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I) Since y=x(x+1)/5 is an integer, x(x+1) must be divisible by 5 to cancel out the 5 in the denominator. x is a prime number greater than 5, so it is definitely NOT divisible by 5. It follows that (x+1) must be divisible by 5. You're looking for a prime number x that is 1 less than a multiple of 5: x=19 fits the bill, since x+1=20 is divisible by 5.
In that case, y = 19·20/5 = 19·4. This is not divisible by 5, so y doesn't HAVE to be divisible by 5.
II) y=x(x+1)/5 is definitely divisible by x, but is y divisible by 2? Remember that x(x+1) is the product of two consecutive integers. x is a prime number greater than 5, therefore it is odd, and thus x+1 is even. Since x+1 is even, 5y is also even, and therefore y is even, i.e. divisible by 2. It follows that y is divisible by 2x.
III) With the same plug in that was used for (I) you can see that y=19·4 is not divisible by x+1=20. Since you have found a single example where y is not divisible by x+1, it follows that y does not HAVE to be divisible by x+1.
Is this the best way to do solve this? I'm not so sure I would have reached those exact conclusions in my brain to be able to get the right answer on this one....
Re: If x is a prime number greater than 5, y is a positive
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26 Oct 2012, 04:02
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6
If x is a prime number greater than 5, y is a positive integer, and 5y=x^2+x, then y must be divisible by which of the following?
I. 5 II. 2x III. x+1
A. I only B. II only C. III only D. I and II only E. II and III only
You can solve this question even not analyzing option II at all.
Since \(5y=x(x+1)\), then \(x+1\) must be a multiple of 5, thus the units digit of \(x\) must be 9 (it cannot be 4, because in this case \(x\) would be even and thus not a prime number).
Now, if \(x=19\), then neither I nor III is true. The only option that does not contains I or III is B. So, B must be correct.
Answer: B.
If you interested why II must be true consider this: since \(x\) is a prime number greater than 5, then it's odd --> \(5y=x(x+1)=odd(odd+1)=odd*even=even\) --> \(5y=even\) --> \(y=even\).
Also, \(\frac{y}{x}=\frac{x+1}{5}\). we know that \(x+1\) is a multiple of 5, thus \(\frac{y}{x}=\frac{x+1}{5}=integer\) --> \(y\) is a multiple of \(x\). And since \(y\) is even, then \(y\) is a multiple of \(2x\).
Re: If x is a prime number greater than 5, y is a positive int..
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25 Oct 2012, 22:28
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anon1 wrote:
If x is a prime number greater than 5, y is a positive integer, and 5y=x2+x, then y must be divisible by which of the following?
I) 5
II) 2x
III) x+1
I only II only III only I and II only II and III only
I'm killing myself.... I got this wrong, and the Master GMAT program suggests that this is the correct methodology for solving this...
I) Since y=x(x+1)/5 is an integer, x(x+1) must be divisible by 5 to cancel out the 5 in the denominator. x is a prime number greater than 5, so it is definitely NOT divisible by 5. It follows that (x+1) must be divisible by 5. You're looking for a prime number x that is 1 less than a multiple of 5: x=19 fits the bill, since x+1=20 is divisible by 5.
In that case, y = 19·20/5 = 19·4. This is not divisible by 5, so y doesn't HAVE to be divisible by 5.
II) y=x(x+1)/5 is definitely divisible by x, but is y divisible by 2? Remember that x(x+1) is the product of two consecutive integers. x is a prime number greater than 5, therefore it is odd, and thus x+1 is even. Since x+1 is even, 5y is also even, and therefore y is even, i.e. divisible by 2. It follows that y is divisible by 2x.
III) With the same plug in that was used for (I) you can see that y=19·4 is not divisible by x+1=20. Since you have found a single example where y is not divisible by x+1, it follows that y does not HAVE to be divisible by x+1.
Is this the best way to do solve this? I'm not so sure I would have reached those exact conclusions in my brain to be able to get the right answer on this one....
Any insight would help immensely.......
We can try plugging in numbers. But since this is a "must be true" question, I would recommend that only as a last measure. We need to find an "x" such that \(x^2 + x\) is divisible by 5. We know x is prime and greater than 5. So we can check it out with 7,11,13,17,19. Fortunately, in this case, 19 suits the picture. Substituting 19, we can find the answer that only the second expression is a factor of y. It would also help to memorize the squares of the first ten prime numbers to save time on calculation.
