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At a certain fast-food restaurant, each sandwich costs x

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kevincan
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At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   26 Nov 2006, 07:39
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At a certain fast-food restaurant, each sandwich costs x dollars and each dessert costs y dollars. If $50 is enough to buy 7 sandwiches and 3 desserts, is $50 enough to buy 5 sandwiches and 5 desserts instead?

(1) $50 is enough to buy 6 sandwiches and 4 desserts.

(2) $50 is enough to buy 3 sandwiches and 7 desserts.



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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   26 Nov 2006, 08:00
Should be D


In both cases we get price of sandwitch = dessert
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   26 Nov 2006, 17:57
D for me.

1) and 2) give us simultaneous equations resulting in x or y is less than or equal to 5 dollars.
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   27 Nov 2006, 05:30
I am getting A.

we know: 7x+3y <=50.
They are asking if 5x+5y <=50.

Also from question, it is obvious that x > y.

From 1) 6x + 4y <= 50
= 5x+5y + (x-y) <= 50

If x> y, it is obvious that 5x+5y is also <= 50.

From 2)

3x+7y <= 50

5x+5y - 2x+2y <=50
5x+5y-2(x-y)<=50


Can't really deduce much from here. 5x+5y may or may not be <=50.
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   27 Nov 2006, 06:11
hsampath wrote:
I am getting A.

we know: 7x+3y <=50.
They are asking if 5x+5y <=50.

Also from question, it is obvious that x > y.

From 1) 6x + 4y <= 50
= 5x+5y + (x-y) <= 50

If x> y, it is obvious that 5x+5y is also <= 50.

From 2)

3x+7y <= 50

5x+5y - 2x+2y <=50
5x+5y-2(x-y)<=50


Can't really deduce much from here. 5x+5y may or may not be <=50.


Cool Sampath.. I think we all misinterpreted the problem. It said ENOUGH to buy..that's not necessarily the price
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   27 Nov 2006, 06:16
Yeah, maybe you did misinterpret the question, but I think I am wrong. y > x not x > y, which makes the the answer B, not A.
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   27 Nov 2006, 10:03
At a certain fast-food restaurant, each sandwich costs x dollars and each dessert costs y dollars. If $50 is enough to buy 7 sandwiches and 3 desserts, is $50 enough to buy 5 sandwiches and 5 desserts instead?

(1) $50 is enough to buy 6 sandwiches and 4 desserts.

(2) $50 is enough to buy 3 sandwiches and 7 desserts.

(1) 7S+3D=6S+4D=> S=D sandwitch and desert same price
so 50$ should be wnough to buy 5S and 5D


(2) 3S+7D=7S+3D=> S=D sandwitch and desert same price
so 50$ should be wnough to buy 5S and 5D

Answer D
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   27 Nov 2006, 10:13
I agree with hsampath: Ans B.
Let the price of a sandwich be S and a dessert be D.
From 1 and stem: 7S+3D<=50 (1) and 6S+4D<=50 (2). let (1)-(2) we have S-D <=0 this means that the price of a sandwich is less than that of a dessert. So, we can not tell for sure if we can by 5S and 5D for $50 or less.

From 2 and stem: setting up the 2 equations: we still have S<=D. but this time, since we can use $50 to buy 3 S and 7D and S<=D, we know that we can use the amount we pay for the 2D and buy 2 extra S and have enough money. Thus, Suff.

Ans B
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   27 Nov 2006, 10:20
kevincan wrote:
At a certain fast-food restaurant, each sandwich costs x dollars and each dessert costs y dollars. If $50 is enough to buy 7 sandwiches and 3 desserts, is $50 enough to buy 5 sandwiches and 5 desserts instead?

(1) $50 is enough to buy 6 sandwiches and 4 desserts.

(2) $50 is enough to buy 3 sandwiches and 7 desserts.


7s+3d<=50
Question asks if 5d+5s<=50 or s+d<=10

S1:6s+4d<=50
insuff to find s+d

S2:3s+7d<=50
+7s+3d<=50
=>10s+10d<=100
s+d<=10 Bingo!

My ans B
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   28 Nov 2006, 09:10
The flaw in Damager reasoning is thr is no way to determine
7S+3D=6S+4D=> S=D
and
3S+7D=7S+3D=> S=D

Since it is mentioned tht both the total are enough i.e <= 50 .. tht is no way tht thy r equal ..
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   28 Nov 2006, 09:13
Very Good prob..

Keep posting Kevin ..!!
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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink] New post   28 Nov 2006, 21:41
kevincan wrote:
Sumithra wrote:
kevincan wrote:
At a certain fast-food restaurant, each sandwich costs x dollars and each dessert costs y dollars. If $50 is enough to buy 7 sandwiches and 3 desserts, is $50 enough to buy 5 sandwiches and 5 desserts instead?

(1) $50 is enough to buy 6 sandwiches and 4 desserts.

(2) $50 is enough to buy 3 sandwiches and 7 desserts.


7s+3d10s+10d<=100
s+d<=10 Bingo!

My ans B

OA and OE


I understand why 2 is sufficient. But can someone explain to me why 1 is not sufficient? What am I doing wrong?

From stem:
7x+3y<=50-----(eqn 0)
from 1:
6x +4y<=50----(eqn 1)

multiplying eqn 0 by 6 and eqn 1 by 7 we get

42x+28y<=350
42x+18y<=300 --subtracting we get
-------------------
10y<=50 ---------eqn 3

multiplying eqn 0 by 4 and eqn 1 by 3 we get

28x+12y<=200
18x+12y<=150 --subtracting we get
-------------------
10x <=50 ----------eqn 4

Adding eqn 3 and 4 we get
10x+10y<=100
or 5x+5y<=50


Thanks :)



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Re: At a certain fast-food restaurant, each sandwich costs x [#permalink]
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