Q
[Questions deleted because source was not cited. Due to copyright laws, sources must be cited or the questions will be deleted.]
Disclaimer: -
I do not know the source of this question, so please do not ask it from me. A friend of mine who just wrote his GMAT and got a cool 750 gave me the pdf files containing some challenging questions. I have the answers to these questions but not the official explanations.
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- darden11
- rfernandez
- Course Students
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- relentlesspursuito700plus
You are right. There are tough...here's my shot at them.
These may NOT be correct, but here's my attempt. If I am wrong, please point out where my logic fails. I think this should be a good exercise for me and for someone correcting my potential mistakes (but hopefully I'm right).
I am actually confused by the first question. I am confused about how to interpret the answer choices for a question like this where the statements seems to contradict each other.... you'll see what I mean.
[deleted - see above]
[NOTE: This is where I am unsure of my logic] I'm not sure so I move on and combine what I know and see what happens. Here we have consecutive integers (x-1), x, (x+1), where the first term and third term are divisible by 7. If X+1 and X-1 are all multiples of 7, then you can multiply everything in this set by 7 to get: (7x-7), (7x), (7x+7). Obviously all the terms are divisible by 7, but that's because they are all multiplied by 7. This shows that x can be any of the integeres I listed above, 8, 15, or 20, 27, which all lead me to the same answer. The answer is still "no" x isn't divisible by 7. So I chose D.
[deleted - see above]
We don’t now what x ^ 4 - y ^ 4 is equal to but we can still factor this out. It takes several factoring but you can strip it down to (x^2+y^2)(x^2-y^2). Then you can take X^2-y^2 and strip it down again to (x+y)(x-y). Since statement 1 tells us one of the factors is divisible by three (a prime number), the entire expression must also be divisible by three according to prime factorization rule. Statement two isn’t very helpful since there are many integers that will yield a remainder of 2 when divided by 3, so x+y could be 5 or even 14. We don’t know.
Therefore A.
[deleted - see above]
Statement 1 tell us that x + 29 (a prime number) is divisible by 10. Since 29 isn’t a multiple of 10, x + 29 must be a multiple of 10. This means x could be 1, 11, 21, or any other integer that has a unit’s digit of 1. If x was 21, yes it is divisible by 7, but if it’s 11, no it’s not divisible by 7. Insufficient.
Statement 2 tells us the reverse. X + 10 must be a multiple of 29. This means that x is equal to an integer that is 10 less than 29, meaning it is 19N. If N is 7 then x would be a multiple of 7 if N is not a multiple of 7, then x isn’t a multiple of seven either. So this is insufficient by itself.
But, both combined, if X+10 = 19(9) +10(9)= 29(9). We have a value of x that has a unit’s digit of 1 . So C. Both together are sufficient.
I am actually confused by the first question. I am confused about how to interpret the answer choices for a question like this where the statements seems to contradict each other.... you'll see what I mean.
[deleted - see above]
[NOTE: This is where I am unsure of my logic] I'm not sure so I move on and combine what I know and see what happens. Here we have consecutive integers (x-1), x, (x+1), where the first term and third term are divisible by 7. If X+1 and X-1 are all multiples of 7, then you can multiply everything in this set by 7 to get: (7x-7), (7x), (7x+7). Obviously all the terms are divisible by 7, but that's because they are all multiplied by 7. This shows that x can be any of the integeres I listed above, 8, 15, or 20, 27, which all lead me to the same answer. The answer is still "no" x isn't divisible by 7. So I chose D.
[deleted - see above]
We don’t now what x ^ 4 - y ^ 4 is equal to but we can still factor this out. It takes several factoring but you can strip it down to (x^2+y^2)(x^2-y^2). Then you can take X^2-y^2 and strip it down again to (x+y)(x-y). Since statement 1 tells us one of the factors is divisible by three (a prime number), the entire expression must also be divisible by three according to prime factorization rule. Statement two isn’t very helpful since there are many integers that will yield a remainder of 2 when divided by 3, so x+y could be 5 or even 14. We don’t know.
Therefore A.
[deleted - see above]
Statement 1 tell us that x + 29 (a prime number) is divisible by 10. Since 29 isn’t a multiple of 10, x + 29 must be a multiple of 10. This means x could be 1, 11, 21, or any other integer that has a unit’s digit of 1. If x was 21, yes it is divisible by 7, but if it’s 11, no it’s not divisible by 7. Insufficient.
Statement 2 tells us the reverse. X + 10 must be a multiple of 29. This means that x is equal to an integer that is 10 less than 29, meaning it is 19N. If N is 7 then x would be a multiple of 7 if N is not a multiple of 7, then x isn’t a multiple of seven either. So this is insufficient by itself.
But, both combined, if X+10 = 19(9) +10(9)= 29(9). We have a value of x that has a unit’s digit of 1 . So C. Both together are sufficient.
- Guest
The first question is garbage because the two statements contradict each other. There is no positive integer that you can both add 1 and subtract 1 from and is divisible by 7. Zero is the only number that goes evenly into seven when you add or subtract one from but you said that its a positive integer.
- StaceyKoprince
- ManhattanGMAT Staff
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