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

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Here we bring you all previous year Arithmetic - Profit & Loss omet questions along with detailed solutions.

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**XAT 2024 QADI | Arithmetic - Profit & Loss omet Question**

The cost of running a movie theatre is Rs. 10,000 per day, plus additional Rs. 5000 per show. The theatre has 200 seats. A new movie released on Friday. There were three shows, where the ticket price was Rs. 250 each for the first two shows and Rs. 200 for the late-night show. For all shows together, total occupancy was 80%. What was the maximum amount of profit possible?

- (a)
Rs. 1,20,000

- (b)
Rs. 87,000

- (c)
Rs. 95,000

- (d)
Rs. 91,000

- (e)
Rs. 1,16,000

Answer: Option D

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**Text Explanation** :

There are 3 shows of 200 seats with 80% occupancy, hence 80% of 600 = 480 tickets were sold.

Profit will be maximum when maximum tickets are sold for first two shows (which have higher ticket price) and least tickets are sold for the night show (which has lower ticket price).

To maximise profit, we will assume first two shows sold 200 + 200 = 400 tickets while the late night show sold only 80 tickets

∴ Total revenue = 400 × 250 + 80 × 200 = 1,00,000 + 16,000 = 1,16,000

Also, total cost = 10000 + 3 × 5000 = 25,000

⇒ Maximum profit = 1,16,000 - 25,000 = 91,000

Hence, option (d).

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**XAT 2024 QADI | Arithmetic - Profit & Loss omet Question**

FS food stall sells only chicken biryani. If FS fixes a selling price of Rs. 160 per plate, 300 plates of biriyani are sold. For each increase in the selling price by Rs. 10 per plate, 10 fewer plates are sold. Similarly, for each decrease in the selling price by Rs. 10 per plate, 10 more plates are sold. FS incurs a cost of Rs. 120 per plate of biriyani, and has decided that the selling price will never be less than the cost price. Moreover, due to capacity constraints, more than 400 plates cannot be produced in a day. If the selling price on any given day is the same for all the plates and can only be a multiple of Rs. 10, then what is the maximum profit that FS can achieve in a day?

- (a)
Rs. 25,300

- (b)
Rs. 28,900

- (c)
Rs. 41,400

- (d)
Rs. 52,900

- (e)
None of the remaining options is correct.

Answer: Option B

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**Text Explanation** :

The selling price of each plate will always be a multiple of 10. So, let the number of selling price of a plate be 160 + 10x.

∴ Number of plates sold will be 300 - 10x.

Total revenue = (300 - 10x) × (160 + 10x) = -100x^{2} + 1400x + 48000

Total cost = 120 × (300 - 10x)

⇒ Net profit = Total revenue - Total cost

= -100x^{2} + 1400x + 48000 - 120 × (300 - 10x)

= -100x^{2} + 1400x + 48000 - 36000 + 1200x

= -100x^{2} + 2600x + 12000

Now, net profit is a quadratic expression -100x^{2} + 2600x + 12000.

This will be maximum when x = - (2600/2 × -100) = 13

∴ Maximum profit = -100 × (13)^{2} + 2600 × (13) + 12000

= - 16900 + 33800 + 12000

= 28,900

Hence, option (b).

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**Read the following scenario and answer the THREE questions that follow.**

A store offers a choice of five different discount coupons to its customers, described as follows:

**Coupon A**: A flat discount of Rs. 250 on a minimum spend of Rs. 1200 in one transaction.

**Coupon B**: A 15% discount on a minimum spend of Rs. 500 in one transaction, up to a maximum discount of Rs. 300.

**Coupon C**: A flat discount of Rs. 100 on a minimum spend of Rs. 600 in one transaction.

**Coupon D**: A 10% discount on a minimum spend of Rs. 250 in one transaction, up to a maximum discount of Rs. 100.

**Coupon E**: A flat discount of Rs. 50 on a minimum spend of Rs. 200 in one transaction.

The customers are allowed to use at most one coupon in one transaction, i.e., two or more coupons cannot be combined for the same transaction.

**XAT 2024 QADI | Arithmetic - Profit & Loss omet Question**

Four customers used four different discount coupons for their respective transactions in such a way that they obtained a total discount of Rs. 710.

