Test: Percentages - 2 - Class 10 MCQ

# Test: Percentages - 2 - Class 10 MCQ

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## 15 Questions MCQ Test The Complete SAT Course - Test: Percentages - 2

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Test: Percentages - 2 - Question 1

### A glass of juice contains 5% fruit extract, 25% of pulp and rest of water. Find amount of water that should be added in glass of 450 ml juice to reduce pulp concentration to 15%?

Detailed Solution for Test: Percentages - 2 - Question 1

Juice of 450 ml contains 5% fruit extract, 25% of pulp.

Amount of pulp = 450 × 25/100 = 112.5 ml

Now, let the amount of water added be ‘x’ ml.

Total volume of juice after adding water = 450 + x

New percentage of pulp = 15%

According to the question

⇒ (450 + x) × 15/100 = 112.5

⇒ (450 + x) = 750

∴ x = 300 ml

Test: Percentages - 2 - Question 2

### A solution contains 33g of common salt in 320g of water. Calculate the concentration in terms of mass, by mass percentage of the solution.

Detailed Solution for Test: Percentages - 2 - Question 2

Formula Used:

Mass percentage in solution = Total Salt/Total Solution × 100

Calculation:

Quantity of salt = 33g

Quantity of water = 320 g

The total quantity of solution,

⇒ 33 + 320 = 353 g

Now by using the formula,

Percentage of salt in the solution

⇒ (33/353) × 100 = 9.35% (approx)

∴ Mass percentage of the solution is 9.35 %.

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Test: Percentages - 2 - Question 3

### In an election A got 55% of the total votes and the remaining votes were casted to B. If 10,000 votes of A are given to B, there would have been a tie. Find the number of total votes polled?

Detailed Solution for Test: Percentages - 2 - Question 3

Given:

In an election A got 55% of the total votes and the remaining votes were casted to B. If 10,000 votes of A are given to B, there would have been a tie

Formula used:

Calculation:

Let the total votes be 100x

A Gets = 55x

B gets = 45x

Given that 10,000 votes of A given to B

As per question

⇒ 55x – 10,000 = 45x + 10,000

⇒ x = 2000

⇒ 100 × 2000 = 200000

∴ Total votes polled is 200000

Test: Percentages - 2 - Question 4

There were two candidates in an election, 10% of voters did not vote and 48 votes were found invalid. The winning candidate got 53% of all the voters in the list  and won by 304 votes. Find the total number of votes enrolled.

Detailed Solution for Test: Percentages - 2 - Question 4

Given:

There were two candidates in an election, 10% of voters did not vote and 48 votes were found invalid. The winning candidate got 53% of the total votes and won by 304 votes.

Concept used:

Percentage

Calculation:

Let the total number of voters be 100x

10% of voters did not vote

Number of voters who vote = 100x - 10x = 90x

Valid votes = 90x - 48

Votes gained by the winning candidate =

Votes gained by the loosing candidate = 90x - 48 - 53x

⇒ 37x - 48

As per the question,

⇒ 53x - (37x - 48) = 304

⇒ 16x = 304 - 48

⇒ 16x = 256

⇒ x = 16

∴ Total number of voters = 100x = 1600
Alternate Method:

Let total number of votes be 100 units,

10% voters did not cast their vote

⇒ Votes polled = 90 units

The winning candidate got 53% of all the voters in the list and won by 304 votes,

⇒ Winning candidate got = 53 units votes

⇒ Other candidate got = 37 units votes

⇒ 16 units = 256

∴ Total number of voters = 1600.

Test: Percentages - 2 - Question 5

65% of a number is more than 25% of the number by 120. What is 20% of that number?

Detailed Solution for Test: Percentages - 2 - Question 5

Given:

65% of a number is more than 25% of the number by 120.

Calculations:

Let the number be x

According to question, (65x/100) - (25x/100) = 120

⇒ (65x - 25x)/100 = 120

⇒ 40x/100 = 120

⇒ x = (120 × 100)/40 = 300

But we have to find 20%, so (20/100) × 300 = 60

∴ The required number is 60.
Shortcut Trick:

65% - 25% = 120

⇒ 40% = 120

⇒ 20% = 60

∴ The required number is 60

Test: Percentages - 2 - Question 6

If the price of petrol has increased from Rs. 40 per litre to Rs. 60 per litre, by how much percent a person has to decrease his consumption so that his expenditure remains same.

