# Profit and Loss Questions for CAT

Profit and Loss is one of the most important topics in the **CAT Quants section.** Over the past few years, CAT Profit and Loss questions have made a recurrent appearance in the Quant section.Â You can also check out these Profit and Loss Questions for CAT PDFs from the **CAT Previous year’s papers**.Â

You can expect around 1-2 questions in the new format of the CAT Quant section. If you’re new to this section, you can check out the CAT Profit and Loss Questions from the CAT Previous Year papers. In this article, we will look into some very important Profit and Loss questions PDF(with solutions) for CAT. If you want to practice these questions, you can download this CAT Profit and Loss Questions PDF (most important) along with the detailed solutions below, which is completely Free.

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**Instructions**

Directions for the following two questions: Shabnam is considering three alternatives to invest her surplus cash for a week. She wishes to guarantee maximum returns on her investment. She has three options, each of which can be utilized fully or partially in conjunction with others.

Option A: Invest in a public sector bank. It promises a return of +0.10%.

Option B: Invest in mutual funds of ABC Ltd. A rise in the stock market will result in a return of +5%, while a fall will entail a return of â€“ 3%.

Option C: Invest in mutual funds of CBA Ltd. A rise in the stock market will result in a return of â€“ 2.5%, while a fall will entail a return of + 2%.

**Question 1:Â **The maximum guaranteed return to Shabnam is

a)Â 0.25%

b)Â 0.10%

c)Â 0.20%

d)Â 0.15%

e)Â 0.30%

**1)Â AnswerÂ (C)**

**Solution:**

Let a, b and c be the percentages of amountÂ invested in options A, B and C respectively => a + b + c = 100

Return attained if there is a rise in the stock market => 0.001a + 0.05b – 0.025c

Return attained if there is a fall in the stock market => 0.001a –Â 0.03b +Â 0.02c

Maximum guaranteed return is attained when both are equal because it is indifferent to rise and fall in the market.

0.001a + 0.05b – 0.025c =Â 0.001a –Â 0.03b +Â 0.02c

=> 0.08b = 0.045c => 16b = 9c

Let’s put the values for a, b and c that satisfy the above equation.

b = 9, c = 16,Â a = 75 => return = 0.125

b = 18, c = 32, a = 50 => return = 0.15

b = 27, c = 48, a = 25 => return = 0.175

b = 36, c = 64, a = 0 => return = 0.2

Hence, the maximum guaranteed returnÂ is 0.2%

**Question 2:Â **What strategy will maximize the guaranteed return to Shabnam?

a)Â 100% in option A

b)Â 36% in option B and 64% in option C

c)Â 64% in option B and 36% in option C

d)Â 1/3 in each of the three options

e)Â 30% in option A, 32% in option B and 38% in option C

**2)Â AnswerÂ (B)**

**Solution:**

Let a, b and c be the percentages of amountÂ invested in options A, B and C respectively => a + b + c = 100

Return attained if there is a rise in the stock market => 0.001a + 0.05b – 0.025c

Return attained if there is a fall in the stock market => 0.001a –Â 0.03b +Â 0.02c

Maximum guaranteed return is attained when both are equal because it is indifferent to rise and fall in the market.

0.001a + 0.05b – 0.025c =Â 0.001a –Â 0.03b +Â 0.02c

=> 0.08b = 0.045c => 16b = 9c

Let’s put the values for a, b and c that satisfy the above equation.

b = 9, c = 16,Â a = 75 => return = 0.125

b = 18, c = 32, a = 50 => return = 0.15

b = 27, c = 48, a = 25 => return = 0.175

b = 36, c = 64, a = 0 => return = 0.2

Hence, the maximum guaranteed returnÂ is 0.2% and it is attained when 36% is invested in option B and 64% is invested in option C.

