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PS - Age problem

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PS - Age problem [#permalink]

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New post 26 Mar 2009, 15:46
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The sum of the ages of 22 boys and 24 girls is 160.What is the sum of ages of one boy plus one girl, if all the boys are of the same age and all the girls are of the same age, and only full years are counted?

(a) 5.
(b) 6.
(c) 7.
(d) 8.
(e) 9.

:)
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Re: PS - Age problem [#permalink]

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New post 26 Mar 2009, 17:17
IMO:C.

22x+24y=160

2*11x+3*2^3y=5*2^5;

x=2^2; y=3;

x+y=7;
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Re: PS - Age problem [#permalink]

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New post 26 Mar 2009, 17:47
pmal04 wrote:
IMO:C.

22x+24y=160

2*11x+3*2^3y=5*2^5;

x=2^2; y=3;

x+y=7;


How did you go from the second step to the third ?
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Re: PS - Age problem [#permalink]

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New post 27 Mar 2009, 02:53
x+y=7 solves the equation.
howere I am not clear about the explanation.

Could you please explain it in detail?
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Re: PS - Age problem [#permalink]

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New post 27 Mar 2009, 05:23
I tried this way and not sure it is the best way to solve this one..
22B + 24G = 160
it is 11B + 12G = 80

Started subst G=1 in the eq will not work (B should result in a integer since it is BOYS)
G=2 in the eq will not work
G=3 and B=4 works
so it is 7
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Re: PS - Age problem [#permalink]

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New post 27 Mar 2009, 11:01
in the left hand side, we need to come up with the same prime factors as those in right hand side. This combination only can give you that.
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Re: PS - Age problem [#permalink]

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New post 27 Mar 2009, 12:53
OA : - C.

Here is my approach: -

22B + 24G = 150. (B and G are integers) ==> 11B + 12G = 80. ==> 11(B+G) +G = 80
==> B+G = (80-G) / 11. Since B and G are intergers, then B + G must be an integer ==> 80 - G must be a multiple of 11. This implies G is the reaminder left when 80 is divided by 11. Ie G = 3.

Substituting G in B +G we get (80 - 3) / 11 = 7 which is the ans.

Hum......Disguise form of remainders problem. :)
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Re: PS - Age problem [#permalink]

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New post 14 Oct 2009, 11:32
Even i am getting C.
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Re: PS - Age problem [#permalink]

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New post 14 Oct 2009, 16:50
C it is... nice explanations by those who did :) wow
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Re: PS - Age problem [#permalink]

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New post 14 Oct 2009, 17:24
My approach is slightly different:

22X+24Y = 160
22(X+Y) + 2Y = 160

We know from the answers that Y must be less than or equal to 8.

So, 22(X+Y) must be greater than or equal to 144, and less that 160. Only answer possible is 7.
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Re: PS - Age problem   [#permalink] 14 Oct 2009, 17:24
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