# Arithmetic - Profit & Loss - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Arithmetic - Profit & Loss. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2021 QA Slot 1 | Arithmetic - Profit & Loss**

Amal purchases some pens at ₹ 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at ₹ 12 each. If the remaining pens are sold at ₹ 11 each, then he makes a net profit of ₹ 300, while he makes a net loss of ₹ 300 if the remaining pens are sold at ₹ 9 each. The wage of the employee, in INR, is

Answer: 1000

**Explanation** :

Let the total number of pens Amal bought = x

Also, let the wage of the employee = w

∴ Amal’s total cost price = 8x + w

Total selling price of the first 100 pen = 100 × 12 = 1200

**Case 1**: The remaining pen when each is sold at Rs. 11

Total selling price of the remaining pen = (x - 100) × 11

⇒ 1200 + 11(x - 100) = 8x + w + 300

⇒ 3x - 200 = w …(1)

**Case 2**: The remaining pen when each is sold at Rs. 9

Total selling price of the remaining pen = (x - 100) × 9

⇒ 1200 + 9(x - 100) = 8x + w - 300

⇒ 4x - 600 = w …(2)

Solving (1) and (2) we get

x = 400 and w = 1000.

Hence, 1000.

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**CAT 2021 QA Slot 2 | Arithmetic - Profit & Loss**

A person buys tea of three different qualities at ₹ 800, ₹ 500, and ₹ 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at ₹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is

- A.
692

- B.
688

- C.
675

- D.
653

Answer: Option B

**Explanation** :

Let 2, 3, 5 kgs is bought of each variety respectively.

∴ Total quantity bought = 2 + 3 + 5 = 10 kgs

Total cost of tea = 2 × 800 + 3 × 500 + 5 × 300 = 4600.

Profit on total quantity = 50%

∴ Total selling price for 10 kg tea = 4600 × 1.5 = 6900.

Selling price for 1/6th of 10 kg tea = 10/6 × 700 = 7000/6

⇒ Selling price for remaining 50/6 kg tea = 6900 – 7000/6 = 34400/6

∴ Selling price per kg for the remaining tea = 41300/6 ÷ 50/6 = 688

Hence, option (b).

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**CAT 2020 QA Slot 1 | Arithmetic - Profit & Loss**

A person spent Rs. 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at 20% profit and the laptop at 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is

Answer: 20000

**Explanation** :

Let the price of desktop be Rs. d, and laptop be Rs. (50,000 - d)

Total profit = 2% of 50,000 = Rs. 1,000

⇒ 1000 = 20% of d - 10% of (50000 - d)

⇒ 100000 = 20d - 500000 + 10d

⇒ 30d = 6,00,000

⇒ d = 20,000

Hence, 20000.

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**CAT 2020 QA Slot 2 | Arithmetic - Profit & Loss**

Anil buys 12 toys and labels each with the same selling price. He sells 8 toys initially at 20% discount on the labeled price. Then he sells the remaining 4 toys at an additional 25% discount on the discounted price. Thus, he gets a total of Rs 2112, and makes a 10% profit. With no discounts, his percentage of profit would have been

- A.
50

- B.
54

- C.
60

- D.
55

Answer: Option A

**Explanation** :

Let the cost price be c and marked price be m. Total selling price for 12 toys = Rs. 12c.

Anil sells 8 toys at 20% discount. Hence, total selling price for 8 toys = 8 × 0.8m = 6.4m

He then sells 4 toys at further 25% discount. Hence, total selling price for 8 toys = 4 × 0.75 × 0.8m = 2.4m

Given, 6.4m + 2.4m = 2112

⇒ m = 240

Now, he earns an overall profit of 10%

∴ 12c × 1.1 = 2112

⇒ c = 160

∴ With no discounts his profit % = (240 - 160)/160 × 100 = 50%

Hence, option (a).

