# PE 1 - Percentage | Arithmetic - Percentage

**PE 1 - Percentage | Arithmetic - Percentage**

The daily compensation of a laborer increase by 10% but the number of days worked by him per month drop by 10%. If he was originally getting Rs. 20,000 per month, what will he get per month now?

- A.
Rs. 21,000

- B.
Rs. 20,100

- C.
Rs. 20,000

- D.
Rs. 19,800

Answer: Option D

**Explanation** :

Let the daily compensation of the laborer be c.

Let number of days worked be d.

∴ Total earnings (e) = c × d = 20,000.

According to the conditions given, new earnings (e’) = 1.1c × 0.9d = 0.99 × cd = 0.99 × 20000 = Rs. 19,800.

Alternately,

Total earnings increases by 10% (due to compensation) and then reduces by 10% (due to number of days).

When a number is increased and then decreased by p% successively, overall % change = -p^{2}/100%

∴ net change in salary = -1%

New salary = 20000(1 - 1%) = Rs. 19,800.

Hence, option (d).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

When 10% is lost in grinding wheat, a country has to import 2 million tons; but when only 6% is lost. It has to import only 1 million ton. Find the quantity of wheat which grows in the country:

- A.
24 million tons

- B.
15 million tons

- C.
20 million tons

- D.
25 million tons

Answer: Option D

**Explanation** :

Let the wheat grown in the country be x million tons.

Consumption in the country = (Wheat produced – wheat lost) + Import

∴ Consumption initially = (w – 10% of w) + 2 = 0.9w + 2

∴ Consumption finally = (w – 6% of w) + 1 = 0.94 + 1

⇒ 0.9x + 2 = 0.94x + 1

⇒ 0.04x = 1

⇒ x =1/0.04= 25 million tons

**Alternately**,

Due to reduction in wastage from 10% to 6% (i.e., 4% reduction) the country imports 1 mn tons less.

⇒ 4% of wheat production = 1 mn tons

⇒ 4/100 × wheat production = 1 mn tons

⇒ wheat production = 100/4 = 25 mn tons.

Hence, option (d).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

In an examination of 5 subjects, all with the same maximum marks, a candidate scores 60% aggregate marks and marks in individual subjects are in proportion to 3 : 3 : 4 : 5 : 5. What was the highest score in any subject?

- A.
69%

- B.
72%

- C.
75%

- D.
80%

Answer: Option C

**Explanation** :

Let highest marks in each subject be 100.

Then for 5 subjects max marks obtainable = 500.

Since the candidate has scored 60% in aggregate

⇒ 0.6 × 500 = 300 marks

It is also given that marks scored in five subjects are in the ratio 3 : 3 : 4 : 5 : 5.

Thus, maximum marks scored in any subject = 5/20 × 300 = 75 marks

Hence, option (c).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

A salesman gets a flat commission of 5% on all sales and a bonus of 20% on all sales exceeding Rs. 1,00,000. If in a particular month he earns Rs. 55000, what were his sales worth?

- A.
Rs. 3,00,000

- B.
Rs. 2,25,000

- C.
Rs. 2,30,000

- D.
Rs. 2,50,000

Answer: Option A

**Explanation** :

Let the monthly sales be denoted by S

Then, by conditions in the problem, 0.05S + 0.2(S – 1,00,000)

⇒ 0.25S – 20,000 = 55,000

⇒ S = Rs. 3,00,000

Hence, option (a).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

A shop’s receipts on the first day of the week were Rs. 10,000. On the second day 50% more than this, and on the third day 80% of the sum of the receipts of the first 2 days. What was the average of the receipts during this three day period (in rupees)?

- A.
10533.33

- B.
12300

- C.
15000

- D.
13500

Answer: Option C

**Explanation** :

Receipts on 1^{st} day Rs. 10,000.

On 2^{nd} day Rs. 1.5 × 10000 = 15,000

On 3^{rd} day Rs. 0.8 × (10,000 + 15,000) = 20,000

∴ Average receipt would be (10000 + 15000 + 2000)/3 = Rs. 15000

Hence, option (c).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

There are n questions in an exam. A student answered 3 of the first 4 questions correctly. Of the remaining questions he answered one-third correctly. All the questions have the same marks and there is no negative marking. If the student scored 50% marks, how many different values of n can be there?

- A.
4

- B.
3

- C.
2

- D.
None of these

Answer: Option D

**Explanation** :

To score 50% he must have answered 50% questions correctly.

From the conditions of the problem

3 + (n - 4)/3 = 0.5n

⇒ 9 + n – 4 = 1.5n

⇒ 0.5n = 5

⇒ n = 10 (i.e., only one value)

Hence, option (d).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

A man in business loses in the first year 10% of his capital, but in the second year he gains 12% of what he had at the end of the first year and his capital is now Rs. 400 more than that at the beginning of first year. Find his original capital.

- A.
Rs. 50,000

- B.
Rs. 40,000

- C.
Rs. 30,000

- D.
None of these

Answer: Option A

**Explanation** :

Let c be the initial capital.

At the end of 1^{st} year, capital remaining = 0.9c

At the end of 2^{nd} year, capital remaining = 1.12 × 0.9c = 1.008c

It is further given that 1.008c = c + 400.

⇒ 0.008c = 400

⇒ c = 400/0.008 = Rs. 50,000

Hence, option (a).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

A sum of money is to be divided among three sons in such a way that the first son gets 25% of the whole, the second 40% of the remainder and the third son the rest. If the 3rd son gets Rs. 1,500 more than the 2nd son, what would be the total sum divided?

- A.
9,000

- B.
10,000

- C.
3,000

- D.
None of these

Answer: Option B

**Explanation** :

Let x be the sum to be shared

1^{st} son gets : 0.25x

Remaining amount = x - 0.25x = 0.75x

2^{nd} son gets : 0.4 × 0.75x = 0.3x

3^{rd} son gets : x – 0.25x – 0.3x = 0.45x

Now, third son gets Rs. 1,500 more than the 2^{nd} son.

⇒ 0.45x – 0.3x = 1500

⇒ x = 10,000

∴ 1^{st} son’s share = Rs. 2500

2^{nd} son’s share = Rs. 3000

3^{rd} son’s share = Rs. 4500

Hence, option (b).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

In an examination Prakash secured 30% of the total marks and failed by 20 marks. Kashish secured 52% of the total marks and secured 24 marks more than minimum pass percentage. Find the pass percentage of marks.

- A.
28%

- B.
25%

- C.
40%

- D.
30%

Answer: Option C

**Explanation** :

Let total marks be x, and let pass marks be p.

Prakash: 0.3x = p – 20 …(1)

Kashish: 0.52x = p + 24 …(2)

(2) – (1)

⇒ 0.22x = 44

⇒ x = 200 and p = 80.

Pass % = 80/200 = 0.4 or 40%

Hence, option (c).

Workspace:

**PE 1 - Percentage | Arithmetic - Percentage**

In an election 4% of the votes cast are invalid. A candidate gets 55% of the valid votes and wins the election by 480 votes. Find the total number of voters who voted (i.e., the number of votes cast). (There are only two candidates)

- A.
5000

- B.
5200

- C.
4800

- D.
None of these

Answer: Option A

**Explanation** :

Let the number of votes cast = v

∴ Valid votes = 0.96v

Winner gets 480 votes more than the other candidate.

⇒ 0.55 × 0.96v = 0.45 × 0.96v + 480

⇒ 0.096v = 480

⇒ v = 5000

Hence, option (a).

Workspace:

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