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Questions from MBA.com preptest 2

by pranavc » Wed Jul 16, 2008 7:42 pm
1. A certain jar contains only 'b' black marbles, 'w' white marbles, and 'r' red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen will be red greater than the probability that the marble chosen will be white?

(1) r/(b+w) > w/(b+r)

(2) b -w > r

2. If 'x' and 'y' are positive integers such that x=8y + 12, what is the greatest common divisor of 'x' and 'y'?

(1) x=12u, where 'u' is an integer
(2) y=12z, where 'z' is an integer

3. Store S sold a total of 90 copies of a certain book during the seven days of last week, and it sold different numbers of copies on any two of the days. If for the seven days store S sold the greatest number of copies on Saturday and the second greatest number of copies on Friday, did store S sell more than 11 copies on Friday?

(1) Last week Store S sold 8 copies of the book on Thursday.
(2) Last week Store S sold 38 copies of the book on Saturday.


4. If k is a positive integer, is 'k' the square of an integer?

(1) k is divisible by 4.
(2) k is divisible by exactly four different prime numbers.

5. Juan bought some paperback books that cost $8 each and some hardcover books that cost $25 each. If Juan bought more than 10 paperback books, how many hardcover books did he buy?

(1) The total cost of all the hardcover books that Juan bought was at least $150.
(2) The total cost of all the books that Juan bought was less than $260.

Any input on any or all of the above will be highly appreciated. Thanks in advance.
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by VP_Tatiana » Thu Jul 17, 2008 2:08 pm
I'll tackle this one problem at a time. Here's problem 1:

Starting by cross multiplying the equation in statement 1, we get:
(b+w)w < r(b+r)
bw + w2 < br + r2
bw-br < r2 - w2
b(w-r) <r> w. So, let's assume r > w. Dividing both sides by r-w yields
-b <r> - (r + w)

This statement makes sense. We know b, r, and w must all be positive, so of course b must be bigger than a negative number.

So, let's consider the situation where r < w. Then b <r> w... because otherwise it results in a statement that does not make sense. Statement 1 provides enough information for us to determine whether r > w... and thus that their is a higher probability of getting a red marble than a white one.

Considering statment 2 on its own, b > r+w does not tell us if r > w or not. So, it is not sufficient. Thus, the answer is A.

What strategy did I use here? Brute force. I wasn't sure off the bat if statement 1 gave me enough info, so I tried to manipulate the numbers into another form that might give me more information. When I got to the point that I had to make a decision whether r-w was positive or not, I saw that this was probably a "proof by assumption" kind of problem. For instance, if I suppose so and I get something impossible... then I know it is not true... and if I suppose not and I get something impossible... then I know it is true.
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by VP_Tatiana » Thu Jul 17, 2008 2:38 pm
Problem 2:

Statement 1 tells us that x is a multiple of 12. For that to be true, 8y must also be a multiple of 12. Since 8 = 4*2, y needs to be a multiple of 3 so that 8y = 4*3*2*something = 12*something. Basically, then, statement 1 tells us that y must be a multiple of 3.

However, we can have y=3 satisfy the condition (in which case greatest common divisor is 3), or we can have y=12 satisfy the condition (in which case gcd is 2). So, statement 1 is insufficient.

Statement 2 tells us that y is a multiple of 12. Since x = 8y + 12, we have x = 8*12*something + 12. Let's say g is the gcd of x. g has to divide 8*12*something, and it also has to divide 12. So, g has to be 12. Since we know y is a multiple of 12, 12 is the gcd of both x and y. Statement 2 is sufficient and the answer is B.
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by VP_Tatiana » Thu Jul 17, 2008 2:57 pm
Problem 3:

We know from the problem that:
M + Tu + W + Th + F + Sa + Su = 90

We also know that Th = 8. Let's try to concoct the scenario where the Friday and Saturday sales are as big as possible (later we will try to make one where they are as small as possible, because if we can get two answers for Friday sales, one greater than 11 and one less than or equal to, we show that statement 1 is not sufficient.) We then have 1 + 2 + 3 + 8 + F + sa + 4 = 90. Thus, F + sa = 72. We could have F = 11, Sa = 61, or F = 12, Sa = 60, for example. We actually don't even have to try to make F and Sa as small as possible by manipulating the sales on the other days; we've already come up with two contradicting examples to show Statement 1 is not sufficient.

From statment 2, we know Sa = 38. Thus, the sum of the other days sales is 52. Let's see if it would be possible for us to hit that number with Friday sales of 11 books. The biggest possible sales we could have for the week with Friday sales of 11 books would be 6 + 7 + 8 + 9 + 10 + 11 (remember, the other 5 days' sales must each be smaller, and each days' sales must be unique.) This would only yield sales of 51 books. So, Friday sales must be greater than 11 books. Thus, statement 2 is sufficient and the answer is B.
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by pranavc » Fri Jul 18, 2008 2:00 am
VP_Tatiana wrote: bw-br < r2 - w2
b(w-r) <r> w. So, let's assume r > w. Dividing both sides by r-w yields
-b <r> - (r + w)

This statement makes sense. We know b, r, and w must all be positive, so of course b must be bigger than a negative number..
OK. This is where you lost me. I don't really get the expression -b<r> - (r+w)

Following your train of thought, I reduced the inequality to

-b < (r+w).