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**CAT 2020 QA Slot 3 | Arithmetic - Profit & Loss**

A man buys 35 kg of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg at this price, and 15 kg at a 10% discount. Accidentally, 3 kg of sugar is wasted. He sells the remaining sugar by raising the marked price by p percent so as to make an overall profit of 15%. Then p is nearest to

- A.
25

- B.
35

- C.
22

- D.
31

Answer: Option A

**Explanation** :

Let the cost price of sugar be Rs. 100/kg and marked price will be = Rs. 120/kg

∴ Total cost for the man = 35 × 100 = Rs. 3500

He earns 15% profit on this, hence total selling price = 3500 × 1.15 = Rs. 4025

Total Selling price for 5 kg sugar = 5 × 120 = 600

Total Selling price for 15 kg sugar = 15 × 120 × 0.9 = 1620

Total Selling price for 3 kg sugar = 3 × 0 = 0

Total Selling price for remaining 12 kg sugar = 12 × 120(1 + p%)

∴ 4025 = 600 + 1620 + 0 + 1440(1 + p%)

⇒ 1805 = 1440(1 + p%)

⇒ 1 + p% ≈ 1.25

⇒ p = 25%

Hence, option (a).

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**CAT 2019 QA Slot 2 | Arithmetic - Profit & Loss**

Mukesh purchased 10 bicycles in 2017, all at the same price. He sold six of these at a profit of 25% and the remaining four at a loss of 25%. If he made a total profit of Rs. 2000, then his purchase price of a bicycle, in Rupees, was

- A.
2000

- B.
6000

- C.
8000

- D.
4000

Answer: Option D

**Explanation** :

Let the cost price of one bicycle = Rs. x

Total cost price = Rs. 10x

He made a total profit of 25% on 6 cycles and 25% loss on 4 cycles and made a profit of Rs. 2000

So, 2000 = 6 × x/4 - 4 × x/4

2000 = x/2

x = Rs. 4000

Hence, option (4).

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**CAT 2019 QA Slot 2 | Arithmetic - Profit & Loss**

A shopkeeper sells two tables, each procured at cost price p, to Amal and Asim at a profit of 20% and at a loss of 20%, respectively. Amal sells his table to Bimal at a profit of 30%, while Asim sells his table to Barun at a loss of 30%. If the amounts paid by Bimal and Barun are x and y, respectively, then (x - y) / p equals

- A.
1

- B.
1.2

- C.
0.7

- D.
0.5

Answer: Option A

**Explanation** :

The Shopkeeper procures the table at price 'p'

He gains 20% on the transaction with Amal

So, Amal buys the table at '1.2p'

Amal sells athe table at 30% profit,

So, the Selling Price of Amal = 1.3 × 1.2p = 1.56p

⇒ x = 1.56p

The Shopkeeper loses 20% on the transaction with Asim

So, Asim buys the table at '0.8p'

Asim sells the table at 30% loss,

So, the Selling Price of Asim = 0.7 * 0.8p = 0.56p

⇒ y = 0.56p

(x - y)/p = (1.56p - 0.56p)/p = 1.

Hence, option (1).

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**CAT 2018 QA Slot 1 | Arithmetic - Profit & Loss**

A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?

- A.
86

- B.
84

- C.
98

- D.
96

Answer: Option D

**Explanation** :

Assume that the wholesaler bought peanut at Rs. X per kg.

∴ Cost price of walnuts = Rs. 3X per kg

The wholesaler sold peanuts at Rs. 1.1X per kg and walnuts at Rs. 3.6X per kg.

Total cost of 8 kg peanuts and 16 kg walnuts for the shopkeeper

= 8 × 1.1X + 16 × 3.6X = 66.4X

Shopkeeper overall profit was 25%

Therefore, revenue = 66.4X × 1.25

He sold (8 – 3 =) 5 kg peanuts and (16 – 5 =) 11 kg walnuts after mixing at Rs. 166 per kg

Therefore, 66.4X × 1.25 = 16 × 166

Solving this, we get

X = 32

Cost price of walnuts for the wholesaler = 3 × 32 = Rs. 96

Hence, option 4.

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**CAT 2018 QA Slot 1 | Arithmetic - Profit & Loss**

Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

- A.
17 : 25

- B.
18 : 25

- C.
21 : 25

- D.
19 : 24

Answer: Option D

**Explanation** :

Let cost of tea A and B be a and b respectively.

If 3 kg of tea A is mixed with 2 kg of tea B, (3a + 2b) × 1.1 = 40× 5 = 200

If 2 kg of tea A is mixed with 3 kg of tea B, (2a + 3b) × 1.05 = 40× 5 = 200

Therefore,

(3a + 2b) × 1.1 = (2a + 3b) × 1.05

∴ 3.3a + 2.2b = 2.1a + 3.15b

∴ a/b = 19/24

Hence, option 4.