- (a)
Coupon A

- (b)
Coupon B

- (c)
Coupon C

- (d)
Coupon D

- (e)
Coupon E

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**XAT 2024 QADI | Arithmetic - Profit & Loss omet Question**

Four customers used four different discount coupons for their respective transactions in such a way that nobody used any discount coupon sub-optimally. (A discount coupon is used suboptimally if using another discount coupon could have resulted in a higher discount for the same transaction.)

- (a)
Rs. 2250

- (b)
Rs. 2500

- (c)
Rs. 2350

- (d)
Rs. 2300

- (e)
Rs. 1550

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**XAT 2024 QADI | Arithmetic - Profit & Loss omet Question**

A family wanted to purchase four products worth Rs. 1000 each, and another product worth Rs. 300.

They were told that they could:

I) pay for the five products through one or more transactions in any way they wanted, as long as the purchase amount of any one product would not get split into different transactions, and

II) use the same discount coupon repeatedly for separate transactions, if they opt for more than one transaction.

What was the maximum discount that they could obtain for their purchase?

- (a)
Rs. 600

- (b)
Rs. 645

- (c)
Rs. 650

- (d)
Rs. 700

- (e)
None of the remaining option is correct.

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**XAT 2023 QADI | Arithmetic - Profit & Loss omet Question**

Rajnish bought an item at 25% discount on the printed price. He sold it at 10% discount on the printed price. What is his profit in percentage?

- (a)
10

- (b)
15

- (c)
17.5

- (d)
20

- (e)
None of the above

Answer: Option D

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**Text Explanation** :

Let the printed price be Rs. 100.

Cost price for Rajnish after 25% discount = Rs. 75

Rajnish sold it at 10% discount in printed price i.e., at Rs. 90

∴ Required profit % = $\frac{90-75}{75}\times 100$ = $\frac{15}{75}\times 100$ = 20%

Hence, option (d).

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**SNAP 10th Dec 2023 Quant Slot 1 (Memory Based) | Arithmetic - Profit & Loss omet Question**

Cost price of each item is Rs. 20 while their selling price are Rs. 2, 4, 6 and so on. If minimum profit percentage earned by selling these items is 40%, what is the minimum number of items?

- (a)
25

- (b)
26

- (c)
27

- (d)
None of these

Answer: Option C

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**Text Explanation** :

Cost price of each item is Rs. 20 while their selling price are Rs. 2, 4, 6 and so on. If minimum profit percentage earned by selling these items is 40%, what is the minimum number of items?

Let the total number of items = n

Total cost priceof these items = 20n

Now, selling price of 1st item is 2, 2nd item is 4, 3rd item is 6 and so on hence, nth item will be sold at 2n.

∴ Selling price of these items = 2 + 4 + 6 + ... + 2n

= 2(1 + 2 + 3 + ... + n)

= n(n + 1)

Since 40% profit is earned

⇒ Total Selling Price = Total Cost Price × 1.4

⇒ n(n + 1) = 20n × 1.4

⇒ (n + 1) = 20 × 1.4

⇒ n + 1 = 28

⇒ n = 27

Hence, option (c).

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**SNAP 10th Dec 2023 Quant Slot 1 (Memory Based) | Arithmetic - Profit & Loss omet Question**

An article is sold for Rs. 17,600 after giving a discount of 12% and 10% profit is earned. Find the profit percent when no discount is given.

- (a)
20

- (b)
22

- (c)
27.5

- (d)
25

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**SNAP 10th Dec 2023 Quant Slot 2 (Memory Based) | Arithmetic - Profit & Loss omet Question**

Two persons sold an article at same selling price of Rs. 1200 and calculated 8% profit. One of them calculated his profit % on cost price while the other calculated profit % on selling price. Find the difference between their cost prices.

- (a)
Rs. 9.6

- (b)
Rs. 7.11

- (c)
Rs. 21.21

- (d)
None of these

Answer: Option B

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**Text Explanation** :

Two persons sold an article at same selling price of Rs. 1200 and calculated 8% profit. One of them calculated his profit % on cost price while the other calculated profit % on selling price. Find the difference between their cost prices.

**Person 1**: Calculate his profit of 8% on cost price.

∴ 1200 = 1.08 × CP

⇒ CP = 1200/1.08 = 1111.11

**Person 2**: Calculate his profit of 8% on selling price.

∴ CP = 1200 - 8% of 1200 = 1200 - 96 = 1104

⇒ Required difference = 1111.11 - 1104 = 7.11

Hence, option (b).

