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## Exercise 2A

**1. Compare the fraction:**

**(i) (5/8)and (7/12)**

**Solution:-**

By cross multiplication, we have:

5 × 12 = 60 and 8 × 7 = 56

But,

60 > 56

∴ (5/8) > (7/12)

**(ii) (5/9)and (11/15)**

**Solution:-**

By cross multiplication, we have:

5 × 15 = 75 and 9 × 11 = 99

But,

75 < 99

∴ (5/9) < (11/15)

**(iii) (11/12) and (15/16)**

**Solution:-**

By cross multiplication, we have:

11 × 16 = 176 and 12 × 15 = 180

But,

176 < 180

∴ (11/12) < (15/16)

**2. Arrange the following fraction in ascending order :**

**(i) (3/4), (5/6), (7/9), (11/12)**

**Solution:-**

LCM of 4, 6, 9, 12 = 2× 2× 3× 3=36

Now, let us change each of the given fraction into an equivalent fraction having 36 as the denominator.

[(3/4) × (9/9)] = (27/36) [(5/6) × (6/6)] = (30/36) [(7/9) × (4/4)] = (28/360 [(11/12) × (3/3)] = (33/36)Clearly,

(27/36) < (28/36) < (30/36) < (33/36)

Hence,

(3/4) < (7/9) < (5/6) < (11/12)

Hence, the given fractions in ascending order are (3/4), (7/9), (5/6), (11/12)

**(ii) (4/5), (7/10), (11/15), (17/20)**

**Solution:-**

LCM of 5, 10, 15, 20 = 5 × 2 × 3 × 2 = 60

Now, let us change each of the given fraction into an equivalent fraction having 60 as the denominator.

[(4/5) × (12/12)] = (48/60) [(7/10) × (6/6)] = (42/60) [(11/15) × (4/4)] = (44/60) [(17/20) × (3/3)] = (52/60)Clearly,

(42/60) < (44/60) < (48/60) < (52/60)

Hence,

(7/10) < (11/15) < (4/5) < (17/20)

Hence, the given fractions in ascending order are (7/10), (11/15), (4/5) , (17/20)

**3. Arrange the following fraction in descending order :**

**(i) (3/4), (7/8), (7/12), (17/24)**

Solution:-

LCM of 4, 8, 12, 24 = 4 × 2 × 3 = 24

Now, let us change each of the given fraction into an equivalent fraction having 24 as the denominator.

[(3/4) × (6/6)] = (18/24) [(7/8) × (3/3)] = (21/24) [(7/12) × (2/2)] = (14/24) [(17/24) × (1/1)] = (17/24)Clearly,

(21/24) > (18/24) > (17/24) > (14/24)

Hence,

(7/8) > (3/4) > (17/24) > (7/12)

Hence, the given fractions in descending order are (7/8), (3/4), (17/24), (7/12)

**(ii) (2/3), (3/5), (7/10), (8/15)**

**Solution:-**

LCM of 3, 5, 10, 15 = 3 × 5 × 2 = 30

Now, let us change each of the given fraction into an equivalent fraction having 30 as the denominator.

[(2/3) × (10/10)] = (20/30) [(3/5) × (6/6)]= (18/30) [(7/10) × (3/3) = (21/30) [(8/15) × (2/2)] = (16/30)Clearly,

(21/30) > (20/30) > (18/30) > (16/30)

Hence,

(7/10) > (2/3) > (3/5) > (8/15)

Hence, the given fractions in descending order are (7/10), (2/3), (3/5), (8/15)

**4. Reenu got (2/7) part of an apple while sonal got (4/5) part of it. Who got the larger part and by how much?**

**Solution:-**

From the question,

Reenu got (2/7) part of an apple

Sonal got (4/5) part of an apple

First we have to compare the given fraction (2/7) and (4/5) to know who got the larger part of the apple.

Then,

By cross multiplication, we have

2 × 5 = 10 and 4 × 7 = 28

But,

10 < 28

∴ (2/7) < (4/5)

So, Sonal got the larger part of the apple

Now,

= (4/5)-(2/7)

= [(28-10)/35]

= [18/35]

∴ Sonal got (18/35) part of the apple larger than Reenu.

**5. Find the sum:**

**(i) (5/9)+(3/9)**

**Solution:-**

For adding two like fractions, the numerators are added and the denominator remains the same.