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**CAT 2018 QA Slot 1 | Arithmetic - Profit & Loss**

A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is

- A.
20

- B.
26

- C.
16

- D.
22

Answer: Option A

**Explanation** :

Let the 10 litres of mixture has ‘Y’ litres of A and (10 – Y) litres of B. Let cost of paint B be Rs. X and that of A be Rs. (X + 8).

We know that, Y ≥ (10 – Y) ⇒ Y ≥ 5

The trader makes 10% profit by selling this mixture at Rs. 264.

∴ Cost price of the mixture = $\frac{264}{1.1}$ = Rs. 240

∴ (X + 8) × Y + (10 – Y) × X = 240

∴ 10X + 8Y = 240

∴ X = 24 – 0.8Y

For maximum value of X, we need to consider minimum value of Y.

∴ X = 24 – (0.8 × 5) = Rs. 20

Hence, option 1.

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**CAT 2017 QA Slot 1 | Arithmetic - Profit & Loss**

If a seller gives a discount of 15% on retail price, she still makes a profit of 2%. Which of the following ensures that she makes a profit of 20%?

- A.
Give a discount of 5% on retail price

- B.
Give a discount of 2% on retail price

- C.
Increase the retail price by 2%

- D.
Sell at retail price

Answer: Option D

**Explanation** :

Let the retail price by 100.

Discount = 15

Selling price = 85

Cost price = 85/1.02 = 500/6

In order to make a profit of 20%, the selling price = 500/6 × 1.2 = 100

The seller must sell at the retail price.

Hence, option 4.

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**CAT 2017 QA Slot 1 | Arithmetic - Profit & Loss**

In a market, the price of medium quality mangoes is half that of good mangoes. A shopkeeper buys 80 kg good mangoes and 40 kg medium quality mangoes from the market and then sells all these at a common price which is 10% less than the price at which he bought the good ones. His overall profit is:

- A.
6%

- B.
8%

- C.
10%

- D.
12%

Answer: Option B

**Explanation** :

Let the price of each good mango be g.

Price of each medium quality mango = g/2

Total cost price = 80g + 40(g/2) = 100g

Total selling price = 120(0.9g) = 108g

Overall profit = 8%

hence, option 2.

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**CAT 2017 QA Slot 1 | Arithmetic - Profit & Loss**

If Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are destroyed in fire. While selling the rest, how much discount should be given on the printed price so that she can make the same amount of profit?

- A.
30%

- B.
25%

- C.
24%

- D.
28%

Answer: Option D

**Explanation** :

Let the printed price be p.

If 40% discount is given, selling price = 0.6 × 60p = 36 p

20% profit is then made.

Total cost price = 36p/1.2 = 30p.

Ten toys are destroyed in the fire.

The remaining toys are sold at a price such that the same amount of profit is made as in the conditional case.

Profit made on remaining toys = 6p

Total selling price of remaining toys = 36p

Discount that should be given = 50p – 36p = 14p

Discount% = $\frac{14p}{50p}\times 100$% = 28%

Hence, option 4.

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**CAT 2017 QA Slot 2 | Arithmetic - Profit & Loss**

The manufacturer of a table sells it to a wholesale dealer at a profit of 10%. The wholesale dealer sells the table to a retailer at a profit of 30%. Finally, the retailer sells it to a customer at a profit of 50%. If the customer pays Rs 4290 for the table, then its manufacturing cost (in Rs) is

- A.
1500

- B.
2000

- C.
2500

- D.
3000

Answer: Option B

**Explanation** :

Lets the manufacturers C.P. be 100

As he sells it to the wholesaler at 10% profit, his S.P will be 100 + 10 or 110

Now 110 is also the wholesaler’s C.P. Further, as the wholesaler sells it to a retailer at 30%, profit, his S.P will be

$\frac{130}{100}$ × 100 = 143.

Now 143 also the retailers C.P

As the retailer sells it to the customer at

50% profit, S.P. will be $\frac{150}{100}$ × 143 or 214.5

Now 214.5 is the price the customer pays for the table, when the manufacturer’s cost is 100.

