CBSE Class 7 Mathematics Fractions And Decimals Worksheet Set E

CBSE Class 7 Maths Worksheet - Fractions and Decimals (4)

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CLASS – VII
SUBJECT – MATHS
TOPIC – FRACTION (DIVISION)

Question. Which of the following is a reducible fraction?
a) (105/112)
b) (104/121)
C) (77/72)
d) (46/63)

Answer : A

Question. [(3/10)+(8/15)] = ?
a) (11/10)
b) (11/15)
c) (5/6)
d) none of these

Answer : C

Question. Which of the following is a vulgar fraction?
a) (3/10)
b) (13/10)
c) (10/3)
d) none of these

Answer : C

Question. 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

Answer : A

Question. Which of the following is an improper fraction?
a) (7/10)
b) (7/9)
c) (9/7)
d) none of these

Answer : C

Question. [3(1/4)] – [2(1/3)] =?
a) [1(1/12)]
b) (1/12)
c) [1 (1/11)]
d) (11/12)

Answer : D

Question. (2/3), (4/6), (6/9), (8/12) are
a) Like fractions
b) irreducible fraction
c) equivalent fraction
d) None of these

Answer : A

Question. Reciprocal of [1(3/4)]
a) [1(4/3)]
b) [4(1/3)]
c) [3(1/4)]
d) none of these

Answer : D

Write down the reciprocal of :

Question. 7
Answer : Reciprocal of 7 is (1/7) [∵ ((7/1) × (1/7)) = 1]

Question. (1/12)
Answer : Reciprocal of (1/12) is (12/1) [∵ ((1/12) × (12/1)) = 1]
= 12

Question. (5/8)
Answer : Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]

Question. [12(3/5)]
Answer : Convert mixed fraction into improper fraction,
= (63/5)
Reciprocal of (63/5) is (5/63) [∵ ((63/5) × (5/63)) = 1]

Simplify :

Question. (4/7) ÷ (9/14)
Answer : 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)

Question. (7/10) ÷ (3/5)
Answer : 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)]

Divide:

Question. (11/24) by (7/8)
Answer : 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)

Question. [6(7/8)] by (11/16)
Answer : 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

Short Answer Type Questions :

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

Question. 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?
Answer : 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

Question. 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?
Answer : 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

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

Question. Mangos are sold at ₹ [43(1/2)] per kg. What is the weight of mangoes available for ₹ [326(1/4)]?
Answer : From the question,
Mangos are sold at = ₹ [43(1/2)]+ = (87/2)
The weight of mangos available for = ₹ [26(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 

Please click the below link to access CBSE Class 7 Maths Worksheet - Fractions and Decimals (4)