= (5+3)/9

= (8/9)

**(ii) (8/9) + (7/12)**

**Solution:-**

For addition of two unlike fractions, first change them to the like fractions.

LCM of 9, 12 = 36

Now, let us change each of the given fraction into an equivalent fraction having 36 as the denominator.

[(8/9) × (4/4)] = (32/36) [(7/12) × (3/3)] = (21/36)Now, add the like fractions,

= (32/36) + (21/36)

= (32+21)/36

= (53/36)

= [1(17/36)]

**(iii) [3(4/5)] + [2(3/10)] + [1(1/15)]**

**Solution:-**

First convert each mixed fraction into improper fraction.

We get,

= [3(4/5)] = (18/5)

= [2(3/10)] = (23/10)

= [1(1/15)] = (16/15)

Then,

(19/5) + (23/10) + (16/15)

LCM of 5, 10, 15 = 30

Now, let us change each of the given fraction into an equivalent fraction having 30 as the denominator.

= [(19/5) × (6/6)] = (114/30)

= [(23/10) × (3/3)] = (69/ 30)

= [(16/15) × (2/2)] = (32/30)

Now,

= (114/30) + (69/30) + (32/30)

= [(114+69+32)/30]

= (215/30)

= [7 (5/30)]

= [7 (1/5)]

**6. Find the difference:**

**(i) (5/7) – (2/7)**

**Solution:-**

The subtraction of fraction can be performed in a manner similar to that of addition.

For subtracting two like fractions, the numerators are subtract and the denominator remains the same.

= (5 – 2)/7

= (3/7)

**(ii) (5/6) – (3/4)**

**Solution:-**

For subtraction of two unlike fractions, first change them to the like fractions.

LCM of 6, 4 = 12

Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator.

= [(5/6) × (2/2)] = (10/12)

= [(3/4) × (3/3)] = (9/12)

Now,

= (10/12)-(9/12)

= [(10 – 9)/12]

= (1/12)

**(iii) [3(1/5)] – (7/10)**

**Solution:-**

Convert mixed fraction into improper fraction,

= [3(1/5)] = (16/5)

= (16/5)-(7/10)

For subtraction of two unlike fractions, first change them to the like fractions.

LCM of 5, 10 = 10

Now, let us change each of the given fraction into an equivalent fraction having 10 as the denominator.

= [(16/5) × (2/2)] = (32/10)

= [(7/10) × (1/1)] = (7/10)

Then,

= (32/10) – (7/10)

= (32-7)/10

= (25/10) … [÷ by 5]

= (5/2)

= [2(1/2)]

**(iv) 7 – [ 4 (2/3)]**

**Solution:-**

Convert mixed fraction into improper fraction, and then find the difference.

= [4(2/3)] = (14/3)

= 7-(14/3)

= (21-14)/3

= (7/3)

= [2 (1/3)]

**(v) [ 3(3/10)] – [1(7/15)]**

**Solution:-**

Convert mixed fraction into improper fraction, and then find the difference.

= [3(3/10)] = (33/10)

= [1(7/15)] = (22/15)

We get,

= (33/10) – (22/15)

LCM of 10, 15 = 30

Now, let us change each of the given fraction into an equivalent fraction having 10 as the denominator.

= [(33/10) × (3/3)] = (99/30)

= [(22/15) × (2/2)] = (44/30)

Then,

= (99/30) – (44/30)

= (99 -44)/30

= (55/30)

= (11/6)

= [1(5/6)]

**(vi) [2(5/9)] – [1(7/15)]**

**Solution:-**

Convert mixed fraction into improper fraction, and then find the difference.

= [2(5/9)] = (23/9)

= [1(7/15)] = (22/15)

We get,

= (23/9) – (22/15)

LCM of 9, 15 = 45

Now, let us change each of the given fraction into an equivalent fraction having 45 as the denominator.

= [(23/9) × (5/5)] = (115/45)

= [(22/15) × (3/3)] = (66/45)

Then,

= (115/45) – (66/45)

= (115 -66)/45

= (49/45)

= [1(4/45)]

**7. Simplify:**

**(i) (2/3) + (5/6) – (1/9)**

**Solution:-**

LCM of 3, 6, 9 = 18

Now, let us change each of the given fraction into an equivalent fraction having 18 as the denominator.

