CAT Profit and Loss Questions & Solutions

A sample of real CAT Profit and Loss past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 62 Profit and Loss questions in all — sign up free to practise them timed.

Q1.CAT 2025
A shopkeeper offers a discount of 22% on the marked price of each chair, and gives 13 chairs to a customer for the discounted price of 12 chairs to earn a profit of 26% on the transaction. If the cost price of each chair is Rs 100, then the marked price, in rupees, of each chair is
Answer: 175
Show solution
Cost price of each chair = 100

For 13 chairs, total cost = $13 \times 100 = 1300$

Profit = 26%, so total revenue

$1.26 \times 1300 = 1638$

We were told that this amount is equal to the discounted price of 12 chairs. So the discounted selling price per chair = $\text{SP}_{\text{disc}} = \frac{1638}{12} = 136.5$

Discount offered = 22%, so:

MP = $\frac{136.5}{0.78}$ = 175
Q2.CAT 2024
Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is
- 10
- 25
- 15
- 20
Show solution
Let us say the cost price of an item is X
It is said that it is marked to make a profit of 20%.
That means it is marked at 1.2X

Ravi gets a 10% discount on the marked price,
$0.9\left(1.2X\right)=1.08X$

Saves 15 rupees, so 1.2X-1.08X
0.12X=15
X=125

Profit made by Gopi is 0.08(125)=10 rupees.
Q3.CAT 2023
Jayant bought a certain number of white shirts at the rate of Rs 1000 per piece and a certain number of blue shirts at the rate of Rs 1125 per piece. For each shirt, he then set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10% and made a total profit of Rs.51000. If he bought both colors of shirts, then the maximum possible total number of shirts that he could have bought is
Answer: 407
Show solution
Let the number of white shirts be m, and the number of blue shirts be n. Hence, the total cost of the shirts = (1000m+1125n), and the number of shirts is (m+n)

The average price of the shirts is $\ \frac{\ 1000m+1125n}{m+n}$ .
It is given that he set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10%.

Hence, the average selling price of the shirts = $\left(\ \frac{\ 1000m+1125n}{m+n}\right)\times\ \frac{5}{4}\times\ \frac{9}{10}=\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)$

The average profit of the shirts = $\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)-\frac{\ 1000m+1125n}{m+n}=\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)$

The total profit of the shirts = $\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)\times\ \left(m+n\right)\ =\ \frac{1}{8}\left(1000m+1125n\right)$

Now, $=>\frac{1}{8}\left(1000m+1125n\right)=51000$

$=>1000m+1125n=51000\times\ 8=408000$

Now to get the maximum number of shirts, we need to minimize n (since the coefficient of n is greater than the coefficient of m), but it can't be zero. Therefore, m has to be maximum.

$m\ =\ \ \frac{\ 408000-1125n}{1000}$

The maximum value of m such that m, and both are integers is m = 399, and n = 8 (by inspection)

Hence, the maximum number of shirts = m+n = 399+8 = 407
Q4.CAT 2020
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
Show solution
Let the price of desktop and laptop be x,y respectively.

Given,

x+y=50000...(i)

1.2x+0.9y=50000(1.02)=51000...(ii)

(ii)-0.9(i) gives

0.3x=6000=> x=20000.
Q5.CAT 2019
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
- 1
- 1.2
- 0.50
- 0.7
Show solution
CP of the table at which the shopkeeper procured each table = p

It is given that shopkeeper sold the tables to Amal and Asim at a profit of 20% and at a loss of 20%, respectively

The selling price of the tables = 1.2p and 0.8p to Amal and Asim respectively.

Amal sells his table to Bimal at a profit of 30%

So, CP of the table by Bimal (x)= 1.2p*1.3 = 1.56p

Asim sells his table to Barun at a loss of 30%

So, CP of the table by Barun (y)= 0.7*0.8p = 0.56p

(x-y)/p = (1.56p-0.56p)/p = p/p=1
Q6.CAT 2017
Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi's savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59500, then Ravi's total monthly savings (in Rs) is
Answer: 70000
Show solution
Let his total savings be 100x.

