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Is x^2*y^4 divisible by 49? (1) x is an integer divisible by

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Is x^2*y^4 divisible by 49? (1) x is an integer divisible by [#permalink] New post 20 Jan 2006, 14:14
00:00
A
B
C
D
E

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(N/A)

Question Stats:

50% (00:00) correct 50% (00:00) wrong based on 2 sessions
Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.
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 [#permalink] New post 20 Jan 2006, 14:23
C.

S1: we don't know about y. Insufficient.

S2:
x = 1, y = 0, ans = yes
x = 2, y = 49/2, ans = no
Insufficient

together:
sufficient
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Re: DS-Divisible [#permalink] New post 20 Jan 2006, 17:04
cool_jonny009 wrote:
Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.


I think this is E.

St1: Clearly INSUFF. We don't knwo value of y

St2: INSUFF. We don't knwo about y. x*y divisible by 49 does not say about value of y.

Combined:

Take examples which satisfies both the statements
(St2 can be written as (x*y)^2 * y^2)
1. x = 98, y =1/2 then Ans to main stem is NO
2. x = 98, y = 2 then ans to main stem is YES
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 [#permalink] New post 20 Jan 2006, 18:59
I think it is C

From 1) we do not know anything about Y
From 2) X can by 5 and Y can be 9.8 or 7 and 7

From 1 and 2) we know for sure that Y is a multiple of 7.
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 [#permalink] New post 20 Jan 2006, 21:11
Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.

Well just a possibility

taking 1)--> suppose x=14 , in stem 14*14*y*y*y*y/49= 4*y^4 which makes stem divisible

taking 2) --> x*y = 14 ( xy*xy*y*y*)/49 = 4*y^2 which makes stem divisible

so cant it be D .

OA ? :roll:
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 [#permalink] New post 20 Jan 2006, 21:43
(1) x is a multiple of 7. Can be 7, 14, 21...etc

If x = 7, x^2*y^4 is divisible by 49.
If x = 14, x^2*y^4 = (2*7)^2*y^4 = 49(4)(y^4) -> divisible 49
If x = 21, x^2*y^4 = (3*7)^2*y^4 = 48(9)(y^4) -> divisible by 49

You can see that a 7 can always be extracted and once sqaured will cancel out the 49 that is in the denominator. Sufficient.

(2) x*y = 49 -> can only be x=7 and y=7. Sufficient.

Ans D
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 [#permalink] New post 20 Jan 2006, 21:43
@ razrulz

it isn't stated in A or in B, for their own, that y is an integer.

C should be right.
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 [#permalink] New post 20 Jan 2006, 22:31
I think it's D


1. x=1*7, 2*7, 3*7; x^2=1*7^2, 2^2*7^2,

So 1. is sufficient

2. x^2*y^4= (xy^2)^2

==>x*y*y/x*y

For example, 7*7*7 is divisible by 49 (7*7)

or 9.8*5*5 is also divisible by 49 (9.8*5)
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Re: DS-Divisible [#permalink] New post 20 Jan 2006, 22:49
cool_jonny009 wrote:
Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.


Well revision again:

1) x is a multiple of 7

But y is the critical value. If it is, for instance, 1/7 then (x^2)*(y^4) is must not be divisible by 49;
if it is 1 is is divisible by 49

2) x*y is an integer divisible by 49

Assume x=7^3 and y=1/7, then x*y is 49

and (7^6)*(7^-4)=7^2, but, as dahya mentioned if x=98 and y=1/2, then isn't divisible by 49


Both combined aren't sufficient too, E
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Re: DS-Divisible [#permalink] New post 20 Jan 2006, 23:49
[quote="allabout"][quote="cool_jonny009"]Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.[/quote]

Well revision again:

1) x is a multiple of 7

But y is the critical value. If it is, for instance, 1/7 then (x^2)*(y^4) is must not be divisible by 49;
if it is 1 is is divisible by 49

2) x*y is an integer divisible by 49

Assume x=7^3 and y=1/7, then x*y is 49

and (7^6)*(7^-4)=7^2, but, as dahya mentioned if x=98 and y=1/2, then isn't divisible by 49


Both combined aren't sufficient too, E[/quote]

But it is given that X is an integer, so it cant be 1/7. Hence I think the answer is A.
Statement 2 is not suff
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 [#permalink] New post 20 Jan 2006, 23:55
ywilfred wrote:
(1) x is a multiple of 7. Can be 7, 14, 21...etc

If x = 7, x^2*y^4 is divisible by 49.
If x = 14, x^2*y^4 = (2*7)^2*y^4 = 49(4)(y^4) -> divisible 49
If x = 21, x^2*y^4 = (3*7)^2*y^4 = 48(9)(y^4) -> divisible by 49

You can see that a 7 can always be extracted and once sqaured will cancel out the 49 that is in the denominator. Sufficient.

(2) x*y = 49 -> can only be x=7 and y=7. Sufficient.

Ans D


I think you are assuming that y is also integer. But there is no such thing mentioned.
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 [#permalink] New post 21 Jan 2006, 22:15
E it is

a) Insufficient we dont know if y is int or not
so it can be x=7 and y=1/2 or x=7 and y =1

b)Insuff
x2y4 is xy * xy * y2
Here again we dnt know y, only thing we know is that xy divisible by 49
so it can be xy=49 and y=1 or 1/2

c)Insuff

xyis divisible by 49 and x is divisible by 7
but xy=x*y=7*7=49 Yes
xy=x*y=98*1/2 No

so E
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 [#permalink] New post 21 Jan 2006, 22:18
cool_jonny009

Please supply the OA?
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 [#permalink] New post 21 Jan 2006, 22:48
I also vote for E.

1) x = 7k (k - integer)
=> x^2*y^4 = 49k^2 * y^4 -- depends on value of y (ycould be fraction also) INSUFF.

2) x*y = 7m (m integer)
=> x^2*y^4 = (xy)^2*y^2 = 49*m^2*y^2 -- again depends on the value of y. INSUFF.

(1) & (2)
No additional conclusion can be drawn. Hence E!
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 [#permalink] New post 25 Jan 2006, 02:50
wow ! cool expln duttsit and anand
  [#permalink] 25 Jan 2006, 02:50
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