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**SNAP 10th Dec 2023 Quant Slot 2 (Memory Based) | Arithmetic - Profit & Loss omet Question**

Usa sells an article for 10% profit to Anil and Anil sells the article to Rahul for Rs. 46,200. If profit earned by Anil is 5%, then at what price Usha purchased the article?

- (a)
Rs. 40,000

- (b)
Rs. 42,000

- (c)
Rs. 40,800

- (d)
None of these

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**SNAP 17th Dec 2023 Quant (Memory Based) | Arithmetic - Profit & Loss omet Question**

A fruit seller purchases 11 fruits at Rs. 10 and sells fruits at Rs. 11. Find his profit or loss percentage?

- (a)
34.5%

- (b)
36.5%

- (c)
38.6%

- (d)
39.5%

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**XAT 2022 QADI | Arithmetic - Profit & Loss omet Question**

Sheela purchases two varieties of apples - A and B - for a total of Rupees 2800. The weights in kg of A and B purchased by Sheela are in the ratio 5 : 8 but the cost per kg of A is 20% more than that of B. Sheela sells A and B with profits of 15% and 10% respectively. What is the overall profit in Rupees?

- (a)
340

- (b)
600

- (c)
240

- (d)
480

- (e)
380

Answer: Option A

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**Text Explanation** :

The two types of apples sold A and B are bought in the ratio of 5: 8.

Considering the weights to be 5x and 8x.

The cost price of A is 20 percent higher than that of B.

Considering the cost price of B = y, A = 6y/5.

The total cost price of A = (5x) ∙ $\left(\frac{6y}{5}\right)$

The total cost price of B = (8x) ∙ (y)

THe total cost price = 8xy + 6xy = 14xy

14xy = 2800.

xy = 200.

THe cost price of A = 1200.

THe cost price of B = 1600.

A is sold a profit of 15 percent. 15 percent of 1200 = 180.

B is sold at a profit of 10 percent. 10 percent of 1600 = 160.

The total profit is 180 + 160

= 340

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**IIFT 23 Dec 2021 QA | Arithmetic - Profit & Loss omet Question**

A shopkeeper marks up the price of the Toor dal by 20% and gives a discount of 10% to thecustomer. Besides, he also tricks 100 grams to his dealer and his customer respectively while buying orselling 1 kilogram of Toor dal. Find the profit percentage of the shopkeeper.

- (a)
22%

- (b)
20%

- (c)
32%

- (d)
27%

Answer: Option C

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**Text Explanation** :

Let the cost of 1000gm of dal be rupees 1000

Now the shopkeeper is buying 1100 gm of dal for rupees 1000

While selling 900 gm of dal the shopkeeping is charging the price of 1000 gm

Therefore while selling 1100 gm of dal the shopkeeper will charge the price of (1000/900)*1100 gm = 11000/9 gm

He has marked up the price by 20% and then given a discount of 10%

So price charged by shopkeeper = (11000/9)*1.2*0.9 = 1320

So the shopkeeper is spending rupees 1000 to buy the dal and is selling the same quantity of dal at rupees 1320. Therefore, profit percentage is (320/1000)*100 = 32%

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**IIFT 23 Dec 2021 QA | Arithmetic - Profit & Loss omet Question**

Same item is sold for Rs. 600 and Rs. 175, respectively. The profit earned on the first sale is 20times the loss incurred on the second sale. To make an overall profit of 30% in the whole transaction, atwhat price the second sale should happen:

- (a)
Rs. 310 approx

- (b)
Rs. 238 approx

- (c)
Rs. 254 approx

- (d)
Rs. 357 approx

Answer: Option C

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**Text Explanation** :

Let the CP be x

Now as per question, 600 - x = 20(x - 175)

21x = 4100

x = 195.24

Therefore, to make a profit of 30% selling price of second article should be = 1.3 × 195.24 = 253.81

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**IIFT 05 Dec 2021 QA | Arithmetic - Profit & Loss omet Question**

A granite stone has been purchased to build the kitchen platform and dining area of Mr. Kumar's bungalow. The cost of the stone varies directly with square of its weight. The stone broke into four parts whose weights are in the ratio of 2 : 4 : 7 : 11. If the granite stone had broken into four equal parts of weight then it would have led to a loss of Rs. 73600. What is the actual cost of the original granite stone (unbroken)?

