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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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14 Jul 2016, 01:59
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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
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14 Jul 2016, 02:09
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/x3 = 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
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14 Jul 2016, 06:35
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,\(\frac{F}{X3}\)=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 SufficientCorrect Answer C
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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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Updated on: 14 Jul 2016, 06:58
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,\(\frac{F}{X3}\)=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 SufficientCorrect Answer CCould 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
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14 Jul 2016, 06:57
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= (X3) (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. C is the answer
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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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14 Jul 2016, 11:51
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,\(\frac{F}{X3}\)=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 SufficientCorrect Answer CCould 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 BookWe can put the value of F from (a) to Statement (1) and (2),and find \(\frac{XY}{X3}\)=Y+1.2,or XY=(Y+1.2)(X3),or XY=XY3Y+1.2X1.2*3,or 3Y=1.2X1.2*3,or Y=\(\frac{6X18}{15}\)......(b) and \(\frac{XY}{X/2}\)=Y+3,or 2Y=Y+3,or Y=3.......(c) (b)+(c) we can write,3=\(\frac{6X18}{15}\),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
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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 \(\frac{XY}{X3}\)=Y+1.2,or XY=(Y+1.2)(X3),or XY=XY3Y+1.2X1.2*3,or 3Y=1.2X1.2*3,or Y=\(\frac{6X18}{15}\)......(b) and \(\frac{XY}{X/2}\)=Y+3,or 2Y=Y+3,or Y=3.......(c) (b)+(c) we can write,3=\(\frac{6X18}{15}\),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=? STATE2: Y=3 Insufficient STATE1: 3 dogs eat (X3)*1.2, each eats =(X3)*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
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15 Jul 2016, 04:24
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 \(\frac{XY}{X3}\)=Y+1.2,or XY=(Y+1.2)(X3),or XY=XY3Y+1.2X1.2*3,or 3Y=1.2X1.2*3,or Y=\(\frac{6X18}{15}\)......(b) and \(\frac{XY}{X/2}\)=Y+3,or 2Y=Y+3,or Y=3.......(c) (b)+(c) we can write,3=\(\frac{6X18}{15}\),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=? STATE2: Y=3 Insufficient STATE1: 3 dogs eat (X3)*1.2, each eats =(X3)*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/therearexdogsinthedoghoundeachdogeatsykgoffoodeveryda222006.html#p1709675Please see my approach to this question here: http://gmatclub.com/forum/therearexdogsinthedoghoundeachdogeatsykgoffoodeveryda222006.html#p1709667Now,Please give me your valuable feedback
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Re: There are X dogs in the dog hound, each dog eats Y Kg of food every da
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Re: There are X dogs in the dog hound, each dog eats Y Kg of food every da
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