Cakes

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tonebeeze: Master | Next Rank: 500 Posts; Posts: 158; Joined: Mon Nov 02, 2009 5:49 pmFollowed by: 3 members; Thanked: 2 times

by tonebeeze » Thu Apr 07, 2011 10:58 am

Eunice sold several cakes. If each cake sold for either exactly 17 or exactly 19 dollars, how many 19 dollar cakes did Eunice sell?

1. Eunice sold a total of 8 cakes.
2. Eunice made 140 dollars in total revenue from her cakes.

OA = b

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Source: Beat The GMAT — Data Sufficiency | Thu Apr 07, 2011 10:58 am

clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Thu Apr 07, 2011 11:41 am

for sure the answer is B, but to me this problem is too time consuming,
it will be great if somebody shares any shortcut

17a+19b=?, find b-?
(1) a+b=8 not suff
(2) 17a+19b=140, we must prove the only integers for a and b exist
here b=2,a=6 (17*6+19*2=140)
and there are no others integers for a and b that give proper result

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tonebeeze: Master | Next Rank: 500 Posts; Posts: 158; Joined: Mon Nov 02, 2009 5:49 pmFollowed by: 3 members; Thanked: 2 times

by tonebeeze » Thu Apr 07, 2011 11:56 am

I agree!

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ldoolitt: Master | Next Rank: 500 Posts; Posts: 184; Joined: Sat Apr 14, 2007 9:23 am; Location: Madison, WI; Thanked: 17 times

by ldoolitt » Thu Apr 07, 2011 1:00 pm

clock60 wrote:for sure the answer is B, but to me this problem is too time consuming,
it will be great if somebody shares any shortcut

17a+19b=?, find b-?
(1) a+b=8 not suff
(2) 17a+19b=140, we must prove the only integers for a and b exist
here b=2,a=6 (17*6+19*2=140)
and there are no others integers for a and b that give proper result

If you wanted to SOLVE the problem (ie a PS problem) first I would subtract 19b from both sides, then divide by 17

(17b/17) = (140/17) - (19a/17)

You know that the left side is an integer so the right side must also be an integer

integer = (136/17) + (4/17) - (17/17) * a - (2/17) * a
integer = integer + 4/17 - integer - (2/17) * a

therefore 4/17 - (2/17) * a must be an integer to remove fractions. a = 2 is the only positive solution that could also satisfy the stem equation (19, etc would work but couldn't possibly satisfy the original equation)

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Thu Apr 07, 2011 1:07 pm

hi Idoolitt
thanks for sharing, but can you elaborate i little bit more
why first I would subtract 19b from both sides, then divide by 17
it sounds great but why in that order
what if substract 17a from both parts and then to divide by 19?
or it does not matter?

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ldoolitt: Master | Next Rank: 500 Posts; Posts: 184; Joined: Sat Apr 14, 2007 9:23 am; Location: Madison, WI; Thanked: 17 times

by ldoolitt » Thu Apr 07, 2011 1:14 pm

clock60 wrote:hi Idoolitt
thanks for sharing, but can you elaborate i little bit more
why first I would subtract 19b from both sides, then divide by 17
it sounds great but why in that order
what if substract 17a from both parts and then to divide by 19?
or it does not matter?

Does not matter

(19a/19) = (140/19) - (17b/19)
integer = (133/19) + (7/19) - (19/19) * b + (2/19) * b
integer = integer + 7/19 - integer + (2/19) * b
7/19 + (2/19)*b = integer

b=6 works there

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Thu Apr 07, 2011 1:17 pm

great reply!!!
thanks ldoolitt

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