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# There are X dogs in the dog hound, each dog eats Y Kg of food every da

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There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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14 Jul 2016, 01:59
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66% (02:03) correct 34% (02:44) wrong based on 73 sessions

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There are X dogs in the dog hound, each dog eats Y Kg of food every day. What percent of the total food weight does each dog eat?

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.

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Re: There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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14 Jul 2016, 02:09
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Bunuel wrote:
There are X dogs in the dog hound, each dog eats Y Kg of food every day. What percent of the total food weight does each dog eat?

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.

we need to find : 100y/xy ==> 100/x ( we just need to find the value of x)

1) 3y/x-3 = 13/10

2) (x/2)*y / (x/2) = 3 ==> y=3

from 1 and 2 combined , we can find x

C
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There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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14 Jul 2016, 06:35
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Bunuel wrote:
There are X dogs in the dog hound, each dog eats Y Kg of food every day. What percent of the total food weight does each dog eat?

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.

Let, total food weight=F.

so ,according to the question,$$\frac{F}{X}$$=Y .........(a)

(1) we can write,$$\frac{F}{X-3}$$=Y+1.2 ,along with (a) ,we can say that two equation but three variable,not sufficient to get value of each variable including total food weight(F),Insufficient
(2)we can write,$$\frac{F}{X/2}$$=Y+3,along with (a) ,we can say that two equation but three variable,not sufficient to get value of each variable including total food weight(F),Insufficient

(1)+(2)=Now we have three equation to solve each individual variable including F and Y,so,we can find percent of the total food weight does each dog eat Sufficient

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There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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Updated on: 14 Jul 2016, 06:58
1
AbdurRakib wrote:
Bunuel wrote:
There are X dogs in the dog hound, each dog eats Y Kg of food every day. What percent of the total food weight does each dog eat?

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.

Let, total food weight=F.

so ,according to the question,$$\frac{F}{X}$$=Y .........(a)

(1) we can write,$$\frac{F}{X-3}$$=Y+1.2 ,along with (a) ,we can say that two equation but three variable,not sufficient to get value of each variable including total food weight(F),Insufficient
(2)we can write,$$\frac{F}{X/2}$$=Y+3,along with (a) ,we can say that two equation but three variable,not sufficient to get value of each variable including total food weight(F),Insufficient

(1)+(2)=Now we have three equation to solve each individual variable including F and Y,so,we can find percent of the total food weight does each dog eat Sufficient

Could u pls solve the 2 equations? I'm not sure whether the answer is c
chetan2u pls assist

Originally posted by rahulkashyap on 14 Jul 2016, 06:57.
Last edited by rahulkashyap on 14 Jul 2016, 06:58, edited 1 time in total.
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Re: There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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14 Jul 2016, 06:57
1
Bunuel wrote:
There are X dogs in the dog hound, each dog eats Y Kg of food every day. What percent of the total food weight does each dog eat?

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.

Total food X dogs eat= XY

We need to find the % of total food each dog eats i.e = 100*Y/XY= 100/X

If we know the value of X, we can find out the % asked.

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
XY= (X-3) (Y+1.2)

This equation will have both X and Y and hence is not sufficient to find the value of X

(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.
XY= X/2 (Y+3)
Y =3
We need value of X. Not sufficient.

Combining both statements and putting the value of Y in equation from statement A, we will get the value of X.
Sufficient together.

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There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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14 Jul 2016, 11:51
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1
rahulkashyap wrote:
AbdurRakib wrote:
Bunuel wrote:
There are X dogs in the dog hound, each dog eats Y Kg of food every day. What percent of the total food weight does each dog eat?

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.

Let, total food weight=F.

so ,according to the question,$$\frac{F}{X}$$=Y .........(a)

(1) we can write,$$\frac{F}{X-3}$$=Y+1.2 ,along with (a) ,we can say that two equation but three variable,not sufficient to get value of each variable including total food weight(F),Insufficient
(2)we can write,$$\frac{F}{X/2}$$=Y+3,along with (a) ,we can say that two equation but three variable,not sufficient to get value of each variable including total food weight(F),Insufficient

(1)+(2)=Now we have three equation to solve each individual variable including F and Y,so,we can find percent of the total food weight does each dog eat Sufficient

Could u pls solve the 2 equations? I'm not sure whether the answer is c
chetan2u pls assist

Thanks,I am happy that you asked.

