# PE 2 - Percentage | Arithmetic - Percentage

**PE 2 - Percentage | Arithmetic - Percentage**

Mr. Manoj saves 30% of his salary every month. But in April, his expenses increase by 20% of his original expenses and he saves $280 less than the amount he saves normally. What is his salary?

- (a)
$2000

- (b)
$1500

- (c)
$2400

- (d)
$3000

- (e)
$1550

Answer: Option A

**Explanation** :

Since, savings is 30%, that means Mr. Manoj’s expenses are 70% of his salary.

Let 'x' be the salary and 0.7x be the expenses.

In April, the expenses increase by 20% and become 1.2 × 0.7x = 0.84x

∴ The savings decrease by 0.84x – 0.7x = 0.14x

⇒ 0.14x = 280

⇒ x = $2000

Hence, option (a).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

In every box of 120 bulbs, 5% are defective. If a company needs 4,000 non-defective bulbs, what is the minimum number of boxes that it has to purchase?

- (a)
32

- (b)
34

- (c)
35

- (d)
36

- (e)
None of these

Answer: Option D

**Explanation** :

5% bulb are defective, i.e., 95% bulb are non-defective.

∴ Each box contains 120 × 95% = 114 non-defective bulbs.

⇒ Minimum number of boxes required = 4000/114 = 35.08

∴ The company has to purchase a minimum of 36 boxes to meet its requirement of bulbs.

Hence, option (d).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

Gurmeet purchased a box of chocolates. She ate 20% of them and gave 25% of the remaining chocolates to her friends. Of the remaining chocolates, she ate 20% of it and gave 25% of the remaining chocolates to her friends. Then, she gave 20 chocolates to a guest. Finally, she is left with 35% of the chocolates that she initially had. How many chocolates did Gurmeet eat?

- (a)
320

- (b)
500

- (c)
640

- (d)
1250

Answer: Option C

**Explanation** :

Assuming that the box contains x number of chocolates.

Remaining = 0.36x

After giving chocolates to friends, she gives 20 chocolates to guests also.

∴ Remaining chocolates with her = 0.36x – 20 = 0.35x

⇒ 0.01x = 20

⇒ x = 2000

∴ She had 2000 chocolates at the start.

She ate 0.2x + 0.12x = 0.32x = 640 chocolates

Hence, option (c).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

It is given that k = x^{2} - 2x. If x is doubled, what is the percentage change in k?

- (a)
100%

- (b)
200%

- (c)
121%

- (d)
0%

- (e)
Cannot be determined

Answer: Option E

**Explanation** :

If x doubles, it becomes 2x.

∴ New value of k = k’ = 4x^{2} - 4x.

∴ Percent change in k = (k’ - k)/k × 100, which cannot be determined without knowing the value of x.

Hence, option (e).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

The cost of an air ticket is Rs. 2,000 plus tax. If a discount of 50% is allowed on the cost price of the air ticket (not tax), the cost of the ticket decreases by 40%. What is the amount of tax charged on the purchase of a ticket?

- (a)
Rs. 400

- (b)
Rs. 1000

- (c)
Rs. 500

- (d)
Rs. 800

- (e)
Cannot be determined

Answer: Option C

**Explanation** :

Original cost = Rs. 2,000 + tax

New cost after discount = 1000 + tax

According to the problem,

60% of (Rs. 2000 + tax) = Rs. 1000 + tax

Rs. 1200 + (0.6) × tax = Rs. 1000 + tax

⇒ 0.4 × tax = 200

⇒ Tax = Rs. 500

Hence, option (c).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

Hitesh said to Rohit, if you give me 10% of your pocket money, I will have 100% more than what you will be left with. By how much percent is Hitesh's pocket money more than Rohit's?

- (a)
80%

- (b)
40%

- (c)
50%

- (d)
70%

- (e)
60%

Answer: Option D

**Explanation** :

Let the pocket money of Hitesh and Rohit be P and V, respectively.

Now, $H+\frac{R}{10}=2\left(R-\frac{R}{10}\right)$

⇒ H = $\frac{9}{5}R-\frac{R}{10}=\frac{17}{10}R$

H = R + $\frac{7}{10}$R

⇒ Hitesh’s money is 7/10 × 100 = 70% more

Hence, option (d).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

Two friends x and y have some money with them. ‘x’ gives 40% of his money to ‘y’ and ‘y’ gives 16.67% of the money he now has to ‘x’. Both the friends have the same amounts of money now. If M and N respectively are the amounts of money with ‘x’ and ‘y’ at the beginning, then

- (a)
M = N

- (b)
M > N

- (c)
M < N

- (d)
M + N = 600

- (e)
None of these

Answer: Option B

**Explanation** :

Money left with x = 0.6M + (0.167N + 0.0667M) = 0.167N + 0.667M

Money left with y = N + 0.4M – (0.167N + 0.0667M) = 0.833N + 0.333M

Given, both have the same amount of money after these exchanges.

∴ 0.167N + 0.667M = 0.833N + 0.333M

⇒ 0.333M = 0.667N

⇒ M/3 = 2N/3

⇒ M = 2N

⇒ M > N

Hence, option (b).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

There are 150 students in a class and 60% of them are girls. 60% of the girls in the class have not studied Math. Taking the class as a whole, 50% of the students have not studied Math. What percentage of boys in the class have studied Math?

- (a)
65%

- (b)
50%

- (c)
40%

- (d)
30%

- (e)
None of these

Answer: Option A

**Explanation** :

Given, total number of students = 150

Number of girls who have not studied Math = 60% of 60% of 150 = 54.

Total number of students who have not studied Math = 50% of 150 = 75.

∴ Number of boys who have not studied Math = 75 – 54 = 21.

Total number of boys = 40% of 150 = 60.

∴ percentage of boys who have studied Math = (60 - 21)/60 × 100 = 65%

Hence, option (a).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

If a quantity is decreased by 30% and then increased by 50%, by how much percentage does it have to be decreased to bring it back to the original quantity?

- (a)
4.76%

- (b)
5.85%

- (c)
1.33%

- (d)
9.5%

- (e)
20%

Answer: Option A

**Explanation** :

Assume the quantity as 100.

It is decreased by 30%, then increased by 50%, and finally decreased by let’s say p%.

Given, final number is same as initial number

∴ $N\left(1-\frac{30}{100}\right)\left(1+\frac{50}{100}\right)\left(1-\frac{p}{100}\right)$ = N

⇒ $\left(\frac{7}{10}\right)\left(\frac{3}{2}\right)\left(1-\frac{p}{100}\right)$ = 1

⇒ $\left(1-\frac{p}{100}\right)=\frac{20}{21}$

⇒ p = 100/21% ≈ 4.76%

Hence, option (a).

Workspace:

**PE 2 - Percentage | Arithmetic - Percentage**

x is a positive integer. How many values can 'x' take such that x% of 350 is less than 50?

- (a)
13

- (b)
14

- (c)
15

- (d)
12

- (e)
16

Answer: Option B

**Explanation** :

Given, x% of 350 < 50.

⇒ 350x/100 < 50

⇒ x < 100/7 ≈ 14.28

⇒ x can be any integer from 1 till 14.

⇒ x can take 14 values.

**Alternately,**

Let us calculate 50 is what % of 350.

⇒ 50/350 × 100 = 14.28%

∴ x < 14.28%

⇒ x can be any integer from 1 till 14.

⇒ x can take 14 values.

Hence, option (b).

Workspace:

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