= (2/3) × (6/6) = (12/18)

= (5/6) × (3/3) = (15/18)

= (1/9) × (2/2) = (2/18)

Then,

= (12/18) + (15/18) – (2/18)

= (12+ 15 – 2)/ 18

= (27-2)/ 18

= (25/18)

= [1(7/18)]

**(ii) 8 – [4(1/2)] – [2(1/4)]**

**Solution:-**

Convert mixed fraction into improper fraction, and then find the difference.

= [4(1/2)] = (9/2)

= [2(1/4)] = (9/4)

LCM of 1, 2, 4 = 4

Now, let us change each of the given fraction into an equivalent fraction having 4 as the denominator.

= (9/2) × (2/2) = (18/4)

= (9/4) × (1/1) = (9/4)

= (8/1) × (4/4) = (32/4)

Then,

= (32/4) – (18/4) – (9/4)

= [(32-18-9)/4]

= [(32-27)/4]

= (5/4)

= [1(1/4)]

**(iii) [8(5/6)] – [3(3/8)] + [1(7/12)]**

**Solution:-**

First convert each mixed fraction into improper fraction.

We get,

= [8(5/6)] = (53/6)

= [3(3/8)] = (27/8)

= [1(7/12)] = (19/12)

Then,

(53/6)- (27/8) – (19/12)

LCM of 6, 8, 12 = 24

Now, let us change each of the given fraction into an equivalent fraction having 24 as the denominator.

= [(53/6) × (4/4)] = (212/24)

= [(27/8) × (3/3)] = (81/ 24)

= [(19/12) × (2/2)] = (38/24)

Now,

= (212/24) – (81/ 24) – (38/24)

= [(212- 81+ 38)/24]

= [(250-81/24)]

= (169/24)

= [7(1/24)]

**8. Aneeta bought [3(3/4)] kg apples and [4(1/2)] kg guava. What is the total weight of fruits purchased by her?**

**Solution:-**

The total weight of fruits bought by Aneeta = [3(3/4)] + [4(1/2)]

We have,

First convert each mixed fraction into improper fraction

= [3(3/4)] = (15/4)

= [4(1/2)] = (9/2)

Then,

= (15/4) + (9/2)

LCM of 4, 2 = 4

Now, let us change each of the given fraction into an equivalent fraction having 4 as the denominator

= (15/4) × (1/1) = (15/4)

= (9/2) × (2/2) = (18/4)

= (15/4) + (18/4)

= (15 + 18)/ 4)

= (33/4)

= [8(1/4)]

The total weight of fruits purchased by Aneeta is [8(1/4)] kg

## Exercise 2B

**1. Find the product:**

**(i) (3/5) × (7/11)**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 × 7)/ (5 × 11)

= (21/55)

**(ii) (5/8) × (4/7)**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5 × 4)/ (8 × 7)

= (20/56) … [÷ by 4]

= (5/14)

**(iii) (4/9) × (15/16)**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (4 × 15)/ (9 × 16)

= (60/144) … [ by 12]

= (5/12)

**(iv) (2/5) × 15**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 15)/ (5 × 1)

= (30/5) … [÷ by 5]

= 6

**(v) (8/15) × 20**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (8 × 20)/ (15 × 1)

= (160/15) … [÷ by 5]

= (32/3)

= [10(2/3)]

**(vi) (5/8) × 1000**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5 × 1000)/ (8 × 1)

= (5000/8) … [÷ by 8]

= 625

**(vii)[3(1/8)] × (16)**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (25/8) × (16/1)

= (25 × 16)/ (8 × 1)

= (400/8) … [÷ by 8]

= 50

**(viii)[2(4/15)] × (12)**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (34/15) × (12/1)

= (34 × 12)/ (15 × 1)

= (408/15) … [÷ by 3]

= (136/5)

= [27(1/5)]

**(ix)[3(6/7)] × [4(2/3)]**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (27/7) × (14/3)

= (27 × 14)/ (7 × 3)

= (378/21) … [÷ by 21]

= 18

**(x) [9(1/2)] × [1(9/19)]**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (19/2) × (28/19)

= (19 × 28)/ (2 × 19)

= (532/38)

= 14

**(xi) [4(1/8)] × [2(10/11)]**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (33/8) × (32/11)

= (33 × 32)/ (8 × 11)

