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M02 #12

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M02 #12 [#permalink]  04 Feb 2009, 20:12
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When $$x$$ is divided by 5, the remainder is 2. Is $$x$$ divisible by 7?

1. $$x$$ is a prime number
2. $$x + 3$$ is a multiple of 10

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

I got B, the official answer is E. I can see why the official answer is correct, but I'm not sure where my approach went wrong. Below is what I did:

I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any.

1. I simply plugged in different values of x and see if there're more than 1 prime # that fits the bill. Starting w/a=1, I got x=7, 12, 17. Since both 7 and 17 are prime # and 7 is divisible by 7 but 17 isn't. 1 is insuff.

2. I got the equation: x+3=10c, so x=10c-3, where c is a constant.
since we already know x=5a+2 from the question, we can set up the eqn as follows:
10c-3=5a+2
10c=5a+5
2c=a+1

Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly?
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Re: M02, 12 [#permalink]  04 Feb 2009, 21:02
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wcgmatclub wrote:
When $$x$$ is divided by 5, the remainder is 2. Is $$x$$ divisible by 7?

1. $$x$$ is a prime number
2. $$x + 3$$ is a multiple of 10

I got B, the official answer is E. I can see why the official answer is correct, but I'm not sure where my approach went wrong. Below is what I did:

I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any.

1. I simply plugged in different values of x and see if there're more than 1 prime # that fits the bill. Starting w/a=1, I got x=7, 12, 17. Since both 7 and 17 are prime # and 7 is divisible by 7 but 17 isn't. 1 is insuff.

2. I got the equation: x+3=10c, so x=10c-3, where c is a constant.
since we already know x=5a+2 from the question, we can set up the eqn as follows:
10c-3=5a+2
10c=5a+5
2c=a+1

Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly?

The highlighted statements are not correct as a and c are not constants. in fact they are variables.

1. $$x$$ is a prime number: x could be 7 or 17 or 37 or 47 and so on..................
2. $$x + 3$$ is a multiple of 10: x could be again 7 or or 17 or 37 or 47 and so on..................

Togather also same. so E.
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Re: M02, 12 [#permalink]  05 Feb 2009, 20:18
my bad in mislabeling "a" and "c" as constants instead of variables, but are there any mistakes in terms of my logic and approach in solving this type of problems? I understand there're multiple solutions for both 1) and 2). But I'm trying to understand how to reach that conclusion outside of brutally plugging in different #'s until I find the answer. This doesn't seem very efficient use of time on the GMAT.

Thanks in advance.
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Re: M02, 12 [#permalink]  06 Feb 2009, 18:54
wcgmatclub wrote:
my bad in mislabeling "a" and "c" as constants instead of variables, but are there any mistakes in terms of my logic and approach in solving this type of problems? I understand there're multiple solutions for both 1) and 2). But I'm trying to understand how to reach that conclusion outside of brutally plugging in different #'s until I find the answer. This doesn't seem very efficient use of time on the GMAT.

Thanks in advance.

Your approach for 1 is correct but 2 leads nowhere. Since you are looking for y, not a or c.
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Re: M02 #12 [#permalink]  17 Mar 2010, 05:02
wcgc wrote:
When $$x$$ is divided by 5, the remainder is 2. Is $$x$$ divisible by 7?

1. $$x$$ is a prime number
2. $$x + 3$$ is a multiple of 10

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

I got B, the official answer is E. I can see why the official answer is correct, but I'm not sure where my approach went wrong. Below is what I did:

I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any.

1. I simply plugged in different values of x and see if there're more than 1 prime # that fits the bill. Starting w/a=1, I got x=7, 12, 17. Since both 7 and 17 are prime # and 7 is divisible by 7 but 17 isn't. 1 is insuff.

2. I got the equation: x+3=10c, so x=10c-3, where c is a constant.
since we already know x=5a+2 from the question, we can set up the eqn as follows:
10c-3=5a+2
10c=5a+5
2c=a+1

Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly?

From Statement 1: X is a prime number.
from this we can say that x should be 7 or 17 or 47....
7 is divisible by 7 but not 17 - not sufficient

From Statement 2:
X+3 is multiple of 10 so again numbers should be 7, 17, 27...
again 7 is divisible by 7 but not 17..

for these type of questions, its easier to solve by picking numbers instead of writing euqations..
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Re: M02 #12 [#permalink]  17 Mar 2010, 09:06
wcgc wrote:
When $$x$$ is divided by 5, the remainder is 2. Is $$x$$ divisible by 7?
1. $$x$$ is a prime number
2. $$x + 3$$ is a multiple of 10
[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any.
Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly?

As statement 1 if x is prime number than it cannot be divided by 7 except 7 itself so we cannot say fully that x is divisible by 7 or not.
As per statement 2 is x+3 is multiple of 10 then x can be 7,17,27,37,47,57,67,77 ...
so from this 7,77 are divisible by 7 and others are not so insufficient.

Combininh also no use hence answer is E
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Re: M02 #12 [#permalink]  19 Mar 2010, 06:34
Its E both are not sufficient
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Re: M02 #12 [#permalink]  21 Mar 2011, 05:09
lionslion wrote:
wcgc wrote:
When $$x$$ is divided by 5, the remainder is 2. Is $$x$$ divisible by 7?

1. $$x$$ is a prime number
2. $$x + 3$$ is a multiple of 10

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

I got B, the official answer is E. I can see why the official answer is correct, but I'm not sure where my approach went wrong. Below is what I did:

I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any.