- (a)
5,18,400

- (b)
2,30,400

- (c)
9,21,600

- (d)
4,66,560

Answer: Option C

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**Text Explanation** :

It is given,

cost(c) is directly proportional to square of weight (w^{2}), i.e.

c = 1.w^{2}

Cost when block is divided into four parts whose weights are in ratio 2 : 4 : 7 : 11

Let the weights be 2x, 4x, 7x, 11x

Cost = k (4x^{2} + 16x^{2} + 49x^{2} + 121x^{2}) = 190kx^{2}

Cost when block is divided into four equal parts

Let the weights be 6x, 6x, 6x, 6x

Cost = k (4 * 36x^{2}) = 144kx^{2}

It is given,

190kx^{2} - 144kx^{2} = 73600

kx^{2 }= 1600

Actual weight of granite = 24x

Cost = 576kx^{2 }= 576*1600 = 921600

The answer is option C.

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**IIFT 05 Dec 2021 QA | Arithmetic - Profit & Loss omet Question**

Sui-Dhaga Company has to prepare two special designer dresses for Garba event. Their unique dress designs are always in demand by the young couples who participate in Garba event competitions. The profit contribution of male dress is $75 per unit and the profit contribution of female dress is $80 per unit. Total 7 hours are required to stitch a male-dress and 3 hours are required to stitch a female-dress. Silk required to prepare a male dress and a female dress is 4 meters and 5 meters respectively. To produce the dresses, total available labour hours are 59 and total availability of silk is 60 meters. They would like to produce male and female dresses as per the available resources in such a way so that the total profit gets maximized for the company. What will be total maximum profit earned by the Sui-Dhaga Company?

- (a)
775

- (b)
900

- (c)
1015

- (d)
1240

Answer: Option C

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**Text Explanation** :

Let the number of male dresses and female dresses be x and y respectively.

It is given,

4x + 5y = 60

7x + 3y = 59

Solving, we get x = 5 and y = 8

Profit = 75(5) + 80(8) = 1015

Answer is option C.

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**IIFT 05 Dec 2021 QA | Arithmetic - Profit & Loss omet Question**

Shankar Fertilizer Limited and Shah Fertilizer Limited purchased one packet of Phosphorus each at the same price. Later on GreenP Company purchased both the packets at equal price from Shankar Fertilizer Limited and Shah Fertilizer Limited. But the profit percentage of Shankar Fertilizer Limited was X while that of Shah Fertilizer Limited was Y. Shah Fertilizer Limited calculated his profit on the selling price. Thus Y = 45 $\frac{9}{20}$%. If the GreenP 20 Company sells one of the packets to Mehrauli Nursery at profit, then what is the cost price for Mehrauli Nursery. while GreenP Company purchased each of the Phosphorus packets at Rs. 330?

- (a)
726

- (b)
762

- (c)
526

- (d)
584

Answer: Option A

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**Text Explanation** :

Shankar's profit% = $\frac{P}{C.P}$ × 100 = X

Shah's profit% = $\frac{P}{S.P}$ × 100 = Y

$\frac{S.P}{C.P}$ = $\frac{X}{Y}$

It is given,

Y = $\frac{909}{2000}$X

$\frac{X}{Y}$ = $\frac{2000}{909}$

$\frac{S.P}{C.P}$ = $\frac{2000}{909}$

$\frac{P}{C.P}$ = $\frac{1091}{909}$

It is given,

C.P = Rs. 330

Profit P = $\frac{1091}{909}$ × 330 ≈ 396

The cost price for Mehrauli Nursery = 396 + 330 = Rs 726

Answer is option A.

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**IIFT 2020 QA | Arithmetic - Profit & Loss omet Question**

A fruit seller in a locality uses dishonest practice as follows:

(i) He cheats on weight by 10 percent for every 1kg weight.

(ii) He pushes up the price of fruit by 15 percent and then gives a discount of 8 percent to the buyers for every kg sold.

Find the percentage profit of the fruit seller from sale of 1kg. (Profit is defined as Revenue - Cost)

- (a)
18.95

- (b)
17.56

- (c)
16.04

- (d)
15.00

Answer: Option B

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**Text Explanation** :

Let the cost of fruit be Rs. 100 for 1000gm.

The fruit seller marks the price by 15% and then gives a discount of 8%. Thus, the selling price of fruit = Rs. 105.8

The fruit seller also cheats the customers by 10% of the weight. Thus, he sells 900gm for Rs. 105.8

Cost Price = $\frac{100}{1}$ = Rs. 100/kg

Selling price = $\frac{105.8}{0.9}$ = Rs. 117.56

Percentage profit = $\frac{117.56-100}{100}$ × 100 = 17.56%

Hence, the answer is option B.