Kudos Please... If my post helped.
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I) Since y=x(x+1)/5 is an integer, x(x+1) must be divisible by 5 to cancel out the 5 in the denominator. x is a prime number greater than 5, so it is definitely NOT divisible by 5. It follows that (x+1) must be divisible by 5. You're looking for a prime number x that is 1 less than a multiple of 5: x=19 fits the bill, since x+1=20 is divisible by 5.
In that case, y = 19·20/5 = 19·4. This is not divisible by 5, so y doesn't HAVE to be divisible by 5.
II) y=x(x+1)/5 is definitely divisible by x, but is y divisible by 2? Remember that x(x+1) is the product of two consecutive integers. x is a prime number greater than 5, therefore it is odd, and thus x+1 is even. Since x+1 is even, 5y is also even, and therefore y is even, i.e. divisible by 2. It follows that y is divisible by 2x.
III) With the same plug in that was used for (I) you can see that y=19·4 is not divisible by x+1=20. Since you have found a single example where y is not divisible by x+1, it follows that y does not HAVE to be divisible by x+1.
Is this the best way to do solve this? I'm not so sure I would have reached those exact conclusions in my brain to be able to get the right answer on this one....
Any insight would help immensely.......
Since \(x\) is a prime greater than 5, \(x\) cannot be divisible by 5. It follows that \(x+1\) must be divisible by 5, which means \(x+1=5n\) for some positive \(n\). Then \(x=5n-1\) must be odd (a prime that is greater than 5), so \(n\) must be even, which means \(n=2k\) for some positive integer \(k\). From \(5y=(5n-1)(5\cdot{2k})\) it follow that y is divisible \(2(5n-1)=2x\). Thus II is TRUE.
I isn't necessarily true: for \(x = 19, \,\,5y = 19\cdot{20}\) and \(y=19\cdot{4}\) is not divisible by 5. Neither is III true for \(x=19.\)
Answer B.
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I) Since y=x(x+1)/5 is an integer, x(x+1) must be divisible by 5 to cancel out the 5 in the denominator. x is a prime number greater than 5, so it is definitely NOT divisible by 5. It follows that (x+1) must be divisible by 5. You're looking for a prime number x that is 1 less than a multiple of 5: x=19 fits the bill, since x+1=20 is divisible by 5.
In that case, y = 19·20/5 = 19·4. This is not divisible by 5, so y doesn't HAVE to be divisible by 5.
II) y=x(x+1)/5 is definitely divisible by x, but is y divisible by 2? Remember that x(x+1) is the product of two consecutive integers. x is a prime number greater than 5, therefore it is odd, and thus x+1 is even. Since x+1 is even, 5y is also even, and therefore y is even, i.e. divisible by 2. It follows that y is divisible by 2x.
III) With the same plug in that was used for (I) you can see that y=19·4 is not divisible by x+1=20. Since you have found a single example where y is not divisible by x+1, it follows that y does not HAVE to be divisible by x+1.
Is this the best way to do solve this? I'm not so sure I would have reached those exact conclusions in my brain to be able to get the right answer on this one....
Any insight would help immensely.......
\(5y=x^2+x=x(x+1)\), and \(x\) and \(y\) are positive integers, \(x>5\). Then \(x+1>5\), therefore necessarily \(x+1\) is a factor of \(y\). Thus III is TRUE.
Since \(x\) is a prime greater than 5, \(x\) cannot be divisible by 5. It follows that \(x+1\) must be divisible by 5, which means \(x+1=5n\) for some positive \(n\). Then \(x=5n-1\) must be odd (a prime that is greater than 5), so \(n\) must be even, which means \(n=2k\) for some positive integer \(k\). From \(5y=(5n-1)(5\cdot{2k})\) it follow that y is divisible \(2(5n-1)=2x\). Thus II is also TRUE.
I isn't necessarily true: for \(x = 19, \,\,5y = 19\cdot{20}\) and \(y=19\cdot{4}\) is not divisible by 5.
Answer E.
Answer to the question is B, not E. Consider x=19.
