Problem Solving

In the fraction x/y, where x and y are positive integers, what is
the value of y ?

(1) x is an even multiple of y.
(2) x -y=2

What property is this testing? I feel like its just plug and chug type problem.

yellowho wrote:In the fraction x/y, where x and y are positive integers, what is
the value of y ?

(1) x is an even multiple of y.
(2) x -y=2

What property is this testing? I feel like its just plug and chug type problem.

(1) is enough because accordingly y = 1.

(2) is not enough because many values of x and y can satisfy x - y = 2.

Answer: B

Statement I

(1) x is an even multiple of y.

we can have infinite no of x and y

x=4,y=2
x=14, y=7

2) x -y=2

again we can have an infinite no. of x & y

x=3, y-1
x=4,y=2

Combining 1 & 2,
since y is an even multiple of x assume that x=2my where m can be 1,2,3..

putting in statement 2

2my-y=2
y=2/(2m-1)

now for y to be a +ve integer, 0<2m-1=<2 & 2m-1 has to be an integer
2m-1=1 or 2m-1=2
m=1 or m=1.5
since m is a multiple it cannot be a fraction
hence m has to 1 & y has to be 2

IMO : C

Please post OA along with the source.

I would go with C.

St 1: x=2, y=1 or x=4, y=2 both are correct. Hence St. 1 is insufficient
St 2: it could x=4, y=2, x=6, y=4. Hence St 2 is insufficient

Together, x=4, y=2 is the only valid values.

Therefore answer C.

You are correct. What's your reasoning why 2m-1 has to be an integer? Although I agree with you it does, just wondering why you thought so. From what you wrote it seems like a circular definition. In a vacuum 2m-1 can equal 1/2 in which case y=4, an integer. 2m-1 has to be an integer because M has to be an integer.


[quote="prachich1987"]Statement I

(1) x is an even multiple of y.

we can have infinite no of x and y

x=4,y=2
x=14, y=7

2) x -y=2

again we can have an infinite no. of x & y

x=3, y-1
x=4,y=2

Combining 1 & 2,
since y is an even multiple of x assume that x=2my where m can be 1,2,3..

putting in statement 2

2my-y=2
y=2/(2m-1)

now for y to be a +ve integer, 0<2m-1=<2 & 2m-1 has to be an integer
2m-1=1 or 2m-1=2
m=1 or m=1.5
since m is a multiple it cannot be a fraction
hence m has to 1 & y has to be 2

IMO : C

Please post OA along with the source.[/quote]

yellowho wrote:You are correct. What's your reasoning why 2m-1 has to be an integer? Although I agree with you it does, just wondering why you thought so. From what you wrote it seems like a circular definition. In a vacuum 2m-1 can equal 1/2 in which case y=4, an integer. 2m-1 has to be an integer because M has to be an integer.

2m-1 has to be an integer because m has to be an integer.
since m represents multiple ,m has to be an integer.
for example
3=2*1.5
can we say 3 is multiple of 2..no we can't because 1.5 (m) is not an integer
& 2m-1=Integer-1=integer

hope it helps

yellowho wrote:If x and y are positive integers, what is
the value of y ?

(1) x is an even multiple of y.
(2) x-y=2

Hi there!

[I´ve slightly modified the question stem, because the fraction mentioned in the original problem has no contribution/restriction involved.]

(1) Insufficient

Take x = 2 and y = 1
Take x = 4 and y = 2

Important: please note that from sttm (1) we may say that x = My, where M is necessarily even whenever y is odd. (When y is even, we just know that M is an integer, but it can be odd or even.)

(2) Insufficient

Take x = 4 and y = 2
Take x = 6 and y = 4

(1+2) Sufficient (This one is the hard one to justify, isn´t it?)

Please note that My = x = y+2 implies (*) y(M-1) = 2, where y and M-1 are integers. That means both are DIVISORS of 2, therefore we should look at -2, 2, -1 and 1 only as candidates for them!

From the fact that y>0, there are only two possibilities for y: 1 and 2.

If y =1 , from the sentence in bold we would have M-1 also odd, therefore y(M-1) could not be even (nor equal to 2, for sure).

The only possible solution for y is therefore 2, and we are done.

POST-MORTEM: take y = 2 in (*) to realize that x = My = 2*2 = 4, therefore (x,y) = (4,2) is the only pair that satisfies the question stem and both statements taken together.

Regards,
Fabio.

Guys, the OA for this is E. Can an Expert help please?Thanks!

Thouraya wrote:Guys, the OA for this is E. Can an Expert help please?Thanks!

Hi Thouraya!

We have two possibilities here:

(i) There is a flaw in my argument (and I got the wrong answer because of that) ;
(ii) My solution is perfect and the OA *you mentioned* is simply wrong.

Please study my solution carefully to see if you understand it fully or if you detect any mistake. If you don´t (as I suspect nobody else did yet), I suggest you believe in (ii)...

Regards,
Fabio.

Hi,

Mr. Fabio explanation is perfect.

Its C, there shouldn't be any doubt in this.

Thanks, m.abdulk!

@Thouraya: feel free to ask further explanations on any of my statements/arguments.

(1) x=2y (because in the prime factorization of y, it will have a 2 somewhere because x is even) --> x/2=y
(2) y=x-2


combined...substitute ---> x/2=x-2 --> x=4.

4/2=y --> y=2.

(C)

In the fraction x/y, where x and y are positive integers, what is
the value of y ?

(1) x is an even multiple of y.
(2) x -y=2

a)insufficient
b)x-y=2
insufficient
a&b) x=2*m*y (m is any integer)
x=2+y -> 2*m*y = 2+y
it means y is even.
y=2,m=1
IMO c