He invests 50x in fixed deposits. 30% of 50x, which is 15x is invested in stocks and 35x goes to savings bank.

It is given 85x = 59500

x = 700

Hence, 100x = 70000
Q7.CAT 2017
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?
- 50
- 30
- 25
- 45
Show solution
Let the number of dozens of candies he bought of each variety be x
Hence total cost = 12x + 15x = 27x
Total selling price = 16.50*2x = 33x
Profit = 33x - 27x = 6x
Given 6x = 150 => x = 25
Hence he bought 50 dozens of candies in total
Passage
Directions for the following two questions:

Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is $240 + bx + cx^2$ , where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.67%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.
Q8.CAT 2007
How many units should Mr. David produce daily?
- 130
- 100
- 70
- 150
- Cannot be determined
Show solution
Cost of 20 units = 240+20b+400c

Cost of 40 units = 240+40b+1600c = 5/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c

=> 480 = 20b + 2800c => 120 = 5b + 700c

Cost of 60 units = 240+60b+3600c = 3/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c

=> 240 = 2400c => c = 1/10 and b = 10

Let the number of items needed for max profit be k

CP = $240+10k+k^2/10$

SP = 30k

Profit = SP - CP = $30k - 240 - 10k - k^2/10$ = $20k - 240 - k^2/10$

or Profit = $\frac{1}{10} (-k^2 + 200k - 2400)$

or, Profit = $\frac{1}{10} (-(k^2 - 200k + 2400))$

or, Profit = $\frac{1}{10} (-(k^2 - 200k + 2400 + 7600 - 7600))$

or, Profit = $\frac{1}{10} (-(k^2 - 200k + 10000) + 7600)$

or, Profit = $\frac{1}{10} (-(k - 100)^2 + 7600)$

To maximise the value of Profit, $-(k - 100)^2$ must be 0.

So, $k$ must be equal to 100.

Hence, option B is the correct answer.
Passage
DIRECTIONS for questions:

These questions are based on the situation given below:

Recently, Ghosh Babu spent his winter vacation on Kyakya Island. During the vacation, he visited the local casino where he came across a new card game. Two players, using a normal deck of 52 playing cards, play this game. One player is called the Dealer and the other is called the Player. First, the Player picks a card at random from the deck. This is called the base card. The amount in rupees equal to the face value of the base card is called the base amount. The face values of Ace, King, Queen and Jack are ten. For other cards, the face value is the number on the card. Once, the Player picks a card from the deck, the Dealer pays him the base amount. Then the dealer picks a card from the deck and this card is called the top card. If the top card is of the same suit as the base card, the Player pays twice the base amount to the Dealer. If the top card is of the same colour as the base card (but not the same suit) then the Player pays the base amount to the Dealer. If the top card happens to be of a different colour than the base card, the Dealer pays the base amount to the Player. Ghosh Babu played the game 4 times. First time he picked eight of clubs and the Dealer picked queen of clubs. Second time, he picked ten of hearts and the dealer picked two of spades. Next time, Ghosh Babu picked six of diamonds and the dealer picked ace of hearts. Lastly, he picked eight of spades and the dealer picked jack of spades. Answer the following questions based on these four games.
Q9.CAT 1999
If the final amount of money that Ghosh Babu had with him was Rs. 100, what was the initial amount he had with him?
- 120
- 8
- 4
- 96
Show solution
After the first game, Ghosh Babu picked 8 of clubs => He gets Rs 8. Then the dealer picked Queen of clubs => Ghosh Babu pays Rs 16 => Ghosh Babu is at a loss of Rs 8 after 1st game.
After the second game, Ghosh Babu picked 10 of hearts => He gets Rs 10. Then the dealer picked 2 of spades => Ghosh Babu gets another Rs 10 => Ghosh Babu is now at a profit of Rs 12.
After the third game, Ghosh Babu picked six of diamonds => He gets Rs 6. Then the dealer picked ace of hearts => Ghosh Babu pays Rs 6 to dealer => Ghosh Babu is still at a profit of Rs 12.
In the fourth game, Ghosh Babu picks 8 of spades => He gets Rs 8. The the dealer picks jack of spades => Ghosh Babu pays Rs 16 to dealer => Ghosh Babu is at a profit of Rs 4.
As Ghosh Babu earned a profit of Rs 4 and now he has Rs 100, he initially would have had 100 - 4 = Rs 96.
Q10.CAT 1997
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 fruits?
- Rs. 80
- Rs. 100
- Rs. 95
- None of these
Show solution
Let's say he buy fruits of weights 3 kg., 4kg., 5 kg.
Total kilograms of dry fruits $=3+4+5=12$