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I) Since y=x(x+1)/5 is an integer, x(x+1) must be divisible by 5 to cancel out the 5 in the denominator. x is a prime number greater than 5, so it is definitely NOT divisible by 5. It follows that (x+1) must be divisible by 5. You're looking for a prime number x that is 1 less than a multiple of 5: x=19 fits the bill, since x+1=20 is divisible by 5.
In that case, y = 19·20/5 = 19·4. This is not divisible by 5, so y doesn't HAVE to be divisible by 5.
II) y=x(x+1)/5 is definitely divisible by x, but is y divisible by 2? Remember that x(x+1) is the product of two consecutive integers. x is a prime number greater than 5, therefore it is odd, and thus x+1 is even. Since x+1 is even, 5y is also even, and therefore y is even, i.e. divisible by 2. It follows that y is divisible by 2x.
III) With the same plug in that was used for (I) you can see that y=19·4 is not divisible by x+1=20. Since you have found a single example where y is not divisible by x+1, it follows that y does not HAVE to be divisible by x+1.
Is this the best way to do solve this? I'm not so sure I would have reached those exact conclusions in my brain to be able to get the right answer on this one....
Any insight would help immensely.......
\(5y=x^2+x=x(x+1)\), and \(x\) and \(y\) are positive integers, \(x>5\). Then \(x+1>5\), therefore necessarily \(x+1\) is a factor of \(y\). Thus III is TRUE.
Since \(x\) is a prime greater than 5, \(x\) cannot be divisible by 5. It follows that \(x+1\) must be divisible by 5, which means \(x+1=5n\) for some positive \(n\). Then \(x=5n-1\) must be odd (a prime that is greater than 5), so \(n\) must be even, which means \(n=2k\) for some positive integer \(k\). From \(5y=(5n-1)(5\cdot{2k})\) it follow that y is divisible \(2(5n-1)=2x\). Thus II is also TRUE.
I isn't necessarily true: for \(x = 19, \,\,5y = 19\cdot{20}\) and \(y=19\cdot{4}\) is not divisible by 5.
Answer E.
Answer to the question is B, not E. Consider x=19.
I'm missing something !! Why isn't E an option. y = x(x + 1)/5 .. Clearly x+1 is a factor of y. Even if we take x = 19 --> y is still divisible by x +1 ... Can someone please infuse a dose of intelligence to my brain ... what's the flaw in my logic
I) Since y=x(x+1)/5 is an integer, x(x+1) must be divisible by 5 to cancel out the 5 in the denominator. x is a prime number greater than 5, so it is definitely NOT divisible by 5. It follows that (x+1) must be divisible by 5. You're looking for a prime number x that is 1 less than a multiple of 5: x=19 fits the bill, since x+1=20 is divisible by 5.
In that case, y = 19·20/5 = 19·4. This is not divisible by 5, so y doesn't HAVE to be divisible by 5.
II) y=x(x+1)/5 is definitely divisible by x, but is y divisible by 2? Remember that x(x+1) is the product of two consecutive integers. x is a prime number greater than 5, therefore it is odd, and thus x+1 is even. Since x+1 is even, 5y is also even, and therefore y is even, i.e. divisible by 2. It follows that y is divisible by 2x.
III) With the same plug in that was used for (I) you can see that y=19·4 is not divisible by x+1=20. Since you have found a single example where y is not divisible by x+1, it follows that y does not HAVE to be divisible by x+1.
Is this the best way to do solve this? I'm not so sure I would have reached those exact conclusions in my brain to be able to get the right answer on this one....
Any insight would help immensely.......
\(5y=x^2+x=x(x+1)\), and \(x\) and \(y\) are positive integers, \(x>5\). Then \(x+1>5\), therefore necessarily \(x+1\) is a factor of \(y\). Thus III is TRUE.
Since \(x\) is a prime greater than 5, \(x\) cannot be divisible by 5. It follows that \(x+1\) must be divisible by 5, which means \(x+1=5n\) for some positive \(n\). Then \(x=5n-1\) must be odd (a prime that is greater than 5), so \(n\) must be even, which means \(n=2k\) for some positive integer \(k\). From \(5y=(5n-1)(5\cdot{2k})\) it follow that y is divisible \(2(5n-1)=2x\). Thus II is also TRUE.
I isn't necessarily true: for \(x = 19, \,\,5y = 19\cdot{20}\) and \(y=19\cdot{4}\) is not divisible by 5.