Let the manufacturer’s cost of the table be Rs. ‘x’ when the price paid by the customer is Rs. 4290

By unitary method.

$\therefore \frac{100}{x}=\frac{214.5}{4290}$ ⇒ x = 2000

So Rs. 2000 is the manufacturer’s cost

Hence, option 3.

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**CAT 2017 QA Slot 2 | Arithmetic - Profit & Loss**

Mayank buys some candies for Rs 15 a dozen and an equal number of different candies for Rs 12 a dozen. He sells all for Rs 16.50 a dozen and makes a profit of Rs 150. How many dozens of candies did he buy altogether?

- A.
50

- B.
30

- C.
25

- D.
45

Answer: Option A

**Explanation** :

Supposing Mayank buys ‘x’ dozen candies at Rs. 15 per dozen and another ‘x’ dozen candies at Rs. 12 per dozen.

Total C.P =15(x) + 12(x) per dozen.

Now he sells these ‘x + x’ or ‘2x’ dozen candies at Rs. 16.50 per dozen.

So total S.P = 16.50 × 2x = 33x

Profit = 33x -27x = 6x

Now 6x = 150 or x = 25

As Mayank buys ‘2x’ dozen candies, number of dozens of candies bought by him

= 2 × 25 or 50

Hence, option 1.

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**CAT 2003 QA - Leaked | Arithmetic - Profit & Loss**

A leather factory produces two kinds of bags, standard and deluxe. The profit margin is Rs. 20 on a standard bag and Rs. 30 on a deluxe bag. Every bag must be processed on machine A and on machine B. The processing times per bag on the two machines are as follows:

The total time available on machine A is 700 hours and on machine B is 1250 hours. Among the following production plans, which one meets the machine availability constraints and maximizes the profit?

- A.
Standard 75 bags, Deluxe 80 bags

- B.
Standard 100 bags, Deluxe 60 bags

- C.
Standard 50 bags, Deluxe 100 bags

- D.
Standard 60 bags, Deluxe 90 bags

Answer: Option A

**Explanation** :

Total time available is 700 hrs on machine A and 1250 hrs on machine B.

Let the number of Standard Bags be s and the number of Deluxe Bags be d.

Here we have to maximize the profit margin i.e. 20s + 30d, subject to the constraints,

4s + 5d ≤ 700 and 6s + 10d ≤ 1250

Now consider options.

1. s = 75 and d = 80

∴ The profit = 75 × 20 + 30 × 80 = 3900

4s + 5d = 700 and 6s + 10d = 1250

∴ the constraints are satisfied.

2. s = 100 and d = 60

∴ The profit = 100 × 20 + 60 × 30 = 3800

The profit is less than in option 1.

∴ Option 2 is not the answer.

3. s = 50 and d = 100

∴ The profit = 50 × 20 + 100 × 30 = 4000

4s + 5d = 700 and 6s + 10d = 1300

∴ The second constraint is not satisfied.

∴ Option 3 cannot be the answer.

4. s = 60 and d = 90

∴ The profit = 60 × 20 + 90 × 30 = 3900

4s + 5d = 690 and 6s + 10d = 1260

∴ The second constraint is not satisfied.

∴ Option 4 cannot be the answer.

As only option 1 satisfies the constraints and also maximizes the profit, option 1 is the answer.

Hence, option 1.

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**CAT 2002 QA | Arithmetic - Profit & Loss**

**Each question is followed by two statements A and B. Answer each question using the following instructions:**

**Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) **if the question can be solved by using either of the two statements.

**Answer (3)**if the question can be solved by using both the statements together and not by any one of them.

**Answer (4)**if the question cannot be solved with the help of the given data and more data is required.

A dress was initially listed at a price that would have fetched the store a profit of 20% on the wholesale cost. What was the wholesale cost of the dress?

A. After reducing the listed price by 10% the dress was sold for a net profit of 10 dollars.

B. The dress was sold for 50 dollars.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Let 100x be the wholesale cost of the dress.

∴ List price of the dress = 120x

**Consider statement A:**

Selling Price = 0.9 × List Price = 0.9 × 120x = 108x

Now, Selling Price – Cost Price = Profit

∴ (108x) − (100x) = 10

∴ x = 10/8

∴ Wholesale Cost = 100x = Rs. 125

∴ Statement A alone is sufficient.