Overall cost price $=3\cdot100+4\cdot80+5\cdot60=300+320+300=920$

So cost price per kg. $=\dfrac{300+320+300}{12} = \dfrac{920}{12}$

Selling price = $\dfrac{920}{12} \times \dfrac{3}{2}$ = 115 per kg (Since Profit is 50%)

Hence answer will be D.
Q11.CAT 1996
The price of a Maruti car rises by 30% while the sales of the car come down by 20%. What is the percentage change in the total revenue?
- -4%
- -2%
- +4%
- +2%
Show solution
let's say price of maruti car is x rs.
Sales = y
revenue = xy
Changed price = 1.3x
changed value of sales = 0.8y
new revenue = 1.04 xy
Percentage change in revenue = 4%
Passage
A dealer deals only in colour TVs and VCRs. He wants to spend up to Rs.12 lakhs to buy 100 pieces. He can purchase a colour TV at Rs.10,000 and a VCR at Rs.15,000. He can sell a colour TV at Rs.12,000 and a VCR at Rs.17,500. His objective is to maximize profits. Assume that he can sell all the items that he stocks.
Q12.CAT 1990
For the maximum profit, the number of colour TVs and VCRs that he should respectively stock are
- 80, 20
- 20, 80
- 60, 40
- None of these
Show solution
It is given that the maximum cost is 12 lakhs
$1200000 <= T \times 10000 + V \times 15000$ (Where T is number of TV's and V is number of VCR's)
and as we know $T+V = 100$
By solving the above two equation, we will get v<=40

We have to keep the no. of VCRs as maximum, as they fetch higher profit than TV, so

$T = 60$
$V = 40$

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CAT Profit and Loss Questions & Solutions

Q1.CAT 2025

A shopkeeper offers a discount of 22% on the marked price of each chair, and gives 13 chairs to a customer for the discounted price of 12 chairs to earn a profit of 26% on the transaction. If the cost price of each chair is Rs 100, then the marked price, in rupees, of each chair is

Answer: 175

Show solution

Cost price of each chair = 100

For 13 chairs, total cost = $13 \times 100 = 1300$

Profit = 26%, so total revenue

$1.26 \times 1300 = 1638$

We were told that this amount is equal to the discounted price of 12 chairs. So the discounted selling price per chair = $\text{SP}_{\text{disc}} = \frac{1638}{12} = 136.5$

Discount offered = 22%, so:

MP = $\frac{136.5}{0.78}$ = 175

Q2.CAT 2024

Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is

Show solution

Let us say the cost price of an item is X
It is said that it is marked to make a profit of 20%.
That means it is marked at 1.2X

Ravi gets a 10% discount on the marked price,
$0.9\left(1.2X\right)=1.08X$

Saves 15 rupees, so 1.2X-1.08X
0.12X=15
X=125

Profit made by Gopi is 0.08(125)=10 rupees.

Q3.CAT 2023

Jayant bought a certain number of white shirts at the rate of Rs 1000 per piece and a certain number of blue shirts at the rate of Rs 1125 per piece. For each shirt, he then set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10% and made a total profit of Rs.51000. If he bought both colors of shirts, then the maximum possible total number of shirts that he could have bought is

Answer: 407

Show solution

Let the number of white shirts be m, and the number of blue shirts be n. Hence, the total cost of the shirts = (1000m+1125n), and the number of shirts is (m+n)

The average price of the shirts is $\ \frac{\ 1000m+1125n}{m+n}$ .

It is given that he set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10%.

