# Aptitude Percentage MCQ, Free Aptitude Percentage Online Test Series

## Aptitude Percentage Online Test, Free Aptitude Quiz

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Aptitude Percentage MCQ, Free Aptitude Percentage Online Test, Free Aptitude Quiz, Online Aptitude Percentage Test. Aptitude Percentage Question and Answers 2018. Aptitude Percentage Quiz. Aptitude Percentage Free Mock Test 2018. **Aptitude Percentage Question and Answers in PDF. **The Aptitude Percentage online mock test paper is free for all students. The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence.** Aptitude MCQ** is very useful for exam preparation and getting for Rank. Aptitude Percentage Question and Answers in Hindi and English. Aptitude Percentage Mock test for topic via Online Mode. Here we are providing** Aptitude Percentage Mock Test in Hindi**. Now Test your self for “**Aptitude Percentage MCQ in Hindi**” Exam by using below 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?

CorrectNumber of runs made by running = 110 – (3 x 4 + 8 x 6)

= 110 – (60)

= 50.

Required percentage = (50/110 × 100)% =IncorrectNumber 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:

CorrectLet 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.IncorrectLet 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:

CorrectSuppose originally he had x apples.

Then, (100 – 40)% of x = 420.

⇒ 60/100 x x = 420

⇒ x = 420×100 / 60) = 700.IncorrectSuppose 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?

CorrectClearly, 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%.IncorrectClearly, 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?

Correctx% of y = (x/100 × y) = (y/100 × x) = y% of x

∴ A = B.Incorrectx% 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:

Correct20% 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.Incorrect20% 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?

CorrectLet 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.IncorrectLet 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 two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.

CorrectIncorrect - 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?CorrectLet the number be x.

Then, error = 5/3 x – 3/5 x = 16/15 x

Error% = 16x/15 × 3/5x × 100)% = 64%IncorrectLet 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:

CorrectNumber of valid votes = 80% of 7500 = 6000.

∴ Valid votes polled by other candidate = 45% of 6000

= (45/100 x 6000) = 2700.IncorrectNumber 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?

CorrectTotal number of votes polled = (1136 + 7636 + 11628) = 20400.

Required percentage = (11628/20400 × 100)% = 57%IncorrectTotal 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?

CorrectLet 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.IncorrectLet 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?

CorrectLet 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. 19.70IncorrectLet 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.

CorrectRebate = 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. 6876.10IncorrectRebate = 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:

CorrectIncrease in 10 years = (262500 – 175000) = 87500.

Increase% = (87500/175000) x 100 % = 50%.

Required average = (50/10) % = 5%.IncorrectIncrease in 10 years = (262500 – 175000) = 87500.

Increase% = (87500/175000) x 100 % = 50%.

Required average = (50/10) % = 5%.