Aptitude Percentage Test Paper 5

21) If 16(2/3) % of a number is added to itself, the number becomes 1400. Find the original number.

1100
1356
1000
1200

Answer: d

Explanation:
Let the number is = x
Now, 16(2/3) % = 16(2/3) * (1/100) = 1/6
So, we have, x + x/6 = 7x/6
As per question, 7x/6 = 1400
So, x = 6* 1400/7 = 1200

22) When the numerator is increased by 200%, and the denominator is increased by 350%, the resultant fraction is 5/12. Find the original fraction.

5/9
11/13
8/9
5/8

Answer: d

Explanation:
Let the original fractions = x/y
So, as per question, (x+ (200/100) * x)/ (y+ (350/100) * y) = 5/12
(300x/100)/ (450y/100) = 5/12
After solving this equation, we get (x/y) = 5/8
Hence, the original fraction was 5/8

23) A number is supposed to multiply by ¾ but by mistake, it is divided by ¾. Find the % error.

77.77% error
36% error
10% error
No error

Answer: a

Explanation:

Note: A number divided by ¾ = multiplied by 4/3.

Now, take LCM of denominators of ¾, and 4/3
i.e., LCM of 4 and 3 = 12
i.) (3/4) * 12 = 9
ii.) (4/3) * 12 = 16
Difference = 16-9 = 7, and original = 9
(7/9) * 100= 77.77 % error.

24) An exam comprises 80 questions carrying one mark each, Parvez answers 65% of the first 40 questions correctly. How much percent of the remaining 40 questions does he need to answer correctly to score 75% in this exam?

Answer: b

Explanation:
Total marks = 80
65% of 40 = (65/100) * 40 = 26
And, 75% of 80 = (75/100) * 80 = 60
i.e., the required marks 60-26 = 34
So, he needs to answer 34 questions correctly out of the remaining 40 questions.
The required percentage = 34/40 * 100 = 85%

25) If the radius of a circle is increased by 50%, what will be the percentage increase in its area?

150%
50%
100%
125%

Answer: d

Explanation:
Solution 1:
Concept: 50% increment Percentage aptitude
The above concept shows that when the old value is 2 units then a 50% increment means an increase of one unit, i.e. the old value becomes 3 after a 50% increment.
I.e. old value = 2 new value after 50% increment = 2+1 + 3
Let old radius= 2, and after the increment, it becomes 3.
i.e., Old area: new area
pi * (22) : pi * (32)
Cancel the constant term, we get
4: 9
The difference between old and new is 5 (4-5).
i.e., required % = (diff/original) * 100
(5/4) * 100 = 125%
So, the area of the circle increases by 125%.

Solution 2:
Area = pi * r * r
Pi is constant so it can be neglected.
R * R = A
Old: 2 * 2 = 4
New: 3* 3 = 9
Difference between area = 5, and original = 4
So, the percentage increase in area = (5/4) * 100 = 125%