In general for n variables you will need at least n equations to get a unique solution.
For details,Please see equations with More than 2 variables at GMAT Math Book

We can put the value of F from (a) to Statement (1) and (2),and find

$$\frac{XY}{X-3}$$=Y+1.2,or XY=(Y+1.2)(X-3),or XY=XY-3Y+1.2X-1.2*3,or 3Y=1.2X-1.2*3,or Y=$$\frac{6X-18}{15}$$......(b)
and
$$\frac{XY}{X/2}$$=Y+3,or 2Y=Y+3,or Y=3.......(c)

(b)+(c) we can write,3=$$\frac{6X-18}{15}$$,or X=$$\frac{33}{6}$$

So F=XY=$$\frac{33}{6}$$*3=$$\frac{33}{2}$$

So,X=$$\frac{33}{6}$$,Y=3 ,and F=$$\frac{33}{2}$$

Now we can calculate,total food weight each dog eat=$$\frac{Y}{F}$$=$$\frac{3}{33/2}$$*100=$$\frac{200}{11}$$=18.18%
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There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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15 Jul 2016, 01:54
AbdurRakib wrote:
Bunuel wrote:
There are X dogs in the dog hound, each dog eats Y Kg of food every day. What percent of the total food weight does each dog eat?

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.

F/X=Y........(a)
We can put the value of F from (a) to Statement (1) and (2),and find
$$\frac{XY}{X-3}$$=Y+1.2,or XY=(Y+1.2)(X-3),or XY=XY-3Y+1.2X-1.2*3,or 3Y=1.2X-1.2*3,or Y=$$\frac{6X-18}{15}$$......(b)
and
$$\frac{XY}{X/2}$$=Y+3,or 2Y=Y+3,or Y=3.......(c)

(b)+(c) we can write,3=$$\frac{6X-18}{15}$$,or X=$$\frac{33}{6}$$

So F=XY=$$\frac{33}{6}$$*3=$$\frac{33}{2}$$

So,X=$$\frac{33}{6}$$,Y=3 ,and F=$$\frac{33}{2}$$

Now we can calculate,total food weight each dog eat=$$\frac{Y}{F}$$=$$\frac{3}{33/2}$$*100=$$\frac{200}{11}$$=18.18%

Hi AbdurRakib, I am convinced with your solution while you substitute F with XY. You can go through this way and finally get the answer. I can suggest you another way that might be easier and economical.
STEM:Total food=XY. Each dog eats= Y. Y*100/XY=? Ultimately X=?
STATE-2: Y=3 Insufficient
STATE-1: 3 dogs eat (X-3)*1.2, each eats =(X-3)*1.2/3=Y Insufficient
combining 1&2, value of X can be found out.
ANS: C
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There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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15 Jul 2016, 04:24
Top Contributor
jasimuddin wrote:
AbdurRakib wrote:
Bunuel wrote:
There are X dogs in the dog hound, each dog eats Y Kg of food every day. What percent of the total food weight does each dog eat?

(1) If there were 3 dogs less then each dog could eat 1.2 Kg more than he is does now.
(2) If there were half the dogs, each dog could eat 3 Kg more than he is does now.

F/X=Y........(a)
We can put the value of F from (a) to Statement (1) and (2),and find
$$\frac{XY}{X-3}$$=Y+1.2,or XY=(Y+1.2)(X-3),or XY=XY-3Y+1.2X-1.2*3,or 3Y=1.2X-1.2*3,or Y=$$\frac{6X-18}{15}$$......(b)
and
$$\frac{XY}{X/2}$$=Y+3,or 2Y=Y+3,or Y=3.......(c)

(b)+(c) we can write,3=$$\frac{6X-18}{15}$$,or X=$$\frac{33}{6}$$

So F=XY=$$\frac{33}{6}$$*3=$$\frac{33}{2}$$

So,X=$$\frac{33}{6}$$,Y=3 ,and F=$$\frac{33}{2}$$

Now we can calculate,total food weight each dog eat=$$\frac{Y}{F}$$=$$\frac{3}{33/2}$$*100=$$\frac{200}{11}$$=18.18%

Hi AbdurRakib, I am convinced with your solution while you substitute F with XY. You can go through this way and finally get the answer. I can suggest you another way that might be easier and economical.
STEM:Total food=XY. Each dog eats= Y. Y*100/XY=? Ultimately X=?
STATE-2: Y=3 Insufficient
STATE-1: 3 dogs eat (X-3)*1.2, each eats =(X-3)*1.2/3=Y Insufficient
combining 1&2, value of X can be found out.
ANS: C

Thanks.

But actually,I replied at the quoted post for the question asked in the below post:http://gmatclub.com/forum/there-are-x-dogs-in-the-dog-hound-each-dog-eats-y-kg-of-food-every-da-222006.html#p1709675

Please see my approach to this question here:http://gmatclub.com/forum/there-are-x-dogs-in-the-dog-hound-each-dog-eats-y-kg-of-food-every-da-222006.html#p1709667

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Re: There are X dogs in the dog hound, each dog eats Y Kg of food every da  [#permalink]

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18 Jan 2019, 02:58
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Re: There are X dogs in the dog hound, each dog eats Y Kg of food every da   [#permalink] 18 Jan 2019, 02:58
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