**Question 3:Â **The owner of an art shop conducts his business in the following manner: every once in a while he raises his prices by X%, then a while later he reduces all the new prices by X%. After one such updown cycle, the price of a painting decreased by Rs. 441. After a second up-down cycle the painting was sold for Rs. 1,944.81. What was the original price of the painting?

a)Â Rs. 2,756.25

b)Â Rs. 2,256.25

c)Â Rs. 2,500

d)Â Rs. 2,000

**3)Â AnswerÂ (A)**

**Solution:**

Let the price of the painting be P

One cycle of price increase and decrease reduces the price by $x^2/100 * P = 441$

Let the new price be N =>Â $P – x^2/100 * P = N$

Price after the second cycle =Â $N – x^2/100 * N$Â = 1944.81

=>Â $(P – x^2/100 * P)(1 – x^2/100) = 1944.81$

=>Â $(P – 441)(1 – 441/P) = 1944.81$

=>Â $P – 441 – 441 + 441^2/P = 1944.81$

=>Â $P^2 – (882 + 1944.81)P + 441^2 = 0$

=>Â $P^2 – 2826.81P + 441^2 = 0$

From the options, the value 2756.25 satisfies the equation.

So, the price of the article is Rs 2756.25

**Question 4:Â **In a stockpile of products produced by three machines M1, M2 and M3, 40% and 30% were manufactured by M1 and M2 respectively. 3% of the products of M1 are defective, 1% of products of M2 defective, while 95% of the products of M3 are not defective. What is the percentage of defective in the stockpile?

a)Â 3%

b)Â 5%

c)Â 2.5%

d)Â 4%

**4)Â AnswerÂ (A)**

**Solution:**

Let’s say total products maufactured by M1, M2 and M3 are 100.

So M1 produced 40, M2 produced 30 and M3 produced 30

Defective pieces for M1 = $\frac{120}{100}$

Defective pieces for M2 = $\frac{30}{100}$

Defective pieces for M3 = $\frac{150}{100}$

So total defective pieces are $\frac{150+30+120}{100}$ = $\frac{300}{100}$ = 3% of total products.

**Question 5:Â **Gopal went to a fruit market with certain amount of money. With this money he can buy either 50 oranges or 40 mangoes. He retains 10% of the money for taxi fare. If he buys 20 mangoes, then the number of oranges he can buy is

a)Â 25

b)Â 18

c)Â 20

d)Â None of these

**5)Â AnswerÂ (C)**

**Solution:**

Let’s say total money was $x$ rs.

So cost price of 40 mango will be = $x$ ;

Hence cost price of 20 mangoes will be = $\frac{x}{2}$

Taxi fare = $\frac{10x}{100}$

Total expense =Â $\frac{x}{2}$ +Â Â $\frac{10x}{100}$ = $\frac{6x}{10}$

Remaining money =$ \frac{4x}{10}$

Cost price of 1 orange will be = $\frac{x}{50}$

Hence inÂ $\frac{4x}{10}$ rs. 20 oranges can be purchased.

Checkout: **CAT Free Practice Questions and Videos**

**Instructions**

Ghosh Babu has a certain amount of property consisting of cash, gold coins and silver bars. The cost of a gold coin is Rs. 4000 and the cost of a silver bar is Rs. 1000. Ghosh Babu distributed his property among his daughters equally. He gave to his eldest daughter gold coins worth 20% of the total property and Rs. 25000 in cash. The second daughter was given silver bars worth 20% of the remaining property and Rs. 50000 cash. Among the third and fourthÂ daughters, he distributed the remaining gold and silver bars equallyÂ both together accounting each for 20% of the property remaining after the previous distribution. He also gave the third and fourth daughtersÂ Rs. 25000 more than what the second daughter had received in cash.

**Question 6:Â **The amount of property in gold and silver possessed by Ghosh Babu is

a)Â 2,25,000

b)Â 2,75,000

c)Â Rs. 4,25,000

d)Â None of these

**6)Â AnswerÂ (B)**

**Solution:**

The total property consists of cash, gold coins and silver bars.

And ghosh babu gave equal parts to 4 daughters, hence they should have 25% of total property each.

As eldest daughter possess gold coins as 20% worth of total property, so 25000 cash should be equal to 5% of total property.

So total property will be =$\frac{ 25000 \times 100}{5}$ = 500000

Hence property amounting only gold coins and silver bars will be = Total property – Total Cash

i.e. = 500000 – (25000 + 50000 + 75000 + 75000)

= 2,75,000

**Question 7:Â **Total property of Ghosh Babu (in Rs.lakh) is

a)Â 5.0

b)Â 7.5

c)Â 10.0

d)Â 12.5.

**7)Â AnswerÂ (A)**

**Solution:**

The total property consists of cash, gold coins and silver bars.

And ghosh babu gave equal parts to 4 daughters, hence they should have 25% of total property each.

