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A certain fraction is divisible by 2/5. If the numerator of 04 Jul 2007, 01:36
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A certain fraction is divisible by 2/5. If the numerator of the fraction is increased by 4 and the denominator doubled, the new fraction is divisible by 1/3. What is the sum of the numerator and denominator of the original fraction?

a) 49
b) 35
c) 28
d) 26
e) 21

Pls. explain your answer.



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A certain fraction is divisible by 2/5. If the numerator of 04 Jul 2007, 02:52
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at least time -cinsuming question.

I found E. Let's assume a/b is original fraction.

a/b=2/5*x and (a+4)/2b=1/3*y ---> where a,b,x,y -integers

1 -statement: x=5a/2b in order X to be Integer a must be divisible by 2b if b=1 then a=2,4,6,.....20,......

2-statement: y=(3a+4)/2b also satisfies with a=20, b=1

If you find 1 possible answer No need to search dor another one.
Thus a+b=21

Maybe I am not right.
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A certain fraction is divisible by 2/5. If the numerator of 04 Jul 2007, 04:09
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at least time -cinsuming question.

I found E. Let's assume a/b is original fraction.

a/b=2/5*x and (a+4)/2b=1/3*y ---> where a,b,x,y -integers

1 -statement: x=5a/2b in order X to be Integer a must be divisible by 2b if b=1 then a=2,4,6,.....20,......

2-statement: y=(3a+4)/2b also satisfies with a=20, b=1

If you find 1 possible answer No need to search dor another one.
Thus a+b=21

Maybe I am not right.

there is a slight calculation error
In statement 2 when u multiply over 3 with "(a+4)" the resultant ans is 3a +12. Hence following the same methodology u used i get the answer 28 i.e.C
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A certain fraction is divisible by 2/5. If the numerator of 04 Jul 2007, 22:59
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UMB
at least time -cinsuming question.

I found E. Let's assume a/b is original fraction.

a/b=2/5*x and (a+4)/2b=1/3*y ---> where a,b,x,y -integers

1 -statement: x=5a/2b in order X to be Integer a must be divisible by 2b if b=1 then a=2,4,6,.....20,......

2-statement: y=(3a+4)/2b also satisfies with a=20, b=1

If you find 1 possible answer No need to search dor another one.
Thus a+b=21

Maybe I am not right.
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UMB can u pls. explain how u get a=20 b=1 from stat 1 (or stat 2)?
If x=5a/2b, and x is an integer, then why can't a be let's say 22 and b be 11?

I guess I'm missing the logic here. Can you pls. explain? Thanks.
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A certain fraction is divisible by 2/5. If the numerator of 05 Jul 2007, 08:10
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GK_Gmat
UMB
at least time -cinsuming question.

I found E. Let's assume a/b is original fraction.

a/b=2/5*x and (a+4)/2b=1/3*y ---> where a,b,x,y -integers

1 -statement: x=5a/2b in order X to be Integer a must be divisible by 2b if b=1 then a=2,4,6,.....20,......

2-statement: y=(3a+4)/2b also satisfies with a=20, b=1

If you find 1 possible answer No need to search dor another one.
Thus a+b=21

Maybe I am not right.

UMB can u pls. explain how u get a=20 b=1 from stat 1 (or stat 2)?
If x=5a/2b, and x is an integer, then why can't a be let's say 22 and b be 11?

I guess I'm missing the logic here. Can you pls. explain? Thanks.
Show more



a=20 and b=1 is from statement 1.
Be careful any chosen number must satisfy statements1 &2 at the same time.. In other words X and Y must be iNTEGERS. 22 and 11 couldn't be solution as they don't satisfy Y (2nd statement). It is right any chosen #s other than 20 and 1 can satisfy Statements 1& 2 at the same time BUT Their sum is not given in the answer choices.
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A certain fraction is divisible by 2/5. If the numerator of 05 Jul 2007, 08:38
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when I do this, I feel like both D & E can be the answer, which is why I did not bother to check the rest. what am I doing wrong here?

starting off with E, 21 can be written as 20 + 1

2/5 * n = a/b
n= 5a/2b ....................... (1)

1/3 * m = a+4/2b
m= 3a + 12/2b .................... (2)

looking at 1, 5a is divisible by 2b, hence a is divisible by a factor of 2
factors of 2 >0 = 2, 4, 6, 8, 10,......18, 20

lets say a = 20 & b = 1 (since 21= 20 + 1 from E)
n= 100/2 = 50
m = 72/ 2= 36

E works.

now D, 26 = 24 + 2
lets say a= 24 & b= 2
n = 120/4 = 30
m= 84/ 4 = 21

seems like D works as well.



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