What exactly does the <r> bit mean and where does it come from? I appreciate you taking the time to answer my question. It'd be even better if you could clarify this bit for me. Thanks in advance.
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by pranavc » Fri Jul 18, 2008 2:05 am
VP_Tatiana wrote:Problem 2:

Statement 1 tells us that x is a multiple of 12. For that to be true, 8y must also be a multiple of 12. Since 8 = 4*2, y needs to be a multiple of 3 so that 8y = 4*3*2*something = 12*something. Basically, then, statement 1 tells us that y must be a multiple of 3.

However, we can have y=3 satisfy the condition (in which case greatest common divisor is 3), or we can have y=12 satisfy the condition (in which case gcd is 2). So, statement 1 is insufficient.

Statement 2 tells us that y is a multiple of 12. Since x = 8y + 12, we have x = 8*12*something + 12. Let's say g is the gcd of x. g has to divide 8*12*something, and it also has to divide 12. So, g has to be 12. Since we know y is a multiple of 12, 12 is the gcd of both x and y. Statement 2 is sufficient and the answer is B.
Makes perfect sense!!! Thank you very much.
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by pranavc » Fri Jul 18, 2008 2:12 am
VP_Tatiana wrote:Problem 3:

We know from the problem that:
M + Tu + W + Th + F + Sa + Su = 90

We also know that Th = 8. Let's try to concoct the scenario where the Friday and Saturday sales are as big as possible (later we will try to make one where they are as small as possible, because if we can get two answers for Friday sales, one greater than 11 and one less than or equal to, we show that statement 1 is not sufficient.) We then have 1 + 2 + 3 + 8 + F + sa + 4 = 90. Thus, F + sa = 72. We could have F = 11, Sa = 61, or F = 12, Sa = 60, for example. We actually don't even have to try to make F and Sa as small as possible by manipulating the sales on the other days; we've already come up with two contradicting examples to show Statement 1 is not sufficient.

From statment 2, we know Sa = 38. Thus, the sum of the other days sales is 52. Let's see if it would be possible for us to hit that number with Friday sales of 11 books. The biggest possible sales we could have for the week with Friday sales of 11 books would be 6 + 7 + 8 + 9 + 10 + 11 (remember, the other 5 days' sales must each be smaller, and each days' sales must be unique.) This would only yield sales of 51 books. So, Friday sales must be greater than 11 books. Thus, statement 2 is sufficient and the answer is B.
Great approach!!! Thanks a lot.
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by VP_Tatiana » Sat Jul 19, 2008 8:13 pm
So sorry... the site kept thinking my < and > signs were HTML brackets. It took me about 10 tries to figure out I had to disable HTML. Here is it posted correctly. (I'm sorry if this spammed anyone with an RSS reader!) I also clarified some parts further. Here's problem 1 again:

Starting by cross multiplying the equation in statement 1, we get:

(b+w)w < r(b+r)
bw + w2 < br + r2
bw-br < r2 - w2
b(w-r) < (r +w)(r-w). So, let's assume r > w. Dividing both sides by r-w yields
-b < (r + w) or
b>-(r+w)

This statement makes sense. We know b, r, and w must all be positive, so of course b must be bigger than a negative number.

So, let's consider the situation where r < w. Then -b < r+w. Thus, it is not possible for r<w... because otherwise it results in a statement that does not make sense. Statement 1 provides enough information for us to determine whether r > w... and thus that there is a higher probability of getting a red marble than a white one.

Considering statment 2 on its own, b > r+w does not tell us if r > w or not. So, it is not sufficient. Thus, the answer is A.

What strategy did I use here? Brute force. I wasn't sure off the bat if statement 1 gave me enough info, so I tried to manipulate the numbers into another form that might give me more information. When I got to the point that I had to make a decision whether r-w was positive or not, I saw that this was probably a "proof by assumption" kind of problem. For instance, if I suppose so and I get something impossible... then I know it is not true... and if I suppose not and I get something impossible... then I know it is true.
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by VP_Tatiana » Sat Jul 19, 2008 8:39 pm
4. If k is a positive integer, is 'k' the square of an integer?

(1) k is divisible by 4.
(2) k is divisible by exactly four different prime numbers.

Problem 4:

Considering statement 1, k could be 8 (not the sq of an integer) or 16 (the sq of an integer.) Thus, statement 1 does not give us sufficient information to determine if k is the square of an integer.

