Mathematics – Class 7 – Chapter 9 – Rational Numbers – Exercise 9.2 – NCERT Exercise Solution

1. Find the sum:

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

Solution:

= (5/4) – (11/4)

= [(5 – 11)/4]

= (-6/4)

Dividing both numerator and denominator by 3

= -3/2

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

Solution:

First, we have to find the LCM of the denominators of the given rational numbers.

LCM of 3 and 5 = 15

Now, for common denominator:

(5/3) = [(5×5)/ (3×5)] = (25/15)

(3/5) = [(3×3)/ (5×3)] = (9/15)

Then,

= (25/15) + (9/15)

= (25 + 9)/15

= 34/15

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

Solution:

First, we find the LCM of the denominators of the given rational numbers.

LCM of 10 and 15 = 30

Now, for common denominator:

(-9/10) = [(-9×3)/ (10×3)] = (-27/30)

(22/15) = [(22×2)/ (15×2)] = (44/30)

Then,

= (-27/30) + (44/30)

= (-27 + 44)/30

= (17/30)

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

Solution:

First, we have to find the LCM of the denominators of the given rational numbers.

LCM of 11 and 9 = 99

Now, for common denominator:

(3/11) = [(3×9)/ (11×9)] = (27/99)

(5/9) = [(5×11)/ (9×11)] = (55/99)

Then,

= (27/99) + (55/99)

= (27 + 55)/99

= (82/99)

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

Solution:

First, we have to find the LCM of the denominators of the given rational numbers.

LCM of 19 and 57 = 57

Now, for common denominator:

(-8/19) = [(-8×3)/ (19×3)] = (-24/57)

(-2/57) = [(-2×1)/ (57×1)] = (-2/57)

Then,

= (-24/57) – (2/57)

= (-24 – 2)/57

= (-26/57)

(vi) -2/3 + 0

Solution:

We know that if we add any number or fraction to zero then the answer will be the same number or fraction.

Hence,

= -2/3 + 0

= -2/3

Mathematics - Class 7 - Chapter 9 - Rational Numbers - Exercise 9.2 - NCERT Exercise Solution

We have, -7/3 + 23/5

We have to find the LCM of the denominators of the given rational numbers.

LCM of 3 and 5 = 15

Now, for making common denominator:

(-7/3) = [(-7×5)/ (3×5)] = (-35/15)

(23/5) = [(23×3)/ (15×3)] = (69/15)

Then,

= (-35/15) + (69/15)

= (-35 + 69)/15

= (34/15)

2. Find

(i) 7/24 – 17/36

Solution:

First, we have to find the LCM of the denominators of the given rational numbers.

LCM of 24 and 36 = 72

Now, for express into common denominator

(7/24) = [(7×3)/ (24×3)] = (21/72)

(17/36) = [(17×2)/ (36×2)] = (34/72)

Then,

= (21/72) – (34/72)

= (21 – 34)/72

= (-13/72)

(ii) 5/63 – (-6/21)

Solution:

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

Now, we have

5/63 – (-2/7)

 5/63 + 2/7

We have to find the LCM of the denominators of the given rational numbers.

LCM of 63 and 7 = 63

Now, express into common denominator

(5/63) = [(5×1)/ (63×1)] = (5/63)

(2/7) = [(2×9)/ (7×9)] = (18/63)

Then,

= (5/63) + (18/63)

= (5 + 18)/63

= 23/63

(iii) -6/13 – (-7/15)

Solution:

According to the question,

LCM of 13 and 15 = 195

Now,

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

(7/15) = [(7×13)/ (15×13)] = (91/195)

Then,

= (-90/195) + (91/195)

= (-90 + 91)/195

= (1/195)

(iv) -3/8 – 7/11

Solution:

According to the question,

LCM of 8 and 11 = 88

Now,

(-3/8) = [(-3×11)/ (8×11)] = (-33/88)

(7/11) = [(7×8)/ (11×8)] = (56/88)

Then,

= (-33/88) – (56/88)

= (-33 – 56)/88

= (-89/88)

Mathematics - Class 7 - Chapter 9 - Rational Numbers - Exercise 9.2 - NCERT Exercise Solution

We have, -19/9 – 6

We have to find the LCM of the denominators of the given rational numbers.

LCM of 9 and 1 = 9

Now,

(-19/9) = [(-19×1)/ (9×1)] = (-19/9)

(6/1) = [(6×9)/ (1×9)] = (54/9)

Then,

= (-19/9) – (54/9)

= (-19 – 54)/9

= (-73/9)

3. Find the product:

(i) (9/2) × (-7/4)

Solution:

We have, (9/2) × (-7/4)

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (9×-7)/ (2×4)

= -63/8

(ii) (3/10) × (-9)

Solution:

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

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (3×-9)/ (10×1)

= -27/10

(iii) (-6/5) × (9/11)

Solution:

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (-6×9)/ (5×11)

= -54/55

(iv) (3/7) × (-2/5)

Solution:

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (3×-2)/ (7×5)

= -6/35

(v) (3/11) × (2/5)

Solution:

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (3×2)/ (11×5)

= 6/55

(vi) (3/-5) × (-5/3)

Solution:

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (3×-5)/ (-5×3)

On simplifying,

= (-15)/ (-15)

= 1

4. Find the value of:

(i) (-4) ÷ (2/3)

Solution:

Reciprocal of (2/3) is (3/2)

Now,

= (-4/1) × (3/2)

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (-4×3) / (1×2)

= (-2×3) / (1×1)

= -6

(ii) (-3/5) ÷ 2

Solution:

Reciprocal of (2/1) is (1/2)

Now,

= (-3/5) × (1/2)

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (-3×1) / (5×2)

= -3/10

(iii) (-4/5) ÷ (-3)

Solution:

Reciprocal of (-3) is (1/-3)

Now,

= (-4/5) × (1/-3)

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (-4× (1)) / (5× (-3))

= -4/-15

= 4/15

(iv) (-1/8) ÷ 3/4

Solution:

Reciprocal of (3/4) is (4/3)

Now,

= (-1/8) × (4/3)

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (-1×4) / (8×3)

= (-1×1) / (2×3)

= -1/6

(v) (-2/13) ÷ 1/7

Solution:

Reciprocal of (1/7) is (7/1)

Now,

= (-2/13) × (7/1)

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (-2×7) / (13×1)

= -14/13

(vi) (-7/12) ÷ (-2/13)

Solution:

Reciprocal of (-2/13) is (13/-2)

Now,

= (-7/12) × (13/-2)

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (-7× 13) / (12× (-2))

= -91/-24

= 91/24

(vii) (3/13) ÷ (-4/65)

Solution:

Reciprocal of (-4/65) is (65/-4)

Now,

= (3/13) × (65/-4)

Multiplying numerator by the numerator and denominator by the denominator of both rational numbers.

= (3×65) / (13× (-4))

= 195/-52

= -15/4

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