Answer E.
Answer to the question is B, not E. Consider x=19.
I'm missing something !! Why isn't E an option. y = x(x + 1)/5 .. Clearly x+1 is a factor of y. Even if we take x = 19 --> y is still divisible by x +1 ... Can someone please infuse a dose of intelligence to my brain ... what's the flaw in my logic
I erased my wrong answer quoted by Bunuel.
See my original corrected post.
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PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: If x is a prime number greater than 5, y is a positive
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17 Jun 2013, 08:49
Bunuel wrote:
If x is a prime number greater than 5, y is a positive integer, and 5y=x^2+x, then y must be divisible by which of the following?
I. 5 II. 2x III. x+1
A. I only B. II only C. III only D. I and II only E. II and III only
You can solve this question even not analyzing option II at all.
Since \(5y=x(x+1)\), then \(x+1\) must be a multiple of 5, thus the units digit of \(x\) must be 9 (it cannot be 4, because in this case \(x\) would be even and thus not a prime number).
Now, if \(x=19\), then neither I nor III is true. The only option that does not contains I or III is B. So, B must be correct.
Answer: B.
If you interested why II must be true consider this: since \(x\) is a prime number greater than 5, then it's odd --> \(5y=x(x+1)=odd(odd+1)=odd*even=even\) --> \(5y=even\) --> \(y=even\).
Also, \(\frac{y}{x}=\frac{x+1}{5}\). we know that \(x+1\) is a multiple of 5, thus \(\frac{y}{x}=\frac{x+1}{5}=integer\) --> \(y\) is a multiple of \(x\). And since \(y\) is even, then \(y\) is a multiple of \(2x\).
Hope it helps.
As per the above highlighted part-how can we say that y is multiple of 2x ?
Re: If x is a prime number greater than 5, y is a positive
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17 Jun 2013, 09:09
debayan222 wrote:
Bunuel wrote:
If x is a prime number greater than 5, y is a positive integer, and 5y=x^2+x, then y must be divisible by which of the following?
I. 5 II. 2x III. x+1
A. I only B. II only C. III only D. I and II only E. II and III only
You can solve this question even not analyzing option II at all.
Since \(5y=x(x+1)\), then \(x+1\) must be a multiple of 5, thus the units digit of \(x\) must be 9 (it cannot be 4, because in this case \(x\) would be even and thus not a prime number).
Now, if \(x=19\), then neither I nor III is true. The only option that does not contains I or III is B. So, B must be correct.
Answer: B.
If you interested why II must be true consider this: since \(x\) is a prime number greater than 5, then it's odd --> \(5y=x(x+1)=odd(odd+1)=odd*even=even\) --> \(5y=even\) --> \(y=even\).
Also, \(\frac{y}{x}=\frac{x+1}{5}\). we know that \(x+1\) is a multiple of 5, thus \(\frac{y}{x}=\frac{x+1}{5}=integer\) --> \(y\) is a multiple of \(x\). And since \(y\) is even, then \(y\) is a multiple of \(2x\).
Hope it helps.
As per the above highlighted part-how can we say that y is multiple of 2x ?
Can you please explain.
y is a multiple of x. Also, we know that y is even and x is odd. So, y=x*even=(2x)*k.
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Re: If x is a prime number greater than 5, y is a positive
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13 Apr 2014, 06:41
I got option B by plugging numbers:
We have:\(y=x*(x+1)/5\) Since x is a prime,it cannot be a multiple of \(5\);so\(x+1\) is a multiple of \(5\) The first value that came to mind for\(x\) was \(19\). Putting \(x=19\);we get \(y=19*4\) Only the II statement is satisfying the relation.
I) Since y=x(x+1)/5 is an integer, x(x+1) must be divisible by 5 to cancel out the 5 in the denominator. x is a prime number greater than 5, so it is definitely NOT divisible by 5. It follows that (x+1) must be divisible by 5. You're looking for a prime number x that is 1 less than a multiple of 5: x=19 fits the bill, since x+1=20 is divisible by 5.
In that case, y = 19·20/5 = 19·4. This is not divisible by 5, so y doesn't HAVE to be divisible by 5.