**Consider statement B:**

This gives the selling price of the dress but it is not mentioned whether any discount is provided on the list price or not.

∴ Statement B alone is not sufficient.

Hence, option 1.

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**CAT 2000 QA | Arithmetic - Profit & Loss**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

Harshad bought shares of a company on a certain day, and sold them the next day. While buying and selling he had to pay to the broker one percent of the transaction value of the shares as brokerage. What was the profit earned by him per rupee spent on buying the shares?

- The sales price per share was 1.05 times that of its purchase price.
- The number of shares purchased was 100.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

From Statement A

The cost of buying the shares for Harshad is

CP + 0.01 CP = 1.01 CP

The cost of selling the shares for Harshad is

SP – 0.01 SP = 0.99 SP

∴ Profit = Cost of selling – Cost of buying

= 0.99 SP – 1.01 CP

∵ SP = 1.05 CP

Profit = (0.99 × 1.05 CP) – 1.01 CP

= 0.0295 CP

∴ Profit earned per rupee spent on buying the shares is Rs 0.0295.

From the number of shares given in the statement B, we cannot conclude anything.

Hence, option 1.

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**Direction: **Answer the questions based on the following information.

A company purchases components A and B from Germany and USA respectively. A and B form 30% and 50% of the total production cost. Current gain is 20%. Due to change in the international scenario, cost of the German mark increased by 30% and that of USA dollar increased by 22%. Due to market conditions, the selling price cannot be increased beyond 10%.

**CAT 1998 QA | Arithmetic - Profit & Loss**

What is the maximum current gain possible?

- A.
10%

- B.
12.5%

- C.
0%

- D.
7.5%

Answer: Option A

**Explanation** :

Let CP = 100

Current gain = 20

⇒ SP = 120

CP = Cost of A + Cost of B + other

= 30% + 50% + 20%

Since cost of German mark (A) increase by 30%

New cost of A = 30 + 30% of 30 = 39

Similarly

New cost of B = 50 + 22% of 50 = 61

New CP = 39 + 61 + 20 = 120

New SP = 120 + 10% of 120 = 132

Maximum profit = $\frac{12}{20}=10\%$

Workspace:

**CAT 1998 QA | Arithmetic - Profit & Loss**

If the USA dollar becomes cheap by 12% over its original cost and the cost of German mark increased by 20%, what will be the gain? (The selling price is not altered.)

- A.
10%

- B.
20%

- C.
15%

- D.
7.5%

Answer: Option B

**Explanation** :

Let CP = 100

Current gain = 20

⇒ SP = 120

CP = Cost of A + Cost of B + other

= 30% + 50% + 20%

New cost of A = 30 + 20% of 30 = 36

New cost of B = 50 – 12% of 50 = 44

New CP = 36 = 44 + 20 = 100

Gain = 20%

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**CAT 1997 QA | Arithmetic - Profit & Loss**

After allowing a discount of 11.11%, a trader still makes a gain of 14.28%. At how many percentage above the cost price does he mark on his goods?

- A.
28.56%

- B.
35%

- C.
22.22%

- D.
None of these

Answer: Option A

**Explanation** :

Hint: Students please note that the percentages that are given are the basic percentages derived from basic fractions. e.g. 11.11% = $\frac{1}{9}$ and 14.28 = $\frac{1}{7}.$

Hence, you should make use of the most of this kind of knowledge. So let the CP be Re 1. Since he makes a profit of $\frac{1}{7},$ his SP = $\left(1+\frac{1}{7}\right)=\mathrm{Rs}.\frac{8}{7}$.

His marked price should be $\frac{1}{9}$ above this. So if we subtract $\frac{1}{9}$ of MP from the MP, we will get the SP.