As eldest daughter possess gold coins as 20% worth of total property, so 25000 cash should be equal to 5% of total property.

So total property will be =$\frac{ 25000 \times 100}{5}$ = 500000

**Question 8:Â **If the ratio of the gold to silver bars thatÂ Ghosh Babu had is 7:27,Â the number of silver bars he has is

a)Â 90

b)Â 60

c)Â 75

d)Â 135

**8)Â AnswerÂ (D)**

**Solution:**

Let the number of gold bars be 7a and the number of silver bars be 27a.

The total value of the gold and silver bars is 500000 – 225000 = 275000

Therefore, 4000*7a + 1000*27a = 275000

Or, 55000a = 275000

Or, a = 5

Therefore the number of silver bars with Ghosh Babu is 27a = 135

**Question 9:Â **A sum of money compounded annually becomes Rs.625 in two years and Rs.675 in three years. The rate of interest per annum is

a)Â 7%

b)Â 8%

c)Â 6%

d)Â 5%

**9)Â AnswerÂ (B)**

**Solution:**

As we know, formulae of compound interest for 2 years Â will be:

$P(1+\frac{r}{100})^{2}$ = 625Â (Where r is rate, P is principal amount)

For 3 years:

$P(1+\frac{r}{100})^{3}$ = 675

Dividing above two equations we will get r=8%

**Question 10:Â **Instead of a metre scale, a cloth merchant uses a faulty 120 cm scale while buying, but uses a faultyÂ 80 cm scale while selling the same cloth. If he offers a discount of 20%, what is his overall profit percentage?

a)Â 20%

b)Â 25%

c)Â 40%

d)Â 15%

**10)Â AnswerÂ (A)**

**Solution:**

Let’s say the cost of the cloth is x rs per metre. Because of the faulty meter, he is paying x for 120 cms when buying.

So cost of 100 cms = 100x/120.

He is selling 80 cms for x, so selling price of 100cms of clothÂ is 100x/80.

discount = 20%

so the effectiveÂ selling price is .8*100x/80= x

profit = SP-CP= x – 100x/120 = x/6

Profit % = x/6 divided by 100x/120 = 20%

**Question 11:Â **The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.

a)Â Rs. 1.4 lakh

b)Â Rs. 2 lakh

c)Â Rs. 1 lakh

d)Â Rs. 2.1 lakh

**11)Â AnswerÂ (C)**

**Solution:**

Let the original weight of the diamond be equal to $10k$. So, after breaking into 4 pieces, the parts of the diamond weight $k, 2k, 3k,4k$

The price of the diamond varies directly in proportion to the weight. Let us assume, the $P=C*W^2$ where $C$ is a constant and $W$ is the weight of the diamond.

Therefore, the original price is $C*10k*10k = 100k^2*C$

The new weight is $Ck^2 + C(2k)^2 + C(3k)^2 + C(4k)^2 = 30k^2C$

The decrease in the price equals 70,000. So, $100k^2C-30k^2C = 70000$

Or, $k^2C = 1000$

Therefore the original price = $100k^2C = 100000$

**Question 12:Â **Once I had been to the post office to buy five-rupee, two-rupee and one-rupee stamps. I paid the clerk Rs. 20, and since he had no change, he gave me three more one-rupee stamps. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I had when I left the post office?

a)Â 10

b)Â 9

c)Â 12

d)Â 8

**12)Â AnswerÂ (A)**

**Solution:**

As shopkeeper gave 3 one-rupee change for 20 rs. change, Buyer must have ordered for a total of 17 rs. stamps.

Now buyer ordered for at leastÂ more than 1 stamp for each type

Hence the minimum he bought was:

2 stamp for 5 rupees = 10 rs.

2 stamp for 2 rupees = 4 rs.

2 stamp for 1 rupee = 2 rs.

For the total to be seventeen, the buyer must have purchased 3 one rupee stamps.

And 3 one rupee stamps were also there as changes given by shopkeeper.

