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GMAT Quantitative: Sample Problems II

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» GMAT Sample Problems and Solutions: Part II

To help you crack the Quantitative Section, here are some sample problems and their solutions, step by step. These problems are based on real GMAT questions, so they may be the ones you'll get stuck on.

Before you begin, be sure to review the Quantitative Section lessons and the Basics of Algebra lessons.

Think you know your stuff? Practice using the Question Bank or the GMAT Tests.


» Question 1:

One day a car rental agency rented 2/3 of its cars, including 3/5 of its cars with CD players. if 3/4 of its cars have CD players, what percentage of the cars that were not rented had CD players?

Answer: This looks like a tough question. So what do we do? We cheat a bit. Instead of working with annoying fractions and percentages, let's just make up real numbers and work with them. There's nothing wrong with that and it will help you beat the GMAT.

So, say we have 120 cars. This means we rented 80 and didn't rent 40 cars. Now, in total, 90 of the cars have CD players, and we rented out 3/5 of them which is 54 cars. We're left with 36 CD cars not rented out of 40. 36 out of 40 is 90%.

See? Working with real numbers got us the answer easy- it's 90%.

Question 2: If -2<a<11 and 3<b<12, then which of the following is NOT true?
(A) 1<a+b<23
(B) -14<a-b<8
(C) -7<b-a<14
(D) 1<b+a<23
(E) -24 < ab < 132

Answer: Take a quick scan of the answers first. You should notice a few things. First, a and d are the same, as a+b=b+a. So we can cancel them out right away. Also, a-b=-(b-a), but b and c are not the same numbers. Thus, either b or c has to be wrong. Great, we've already narrowed it down a lot without thinking much.

Which one is it though? A simple check using the end points is sufficient to find out the answer. Let's try b first. We can get to -2-3=-5, 11-12=-1, 11-3=8 or -2-12=-14 using all the endpoints. Our range is clearly -14 to 8. So b is right. Without doing anything else, we already the know the answer is C.

Question 3: In this figure, we have a circle inside a square. What's the ratio of the area of the circle to the area of the square?

Photo 3909


(A) \frac{\pi}{2}
(B) \frac{4}{\pi}
(C) \frac{\pi}{3}
(D) \frac{\pi}{4}
(E) \frac{\pi}{6}

Answer: First of all, take a look at the answers. Keep in mind thet \pi is approximately 3.14. We're looking for a ratio: the area of the circle to the square. Clearly, the circle is smaller than the square, so the ratio has to be a value less than 1. Thus, answer choices A, B and C can't possible be right. It's either D or E.

Now, what do we do? Let's plug in values again. We know that for the area of the circle we need a radius, so let's say the radius is 2. Then the area of the circle is 4\pi. The square on the other hand would have a side that's twice the radius- 4, and so its area would be 16.

Now we just divide for the ratio: \frac{4\pi}{16}=\frac{\pi}{4}. Our answer is just D.

That's it for now! Check back for more sample problems with solutions.



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  1. samim saidSun, 28 Sep 2008 05:29:39 -0000 ( Link )

    that was nice

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  2. Sanchit kshirsagar saidMon, 27 Oct 2008 05:30:04 -0000 ( Link )

    Sir,

    That was pretty explanatory but I am not convinced with the first question.

    Allow me to explain my point.

    I think we must know the principle behind solving the first problem. The trick you used is still nice but numbers might not be same as this always. YOu went by LCM of three numbers 3,4,5,2 and decided to go by 120 as a total number of cars but what if the numbers were 2/7, 3/8, 1/9, .... LCM witll be not a good figure to work with.

    Please let me know the principle used in 1st question

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  3. oLahav saidTue, 28 Oct 2008 14:29:19 -0000 ( Link )

    I believe that this type of method will always work, even with uglier fraction. You just need to identify which “magic number” will make the calculation nicer. For example, if you have fractions like 2/7 and 5/9, then 63 will be a good thing to use since it has a factor of both 7 and 9. Using 120 will not be of much use this time though… so if you go with a number that fits, just by looking at the denominator, this method is useful in most cases.

    Of course, if you find that working the actual numbers and fractions works better for you, don’t waste your time on examples and just do the straight calculation. I hope this helps.

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  4. abhi_221984 saidWed, 24 Dec 2008 09:52:30 -0000 ( Link )

    that wa nice but ans 2 was not satisfactorily understood.

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  5. coolvenky9 saidTue, 03 Feb 2009 08:28:02 -0000 ( Link )

    hey i think the second one is wrong here how can u take -2 to for a and 3 for b when it clearly says a is greater than -2 and B is greater than 3. pls let me know

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  6. oLahav saidTue, 03 Feb 2009 14:46:39 -0000 ( Link )

    We’re trying to check the boundary. We know that the boundary for a is – 2, since we’re given that. It’s true that a could never be – 2, since we have a strict inequality, but just to look at the range it’s ok to use – 2 as long as we keep the strict inequality.

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