II) y=x(x+1)/5 is definitely divisible by x, but is y divisible by 2? Remember that x(x+1) is the product of two consecutive integers. x is a prime number greater than 5, therefore it is odd, and thus x+1 is even. Since x+1 is even, 5y is also even, and therefore y is even, i.e. divisible by 2. It follows that y is divisible by 2x.
III) With the same plug in that was used for (I) you can see that y=19·4 is not divisible by x+1=20. Since you have found a single example where y is not divisible by x+1, it follows that y does not HAVE to be divisible by x+1.
Is this the best way to do solve this? I'm not so sure I would have reached those exact conclusions in my brain to be able to get the right answer on this one....
Any insight would help immensely.......
Since \(x\) is a prime greater than 5, \(x\) cannot be divisible by 5. It follows that \(x+1\) must be divisible by 5, which means \(x+1=5n\) for some positive \(n\). Then \(x=5n-1\) must be odd (a prime that is greater than 5), so \(n\) must be even, which means \(n=2k\) for some positive integer \(k\). From \(5y=(5n-1)(5\cdot{2k})\) it follow that y is divisible \(2(5n-1)=2x\). Thus II is TRUE.
I isn't necessarily true: for \(x = 19, \,\,5y = 19\cdot{20}\) and \(y=19\cdot{4}\) is not divisible by 5. Neither is III true for \(x=19.\)
Answer B.
I don't quite get it. Do we need y to be divisible by 5? If x is 29, y would still not be divisible by 5.
I) Since y=x(x+1)/5 is an integer, x(x+1) must be divisible by 5 to cancel out the 5 in the denominator. x is a prime number greater than 5, so it is definitely NOT divisible by 5. It follows that (x+1) must be divisible by 5. You're looking for a prime number x that is 1 less than a multiple of 5: x=19 fits the bill, since x+1=20 is divisible by 5.
In that case, y = 19·20/5 = 19·4. This is not divisible by 5, so y doesn't HAVE to be divisible by 5.
II) y=x(x+1)/5 is definitely divisible by x, but is y divisible by 2? Remember that x(x+1) is the product of two consecutive integers. x is a prime number greater than 5, therefore it is odd, and thus x+1 is even. Since x+1 is even, 5y is also even, and therefore y is even, i.e. divisible by 2. It follows that y is divisible by 2x.
III) With the same plug in that was used for (I) you can see that y=19·4 is not divisible by x+1=20. Since you have found a single example where y is not divisible by x+1, it follows that y does not HAVE to be divisible by x+1.
Is this the best way to do solve this? I'm not so sure I would have reached those exact conclusions in my brain to be able to get the right answer on this one....
Any insight would help immensely.......
Since \(x\) is a prime greater than 5, \(x\) cannot be divisible by 5. It follows that \(x+1\) must be divisible by 5, which means \(x+1=5n\) for some positive \(n\). Then \(x=5n-1\) must be odd (a prime that is greater than 5), so \(n\) must be even, which means \(n=2k\) for some positive integer \(k\). From \(5y=(5n-1)(5\cdot{2k})\) it follow that y is divisible \(2(5n-1)=2x\). Thus II is TRUE.
I isn't necessarily true: for \(x = 19, \,\,5y = 19\cdot{20}\) and \(y=19\cdot{4}\) is not divisible by 5. Neither is III true for \(x=19.\)
Answer B.
I don't quite get it. Do we need y to be divisible by 5? If x is 29, y would still not be divisible by 5.
Re: If x is a prime number greater than 5, y is a positive
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15 Aug 2015, 05:53
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If x is a prime number greater than 5, y is a positive integer, and 5y=x^2+x, then y must be divisible by which of the following?
I. 5 II. 2x III. x+1
B is de answer. In other words: is x(x+1)/5 is divisble by which of de following X is > 5 and is a prime. 1) For de above number to be divisble by 5, X must end in 9. So, X+1 will be divisble by 5 from equation and still X is not divisble by 5, hence cannot be I
3) x(x+1)/5 is always divisble by x+1, but we are sure that remaining x/5 cannot be an integer, since X being prime, must end in 9. Hence III cannot be true
2) x(x+1)/5 by 2x must be true because X ends in 9, making x+1 ending in 0 making it divisble by 5 and 2, and X will be cancelld with X in numerator. Hence II must be true.
Re: If x is a prime number greater than 5, y is a positive
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