So $\left(\mathrm{MP}-\frac{1}{9}\mathrm{MP}\right)=\mathrm{SP}=\frac{8}{7}$

Hence, MP = $\frac{9}{7}$

Therefore, percentage of mark-up on CP = (MP – CP)/CP

$=\left(\frac{9}{7}-1\right)/1=\frac{2}{7}=2\left(\frac{1}{7}\right)$ = 2 × 14.28 = 28.56%

**Alternative method:**

We can use the formula z = x – y – $\frac{xy}{100},$ where

z = Gain percentage

x = Percentage above CP

y = Discount percentage

∴ 14.28% = x – 11.11% – $\frac{11.11x}{100}$

or 14.28 = $\frac{100x-1111-11.11x}{100}$

or 1428 - 1111 = 88.89x

or x = 28.56% (Approximately)

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**CAT 1997 QA | Arithmetic - Profit & Loss**

A dealer buys dry fruits at Rs. 100, Rs. 80 and Rs. 60 per kilogram. He mixes them in the ratio 3 : 4 : 5 by weight, and sells at a profit of 50%. At what price per kilogram does he sell the dry fruit?

- A.
Rs. 80

- B.
Rs. 100

- C.
Rs. 95

- D.
None of these

Answer: Option D

**Explanation** :

Let he mix 3 kg, 4 kg and 5 kg of dry fruits at Rs. 100, Rs. 80 and at Rs. 60 per kilogram respectively. Hence, his effective cost of the dry fruits per kilogram should be the weighted average

$=\left(\frac{3\times 100+4\times 80+5\times 60}{3+4+5}\right)=\frac{920}{12}$

In order to make a 50% profit, he will have to sell it at $\left(\frac{920}{12}\times 1.5\right)=\frac{920}{12}\times \frac{3}{2}=\frac{920}{8}$ = Rs. 115 per kg.

Since none of the answer-choices confirms this, the answer is (d).

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**CAT 1996 QA | Arithmetic - Profit & Loss**

Instead of a metre scale, a cloth merchant uses a 120 cm scale while buying, but uses an 80 cm scale while selling the same cloth. If he offers a discount of 20% on cash payment, what is his overall profit percentage?

- A.
20%

- B.
25%

- C.
40%

- D.
15%

Answer: Option A

**Explanation** :

Let the price per metre of cloth be Re 1. The shopkeeper buys 120 cm, but pays for only 100 cm. In other words, he buys 120 cm for Rs. 100. So his CP = $\left(\frac{100}{120}\right)$ = Re 0.833 per metre. Now he sells 80 cm, but charges for 100 cm. In other words, he sells 80 cm for Rs. 100. On this he offers a 20% discount on cash payment. So he charges Rs. 80 for 80 cm cloth. In other words, his SP = $\left(\frac{80}{80}\right)$ = Re 1 per metre. So his percentage profit in the overall transaction = $\frac{(1-0.833)}{0.833}=20\%.$

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**Direction: Answer the questions based on the following information.**

A watch dealer incurs an expense of Rs. 150 for producing every watch. He also incurs an additional expenditure of Rs. 30,000, which is independent of the number of watches produced. If he is able to sell a watch during the season, he sells it for Rs. 250. If he fails to do so, he has to sell each watch for Rs. 100.

**CAT 1996 QA | Arithmetic - Profit & Loss**

If he is able to sell only 1,200 out of 1,500 watches he has made in the season, then he has made a profit of

- A.
Rs. 90,000

- B.
Rs. 75,000

- C.
Rs. 45,000

- D.
Rs. 60,000

Answer: Option B

**Explanation** :

Total expense incurred in making 1,500 watches = (1500 x 150) + 30000 = Rs. 2,55,000.

Total revenue obtained by selling 1,200 of them during the season = (1200 × 250) = Rs. 3,00,000. The remaining

300 of them has to be sold by him during off season. The total revenue obtained by doing that = (300 × 100)

= Rs. 30,000. Hence, total revenue obtained

= (300000 + 30000) = Rs. 3,30,000. Hence, total profit

= (330000 – 255000) = Rs. 75,000.

Workspace:

**CAT 1996 QA | Arithmetic - Profit & Loss**

If he produces 1,500 watches, what is the number of watches that he must sell during the season in order to break-even, given that he is able to sell all the watches produced?

- A.
500

- B.
700

- C.
800

- D.
1,000

Answer: Option B

**Explanation** :

From the previous solution, we can see that the total expense incurred by him in manufacturing 1,500 watches

= Rs.2,55,000. In order to break-even, he has to make a minimum revenue in order to recover his expenditure. He gets Rs. 250 per watch sold and Rs. 100 on every watch not sold. Let him sell x watches to break-even. So our equation will be 250x + 100(1500 – x) = 255000. Solving this, we get x = 700 watches

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