Percentage Quantitative Aptitude Questions Page-3

17) In an examination a candidate has to secure 40% marks to pass. If a candidate secures 190 marks and fails by 10 marks. The total number of marks in the examination is?
(A) 400
(B) 360
(C) 500
(D) 600

View Answer

Answer: Option C

Solution:
Passing Marks = 40%
Candidate secures 190 marks and fails by 10. It means passing marks = 40% = 200
x =
200 * 100/40

= 500

18) If 50% of y is 250, then y is 50% of
(A) 250
(B) 500
(C) 750
(D) 1000

View Answer

Answer: Option D

Solution:
50/100
* y = 250
y =
250 * 100/50

y = 500
y =
50/100
* ?
500 * 100/50

= 1000

19) Three teams scored in the tournament as 125, 375 and 250 respectively. What percentage of the total score did the winning team get?
(A) 50%
(B) 60%
(C) 65%
(D) 90%

View Answer

Answer: Option A

Solution:
Total score = 125 + 375 + 250 = 750
Highest score = 375
=
375/750
* 100
= 50%

20) In an examination 33% of the students failed in Mathematics, 27% failed in English, while 15% of the students failed in both the subjects. Find the percentage of the students, who passed in both the subjects.
(A) 60%
(B) 55%
(C) 85%
(D) 45%

View Answer

Answer: Option B

Solution:
% of students passed in mathematics = 100 - 33 = 67
% of students passed in english = 100 - 27 = 73
Total students passed in both subjects = 100 - 15 = 85
Passed in Mathematics + Passed in English - Passed in Both subjects
67 + 73 - 85 = 55%

21) Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are
(A) 21, 30
(B) 45, 36
(C) 33, 45
(D) 42, 33

View Answer

Answer: Option D

Solution:
If one student get x marks, Second student gets x + 9
Marks of second student were 56% of the sum of their marks
x + 9 =
56/100
(x + x+9)
100(x + 9) = 56 (2x + 9)
100x + 900 = 112x + 504
900 - 504 = 112x - 100x
396 = 12x
x =
396/12
=> 33
Marks of second student = 33 + 9 = 42

22) After deducting 20% from a certain sum and then 30% from the remainder, there is 3500 left. Find the original sum?
(A) 6500
(B) 3175
(C) 6250
(D) 12500

View Answer

Answer: Option C

Solution:
=
3500 * 100 * 100/(100 - 20) (100 - 30)

=
3500 * 100 * 100/(80) (70)

= 6250

23) If N = X% of Y and M = Y% of X, then which of the following is true?
(A) N is equal to M
(B) N is smaller than M
(C) N is greater than M
(D) Relationship between N and M cannot be determined

View Answer

Answer: Option A

Solution:
Let X = 100 and Y = 100.
N = X% of Y
Substitute the value of X and Y => N = 100/100 * 100 => N = 100
M = Y% of X
M = 100/100 * 100 => M = 100
Now N = 100 and M = 100 therefore you can say that N = M

24) Evaluate: 15% of 80% of 90
(A) 11.7
(B) 13.5
(C) 10.8
(D) 19.4

View Answer

Answer: Option C

Solution:
80/100
* 90 = 72
15/100
* 72 = 10.8