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Olympiad Test: Percentage -1 - Class 7 MCQ


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

Olympiad Test: Percentage -1 for Class 7 2024 is part of Mathematics Olympiad Class 7 preparation. The Olympiad Test: Percentage -1 questions and answers have been prepared according to the Class 7 exam syllabus.The Olympiad Test: Percentage -1 MCQs are made for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Percentage -1 below.
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Olympiad Test: Percentage -1 - 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?

Detailed Solution for Olympiad Test: Percentage -1 - Question 1

Olympiad Test: Percentage -1 - 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:

Detailed Solution for Olympiad Test: Percentage -1 - Question 2

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.

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Olympiad Test: Percentage -1 - Question 3

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

Detailed Solution for Olympiad Test: Percentage -1 - Question 3

Suppose originally he had x apples.
Then, (100 – 40) % of x = 420
⇒ (60/100) × x = 420
⇒ x = (420 × 100)/60 = 700

Olympiad Test: Percentage -1 - Question 4

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

Detailed Solution for Olympiad Test: Percentage -1 - Question 4

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%.

Olympiad Test: Percentage -1 - Question 5

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

Detailed Solution for Olympiad Test: Percentage -1 - Question 5

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

Olympiad Test: Percentage -1 - Question 6

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

Detailed Solution for Olympiad Test: Percentage -1 - Question 6

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

Olympiad Test: Percentage -1 - 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?

Detailed Solution for Olympiad Test: Percentage -1 - Question 7

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

Olympiad Test: Percentage -1 - 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.

Detailed Solution for Olympiad Test: Percentage -1 - Question 8

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

Olympiad Test: Percentage -1 - Question 9

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

Detailed Solution for Olympiad Test: Percentage -1 - Question 9

Let the number be x.
Then, error = (5/3) x – (3/5) x = (16/15) x.
Error% = [(16x/15) × (3/5x) × 100] %
= 64%

Olympiad Test: Percentage -1 - 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:

Detailed Solution for Olympiad Test: Percentage -1 - Question 10

Number of valid votes = 80% of 7500
= 6000
∴ Valid votes polled by other candidates
= 45% of 6000 = (45/100 × 6000) = 2700

Olympiad Test: Percentage -1 - 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?

Detailed Solution for Olympiad Test: Percentage -1 - Question 11

Total number of votes polled
= (1136 + 7636 + 11628) = 20400
∴ Required percentage
= [(11628/20400)× 100] % = 57%

Olympiad Test: Percentage -1 - 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?

Detailed Solution for Olympiad Test: Percentage -1 - Question 12

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

Olympiad Test: Percentage -1 - 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?

Detailed Solution for Olympiad Test: Percentage -1 - Question 13

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

Olympiad Test: Percentage -1 - 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.

Detailed Solution for Olympiad Test: Percentage -1 - Question 14

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

Olympiad Test: Percentage -1 - 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:

Detailed Solution for Olympiad Test: Percentage -1 - Question 15

Increase in 10 years = (262500 – 175000) = 875 00
Increase% = [(87500/175000) × 100] % = 50%
∴ Required average = (50/10) % = 5%

Olympiad Test: Percentage -1 - 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:

Detailed Solution for Olympiad Test: Percentage -1 - Question 16

C.P. = Rs 3000
S.P. = Rs [(3600 × 100)/ 100 + (10 × 2)] 
= Rs 3000
Gain = 0%

Olympiad Test: Percentage -1 - Question 17

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

Detailed Solution for Olympiad Test: Percentage -1 - Question 17

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%

Olympiad Test: Percentage -1 - 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?

Detailed Solution for Olympiad Test: Percentage -1 - Question 18

Required money
= P.W. of  Rs 10028 due 9 months hence
= Rs [(10028 × 100)/ (100 + 12 × 9/12)]
= Rs 9200

Olympiad Test: Percentage -1 - 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?

Detailed Solution for Olympiad Test: Percentage -1 - Question 19

P.W. of  Rs12,880 due 8 months hence
= Rs [(12880 × 100)/ (100 +18 × 8/12)]
= Rs (12880 × 100)/112 = Rs 11500

Olympiad Test: Percentage -1 - 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:

Detailed Solution for Olympiad Test: Percentage -1 - Question 20

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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