Are you desperately looking for practice questions on the GMAT club forum? How many questions did you find for the topic you were looking for? Are they any good?

Well, if you are scavenging for good practice questions you’ve just hit a gold mine! Problem solving comprises roughly 50% of the total questions in the GMAT quant section and in this blog, we will show you how to solve GMAT profit and loss questions through some quality practice questions and their detailed explanations. Let’s go!

## GMAT Profit and Loss Problem 1.

By selling 5 articles for INR 15, a man makes a profit of 20%. Find his gain or loss percentage if he sells 8 articles for INR 18.4?

**Solution:** Questions of this type normally appear as part of a more complex problem in an exam like the GMAT.

Remember, such a question should be solved by you as soon as you finish reading the question by solving-while reading process, as follows.

By selling 5 articles for INR 15, a man makes a profit of 20% – SP = 3.

Hence, CP = 2.5, if he sells 8 articles for INR 18.4 – SP = 2.3.

Hence percentage loss = 8%.

## GMAT Profit and Loss Problem 2.

RFO Tripathi bought some oranges in Nagpur for 32 Rupees. He has to sell it off in Yeotmal. He is able to sell off all the oranges in Yeotmal and on reflection finds that he has made a profit equal to the cost price of 40 oranges. How many oranges did RFO Tripathi buy?

**Solution: ** Suppose we take the number of oranges bought as x. Then, the cost price per orange would be Rupees 32/x, and his profit would be

40 x 32/x = 1280/x.

To solve for x, we need to equate this value with some value on the other side of the equation. But, we have no information provided here to find out the value of the variable x. Hence, we cannot solve this

## GMAT Profit and Loss Problem 3.

A dishonest businessman professes to sell his articles at cost price but he uses false weights with which he cheats by 10% while buying and by 10% while selling. Find his percentage profit.

**Solution:** Assume that the businessman buys and sells 1 kg of items. While buying the cheats by 10%, which means that when he buys 1 kg he actually takes 1100 grams. Similarly, he cheats by 10% while selling that is, he gives only 900 grams when he sells a kilogram. Also, it must be understood that since he purportedly buys and sells the same amount of goods and he is trading at the same price while buying and selling, money is already equated in this case.

Hence, we can directly use:

% Profit = (Goods left x 100/Goods sold)

= 200 x 100/900 = 22.22%

Note that you should not need to do this calculation since this value comes from the fraction to percentage conversion table.

If you are looking at a 680+ plus score in quantitative ability you should be able to come to this solution under 90 seconds inclusive of problem reading time. And the calculation should go like this:

Money is equated – % profit = 2/9 = 22.22%

Have you taken the GMAT before?

**GMAT Profit and Loss Problem 4.**

In order to maximize its profits, AMS Corporate defines a function. Its unit sales price is Rs. 700 and the function representing the cost of production = 300 + 2p^{2}, where p is the total units produced or sold. Find the most profitable production level. Assume that everything produced is necessarily sold.

**Solution: **The function for profit is a combination of revenue and costs. It is given by Profit = Revenue – Costs = 700p – (300 + 2p^{2}) = – 2p^{2} + 700p – 300.

In order to find the maxima or minima of any quadratic function, we differentiate it and equate the differentiated equation to zero.

Thus, the differentiated profit function is -4p + 700 = 0 = p = 175. This value of production will yield the maximum profits in this case.

## GMAT Profit and Loss Problem 5. *(based on Question 4)*

What is the value of the maximum profits for AMS Corporate?

**Solution:** For this, continuing from the previous question’s we just put the value of p = 175 in the equation for profit. Thus, substitute p = 175 in the equation. Profit – 2p^{2} + 700p – 300 and get the answer.

## GMAT Profit and Loss Problem 6.

A shopkeeper allows a rebate of 25% to the buyer. He sells only smuggled goods and as a bribe, pays 10% of the cost of the article. If his cost price is 2500, then find what should be the marked price if he to make a profit of 9.09%.

**Solution:** Use solving-while-reading as follows:

Cost Price (= 2500) + Bribe (= 10% of cost of article = 250)

= Total cost to the shopkeeper (2500 + 250 = 2750).

He wants a profit of 9.09 percent on this value – Using fraction to percentage change table we get 2750 + 9.09% of 2750 = 2750 + 250 = Rs. 3000.

But this Rs.3000 is got after a rebate of 25%. Since we do not have the value of the marked price on which the 25% rebate is to be calculated, it would be a good idea to work reverse through the percentage change graphic:

Going from the marked price to Rs. 3000 requires a 25% rebate. Hence the reverse process will be got by increasing Rs. 3000 by 33.33% and getting Rs. 4000.

## GMAT Profit and Loss Problem 7.

A man sells three articles, one at a loss of 10%, another at a profit of 20%, and the third one at a loss of 25%. If the selling price of all the three is the same, find by how much percent is their average CP lower than or higher than their SP.

**Solution:** We have to calculate: (average CP – average SP)/average SP.

Here, the selling price is equal in all three cases. Since the maximum number of calculations is associated with the SP, we assume it to be 100. This gives us an average SP of 100 for the three articles. Then, the first article will be sold at 111.11, the second at 83.33, and the third at 133.33. (The student is advised to be fluent at these calculations)

Further, the CP of the three articles is 111.11 + 83.33 + 133.33 = 327.77.

The average CP of the three articles is 327.77/3

= 109.2566.

Hence, (average CP- average SP)/average SP = 9.2566% higher.

Any other process adopted for this problem is likely to require much more effort and time.

Note that it is always convenient to solve questions involving percentages by using the number 100. The reason for this is that it **reduces the amount of effort required in calculating the solution**. Hence, it goes without saying that the variable to be fixed at 100 should be the one with the highest number of calculations associated with it. Another thumb rule for this is that the variable to be fixed at 100 should be the one with which the most difficult calculation set is associated.

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