)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-7× 13) / (12× (-2))

= -91/-24

= 91/24

(vii) (3/13) ÷ (-4/65)

Solution:-

We have,

= (3/13) × (65/-4) … [∵ reciprocal of (-4/65) is (65/-4)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (3×65) / (13× (-4))

= 195/-52

= -15/4


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