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# Practice Test: Percentages - 1 - UPSC

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## 10 Questions MCQ Test CSAT Preparation - Practice Test: Percentages - 1

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Practice Test: Percentages - 1 - Question 1

### A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest. What percent of his apples does the vendor throw?

Detailed Solution for Practice Test: Percentages - 1 - Question 1

Let the number of apples be 100.

On the first day, he sells 60% apples ie.,60 apples. Remaining apples =40.

He throws 15% of the remaining i.e., 15% of 40 = 6.Now he has 40 - 6 = 34 apples

The next day he throws 50% of the remaining 34 apples i.e., 17.

Therefore, in all, he throws 6 + 17 =23 apples.

Practice Test: Percentages - 1 - Question 2

### If X% of Y is 100 and Y% of Z is 200, find a relation between X and Z.

Detailed Solution for Practice Test: Percentages - 1 - Question 2 Practice Test: Percentages - 1 - Question 3

### In an election between 2 candidates, Walter gets 80% of the total valid votes. If the total votes were 12000, what is the number of valid votes that the other candidate Jesse Pinkman gets if 15% of the total votes were declared invalid?

Detailed Solution for Practice Test: Percentages - 1 - Question 3

Total Valid votes = 85% of 12000 = 10200.
Jesse Pinkman would get 10200 – 8160 = 2040 Votes.

Practice Test: Percentages - 1 - Question 4

30% of a% of b is 25% of b% of c. Which of the following is c?

Detailed Solution for Practice Test: Percentages - 1 - Question 4

Given:
a% of b = (a/100) x b = ab/100
30% of a% of b = 30% of ab/100
= (30/100) x ab/100
= 3ab/1000

Also Given:
b% of c = (b/100) x c = bc/100
25% of b% of c = 25% of bc/100
= (25/100) x bc/100
= bc/400

Now ATQ:
(3ab/1000) = (bc/400)

⇒ 3a/10 = c/4
⇒ c = 12a/10
⇒ c = 1.20a

Practice Test: Percentages - 1 - Question 5

A number is mistakenly divided by 2 instead of being multiplied by 2. Find the percentage change in the result due to this mistake.

Detailed Solution for Practice Test: Percentages - 1 - Question 5

Let the number be 100. Then, 200 should be the correct outcome. But instead, the value got is 50.
Change in value = 200 – 50 = 150.
The percentage change in the value = 150 X 100 / 200 = 75%.
Alternatively, you could think of this as the number being ‘x’ and the required result being 2x and the derived result being 0.5x.

Hence, the percentage change in the result is 1.5x X 100 / 2x. Clearly, the value would be 75%. (Note: In this case, the percentage change in the answer does not depend on the value of ‘x’).

Alternate solution :

Let the number be x.

Then actual value should be 2x and the measured value is x/2.

Therefore, the percentage error can be calculated as:

Actual value−Measured value/Actual value×100

=2x−x/2/2x×100

=4x−x/2/2x×100

=3x/2×1/2x×100

=3/4×100

=75%

Practice Test: Percentages - 1 - Question 6

10% of Mexico’s population migrated to South Asia, 10% of the remaining migrated to America and 10% of the rest migrated to Australia. If the female population, which was left in Mexico, remained only 3,64,500, find the population of Mexico City before the migration and its effects if it is given that before the migration the female population was half the male population and this ratio did not change after the migration?

Detailed Solution for Practice Test: Percentages - 1 - Question 6

Let the population of Mexico be X Pop.

After 10% migrated to SA = 1-0.1= 0.9x

Population after 10% migrated to America = 0.9x * 0.9 = 0.81x Population after 10% of the rest migrated to Australia = 0.81x * 0.9 = 0.729x

Let the no of males be a and females be b => a+b = 0.729x Given that b=a/2 ratio didn't change after and before migration so,

b = 364500

=> a = 2*364500 = 729000

=> 729000 + 364500 = 0.729x

=> x = 1500000

Practice Test: Percentages - 1 - Question 7

In a mixture of 100 litres of milk and water, 25% of the mixture is milk. How much water should be added to the mixture so that milk becomes 20% of the mixture?

Detailed Solution for Practice Test: Percentages - 1 - Question 7

From the first statement we get that out of 100 litres of the mixture, 25 litres must be milk.
Since we are adding water to this and keeping the milk constant, it is quite evident that 25 litres of milk should correspond to 20% of the total mixture.
Thus, the amount in the total mixture must be 125, which means we need to add 25 litres of water to make 100 litres of the mixture.

Practice Test: Percentages - 1 - Question 8

If the length, breadth and height of a cube are decreased, decreased and increased by 5%, 5% and 20%, respectively, then what will be the impact on the surface area of the cube (in percentage terms)?

Detailed Solution for Practice Test: Percentages - 1 - Question 8

Let the initial side of the cube be 'x' units.

⇒ Surface area of cube =6×side2=6x2 square units

Given, the length, breadth and height of a cube are decreased, decreased and increased by 5%, 5% and 20% respectively.

This means that the cube is now converted into a cuboid.

Length of cuboid =(100-5)% of x=95% of x=19x/20

Breadth of cuboid =(100-5)% of x=95% of x=19x/20

Height of cuboid =(100+20)% of x=120% of x=24x/20

⇒ Surface area of cuboid =2[(19x/20×19x/20)+(19x/20×24x/20)+(24x/20×19x/20)]

=2[(361x2/400)+(456x2/400)+(456x2/400)]

=2×[1273x2/400]=1273x2/200 square units

⇒ Increase in surface area ={[(1273x2/200)-6x2]  / [6x2]}×100=(73/1200)×100=6.0833%

Hence, the correct answer is 6.0833%.

Practice Test: Percentages - 1 - Question 9

The price of raw materials has gone up by 15%, labour cost has also increased from 25% of the cost of raw material to 30% of the cost of raw material. By how much percentage should there be a reduction in the usage of raw materials so as to keep the cost the same?

Detailed Solution for Practice Test: Percentages - 1 - Question 9

Let the initial price of raw materials be 100.
The new cost of the same raw material would be 115.
The initial cost of labour would be 25 and the new cost would be 30% of 115 = 34.5

The total cost initially would be 125.
The total cost for the same usage of raw material would now be: 115 + 34.5 = 149.5

This cost has to be reduced to 125. The percentage reduction will be given by 24.5 / 149.5 = 16 % approx.

Practice Test: Percentages - 1 - Question 10

There are five containers in a truck hold. The weight of the first container is 100 kg and the weight of the second container is 20% higher than the weight of the third container, whose weight is 25% higher than the first container’s weight. The fourth container at 175 kg is 30% lighter than the fifth container. What is the difference in the average weight of the heaviest three and the lightest three?

Detailed Solution for Practice Test: Percentages - 1 - Question 10

Weight of first container = 100kg

According to the question,

Weight of the 3rd container = 100 × 5/4 = 125 kg

Weight of the 2nd container = 125 × 6/5 = 150 kg

Weight of the 4th container = 175 kg

Weight of the 5th container = 175 × 10/7 = 250 kg

Average of three heaviest container = (250 + 175 + 150)/3 = 191.66 kg

Average of three lightest container = (100 + 125 + 150)/3 = 125 kg

∴ Required answer = 191.66 – 125 = 66.66 kg

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