So total number of stamps = 2+2+(3+3) = 10

**Question 13:Â **I sold two watches for Rs. 300 each, one at the loss of 10% and the other at the profit of 10%. What is the percentage of loss(-) or profit(+) that resulted from the transaction?

a)Â (+)10

b)Â (-)1

c)Â (+)1

d)Â (-)10

**13)Â AnswerÂ (B)**

**Solution:**

Selling price of first watch = 300

Profit = 10%

cost price = $\frac{300}{1.1}$

Selling price of second watch = 300

Loss = 10%

cost price = $\frac{300}{0.9}$

Total selling price of transaction= 600

Total cost price of transaction = $300(\frac{10}{11} + \frac{10}{9}) = 600 (\frac{100}{99})$

Loss = $600 (\frac{100}{99} – 1)$

%loss = $(600 (\frac{100}{99} – 1)) \div (600(\frac{100}{99})) \times 100 = 1$

**Question 14:Â **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)Â 28.56%

b)Â 35%

c)Â 22.22%

d)Â None of these

**14)Â AnswerÂ (A)**

**Solution:**

Let’s say cost price is 100

gain = 14.28

selling price = 114.28

Marked price = x(say)

So $x- \frac{11.11x}{100} = \frac{8x}{9} = 114.28$

Or $x = 128.52$

So marked price is 28.52% more than cost price.

**Question 15:Â **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?

a)Â Rs. 80

b)Â Rs. 100

c)Â Rs. 95

d)Â None of these

**15)Â AnswerÂ (D)**

**Solution:**

Let’s say he buy fruits of weights 3 kg., 4kg., 5 kg.

Total kilograms of dry fruitsÂ $=3+4+5=12$

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.

**Question 16:Â **A man earns x% on the first Rs. 2,000 and y% on the rest of his income. If he earns Rs. 700 from income of Rs. 4,000 and Rs. 900 from Rs. 5,000 of income, find x%.

a)Â 20%

b)Â 15%

c)Â 25%

d)Â None of these

**16)Â AnswerÂ (B)**

**Solution:**

He earns x% on first 2000 and y% on rest of his income.

So on 4000 rs. , he will earn as follows:

$2000 \frac{x}{100} + 2000 \frac{y}{100}$ = 700

Or $x+y = 35$

Similarly on 5000 rs. ,he will earn 900 as follows:

$2000 \frac{x}{100} + 3000 \frac{y}{100} = 900$

Or $20x + 30y = 900$

On solving above equations, we will get value of x = 15

**Question 17:Â **A yearly payment to the servant is Rs. 90 plus one turban. The servant leaves the job after 9 months and receives Rs. 65 and a turban. Then find the price of the turban.

a)Â Rs. 10

b)Â Rs. 15

c)Â Rs. 7.50

d)Â Cannot be determined

**17)Â AnswerÂ (A)**

**Solution:**

Let’s say price of turban is x.

So total price for 12 months will be = $90+x$

total price for 9 months = $\frac{(90+x) \times 9}{12} = (65+x) $

By solving above equation, we will get value of x= 10.

**Question 18:Â **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

**18)Â Answer:Â 70000**

**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

**Question 19:Â **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)Â Give a discount of 5% on retail price.

b)Â Give a discount of 2% on retail price.

c)Â Increase the retail price by 2%.

d)Â Sell at retail price.

**19)Â AnswerÂ (D)**

**Solution:**

Let the retail price be M and cost price be C.

Given,

0.85 M = 1.02 C

M = 1.2 C

If he wants 20% profit he has to sell at 1.2C, which is nothing but the retail price.

**Question 20:Â **Suppose, C1, C2, C3, C4, and C5 are five companies. The profits made by Cl, C2, and C3 are in the ratio 9 : 10 : 8 while the profits made by C2, C4, and C5 are in the ratio 18 : 19 : 20. If C5 has made a profit of Rs 19 crore more than C1, then the total profit (in Rs) made by all five companies is

a)Â 438 crore

b)Â 435 crore

c)Â 348 crore

d)Â 345 crore

**20)Â AnswerÂ (A)**

**Solution:**

Given,

C1 : C2 : C3 = 9 : 10 : 8 … i

C2 : C4 : C5 = 18 : 19 : 20 … ii

Let’s multiply i by 9 and ii by 5

C1 : C2 : C3 = 81 : 90 : 72

C2 : C4 : C5 = 90 : 95 : 100

Therefore, C1 : C2 : C3 : C4 : C5 = 81 : 90 : 72 : 95 : 100

Given,

100x – 81x = 19

x = 1

Hence, total profit = 100 + 95 + 72 + 90 + 81 = 438

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