Aptitude Percentage Online Test, Free Aptitude Quiz
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Question 1 of 15
1. Question
A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
Correct
Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage = (50/110 × 100)% =Incorrect
Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage = (50/110 × 100)% = 
Question 2 of 15
2. Question
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:
Correct
Let their marks be (x + 9) and x.
Then, x + 9 = 56/100 (x + 9 + x)
⇒ 25(x + 9) = 14(2x + 9)
⇒ 3x = 99
⇒ x = 33
So, their marks are 42 and 33.Incorrect
Let their marks be (x + 9) and x.
Then, x + 9 = 56/100 (x + 9 + x)
⇒ 25(x + 9) = 14(2x + 9)
⇒ 3x = 99
⇒ x = 33
So, their marks are 42 and 33. 
Question 3 of 15
3. Question
A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
Correct
Suppose originally he had x apples.
Then, (100 – 40)% of x = 420.
⇒ 60/100 x x = 420
⇒ x = 420×100 / 60) = 700.Incorrect
Suppose originally he had x apples.
Then, (100 – 40)% of x = 420.
⇒ 60/100 x x = 420
⇒ x = 420×100 / 60) = 700. 
Question 4 of 15
4. Question
What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
Correct
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number = 14
Required percentage = (14/70 × 100)% = 20%.Incorrect
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number = 14
Required percentage = (14/70 × 100)% = 20%. 
Question 5 of 15
5. Question
If A = x% of y and B = y% of x, then which of the following is true?
Correct
x% of y = (x/100 × y) = (y/100 × x) = y% of x
∴ A = B.Incorrect
x% of y = (x/100 × y) = (y/100 × x) = y% of x
∴ A = B. 
Question 6 of 15
6. Question
If 20% of a = b, then b% of 20 is the same as:
Correct
20% of a = b ⇒ 20/100 a = b.
b% of 20 = (b/100 × 20) = (20/100 a × 1/100 × 20) = 4/100 a = 4% of a.Incorrect
20% of a = b ⇒ 20/100 a = b.
b% of 20 = (b/100 × 20) = (20/100 a × 1/100 × 20) = 4/100 a = 4% of a. 
Question 7 of 15
7. Question
In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
Correct
Let the number of students be x. Then,
Number of students above 8 years of age = (100 – 20)% of x = 80% of x.
∴ 80% of x = 48 + 2/3 of 48
⇒ 80/100 x = 80
⇒ x = 100.Incorrect
Let the number of students be x. Then,
Number of students above 8 years of age = (100 – 20)% of x = 80% of x.
∴ 80% of x = 48 + 2/3 of 48
⇒ 80/100 x = 80
⇒ x = 100. 
Question 8 of 15
8. Question
Two numbers A and B are such that the sum of 5% of A and 4% of B is twothird of the sum of 6% of A and 8% of B. Find the ratio of A : B.
Correct
Incorrect

Question 9 of 15
9. Question
A student multiplied a number by 3/5 instead of 5/3 .
What is the percentage error in the calculation?Correct
Let the number be x.
Then, error = 5/3 x – 3/5 x = 16/15 x
Error% = 16x/15 × 3/5x × 100)% = 64%Incorrect
Let the number be x.
Then, error = 5/3 x – 3/5 x = 16/15 x
Error% = 16x/15 × 3/5x × 100)% = 64% 
Question 10 of 15
10. Question
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:
Correct
Number of valid votes = 80% of 7500 = 6000.
∴ Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700.Incorrect
Number of valid votes = 80% of 7500 = 6000.
∴ Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700. 
Question 11 of 15
11. Question
Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
Correct
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 × 100)% = 57%Incorrect
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 × 100)% = 57% 
Question 12 of 15
12. Question
Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
Correct
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
⇒ z + 120/100 z = 550
⇒ 11/5 z = 550
⇒ z = (550 × 5/11) = 250.Incorrect
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
⇒ z + 120/100 z = 550
⇒ 11/5 z = 550
⇒ z = (550 × 5/11) = 250. 
Question 13 of 15
13. Question
Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
Correct
Let the amount taxable purchases be Rs. x.
Then, 6% of x = 30/100
x = (30/100 × 100/6) = 5.
∴ Cost of tax free items = Rs. [25 – (5 + 0.30)] = Rs. 19.70Incorrect
Let the amount taxable purchases be Rs. x.
Then, 6% of x = 30/100
x = (30/100 × 100/6) = 5.
∴ Cost of tax free items = Rs. [25 – (5 + 0.30)] = Rs. 19.70 
Question 14 of 15
14. Question
Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.
Correct
Rebate = 6% of Rs. 6650 = Rs. (6/100 x 6650) = Rs. 399.
Sales tax = 10% of Rs. (6650 – 399) = Rs. (10/100 x 6251) = Rs. 625.10
∴ Final amount = Rs. (6251 + 625.10) = Rs. 6876.10Incorrect
Rebate = 6% of Rs. 6650 = Rs. (6/100 x 6650) = Rs. 399.
Sales tax = 10% of Rs. (6650 – 399) = Rs. (10/100 x 6251) = Rs. 625.10
∴ Final amount = Rs. (6251 + 625.10) = Rs. 6876.10 
Question 15 of 15
15. Question
The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:
Correct
Increase in 10 years = (262500 – 175000) = 87500.
Increase% = (87500/175000) x 100 % = 50%.
Required average = (50/10) % = 5%.Incorrect
Increase in 10 years = (262500 – 175000) = 87500.
Increase% = (87500/175000) x 100 % = 50%.
Required average = (50/10) % = 5%.