= (1056/88)

= 12

**(xii) [5(5/6)] × [1(5/7)]**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (35/6) × (12/7)

= (35 × 12)/ (6 × 7)

= (420/42)

= 10

**2. Simplify:**

**(i) (2/3) × (5/44) × (33/35)**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2×5×33)/ (3×44×35)

On simplifying we get,

= (1×1×11) / (1×22×7)

= (11/154)

= (1/14)

**(ii) (12/25) × (15/28) × (35/36)**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (12×15×35)/ (25×28×36)

On simplifying we get,

= (1×3×5) / (5×4×3)

Again simplifying we get,

= (1×1×1)/ (1×4×1)

= (1/4)

**(iii) (10/27) × (28/65) × (39/56)**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (10×28×39)/ (27×65×56)

On simplifying we get,

= (10×1×3) / (27×5×2)

Again simplifying we get,

= (1×1×3)/ (27×1×1)

= (3/27)

=9

**(iv) [1(4/7)] × [1(13/22)] × [1(1/15)]**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

By Converting mixed fraction into improper fraction we get,

= (11/7) × (35/22) × (16/15)

= (11×35×16)/ (7×22×15)

On simplifying we get,

= (1×5×16) / (1×2×15)

Again simplifying we get,

= (1×1×8)/ (1×1×3)

= (8/3)

= [2(2/3)]

**(v) [2(2/17)] × [7(2/9)] × [1(33/52)]**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

By Converting mixed fraction into improper fraction we get,

= (36/17) × (65/9) × (85/52)

= (36×65×85)/ (17×9×52)

On simplifying we get,

= (4×5×5) / (1×1×4)

Again simplifying we get,

= (1×5×5)/ (1×1×1)

= 25

**(vi) [3(1/16)] × [7(3/7)] × [1(25/39)]**

**Solution:-**

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

By Converting mixed fraction into improper fraction we get,

= (49/16) × (52/7) × (64/39)

= (49×52×64)/ (16×7×39)

On simplifying we get,

= (7×4×4) / (1×1×3)

= (112)/ (3)

= [37(1/3)]

**3. Find :**

**(i) (1/3) of 24**

**Solution:-**

We have:

= (1/3) of (24/1)

This can be written as,

= (24/1) × (1/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (24 × 1)/ (1 × 3)

= (24/3)

= 8

**(ii) (3/4) of 32**

**Solution:-**

We have:

= (3/4) of (32/1)

This can be written as,

= (32/1) × (3/4)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (32 × 3)/ (1 × 4)

On simplifying we get,

= (8×3)/ (1×1)

= 24

**(iii) (5/9) of 45**

**Solution:-**

We have:

= (5/9) of (45/1)

This can be written as,

= (45/1) × (5/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (45 × 5)/ (1 × 9)

On simplifying we get,

= (5×5)/ (1×1)

= 25

**(iv) (7/50) of 1000**

**Solution**:-

We have:

= (7/50) of (1000/1)

This can be written as,

= (1000/1) × (7/50)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1000 × 7)/ (1 × 50)

On simplifying we get,

= (20×7)/ (1×1)

= 140

**(v) (3/20) of 1020**

**Solution:-**

We have:

= (3/20) of (1020/1)

This can be written as,

= (1020/1) × (3/20)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1020 × 3)/ (1 × 20)

On simplifying we get,

= (51×3)/ (1×1)

= 153

**(vi) (5/11) of ₹ 220**

**Solution:-**

We have:

= (5/11) of (220/1)

This can be written as,

= (220/1) × (5/11)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (220 × 5)/ (1 × 11)

On simplifying we get,

= (20×5)/ (1×1)

= ₹ 100

**(vii) (4/9) of 54 meters**

**Solution:-**

We have:

= (4/9) of (54/1)

This can be written as,

= (54/1) × (4/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (54 × 4)/ (1 × 9)

On simplifying we get,

= (6×4)/ (1×1)

= 24 meters

**(viii) (6/7) of 35 liters**

**Solution:-**

We have:

= (6/7) of (35/1)

This can be written as,

= (35/1) × (6/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (35 × 6)/ (1 × 7)

On simplifying we get,

= (5×6)/ (1×1)

= 30 liters

**(ix) (1/6) of an hour**

**Solution:-**

The above question can be written as,

= (1/6) of 60 min

We have:

= (1/6) of (60/1)

This can be written as,

= (60/1) × (1/6)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (60 × 1)/ (6 × 1)

On simplifying we get,

= (10×1)/ (1×1)

= 10 min

**(x) (5/6) of an year**

**Solution:-**

The above question can be written as,

= (5/6) of 12 months

We have:

= (5/6) of (12/1)

This can be written as,

= (12/1) × (5/6)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (12 × 5)/ (1 × 6)

On simplifying we get,

= (2×5)/ (1×1)

= 10 months

**(xi) (7/20) of a kg**

**Solution:-**

The above question can be written as,

= (7/20) of 1000 grams

We have:

= (7/20) of (1000/1)

This can be written as,

= (1000/1) × (7/20)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1000 × 7)/ (1 × 20)

On simplifying we get,

= (50×7)/ (1×1)

= 350 grams

**(xii) (9/20) of a meter**

**Solution:-**

The above question can be written as,

= (9/20) of 100 cm

We have:

= (9/20) of (100/1)

This can be written as,

= (100/1) × (9/20)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (100 × 9)/ (1 × 20)

On simplifying we get,

= (5×9)/ (1×1)

= 45 cm

**(xiii) (7/8) of a day**

**Solution:-**

The above question can be written as,

= (7/8) of 24 hours

We have:

= (7/8) of (24/1)

This can be written as,

= (24/1) × (7/8)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (24 × 7)/ (1 × 8)

On simplifying we get,

= (3×7)/ (1×1)

= 21 hours

**(xiv) (3/7) of a week**

**Solution:-**

The above question can be written as,

= (3/7) of 7 days

We have:

= (3/7) of (7/1)

This can be written as,

= (7/1) × (3/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (7 × 3)/ (1 × 7)

On simplifying we get,

= (1×3)/ (1×1)

= 3 days

**(xv) (7/50) of a liter**

**Solution:-**

The above question can be written as,

= (7/50) of 1000 ml

We have:

= (7/50) of (1000/1)

This can be written as,

= (1000/1) × (7/50)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1000 × 7)/ (1 × 50)

On simplifying we get,

= (20×7)/ (1×1)

= 140 ml

**4. Apples are sold at ₹ [48(4/5)] per kg. What is the cost of [3(3/4)] kg of apples?**

**Solution:-**

From the question,

The cost of 1 kg of apples = [48(4/5)] = (244/5)

Therefore, the cost of [3(3/4)] kg of apples = (15/4)

Then,

= (244/5) × (15/4)

= (244 × 15) / (5 × 4)

On simplifying we get,

= (61 × 3) / (1 × 1)

= ₹ 183

Hence, the cost of [3(3/4)] kg is ₹ 183

**5. Cloth is being sold at ₹ [42(1/2)] per meter. What is the cost of [5(3/5)] meters of this cloth?**

**Solution:-**

From the question,

The cost of 1 meter of cloth = ₹ [42(1/2)] = (85/2)

Therefore, the cost of [5(3/5)] meters of cloth = (28/5)

Then,

= (85/2) × (28/5)

= (85 × 28) / (2 × 5)

On simplifying we get,

= (17 × 14) / (1 × 1)

= ₹ 238

Hence, the cost of [5(3/5)] meters of cloth is ₹ 238.

**6. A car covers a certain distance at a uniform speed of [66(2/3)] km per hour. How much distance will it cover in 9 hours?**

**Solution:-**

From the question,

The total distance covered by a car in 1 hour = [66(2/3)] km = (200/3)

Therefore, the distance covered by a car in 9 hour = (200/3) × 9

Then,

= (200/3) × (9/1)

= (200 × 9) / (3 × 1)

On simplifying we get,

= (200× 3) / (1 × 1)

= 600 km

Hence, the distance covered by a car in 9 hour is 600 km.

**7. One tin holds [12(3/4)] liters of oil. How many liters of oil can 26 such tins hold?**

**Solution:-**

From the question,

The total amount of oil in 1 tin = [12(3/4)] liters = (51/4)

Therefore, the amount of oil in 26 such tins = (51/4) × 26

Then,

= (51/4) × (26/1)

= (51 × 26) / (4 × 1)

On simplifying we get,

= (51× 13) / (2 × 1)

= (663/2)

= [331(1/2)] liters

Hence, the amount of oil in 26 such tins is [331(1/2)] liters.

