Percentage Mcqs

A. 1000 kg
B. 1100 kg
C. 1125 kg
D. 1225 kg
Explanation:
Let the weight of fresh grapes be x kg
Quantity of water in it = (80/100 × x)kg = 4x/5 kg
Quantity of pulp in it = (x – 4x/5)kg = x/5 kg
Quantity of water in 250 kg dry grapes
= (10/100 × 250)kg = 25kg
Quantity of pulp in it = (250 – 25)kg = 225 kg
Therefore, x/5 = 225
=> x = 1125

A. 32 kg
B. 40 kg
C. 52 kg
D. 80 kg
Explanation:
Quantity of water in 100kg of fresh fruits =(68/100 × 100)kg
Quantity of pulp in it = (100 – 68)kg = 32 kg
Let the dry fruit be x kg
Water in it = (20/100 × x)kg = x/5 kg
Quantity of pulp in it = (x – x/5)kg = 4x/5 kg
Therefore, 4x/5 = 32 => x = 160/4 = 40 kg

A. 3/5
B. 5/16
C. 2/9
D. None of these

A. 8. 35 %
B. 7.15 %
C. 8.15 %
D. 6. 25%
Explanation:
Let length = x units and breadth = Y units
Then actual area = xy sq.units
Length shown = (105/100 × x)units = 21x/20 units;
Breadth shown = (103/100 × Y)
Calculated area = (21x/20 × 103y/100)sq.units
= 2163xy/2000 sq.units
Error = (2163xy/2000 – xy)
= 163xy/2000
Error % = (163xy/2000 × 1/xy × 100)%
= 163/20 % = 8.15 %

A. 25 %
B. 30 %
C. 50 %
D. 20 %
Explanation:
Reduction % in consumption = {r/(100+ r) × 100}% = (25/125 × 100)% = 20%

A. 8
B. 6
C. 4
D. 2
Explanation:
Let n erasers be available for a rupee
Reduced Price = (75/100 × 1) = Re ¾
¾ rupee fetch n erasers = 1 Rupee will fetch (n × 4/3) erasers
Therefore, 4n/3 = n +2 => 4n = 3n +6 => n =6

A. 40 %
B. 32 %
C. 22 %
D. 8 %
Explanation:
Let A = set of students who play football and
B = set of students play cricket.
Then n(A) = 40, n (B) = 50 and
n(A U B) = (100 – 18) = 82
n(A U B) = n(A) + n(B) – n(A ∩ B)
n(A∩B) = n(A) + n(B) – n(AUB) = (40 + 50 -82) = 8
Percentage of the students who play both = 8%

A. 26
B. 28
C. 30
D. 32
Explanation:
Let A = Set of students who fail in English
B = Set of students who fair in mathematics
Then , n(A) = 50, n(B) = 30 and n(A∩ B) = 12.
n(AUB) = n(A) + n(B)- n(A∩B) = (50 + 30 -12) = 68
Number of students who fail in one or both the students = 68
Number of those who pass in both = (100 – 68) = 32

A. 13 %
B. 11 1/9 %
C. 11.25 %
D. 12 1/9 %
Explanation:
Let the C.P of the article be Rs 1 per gram
C.P of 900gm = Rs 900,
S.P of 1000gm = Rs 1000
Gain % = (100/900 × 100)% = 100/9 % = 11 1/9 %

A. 10 litres
B. 15 litres
C. 66 2/3 litres
D. 20 litres
Explanation:
Let the capacity of the bucket be x liters
(80% of x) – (200/3% of x) = 2
=> (80/100 × x) – (200/300 × x) = 2
=> 4x/5 – 2x/3 = 2
=> (12x -10x) = 30
=> 2x = 30
=> X = 15 Litres