# Profit and Loss Questions for CAT

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Profit and Loss Questions for CAT PDF consists of the important questions on profit and loss for CAT exam. ThisÂ Profit, Loss and Discount questions and answers PDF will be very helpful for Quantitative aptitude section of CAT.

Profit and Loss questions will have some basic formulas and theory to learn. You can download the Profit and LossÂ Tricks/shortcuts for CAT PDF to solve this problems with ease. Speed and accuracy is also important while solving the questions of this topic.

Every year around 3-5 questions are asked in CAT Previous question papers. Take a free CAT mock test to take mock similar to actual CAT exam.

You can download the Profit and Loss Questions for CAT PDF or you can go through the details below.

Profit and Loss Questions for CAT:

Question 1:

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

Question 2:

Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every 6 months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?

A.Â Rs. 93,300

B.Â Rs. 93,200

C.Â Rs. 93,100

D.Â None of these

Question 3:

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%

Question 4:

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 III 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%

You can download the CAT Maths formulas PDF for other Quant formulas for CAT.

Question 5:

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

Answers and Solutions for Profit and Loss Questions for CAT PDF:

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

January 1, 1950 to December 31, 1959 is a period of 10 years or 20 half years.
The person X after 1st year gets Rs. 300Â in next year he gets Rs. 330Â and so on.
So his earning is in AP with 10 300+330+360+…

Similarly earning of Y is in AP with 20 terms 200+215+230+245…. .

So, the total earnings of X equals 12*(300+330+….10 terms) = 52200
The total earnings of Y equals 6*(200+215+230+…20 terms) = 41100

So, the total earnings of the two equals 52200+41100 = 93300

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%

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.

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.