Hence, the average selling price of the shirts = $\left(\ \frac{\ 1000m+1125n}{m+n}\right)\times\ \frac{5}{4}\times\ \frac{9}{10}=\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)$

The average profit of the shirts = $\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)-\frac{\ 1000m+1125n}{m+n}=\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)$

The total profit of the shirts = $\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)\times\ \left(m+n\right)\ =\ \frac{1}{8}\left(1000m+1125n\right)$

Now, $=>\frac{1}{8}\left(1000m+1125n\right)=51000$

$=>1000m+1125n=51000\times\ 8=408000$

Now to get the maximum number of shirts, we need to minimize n (since the coefficient of n is greater than the coefficient of m), but it can't be zero. Therefore, m has to be maximum.

$m\ =\ \ \frac{\ 408000-1125n}{1000}$

The maximum value of m such that m, and both are integers is m = 399, and n = 8 (by inspection)

Hence, the maximum number of shirts = m+n = 399+8 = 407

Q4.CAT 2020

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

Show solution

Let the price of desktop and laptop be x,y respectively.

Given,

x+y=50000...(i)

1.2x+0.9y=50000(1.02)=51000...(ii)

(ii)-0.9(i) gives

0.3x=6000=> x=20000.

Q5.CAT 2019

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

1
1.2
0.50
0.7

Show solution

CP of the table at which the shopkeeper procured each table = p

It is given that shopkeeper sold the tables to Amal and Asim at a profit of 20% and at a loss of 20%, respectively

The selling price of the tables = 1.2p and 0.8p to Amal and Asim respectively.

Amal sells his table to Bimal at a profit of 30%

So, CP of the table by Bimal (x)= 1.2p*1.3 = 1.56p

Asim sells his table to Barun at a loss of 30%

So, CP of the table by Barun (y)= 0.7*0.8p = 0.56p

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

Q6.CAT 2017

Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi's savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59500, then Ravi's total monthly savings (in Rs) is

Answer: 70000

Show solution

Let his total savings be 100x.

He invests 50x in fixed deposits. 30% of 50x, which is 15x is invested in stocks and 35x goes to savings bank.

It is given 85x = 59500

x = 700

Hence, 100x = 70000

Q7.CAT 2017

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?

Show solution

Let the number of dozens of candies he bought of each variety be x
Hence total cost = 12x + 15x = 27x
Total selling price = 16.50*2x = 33x
Profit = 33x - 27x = 6x
Given 6x = 150 => x = 25
Hence he bought 50 dozens of candies in total

Passage

Directions for the following two questions:

Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is $240 + bx + cx^2$ , where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.67%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.

Q8.CAT 2007

How many units should Mr. David produce daily?

130
100
70
150
Cannot be determined

Show solution

Cost of 20 units = 240+20b+400c

Cost of 40 units = 240+40b+1600c = 5/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c

=> 480 = 20b + 2800c => 120 = 5b + 700c

Cost of 60 units = 240+60b+3600c = 3/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c

=> 240 = 2400c => c = 1/10 and b = 10

Let the number of items needed for max profit be k

CP = $240+10k+k^2/10$

SP = 30k

Profit = SP - CP = $30k - 240 - 10k - k^2/10$ = $20k - 240 - k^2/10$

or Profit = $\frac{1}{10} (-k^2 + 200k - 2400)$

or, Profit = $\frac{1}{10} (-(k^2 - 200k + 2400))$

or, Profit = $\frac{1}{10} (-(k^2 - 200k + 2400 + 7600 - 7600))$

or, Profit = $\frac{1}{10} (-(k^2 - 200k + 10000) + 7600)$

or, Profit = $\frac{1}{10} (-(k - 100)^2 + 7600)$

To maximise the value of Profit, $-(k - 100)^2$ must be 0.

So, $k$ must be equal to 100.

Hence, option B is the correct answer.