## Exercise 2C

**1. Write down the reciprocal of:**

**(i) (5/8)**

**Solution:-**

Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]

**(ii) 7**

**Solution:-**

Reciprocal of 7 is (1/7) [∵ ((7/1) × (1/7)) = 1]

**(iii) (1/12)**

**Solution**:-

Reciprocal of (1/12) is (12/1) [∵ ((1/12) × (12/1)) = 1]

= 12

**(iv) [12(3/5)]**

**Solution:-**

Convert mixed fraction into improper fraction,

= (63/5)

Reciprocal of (63/5) is (5/63) [∵ ((63/5) × (5/63)) = 1]

**2. Simplify:**

**(i) (4/7) ÷ (9/14)**

**Solution:-**

We have,

= (4/7) ÷ (9/14)

= (4/7) × (14/9)

(Because reciprocal of (9/14) is (14/9)

= (4 × 14) / (7 × 9)

= (4 × 2) / (1×9)

= (8/9)

**(ii) (7/10) ÷ (3/5)**

**Solution:-**

We have,

= (7/10) ÷ (3/5)

= (7/10) × (5/3)

(Because reciprocal of (3/5) is (5/3)

= (7 × 5) / (10 × 3)

= (7 × 1) / (2 × 3)

= (7/6)

= [1(1/6)]

**(iii) (8/9) ÷ (16)**

**Solution:-**

We have,

= (8/9) ÷ (16/1)

= (8/9) × (1/16)

(Because reciprocal of (16/1) is (1/16)

= (8 × 1) / (9 × 16)

= (1 × 1) / (9 × 2)

= (1/18)

**(iv) (9) ÷ (1/3)**

**Solution:-**

We have,

= (9/1) ÷ (1/3)

= (9/1) × (3/1)

(Because reciprocal of (1/3) is (3/1)

= (9 × 3) / (1 × 1)

= 27

**(v) (24) ÷ (6/7)**

**Solution:-**

We have,

= (24/1) ÷ (6/7)

= (24/1) × (7/6)

(Because reciprocal of (6/7) is (7/6)

= (24 × 7) / (1 × 6)

= (4 × 7) / (1 × 1)

= 28

**(vi) [3(3/5)] ÷ (4/5)**

**Solution:-**

Convert mixed fraction into improper fraction,

= [3(3/5)] = (18/5)

We have,

= (18/5) ÷ (4/5)

= (18/5) × (5/4)

(Because reciprocal of (4/5) is (5/4)

= (18 × 5) / (5 × 4)

= (9 × 1) / (1 × 2)

= (9/2)

= [4(1/2)

**(vii) [3(3/7)] ÷ (8/21)**

**Solution:-**

Convert mixed fraction into improper fraction,

= [3(3/7)] = (24/7)

We have,

= (24/7) ÷ (8/21)

= (24/7) × (21/8)

(Because reciprocal of (8/21) is (21/8)

= (24 × 21) / (7 × 8)

= (3 × 3) / (1 × 1)

= 9

**(viii) [5(4/7)] ÷ [1(3/10)]**

**Solution:-**

Convert mixed fraction into improper fraction,

= [5(4/7)] = (39/7)

= [1(3/10)] = (13/10)

We have,

= (39/7) ÷ (13/10)

= (39/7) × (10/13)

(Because reciprocal of (13/10) is (10/13)

= (39 × 10) / (7 × 13)

= (390) / (91)

= (30 / 7)

= [4(2/7)]

**(ix) [15(3/7)] ÷ [1(23/49)]**

**Solution:-**

Convert mixed fraction into improper fraction,

= [15(3/7)] = (108/7)

= [1(23/49)] = (72/49)

We have,

= (108/7) ÷ (72/49)

= (108/7) × (49/72)

(Because reciprocal of (72/49) is (49/72)

= (108 × 49) / (7 × 72)

= (9×7) / (1 × 6)

= (3×7) / (1×2)

= (21/2)

= [10(1 /2)

**3. Divide:**

**(i) (11/24) by (7/8)**

**Solution:-**

The above question can be written as,

= (11/24) ÷ (7/8)

We have,

= (11/24) × (8/7)

(Because reciprocal of (7/8) is (8/7)

= (11 × 8) / (24 × 7)