1. I simply plugged in different values of x and see if there're more than 1 prime # that fits the bill. Starting w/a=1, I got x=7, 12, 17. Since both 7 and 17 are prime # and 7 is divisible by 7 but 17 isn't. 1 is insuff.

2. I got the equation: x+3=10c, so x=10c-3, where c is a constant.
since we already know x=5a+2 from the question, we can set up the eqn as follows:
10c-3=5a+2
10c=5a+5
2c=a+1

Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly?

From Statement 1: X is a prime number.
from this we can say that x should be 7 or 17 or 47....
7 is divisible by 7 but not 17 - not sufficient

From Statement 2:
X+3 is multiple of 10 so again numbers should be 7, 17, 27...
again 7 is divisible by 7 but not 17..

for these type of questions, its easier to solve by picking numbers instead of writing euqations..

I agree, it is much faster to just plug in numbers then to derive an equation in this case
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Re: M02 #12 [#permalink]  21 Mar 2011, 22:51
lionslion wrote:

From Statement 1: X is a prime number.
from this we can say that x should be 7 or 17 or 47....
7 is divisible by 7 but not 17 - not sufficient

From Statement 2:
X+3 is multiple of 10 so again numbers should be 7, 17, 27...
again 7 is divisible by 7 but not 17..

for these type of questions, its easier to solve by picking numbers instead of writing euqations..

Nicely explained. Going with equations is time consuming.
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Re: M02 #12 [#permalink]  22 Mar 2011, 00:34
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x = 5k + 2, where k is an integer

x = 7,12,19,22,27,32...

So is x = 7n, where n is an integer ?

From (1), x is prime number

x = 7,17.. so (1) is not sufficient

From (2)

x+3 = 10m

x = 7, 17, 27 , So (2) is not sufficient

(1) and (2) are also not sufficient, so answer is E.
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Re: M02 #12 [#permalink]  24 Mar 2012, 14:12
I got E, I simply tried a couple numbers and showed that in some cases it was divisible by 7 and others it wasn't.
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Re: M02 #12 [#permalink]  25 Mar 2012, 05:12
E

When is divided by 5, the remainder is 2. Is divisible by 7?

Stmt says X when divided by 5 gives Remainder (R) = 2...
that means X=(5*a+2) X can be 2, 7, 12, 17, 22... (putting a =0,1,2...)
and asks is X divisible by 7?

1. X is a prime number

this says X is prime both 7 and 17 are prime, not sufficient

2. (x+ 3) is a multiple of 10
again taking 7 and 17, both (7+3) and (17+3) are multiple of 10. not sufficient.

Since same numbers are used for both 1 and 2, answer E
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Re: M02 #12 [#permalink]  26 Mar 2012, 00:13
Its E.....
I got it wrong because i was focusing on statement 2 in wrong way.....i was considering only multiples of 7
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Re: M02 #12 [#permalink]  26 Mar 2012, 06:16
one observation i have made is , when both the options are saying same ( like wht we have now)...
1. If Option A is suffcient, then Option B would be sufficient too making answer D
2. If Option A is insuffcient, then Option B would be insufficient too making answer E....

I go with E for the above question n i agree with the above explantions....
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Re: M02 #12 [#permalink]  26 Mar 2013, 04:15
When is divided by 5, the remainder is 2. Is divisible by 7?

1. is a prime number
2. x+3 is a multiple of 10

The numbers for two sets:
SET A:.7,17,27,37 etc
SET B:12,22,32,42 etc

for Case 1: we cannot draw a conclusion as SET A contains prime and non prime nos.
for Case 2: the statement is false for SET B data.

Combining also we don't get a true correlation.

Hence IMO E.
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Re: M02 #12 [#permalink]  26 Mar 2013, 18:05
qtn says x leaves reminder 2 when divided by 5 => 2,7,12,17,27,32,37,....

1. x is a prime number =>2,3,5,7,11,13,17,...
2. x+3 is a multiple of 10 => 7,17,27,37,...

Checking choice 1: Qtn & 1 => 2,7,11,...
2 is not divisible by 7
7 is divisible by 7
=> choice 1 insuffecient

Checking choice 2: Qtn & 1 => 7,17,27,...
7 is divisible by 7
17 is not divisible by 7
=> choice 2 insuffecient

Combine 1+2:
7,17,..
7 is divisible by 7
17 is not divisible by 7
=> choice 1+2 together is insuffecient

So E, hope it helps
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Re: M02 #12 [#permalink]  27 Mar 2013, 05:01
5a + 2 = x (given)

1) X is a prime number, prime numbers 2,3,5,7,11

a = 0 in given equation makes X = 2.
a = 1 in given equation makes X = 7.

two different answers, insufficient.

2) X+3 is a multiple of 10

If X = 7, then X+3 = 10, is a multiple of 10
If X =17, then X+3 = 20, is a multiple of 10

again two different answers, insufficient.

Combining 1 & 2 insufficient as, X can be 17 & 7

Solution E
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Re: M02 #12 [#permalink]  27 Mar 2013, 10:39
for option 1 > I picked couple of examples like 7,17,37 which are prime numbers and leaves remainder as 2. In all three of them, first one 7 is divisible by 7 but rest 17 and 37 are not. Clearly, we can not confirm the question under investigation "is number divisible by 7 or not?"

for option 2 > Again picked examples like 7, 17, 27, 37 and so on.. In all of them, first one is divisible by 7 not rest are not. So we can not again be pretty sure if the number is divisible by 7 or not.

Therefore, I go with option E.
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Re: M02 #12   [#permalink] 27 Mar 2013, 10:39
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