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The length of each side of square A is increased by

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The length of each side of square A is increased by

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The length of each side of square A is increased by 100 percent to make square B. If the length of the side of square B is increased by 50 percent to make square C, by what percent is the area of square C greater than the sum of the areas of squares A and B?

A. 75%
B. 80%
C. 100%
D. 150%
E. 180%

The OA is the option B.

Experts, how can I solve this PS question. I got confused trying to set the equations.

Can you give me some help, please?

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GMAT/MBA Expert

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Hello vjesus12.

Let's take a look at your question.

Let's suppose that the length of each side of A is equal to 10cm. Then, the area of A is equal to 10*10=100. Now:

each side of square A is increased by 100 percent to make square B implies that the length of each side of B is equal to 10+10*100%=10+10=20cm. Hence, the area of B is equal to 20*20=400.

length of the side of square B is increased by 50 percent to make square C implies that the length of each side of C is equal to 20+20*50%=20+10=30cm. Hence, the area of C is equal to 30*30=900.

Finally, we have to compare the area of C with A+B.

A+B=500 -------> 100%
C=900 -----------> x% ?

x = 900*100/500 = 900/5 = 180.

It implies that the area of C is 80% greater than the area of A+B.

This is why the correct answer is the option B=80%.

I hope this answer can help you.

Feel free to ask me again if you have a doubt.

Regards.

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VJesus12 wrote:
The length of each side of square A is increased by 100 percent to make square B. If the length of the side of square B is increased by 50 percent to make square C, by what percent is the area of square C greater than the sum of the areas of squares A and B?

A. 75%
B. 80%
C. 100%
D. 150%
E. 180%
In any square:
Area = s².

Square A:
Let s = 1, implying that the area of A = 1² = 1.

Square B:
If a value increases by 100%, it DOUBLES.
Thus, each side of B is twice that of A.
Since each side of B = 2*1 = 2, the area of B = 2² = 4.

Square C:
Here, s = 50% more than each side of B = 2 + (1/2)(2) = 3.
Thus, the area of C = 3² = 9.

The area of square C is what percent greater than the sum of the areas of squares A and B?
Since C=9 and A+B = 1+4 = 5, we get:
Percent increase from 5 to 9 = Difference/Smaller * 100 = (9-5)/5 * 100 = 80.

The correct answer is B.

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VJesus12 wrote:
The length of each side of square A is increased by 100 percent to make square B. If the length of the side of square B is increased by 50 percent to make square C, by what percent is the area of square C greater than the sum of the areas of squares A and B?

A. 75%
B. 80%
C. 100%
D. 150%
E. 180%
We can let the length of each side of square A = 2, so the length of each side of B is 4, and the length of each side of square C is 1.5 x 4 = 6.

The sum of the areas of squares A and B is 4 + 16 = 20, and the area of C is 36. Let’s now use the percent change formula:

(36 - 20)/20 = 16/20 = 4/5 = 0.80 = 80%

Answer: B

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Scott Woodbury-Stewart
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scott@targettestprep.com



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