Profit and Loss Practice Problems: CAT Quant Sample Question and Answer

cat quant profit loss

Profit & loss questions are common on the CAT test. These questions are very appropriate for a b-school entrance exam, since they deal with business operations.

This question type will always be presented as a story problem. You’ll read about some kind of business transaction, whether it’s the sale of an individual item, or a large operation such as an investment and a return.

Sometimes you’ll need to calculate the profit or loss generated by the business transaction. In that case, your end solution will be based on data such as cost, revenue, the percentage of profit or loss from one item in a set, etc….

At other times, you’ll be told exactly what the profit or loss was at the end of the business transaction. When you already have the profit or loss, the CAT will ask you to calculate other business-related data. You may need to figure out the original cost of an item, its retail markup for sale, or other similar information, based on the given profit or loss.

Below is an example CAT Quant profit & loss question. While some questions on the exam deal only with profit or only with loss, this question deals with both. The question will be followed by an answer explanation. And we’ll end this article with a tutorial on how to find and plug in variables for this question type.

CAT Quant: Profit & Loss Question

A grocery store purchases a bag of apples and a bag of oranges for a combined price of Rs 70. The two bags are then sold together in a fruit bundle. The apples are sold at a 40% profit and the oranges are sold at a 20% loss. If the grocery store makes a 16 Rs profit on each fruit bundle, what is the store’s cost for the apples?

A) 25
B) 30
C) 50
D) none of the above

Answer and Explanation for the Profit & Loss Question

The answer to this question is C) 50. How do you get to that answer? Read on.

You can solve this problem by coming up with two variables and plugging them into two equations.

First, the variables:
X = the original cost of the bag of apples, and Y = the original cost of the bag of oranges.

Next, the equations:
X + Y = 70.

This makes perfect sense, because 70 is the original cost price (or CP) the store paid for the apples and oranges together. That’s easy enough, right?

The other equation is 1.4X + 0.8Y = 86.

Where this equation comes from is a bit less obvious. But have no fear! I’ll break it down for you.

1.4X represents the selling price of the bag of apples. Since the apples get the store a 40% profit, their selling price must be 100% of the original cost, plus 40% more. This is 140% of the original cost of the apples, or 1.4 times variable X. (Note that 1.4 is the decimal form of 140%.)

Similarly, the oranges represent a proportion of the original cost. As you’ll recall, the bag of oranges was sold at a 20% loss. In other words, the oranges are sold for 20% less than the original cost, or 80% of the original cost. So you want to multiply the original cost of the oranges (variable Y) by 0.8, the decimal expression for 80%.

So the selling price (or SP) for the bag of apples is 1.4X and the selling price for the bag of oranges is 0.8Y. Together, these must add up to the total selling price.

And we know what that total is. It’s 86 Rs. How do we know the two bags were purchased for 86 Rs by a customer? Easy! It’s because the story problem tells us that the grocery store makes 16 Rs in profit from a pair of fruit bags that it originally paid 70 Rs for. That means they sell the bags for 16 Rs more than 70 Rs. 70 Rs + 16 Rs = 86 Rs. Got it?

OK, now that we know where both equations come from, let’s look at the equations again….

X + Y = 70
1.4X + 0.8Y = 86

And let’s solve the equations by solving for X. (Remember, we need to know the store cost for the bag of apples, and that cost is represented by variable X.)

To do this, first we need to solve for Y in the first equation. That way we can get rid of the Y in the second equation. Then we’ll just have X to solve for. Here’s how that works:

  • 1) X + Y = 70 >>>>>> Y = 70 – X (Isolate Y by subtracting X from both sides.)
  • 2) 1.4X + 0.8Y = 86 >>>>>> 1.4X + 0.86(70-X) = 86 (Replace Y with the Y-value from the first equation.)
  • 3) 1.4X + 56 – 0.8X = 86 (In the parenthesis from step 2, mutliply 0.86 by 70 and by -X.)
  • 4) 1.4X – 0.8X = 30 (Get rid of the 56 by subtracting it from both sides of the equation.)
  • 5) 0.6X = 30 (Subtract the two X values to get a single X value.)
  • 6) X = 50 (To isolate X and solve for it, divide both sides by 0.6.)

How to Find the Variables in Any CAT Quant Profit and Loss Question

As you saw above, finding your variables is key to coming up with an equation and solution to any profit/loss problem. I briefly mentioned two variables that always come up in this problem type: cost price (CP) and selling price (SP).

There are a few equations you should be aware of, related to the CP and SP variables.

First, SP is CP + profit. Second, logically this must mean that CP is SP – profit. Knowing these two equations, you can find the profit if you have the SP and CP, you can find the SP if you have the CP and the profit, and you can find the CP of you have the SP and profit.

Logically, then, if follows that if the SP is lower than the CP, then CP – SP = loss. Again, if you have any two of these three variables– CP, SP, or loss– you can find the third variable.

Finally, remember that % of profit = [(profit or loss)*(100)]/CP.

So in the problem above, the apples and oranges were sold together at a 16 Rs profit. This comes out to (16*100)/CP. The CP for the apples and oranges together was 70 Rs, which means the percentage of combined profit for the two bags of fruit was 1600/70. This comes out to 22.86, or roughly a 23% profit. (Conversely, a 16 Rs loss from a cost price of 70 Rs would come out to a 23% loss, using the same formula.)

Sometimes the cost price or sell price seem less clear, because the profit & loss problem isn’t dealing with retail sales. Suppose, for example, that a problem mentions an 70,000 Rs, with a profit of 34,000 Rs from that investment after one year.

Here, there is technically no sale; we’re just dealing with the growth of an investment. But the investment is still the initial cost, so CP is 70,000 Rs. And the growth functions mathematically as sales price. So CP + 34,000 = SP in this case.

Anytime you look at a profit and loss problem in CAT Quantitative, you will be able to find the values for CP, SP, and profit, even when retail sales aren’t involved. So always keep an eye out for the variables; they’re the key to cracking this problem type.

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