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**IIFT 2020 QA | Arithmetic - Profit & Loss omet Question**

An owner of a grocery shop purchases two varieties of grain. The price of first variety is twice the price of the second one. He mixes both the varieties and sells the mixture at the price of Rs. 28 per kg, making a profit of 25%. If the ratio of first variety of grain and the second variety of grain in the mixture is 2 : 3, find the price of first variety of grain.

- (a)
Rs. 16/kg

- (b)
Rs. 24/kg

- (c)
Rs. 32/kg

- (d)
Rs. 64/kg

Answer: Option C

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**Text Explanation** :

Let the mixture so created to be sold was exactly one kg.

Hence, amount of first variety procured to contribute to this mixture = (2/5)*1000 = 400gm or 0.4 kg

Hence, amount of second variety procured to contribute to this mixture = (3/5)*1000 = 600gm or 0.6 kg

Let the cost price of second variety = Rs. a/kg and hence, cost price of first variety = Rs. 2a / kg

Hence, cost price of the entire mixture = 0.6a + 0.4*2a = 1.4a rupees / kg

Also, the mixture is sold at Rs. 28 per kg and it means a profit of 25% ⇒ CP per kg = Rs. (28/1.25) per kg

Equating both, we get a = 16, hence, price of the first variety = 32 rupees per kg

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**IIFT 2020 QA | Arithmetic - Profit & Loss omet Question**

Saudi Aramco and Reliance Industries entered into a joint venture where Saudi Aramco invested 20 billion dollars and Reliance Industries invested 30 billion dollars. The ownership ratio is always equal to the investment ratio. After 1 year, the venture made a profit of 6 billion dollars which they reinvested. Now Reliance Industries wants to increase its ownership to 75%, how much it should pay to Saudi Aramco in billion dollars?

- (a)
12.4

- (b)
8.4

- (c)
10.4

- (d)
14.4

Answer: Option B

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**Text Explanation** :

The starting net worth of the venture is 50 billion dollars. After the first year profit of 6 billion dollars is reinvested, the net worth of the venture becomes 56 billion dollars. This is held in the ratio of 2 : 3 by Saudi Aramco and Reliance respectively.

Of this 56 billion dollars, Reliance has a claim to 3/5 th of it i.e. 33.6 billion dollars and Aramco has a claim to 2/5 th of it i.e. 22.4 billion dollars.

Now, Reliance wants to own 75% of this i.e. 3/4th of 56 billion dollars = 42 billion dollars.

To increase its stake from 33.6 to 42 billion dollars, it should pay the difference amount to Aramco i.e. 8.4 billion dollars.

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**XAT 2019 QADI | Arithmetic - Profit & Loss omet Question**

An article is marked x% above the cost price. A discount of 2/3x% if given on the marked price. If the profit is 4% of the cost price and the value of x lies between 25 and 50, then the value of 50% of x is:

- (a)
15

- (b)
13

- (c)
20

- (d)
16

- (e)
12

Answer: Option A

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**Text Explanation** :

Let the CP of the article be Rs 100.

∴ M.P = 100 + x

As a discount of 2x/3 % is given on the Marked Price.

SP = (100 + x)$\left(1-\frac{2x/3}{100}\right)$ …(1)

Also, since profit is 4%, it means selling price = 100 + 4 = 104 …(2)

Equating (1) and (2)

⇒ (100 + x)$\left(1-\frac{2x/3}{100}\right)$ = 104

⇒ 100 + x – $\frac{2x}{3}$ – $\frac{2{x}^{2}}{300}$ = 104

⇒ $\frac{2{x}^{2}}{300}$ - $\frac{x}{3}$ + 4 = 0

⇒ x^{2} – 50x + 600 = 0

⇒ x = 20 or 30.

Since it is given that x is between 25 and 50 hence x = 30.

∴ 50% of 30 = 15.

Hence, option (a).

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**XAT 2017 QADI | Arithmetic - Profit & Loss omet Question**

A shop, which sold same marked price shirts, announced an offer - if one buys three shirts then the fourth shirt is sold at a discounted price of ₹ 100 only. Patel took the offer. He left the shop with 20 shirts after paying ₹ 20,000. What is the marked price of a shirt?

