6th Grade > Mathematics
FRACTIONS MCQs
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Each part: 1 Mark
(a) LCM of 7 and 5 is 35
37=3×57×5=1535
35=3×75×7=2135
Clearly, 2135 is greater than 1535
Hence, 35 is greater than 37
(b) LCM of 8 and 2 is 8
58=5×18×1=58
12=1×42×4=48
Clearly, 58 is greater than 48
Hence, 58 is greater than 12
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Each option: 1 Mark
(a) Length of the other piece = Total length of the wire - Length of the given piece
= (78−14)m
= (7−28)m
= (58)m
(b) Total length of ribbon = Length of ribbon bought by Sarita + Length of ribbon bought by Lalita
= 25+34
= 8+1520
= 2320
= (1320)m
(c) 814−256 = 334−176
= 198−6824
= 13024
= 6512
= 5512
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Each option: 1 Mark
(a) Given pizza has 4 slices
Pizza slices left in total = 3
Fraction of pizza slices left= 34
(b) 6729×7296=1 ( product of a fraction and its reciprocal is always 1)
In a match of 50 overs, the team batting second was given a target of 310.They lost the match by 10 runs. The team batting second i.e. the losing team, scored one-half of their runs in 30 overs and next five overs they scored one-half of the remaining runs. Later in the next 8 overs, they scored 15th of the remaining runs. In last seven overs, they scored the remaining runs. How many runs did they score in last seven overs?
[3 MARKS]
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Steps: 2 Marks
Answer: 1 Mark
Total runs scored by the team that lost = 300
Now, total no. of overs played by them = 50
That is in 30 overs, this team scored =3002
=150 runs
The balance 300 - 150 = 150 runs were scored by this team in balance 20 overs.
By the given condition,
In the next 5 overs, 12 of 150 runs
= 75 runs were scored.
⇒ 150 runs - 75 runs = 75 runs are still in balance.
Again in the next 8 overs.
15 of 75 = 15 runs were scored.
Now No. of overs left = 20 - (5 + 8)
⇒ 20 - 13 = 7
In seven overs, runs scored will be 75 - 15 = 60.
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Steps: 1 Mark
Difference in fractions: 1 Mark
Solution: 1 Mark
Time taken by Mena = 715 hr
Time taken by Vaibhav
= 25 hr=2×35×3=615
Clearly, 715>615
∴ Vaibhav worked longer.
Time for which Vaibhav has worked more
= 715−25=7−615=115 hrs longer.
(a) Last month, Shakshi and Shashank sold candy to raise money for their debate team. Shashank sold 25 as much candy as Shakshi did. If Shakshi sold 3 boxes of candy, how many boxes of candy did Shashank sell? [4 MARKS]
(b) Naina was given 112 piece of cake and Najma was given 113 piece of cake. Find the total amount of cake that was given to both of them.
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Each part: 2 Marks
(a) Shashank sold 25 times 3 boxes of candy ⇒25×3
Simplify the product 65=115
Shashank sold 115 boxes of candy.
(b) Total mount of cake = Naina's share of cake + Najma's share of cake
= 112+113
= 32+43
= 9+86
= 116
= 156
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Concept: 1 Mark
Application: 3 Marks
Distance covered by car A in 4 litres of petrol is 5500m.
Distance covered by car A in 1 litre of petrol =(5500÷4)= 1375m.
Distance covered by car B in 1 litre of petrol = 1800m.
So car B travels more compared to car A
Difference = 1800-1375 = 425m
Car B travels 425m more than car A
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Definitions: 3 Marks
Example: 1 Mark
Saurabh should answer-
Proper Fraction: The fractions in which the numerator is less than the denominator are called proper fractions.
e.g. 13,34,21793
Improper Fraction: The fractions in which the numerator is greater than (or equal to) the denominator are called proper fractions.
e.g. 43,94,229
Mixed Fraction: A whole number and a proper fraction mixed together is called a mixed fraction.
e.g. 213,134,3293
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Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Let the length and breadth of the bigger rectangle be ′a′m and ′b′m respectively
Then, the length of the shaded rectangle is (0.1 a)m.
Then, the breadth of the shaded rectangle is (0.001 b)m.
Area of the given bigger rectangle = ab metres. = 10000 m2
Area of the shaded rectangle = (0.1a) × (0.001b)
= (0.0001) × (ab)
= (0.0001) × (10000)
= 1m2