Passage

DIRECTIONS for questions:

These questions are based on the situation given below:

Recently, Ghosh Babu spent his winter vacation on Kyakya Island. During the vacation, he visited the local casino where he came across a new card game. Two players, using a normal deck of 52 playing cards, play this game. One player is called the Dealer and the other is called the Player. First, the Player picks a card at random from the deck. This is called the base card. The amount in rupees equal to the face value of the base card is called the base amount. The face values of Ace, King, Queen and Jack are ten. For other cards, the face value is the number on the card. Once, the Player picks a card from the deck, the Dealer pays him the base amount. Then the dealer picks a card from the deck and this card is called the top card. If the top card is of the same suit as the base card, the Player pays twice the base amount to the Dealer. If the top card is of the same colour as the base card (but not the same suit) then the Player pays the base amount to the Dealer. If the top card happens to be of a different colour than the base card, the Dealer pays the base amount to the Player. Ghosh Babu played the game 4 times. First time he picked eight of clubs and the Dealer picked queen of clubs. Second time, he picked ten of hearts and the dealer picked two of spades. Next time, Ghosh Babu picked six of diamonds and the dealer picked ace of hearts. Lastly, he picked eight of spades and the dealer picked jack of spades. Answer the following questions based on these four games.

Q9.CAT 1999

If the final amount of money that Ghosh Babu had with him was Rs. 100, what was the initial amount he had with him?

Show solution

After the first game, Ghosh Babu picked 8 of clubs => He gets Rs 8. Then the dealer picked Queen of clubs => Ghosh Babu pays Rs 16 => Ghosh Babu is at a loss of Rs 8 after 1st game.

After the second game, Ghosh Babu picked 10 of hearts => He gets Rs 10. Then the dealer picked 2 of spades => Ghosh Babu gets another Rs 10 => Ghosh Babu is now at a profit of Rs 12.

After the third game, Ghosh Babu picked six of diamonds => He gets Rs 6. Then the dealer picked ace of hearts => Ghosh Babu pays Rs 6 to dealer => Ghosh Babu is still at a profit of Rs 12.

In the fourth game, Ghosh Babu picks 8 of spades => He gets Rs 8. The the dealer picks jack of spades => Ghosh Babu pays Rs 16 to dealer => Ghosh Babu is at a profit of Rs 4.

As Ghosh Babu earned a profit of Rs 4 and now he has Rs 100, he initially would have had 100 - 4 = Rs 96.

Q10.CAT 1997

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 fruits?

Rs. 80
Rs. 100
Rs. 95
None of these

Show solution

Let's say he buy fruits of weights 3 kg., 4kg., 5 kg.
Total kilograms of dry fruits $=3+4+5=12$

Overall cost price $=3\cdot100+4\cdot80+5\cdot60=300+320+300=920$

So cost price per kg. $=\dfrac{300+320+300}{12} = \dfrac{920}{12}$

Selling price = $\dfrac{920}{12} \times \dfrac{3}{2}$ = 115 per kg (Since Profit is 50%)

Hence answer will be D.

Q11.CAT 1996

The price of a Maruti car rises by 30% while the sales of the car come down by 20%. What is the percentage change in the total revenue?

Show solution

let's say price of maruti car is x rs.
Sales = y
revenue = xy
Changed price = 1.3x
changed value of sales = 0.8y
new revenue = 1.04 xy
Percentage change in revenue = 4%

Passage

A dealer deals only in colour TVs and VCRs. He wants to spend up to Rs.12 lakhs to buy 100 pieces. He can purchase a colour TV at Rs.10,000 and a VCR at Rs.15,000. He can sell a colour TV at Rs.12,000 and a VCR at Rs.17,500. His objective is to maximize profits. Assume that he can sell all the items that he stocks.

Q12.CAT 1990

For the maximum profit, the number of colour TVs and VCRs that he should respectively stock are

80, 20
20, 80
60, 40
None of these

Show solution

It is given that the maximum cost is 12 lakhs
$1200000 <= T \times 10000 + V \times 15000$ (Where T is number of TV's and V is number of VCR's)
and as we know $T+V = 100$
By solving the above two equation, we will get v<=40

We have to keep the no. of VCRs as maximum, as they fetch higher profit than TV, so

$T = 60$
$V = 40$

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