- (a)
Rs. 1260

- (b)
Rs. 1300

- (c)
Rs. 1350

- (d)
Rs. 1400

- (e)
Rs. 1500

Answer: Option B

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**Text Explanation** :

Let the marked price of a shirt = M

∴ If a customer buys 3 shirts then the fourth shirt would be given at a discounted price of 100 rupees.

⇒ Total cost of 4 shirts = 3M + 100

Customer bought a total of 20 shirts, hence he bought 15 shirts and marked price and got 5 shirts at the discounted price.

∴ Total amount paid by the customer = 15M + 500

⇒ 15M + 500 = 20,000

⇒ M = 1300

Hence, option (b).

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**XAT 2016 QADI | Arithmetic - Profit & Loss omet Question**

Rani bought more apples than oranges. She sells apples at Rs. 23 apiece and makes 15% profit. She sells oranges at Rs. 10 apiece and marks 25% profit. If she gets Rs. 653 after selling all the apples and oranges, find her profit percentage.

- (a)
16.8%

- (b)
17.4%

- (c)
17.9%

- (d)
18.5%

- (e)
19.1%

Answer: Option B

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**Text Explanation** :

Let the number of apples sold be ‘a’ and the number of oranges sold be ‘o’.

∴ Total Selling Price = 23a + 10b = 653

In R.H.S., there is a 3 in the unit’s place

∴ 23a + 10b should end with 3.

Now, unit’s digit of 10b will be ‘0’ hence units digit of 13a should be 3.

For this the values possible values of ‘a’ are 1, 11, 21, …

When a = 11, b = 40 [not possible since a should be more than b.]

When, a = 21, b = 17 which is in line with the condition of a > b

When a ≥ 31, b is negative, hence we will not consider those values.

∴ a = 21 and b = 17.

Now, the profit per apple is 15% and profit per orange is 25%

Cost price of each apple = 23/1.15 = Rs. 20

Cost price of each orange = 10/1.25 = Rs. 8

∴ Total Cost price = 20a + 8b = Rs.556

∴ Profit percent = ((653 - 556)/ 556) × 100 = 17.4%

Hence, option (b).

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**XAT 2016 QADI | Arithmetic - Profit & Loss omet Question**

Company ABC starts an educational program in collaboration with Institute XYZ. As per the agreement, ABC and XYZ will share profit in 60 : 40 ratio. The initial investment of Rs. 100,000 on infrastructure is borne entirely by ABC whereas the running cost of Rs. 400 per student is borne by XYZ. If each student pays Rs. 2000 for the program find the minimum number of students required to make the program profitable, assuming ABC wants to recover its investment in the very first year and the program has no seat limits.

- (a)
63

- (b)
84

- (c)
105

- (d)
157

- (e)
167

Answer: Option C

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**Text Explanation** :

Initial investment by ABC = 1,00,000

Let the total number of students be x.

Net profit from a student = 2000 – 400 = Rs. 1600

Net profit from ‘x’ students = 1600x

This will be divided between ABC and XYZ in the ratio of 60 : 40 i.e., 3 : 2.

∴ ABC will receive 1600x × 3/5

Hence, 1600x × 3/5 ≥ 1,00,000

⇒ x ≥ 104.2

∴ Minimum number of students should be 105.

Hence, option (c).

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**IIFT 2016 QA | Arithmetic - Profit & Loss omet Question**

In a local shop, as part of promotional measures, the shop owner sells three different varieties of soap, one at a loss of 13 percent, another at a profit of 23 percent and the third one at a loss of 26 percent. Assuming that the shop owner sells all three varieties of soap at the same price, the approximate percentage by which average cost price is lower or higher than the selling price is

- (a)
10.5 higher

- (b)
12.5 lower

- (c)
14.5 lower

- (d)
8.5 higher

Answer: Option A

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**Text Explanation** :

Let the S.P. of each type be Rs. 100

C.P. of soap sold at 13% loss = 100/0.87 ≅ Rs. 115

C.P. of soap sold at 23% profit = 100/1.23 ≅ Rs. 81

C.P. of soap sold at 26% loss = 100/0.74 ≅ Rs. 135

∴ Average C.P. = (115 + 81 + 135)/3 = Rs. 110.3

∴ % by which average C.P. is higher = (10.3/100) × 100 = 10.3%

The closest value in the options is 10.5%

Hence, option (a).

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