Considering statement 2, k could be 2*3*5*7=210 which is not the sq of an integer. Or, k could be (2*3*5*7)^2 = 210^2, which is the sq of an integer. Thus, statement 2 does not give us sufficient information.

When we have to satisfy both statements (divisible by 4 and divisible by exactly 4 primes), we could have:
k = 2*2*3*5*7 = 410 (not a sq)
or
k = (2*3*5*7)^2 = 210^2 (is a sq)

So, the answer is E.
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by VP_Tatiana » Sat Jul 19, 2008 9:02 pm
5. Juan bought some paperback books that cost $8 each and some hardcover books that cost $25 each. If Juan bought more than 10 paperback books, how many hardcover books did he buy?

(1) The total cost of all the hardcover books that Juan bought was at least $150.
(2) The total cost of all the books that Juan bought was less than $260.

Considering statement 1, we know Juan must have bought at least 6 hardcover books. However, he could have bought more than that, so statement 1 alone is not sufficient.

Considering statement 2, and the fact that Juan spent $88 or more on paperback books (he bought MORE than 10), Juan could have bought anywhere between 0 and 6 hardcover books. So, statement 2 alone is not sufficient.

Putting the two statements together, we know Juan bought at least 6 hardcover books and anywhere between 0 and 6 hardcover books... so he must have bought exactly 6 hardcover books. Thus, both statements together tell us how many hardcover books Juan bought (and thus how much he spent on them.) Since both statements together are sufficient, the answer is C.
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by nepaligirl » Tue Jul 22, 2008 9:30 pm
VP_Tatiana wrote:So sorry... the site kept thinking my < and > signs were HTML brackets. It took me about 10 tries to figure out I had to disable HTML. Here is it posted correctly. (I'm sorry if this spammed anyone with an RSS reader!) I also clarified some parts further. Here's problem 1 again:

Starting by cross multiplying the equation in statement 1, we get:

(b+w)w < r(b+r)
bw + w2 < br + r2
bw-br < r2 - w2
b(w-r) < (r +w)(r-w). So, let's assume r > w. Dividing both sides by r-w yields
-b < (r + w) or
b>-(r+w)

This statement makes sense. We know b, r, and w must all be positive, so of course b must be bigger than a negative number.

So, let's consider the situation where r < w. Then -b < r+w. Thus, it is not possible for r<w... because otherwise it results in a statement that does not make sense. Statement 1 provides enough information for us to determine whether r > w... and thus that there is a higher probability of getting a red marble than a white one.

Considering statment 2 on its own, b > r+w does not tell us if r > w or not. So, it is not sufficient. Thus, the answer is A.

What strategy did I use here? Brute force. I wasn't sure off the bat if statement 1 gave me enough info, so I tried to manipulate the numbers into another form that might give me more information. When I got to the point that I had to make a decision whether r-w was positive or not, I saw that this was probably a "proof by assumption" kind of problem. For instance, if I suppose so and I get something impossible... then I know it is not true... and if I suppose not and I get something impossible... then I know it is true.

i didn't get the above line <in bold>. could you please explain? thanks in advance.
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Re: Questions from MBA.com preptest 2

by Ian Stewart » Wed Jul 23, 2008 12:48 am
pranavc wrote:1. A certain jar contains only 'b' black marbles, 'w' white marbles, and 'r' red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen will be red greater than the probability that the marble chosen will be white?

(1) r/(b+w) > w/(b+r)

(2) b -w > r
There are a few ways to see why 1) is sufficient without using any algebra. r/(b+w) just represents the ratio of red marbles to the other marbles in the jar. If the ratio of the red marbles to the rest in the jar is greater than the ratio of white marbles to the rest in the jar, you certainly have more red marbles than white marbles.

Alternatively, you can look at the inequality:

r/(b+w) > w/(b+r)

Bearing in mind that all values are positive here, could w be larger than r? If w were larger than r, the numerator on the left side of the inequality would be smaller than the numerator on the right, and the denominator on the left would be larger than the denominator on the right. So the inequality can't possibly be true if w > r; and since clearly r and w cannot be equal, we must have r > w.

Or, of course, you can approach the problem algebraically, as was done in one of the posts above.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by kanha81 » Fri Mar 27, 2009 9:06 am
VP_Tatiana wrote:Problem 2:
Statement 2 tells us that y is a multiple of 12. Since x = 8y + 12, we have x = 8*12*something + 12. Let's say g is the gcd of x. g has to divide 8*12*something, and it also has to divide 12. So, g has to be 12. Since we know y is a multiple of 12, 12 is the gcd of both x and y. Statement 2 is sufficient and the answer is B.
For stmt-2: Why does gcd (x) <> 4?
if gcd (x) = 4, then 4 divides 8*12*something + 12
so, then gcd(x,y) = gcd(4,12) = 4

Grateful, if anyone can help me understand this problem better! I understood and agree with the stmt-1. have doubts abt stmt-2

thanks a bunch!
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