= (11 × 1) / (3 × 9)

= (11/21)

**(ii) [6(7/8)] by (11/16)**

**Solution:-**

The above question can be written as,

= [6(7/8)] ÷ (11/16)

Convert mixed fraction into improper fraction,

= [6(7/8)] = (55/8)

We have,

= (55/8) × (16/11)

(Because reciprocal of (11/16) is (16/11)

= (55 × 16) / (8 × 11)

= (5 × 2) / (1 × 1)

= 10

**(iii) [5(5/9)] by [3(1/3)]**

**Solution:-**

The above question can be written as,

= [5(5/9)] ÷ [3(1/3)]

Convert mixed fraction into improper fraction,

= [5(5/9)] = (50/9)

= [3(1/3)] = (10/3)

We have,

= (50/9) × (3/10)

(Because reciprocal of (10/3) is (3/10)

= (50 × 3) / (9 × 10)

= (5 × 1) / (3 × 1)

= (5/3)

= [1(2/3)]

**(iv) 32 by [1(3/5)]**

**Solution:-**

The above question can be written as,

= 32 ÷ [1(3/5)]

Convert mixed fraction into improper fraction,

= [1(3/5)] = (8/5)

We have,

= (32/1) × (5/8)

(Because reciprocal of (8/5) is (5/8)

= (32 × 5) / (1 × 8)

= (4 × 5) / (1 × 1)

= 20

**(v) 45 by [1(4/5)]**

**Solution:-**

The above question can be written as,

= 45 ÷ [1(4/5)]

Convert mixed fraction into improper fraction,

= [1(4/5)] = (9/5)

We have,

= (45/1) × (5/9)

(Because reciprocal of (9/5) is (5/9)

= (45 × 5) / (1 × 9)

= (5 × 5) / (1 × 1)

= 25

**(vi) 63 by [2(1/4)]**

**Solution:-**

The above question can be written as,

= 63 ÷ [2(1/4)]

Convert mixed fraction into improper fraction,

= [2(1/4)] = (9/4)

We have,

= (63/1) × (4/9)

(Because reciprocal of (9/4) is (4/9)

= (63 × 4) / (1 × 9)

= (7 × 4) / (1 × 1)

= 28

**4. A rope of length [13(1/2)] m has been divided into 9 pieces of the same length. What is the length of each piece? **

**Solution:-**

From the question,

Rope length = [13(1/2)] m = (27/2)

Number of equal pieces divided into = 9

Then we have,

= (27/2) ÷ (9/1)

= (27/2) × (1/9)

(Because reciprocal of (9/1) is (1/9)

= (27 × 1) / (2 × 9)

= (3×1) / (2×1)

= (3 / 2)

= [1(1/2)] m

Hence, the length of 9 pieces of rope is [1(1/2)] m

**5. 18 boxes of nails weigh equally and their total weight is [49(1/2)] kg. How much does each box weigh?**

**Solution:-**

From the question,

Total weight of boxes= [49(1/2)] kg = (99/2)

Number of boxes = 18

Then we have,

= (99/2) ÷ (18/1)

= (99/2) × (1/18)

(Because reciprocal of (18/1) is (1/18)

= (99 × 1) / (2 × 18)

= (11×1) / (2×2)

= (11 / 4)

= [2(3/4)] kg

Hence, the weight of each box is [2(3/4)] kg

**6. By selling oranges at the rate of ₹ [6(3/4)] per orange, a man gets ₹ 378. How many oranges does he sell?**

**Solution:-**

From the question,

Cost for 1 orange = ₹ [6(3/4)] = (27/4)

Man gets = ₹ 378

Then we have,

= (378/1) ÷ (27/4)

= (378/1) × (4/27)

(Because reciprocal of (27/4) is (4/27)

= (378 × 4) / (1 × 27)

= (42×4) / (1×3)

= (14×4) / (1×1)

= 56

Hence, the man sold 56 orange.

