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Intern
Joined: 03 Dec 2010
Posts: 43
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Kudos [?]: 23 [0], given: 7

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Hey guys, i was just wondering that the options and answers posted for the PS1000 are correct or have been modified since last time it was posted? When i sit down to solve it, i get answers, which are not even in the options.

Would appreciate if someone could guide me.
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 149

Kudos [?]: 1360 [0], given: 376

Acer86 wrote:
Hey guys, i was just wondering that the options and answers posted for the PS1000 are correct or have been modified since last time it was posted? When i sit down to solve it, i get answers, which are not even in the options.

Would appreciate if someone could guide me.

Most of the answers are right, I believe!! If you find some discrepancy in few of the answers, why don't you just post them, mentioning proper question number and set. Others can have a look and confirm whether there is indeed a discrepancy. It will benefit many.
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Intern
Joined: 03 Dec 2010
Posts: 43
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Kudos [?]: 23 [0], given: 7

This is from Section 1 Question 5

5. At a certain diner, a hamburger and coleslaw cost $3.59, and a hamburger and french fries cost$4.40. If french fries cost twice as much as coleslaw, how much do french fries cost?
(A) $0.30 (B)$0.45
(C) $0.60 (D)$0.75
(E) $0.90 The answer is in the answer sheet is (E). I tried back solving and the only answer that came was close to matching the description was E. But when i assigned variables such as X (hamburger) Y (Coleslaw) and Z (French Fries) I got the following Equations: X + Y = 3.59 X + Z = 4.40 Z = 2Y (FRENCH FRIES twice as expensive as COLESLAW) Solving the above equations i get Y = 0.81 and Z = 1.62, and 1.62 is not even in the options for the cost of french fries.. I dont know know if i have done something wrong in forming the equations. Could someone guide me plzz. Math Expert Joined: 02 Sep 2009 Posts: 31303 Followers: 5363 Kudos [?]: 62507 [0], given: 9457 Re: PS1000 Questions and Answers [#permalink] 11 Mar 2011, 02:09 Expert's post Acer86 wrote: This is from Section 1 Question 5 5. At a certain diner, a hamburger and coleslaw cost$3.59, and a hamburger and french fries cost $4.40. If french fries cost twice as much as coleslaw, how much do french fries cost? (A)$0.30
(B) $0.45 (C)$0.60
(D) $0.75 (E)$0.90

The answer is in the answer sheet is (E). I tried back solving and the only answer that came was close to matching the description was E.

But when i assigned variables such as X (hamburger) Y (Coleslaw) and Z (French Fries) I got the following Equations:

X + Y = 3.59
X + Z = 4.40
Z = 2Y (FRENCH FRIES twice as expensive as COLESLAW)

Solving the above equations i get Y = 0.81 and Z = 1.62, and 1.62 is not even in the options for the cost of french fries..

I dont know know if i have done something wrong in forming the equations. Could someone guide me plzz.

This question is discussed here: ps-section-1-q5-95497.html?hilit=coleslaw#p735081

Given:
$$h+c=3.59$$, $$h+f=4.4$$ and $$f=2c$$. Q: $$f=?$$

Subtract 1 from 2: $$h+f-h-c=4.4-3.59$$ --> $$f-c=0.81$$, as $$f=2c$$, then $$f-\frac{f}{2}=0.81$$ --> $$f=1.62$$.

P.S.
 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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