Percentage mcqs with answers pdf for FPSC NTS PPSC SPSC BPSC, KPPSC and ETEA. Percentage Questions with solutions PDF and mathematics mcqs with answers pdf.
When a number is first increased by 10 % and then reduced by 10 % the number ?
A. decreased by 1 %
B. increased by 1 %
C. does not change
D. None of themIncreased Number = (110% of x)
= (110/100 × x) = (11x/10)
Finally reduced number = (90 % of 11x/10)
= (90/100 × 11x/10) = 99x/100
Decrease = (x – 99x/100) = x/100
Decrease % = (x/100 × 1/x × 100)% = 1 %
A period of 4 hrs 30 min is what percent of a day ?
A. 16 3/4 %
B. 19 %
C. 18 3/4 %
D. 20 %Required % = (9/48 × 100)% = 75/4 % = 18 ¾ %
A bucket contains 2 litres more water when it is filled 80 % in comparison when it is filled 66 2/4 % what is the capacity of the bucket ?
A. 15 litres
B. 20 litres
C. 10 litres
D. 66 2/3 litres(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
A dishonest dealer claims to sell his goods at the cost price but uses a false weight of 900 gm for 1 kg what is his gain percent ?
A. 11 1/9 %
B. 12 1/9 %
C. 13 %
D. 11.25 %C.P of 900gm = Rs 900,
S.P of 1000gm = Rs 1000
Gain % = (100/900 × 100)% = 100/9 % = 11 1/9 %
Out of 100 students, 50 failed in English and 30 in Mathematics. If 12 students fail in both English and Mathematics. Then the number of students who passed in both the subjects is ?
A. 76
B. 32
C. 45
D. 23B = 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
In a school, 40 % of the students play football and 50 % play cricket. If 18 % of the students play neither football nor cricket, the percentage of students playing both is ?
A. 32 %
B. 22 %
C. 8 %
D. 40 %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%
If the price of the eraser is reduced by 25% a person buy 2 more erasers for a rupee. How many erasers available for a rupee ?
A. 6
B. 9
C. 13
D. 17Reduced 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
Price of petrol is increased by 25 %. By how much percent a car owner should reduce his consumption of petrol. So that expenditure on petrol would not be increased ?
A. 20 %
B. 50 %
C. 25 %
D. 30 %Reduction % in consumption = {r/(100+ r) × 100}% = (25/125 × 100)% = 20%
In measuring the sides of a rectangle errors of 5 % and 3 % in excess are made. The error percent in the calculates area is ?
A. 7.15 %
B. 6. 25%
C. 8. 35 %
D. 8.15 %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 %
If the numerator of a fraction is increased by 140 % and the denominator is increased by 150 % the resultant fraction is 4/15 what is the original fraction ?
A. 2/9
B. 3/5
C. 5/16
D. None of them