**7. Mangos are sold at ₹ [43(1/2)] per kg. What is the weight of mangoes available for ₹ [326(1/4)]?**

**Solution:-**

From the question,

Mangos are sold at = ₹ [43(1/2)] = (87/2)

The weight of mangos available for = ₹ [326(1/4)] = (1305/4)

Then we have,

= (1305/4) ÷ (87/2)

= (1305/4) × (2/87)

(Because reciprocal of (87/2) is (2/87)

= (1305 × 2) / (4 × 87)

= (435×1) / (2×29)

= [7(1/2)] kg

Hence, the weight of mangos available for (1305/4) is [7(1/2)] kg

**8. Vikas can cover a distance of [20(2/3)] km in [7(3/4)] hours on foot. How many km per hour does he walk?**

**Solution:-**

From the question,

Distance covered by vikas in [7(3/4)] hours on foot = [20(2/3)] km = (62/3)

Distance covered by vikas in 1 hour = (62/3) ÷ (31/4)

Then we have,

= (62/3) × (4/31)

(Because reciprocal of (31/4) is (4/31)

= (62 × 4) / (3 × 31)

= (2×4) / (3×1)

= (8) / (3)

= [2(2/3) km

Hence, Distance covered by vikas in 1 hour is [2(2/3) km

## Exercise 2D

**Mark against the correct answer in each of the following:**

**1. Which of the following is a vulgar fraction?**

**a) (3/10) b)(13/10) c)(10/3) d)none of these**

**Solution:-**

(10/3). Because, its denominator is a whole number, other than 10, 100, 10000 etc.

**2. Which of the following is an improper fraction?**

**a)(7/10) b)(7/9) c)(9/7) d)none of these**

**Solution:-**

(9/7). Because, its numerator is more than its denominator.

**3. Which of the following is a reducible fraction?**

**a) (105/112) b)(104/121) C)(77/72) d)(46/63)**

**Solution:-**

(105/112). Because the fraction can be reduced by dividing both numerator and denominator by a common factor.

**4. (2/3), (4/6), (6/9), (8/12) are**

**a)Like fractions b)irreducible fraction c)equivalent fraction**

**d) None of these**

**Solution: –**

Like fractions. Because the numerator is less than the denominator.

**5. Which of the following statement is true?**

**a)(9/16) = (13/24) b)(9/16) < (13/4) c)(9/16) > (13/24) d)none of these**

**Solution:-**

(9/16) > (13/24)

Because,

By cross multiplication

(9 × 24) = 216 and (13 × 16) = 208

There for,

216 > 208

Hence,

(9/16) > (13/24)

**6. Reciprocal of [1(3/4)]**

**a)[1(4/3)] b)[4(1/3)] c)[3(1/4)] d)none of these**

**Solution:-**

None of these.

Because, Reciprocal of [1(3/4)] = (7/4) = (4/7)

**7. [(3/10)+(8/15)] = ?**

**a)(11/10) b)(11/15) c)(5/6) d)none of these**

**Solution:-**

(5/6)

Because,

= (3/10) + (8/15)

LCM of 10 and 15 = 30

= (3/10) × (3/3) = (9/30)

= (8/15) × (2/2) = (16/30)

= (9/30) + (16/30)

= (9 + 16)/ 30

= (25/30)

= (5/6)

**8. [3(1/4)] – [2(1/3)] =?**

**a)[1(1/12)] b)(1/12) c)[1 (1/11)] d)(11/12)**

**Solution:-**

(11/12)

Because,

= (13/4) – (5/3)

LCM of 4 and 3 = 12

= (13/4) × (3/3) = (39/12)

= (6/3) × (4/4) = (28/12)

= (39/12) – (28/12)

= (39 – 28)/12

= (11/12)

## RS Aggarwal Solutions for Class 7 Maths Chapter 2 – Fractions

Chapter 2 – Fractions contains 4 exercises, and also RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercise. Now, let us have a look at some of the important concepts discussed in this chapter.

- Various Types of Fractions
- Method of Changing Unlike Fractions To Like Fractions
- Comparing Fractions
- Method of Comparing More Than Two Fractions
- Addition And Subtraction of Fractions
- Addition of Fractions
- Subtraction of Fractions
- Multiplication of Fractions
- Division of Fractions
- Reciprocal of a Fraction

### Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 2 – Fractions

RS Aggarwal Solutions for Class 7 Maths Chapter 2 – Fractions, refers to numbers that represent a part of the whole. When a component or a group of components is divided into equal parts, then each individual part is a fraction. A fraction is usually written as ¾ or ⅝ and so on. Fraction contains numerator and denominator. The solutions are solved in such a way that students will understand clearly and they will never get confused.