Detailed Solution for Test: Percentages - 2 - Question 6

Given:

If the price of petrol has increased from Rs. 40 per litre to Rs. 60 per litre

Calculation:

Let the consumption be 100 litres.

When price is Rs. 40 per litres, then, the expenditure = 100 × 40

⇒ Rs. 4,000.

At Rs. 60 per litre, the 60 × consumption = 4000

Consumption = 4,000/60 = 66.67 litres.

∴ Required decreased % = 100 - 66.67 = 33.33%

Test: Percentages - 2 - Question 7

Out of two numbers, 65% of the smaller number is equal to 45% of the larger number. If the sum of two numbers is 2574, then what is the value of the larger number?

Detailed Solution for Test: Percentages - 2 - Question 7

Given:

Out of two numbers, 65% of the smaller number is equal to 45% of the larger

number. If the sum of two numbers is 2574

Calculation:

Let the smaller number be ‘x’ and the larger number be ‘y’

From the problem, it is given that

65%x = 45%y

⇒ 13x = 9y

⇒ x = (9/13)y    ----(1)

Given the sum of the numbers = 2574

⇒ (x + y) = 2574     ----(2)

Substituting the value of ‘x’ from Equation 1 in Equation 2, we get

(9/13)y + y = 2574

⇒ (9y + 13y) = 2574 × 13

⇒ 22y = (2574 × 13)

⇒ y = (2574 × 13)/22 = 1521

∴ Value of the larger number is 1521

Test: Percentages - 2 - Question 8

If the average, of a given number, 50% of that number and 25% of the same number is 280, then the number is

Detailed Solution for Test: Percentages - 2 - Question 8

Given:

Average is 280.

Formula used:

Average = sum of the observation/number of the observation

Calculation:

Let the number be x

According to the question,

⇒ (x + 50% of x + 25% of x)/3 = 280

⇒ (x + x/2 + x/4)/3 = 280

⇒ 7x/12 = 280

⇒ x =  480

∴ The number is 480.

Test: Percentages - 2 - Question 9

Two students appeared for an examination. One of them secured 22 marks more than the other and his marks were 55% of the sum of their marks. The marks obtained by them are _______.

Detailed Solution for Test: Percentages - 2 - Question 9

Given:

Two students appeared for an examination. One of them secured 22

marks more than the other and his marks were 55% of the sum of

their marks

Calculation:

Let the students be A and B

Let the marks secured by B = x

Marks secured by A = x + 22

Sum of their marks of A & B = x + x + 22 = 2x + 22

Accoring to question,

Marks of A =  55% of the sum of marks

⇒  x + 22 = 0.55 × (2x + 22)

⇒ x + 22 = 1.1x + 12.1

⇒ 0.1x = 9.9

⇒ x = 9.9/0.1 = 99 marks

Marks secured by A = 99 + 22 = 121 marks

Therefore the correct answer is 121.
Shortcut Trick:

Let us go by options.

The difference between the numbers is 22 in all cases.

So, now check the next statement.

121 = 0.55 × (121+ 99)

That is, as the first option itself satisfies the condition and since we do no have a combination of answers, it is the solution.

∴ The required numbers are 121 and 99.

Test: Percentages - 2 - Question 10

A fruit seller sells 45% of the oranges that he has along with one more orange to a customer. He then sells 20% of the remaining oranges and 2 more oranges to a second customer. He then sells 90% of the now remaining oranges to a third customer and is still left with 5 oranges. How many oranges did the fruit seller have initially?

Detailed Solution for Test: Percentages - 2 - Question 10

Calculation:

Let the initial oranges with the fruit seller be x.

1st selling = 0.45x + 1

Remaining = x - (0.45x + 1) = 0.55x - 1

2nd selling = 1/5 × (0.55x - 1) = 0.11x - 0.2 + 2 = 0.11x + 1.8

Remaining after second selling = 0.55x - 1 - (0.11x + 1.8) = 0.55x - 0.11x - 1 - 1.8 = 0.44x - 2.8

3rd selling = 90% × (0.44x - 2.8)

Remaining after 3rd selling = 0.1 × (0.44x - 2.8) = 0.044x - 0.28

According to the question-

⇒ 0.044x - 0.28 = 5

⇒ 0.044x = 5.28

⇒ x = 5.28/0.044 = 120

∴ The number of oranges was 120.

