Olympiad Test: Percentage -1


20 Questions MCQ Test Mathematics Olympiad Class 7 | Olympiad Test: Percentage -1


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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. The solved questions answers in this Olympiad Test: Percentage -1 quiz give you a good mix of easy questions and tough questions. Class 7 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, long questions & short questions for Class 7 on EduRev as well by searching above.
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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