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This mock test of Olympiad Test: Percentage -1 for Class 7 helps you for every Class 7 entrance exam.
This contains 20 Multiple Choice Questions for Class 7 Olympiad Test: Percentage -1 (mcq) to study with solutions a complete question bank.
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students definitely take this Olympiad Test: Percentage -1 exercise for a better result in the exam. You can find other Olympiad Test: Percentage -1 extra questions,
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QUESTION: 1

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?

Solution:

Number of runs made by running

= 110 – (3 × 4 + 8 × 6)

= 110 – (60)

= 50

Required percentage = (50/100 × 100) %

QUESTION: 2

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks were 56% of the sum of their marks. The marks obtained by them are:

Solution:

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

A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:

Solution:

Suppose originally he had x apples.

Then, (100 – 40) % of x = 420

⇒ (60/100) × x = 420

⇒ x = (420 × 100)/60 = 700

QUESTION: 4

What percentage of numbers from 1 to 70 has 1 or 9 in the unit’s digit?

Solution:

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.

The number of these numbers = 14

Required percentage = (14/70 × 100) %

= 20%.

QUESTION: 5

If A = x% of y and B = y% of x, then which of the following is true?

Solution:

x% of y = (x/100 × y) = (y/100 × x) = y% of x Therefore, A = B.

QUESTION: 6

If 20% of a = b, then b% of 20 is the same as:

Solution:

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

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?

Solution:

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

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.

Solution:

5% of A + 4% of B

= 2/3 (6% of A + 8% of B)

(5/100)A + (4/100)B

= 2/3[(6/100)A + (8/100)B]

⇒ (1/20)A + (1/25)B = (1/25)A + (4/75)B

⇒ (1/20 – 1/25)A = (4/75 – 1/25)B

⇒ (1/100)A = (1/75) B

A/B = 100/75 = 4/3

∴ Required ratio = 4 : 3

QUESTION: 9

A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation?

Solution:

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

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:

Solution:

Number of valid votes = 80% of 7500

= 6000

∴ Valid votes polled by other candidates

= 45% of 6000 = (45/100 × 6000) = 2700

QUESTION: 11

Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

Solution:

Total number of votes polled

= (1136 + 7636 + 11628) = 20400

∴ Required percentage

= [(11628/20400)× 100] % = 57%

QUESTION: 12

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?

Solution:

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

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?

Solution:

Let the amount on taxable purchases be Rs x.

Then, 6% of x = 30/100

⇒ x = [(30/100) × (100/6)] = 5

∴ Cost of tax free items

= Rs [2 5 – (5 + 0.30)] = Rs 19.70

QUESTION: 14

Rajeev buys goods 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.

Solution:

Rebate = 6% of Rs 6650 = Rs 399.

Sales tax = 10% of Rs (6650 - 399)

= Rs 625.10

∴ Final amount = Rs (6251 + 625.10)

= Rs 6876.10

QUESTION: 15

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:

Solution:

Increase in 10 years = (262500 – 175000) = 875 00

Increase% = [(87500/175000) × 100] % = 50%

∴ Required average = (50/10) % = 5%

QUESTION: 16

A man purchased a cow for Rs 3000 and sold it the same day for Rs 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:

Solution:

C.P. = Rs 3000

S.P. = Rs [(3600 × 100)/ 100 + (10 × 2)]

= Rs 3000

Gain = 0%

QUESTION: 17

The true discount on Rs 2562 due 4 months hence is Rs 122. The rate percent is:

Solution:

P.W. = Rs (2562 – 122) = Rs 2440

∴ S.I. on Rs 2440 for 4 months is Rs 122.

∴ Rate = [(100 × 122)/ (2440 × 1/3)] %

= 15%

QUESTION: 18

A trader owes a merchant Rs 10,028 due 1 year hence. The trader wants to settle the account after 3 months. If the rate of interest 12% per annum, how much cash should he pay?

Solution:

Required money

= P.W. of Rs 10028 due 9 months hence

= Rs [(10028 × 100)/ (100 + 12 × 9/12)]

= Rs 9200

QUESTION: 19

A man wants to sell his scooter. There are two offers, one at Rs 12,000 cash and the other a credit of Rs 12,880 to be paid after 8 months, money being at 18% per annum. Which is the better offer?

Solution:

P.W. of Rs12,880 due 8 months hence

= Rs [(12880 × 100)/ (100 +18 × 8/12)]

= Rs (12880 × 100)/112 = Rs 11500

QUESTION: 20

If Rs 10 be allowed as true discount on a bill of Rs 110 due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time is:

Solution:

S.I. on Rs (110 – 10) for a certain time = Rs 10

S.I. on Rs 100 for double the time = Rs 20

T.D. on Rs 120 = Rs (120 – 100) = Rs 20

T.D. on Rs 110 = Rs (20/120) × 110

= Rs 18.33

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