Test: Percentages - 2 - Question 11

5%-income of A is equal to 15% income of B and 10% income of B is equal to 20% income of C. If income of C is Rs. 2,000, then the sum of incomes (in Rs.) of A, B and C is:

Detailed Solution for Test: Percentages - 2 - Question 11

C's income = Rs.2000

20% of C's income = Rs.400

10%ofB′sincome = 20%of  C′s  income

or, 10% of B's income = 400

or, B's income = Rs.4000

15% of B's income = 15% of 4000 = Rs.600

5% of A's income = 15% of B income = 600

Thus, A's income = Rs.12000

Total income of A + B + C = 12000 + 4000 + 2000

= Rs.18,000

Test: Percentages - 2 - Question 12

A furniture company imported three types of woods, Wood A worth Rs. 440000, Wood B worth Rs. 230000 and Wood C worth Rs. 190000. Company had to pay 15% duty on Wood A, 9% on Wood B and 7% on Wood C. How much total duty (in rupees) Company had to pay on all items?

Detailed Solution for Test: Percentages - 2 - Question 12

Given:

Wood A worth Rs. 440000, Wood B worth Rs. 230000 and Wood C worth Rs. 190000

Company had to pay 15% duty on Wood A, 9% on Wood B and 7% on Wood C

Calculation:

Total duty on three woods = 440000 × 15% + 230000 × 9% + 190000 × 7%

= 440000 × 0.15 + 230000 × 0.09 + 190000 × 0.07

= 66000 + 20700 + 13300

= 100000

Test: Percentages - 2 - Question 13

If a number x is 10% less than another number y and y is 10% more than 125, then the value of x is:

Detailed Solution for Test: Percentages - 2 - Question 13

Given:

The number is 125.

Solution:

Given number is 125

∵ y is 10% more than 125:

So y = 125 × ( 110 / 100 )

⇒ y = 137.5

∵ x is 10 % lesser than y :

So x = 137.5 × (90 / 100)

⇒ x = 123.75

Hence the correct answer is "123.75".

Test: Percentages - 2 - Question 14

In the beginning of 2020, the population of three cities 'A', 'B', and 'C' was in the ratio of 4 : 5 : 6 respectively. During the year 2021, the population of respective cities increased by a percentage in the ratio of 7 : 5 : 9. In the beginning of 2021, the population of city 'C' was 27000 more than that of the beginning of the year 2020 and in the beginning of the year 2021 the ratio of the population of A : B was 108 : 125 then during the year 2020, what was the sum of the increase in the population of all the cities together?

Detailed Solution for Test: Percentages - 2 - Question 14

Given:

Population of A : B : C = 4 : 5 : 6 (in the beginning of 2020)

Percentage increase in the population of A : B : C = 7 : 5 : 9

Population of city 'C' (in 2021) - Population of city 'C' (in 2020) = 27000

Population of A : B (in the beginning of 2021) = 108 : 125

Calculation:

Let the population of city 'A', 'B' and 'C' was 7x, 5x and 9x (in the beginning of 2020)

And the percentage increase in the population of cities = 7y%, 5y%, and 9y%.

ATQ, 6x × [(100 + 9y)/100] - 6x = 27000 ...(i)

Again (4x) × [(100 + 7y)/100]/(5x) × [(100 + 5y)/100] = 108/125

⇒ (100 + 7y)/(100 + 5y) = 27/25

⇒ 40y = 200

⇒ y = 5

Using (i)

⇒ 6x × 145% - 6x = 27000

⇒ x = 10000

Population of city A (in 2020) = 4x = 40000

Population of city B (in 2020) = 5x = 50000

Population of city C (in 2020) = 6x = 60000

The sum of the increase in the population of all the cities together = 35% × 40000 + 25% × 50000 + 45% × 60000 = 53500

Hence, the correct answer is 53500.

Test: Percentages - 2 - Question 15

Monthly income of a person was Rs. 36,000. if he spent 25% on food, 12 % on his children's education and saves x% in bank account. After the calculation, he found that he was left with Rs. 2160. Find the sum of amount which he saves in his bank.

Detailed Solution for Test: Percentages - 2 - Question 15

Monthly income of a person was Rs. 36,000.

25% on food = (25/100) × 36000 = 9000 Rs.

12 % on his children's education = (12/100) × 36000 = Rs. 4320

x% in bank account = (x/100) × 36000 = 360x

Remaining amount = Rs. 2160.

So, we can write the above as:

9000 + 4320 + 360x + 2160 = 36000

360x = 36000 - 15480 = 20520

⇒ x = (20520/360) = 57%

The sum of amount which he saves in his bank = (57/100) × 36000 = Rs. 20520

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