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# A woman sold 100 oranges at $12.10, some at the rate of 3  Question banks Downloads My Bookmarks Reviews Author Message TAGS: Director Joined: 07 Jun 2004 Posts: 638 Location: PA Followers: 1 Kudos [?]: 60 [2] , given: 22 A woman sold 100 oranges at$12.10, some at the rate of 3 [#permalink]  26 Jul 2010, 08:39
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A woman sold 100 oranges at $12.10, some at the rate of 3 for 35 cents and the rest at 7 for 85 cents. How many were sold at the first rate? A. 45 B. 21 C. 9 D. 15 E. 12 [Reveal] Spoiler: OA _________________ If the Q jogged your mind do Kudos me : ) Manager Joined: 02 Apr 2010 Posts: 103 Followers: 2 Kudos [?]: 76 [8] , given: 18 Re: Oranges sold at different rates [#permalink] 26 Jul 2010, 09:58 8 This post received KUDOS This can be solved like a classical mixture problem but numbers are awkward to deal with. It's easier to just look at the answer choices. You know that a multiple of 3 oranges has to be sold at the first rate, and a multiple of 7 at the second rate. You simple subtract the answer choices for the first rate from 100 and check whether the remainder (i.e. the number of oranges sold at the second rate) is a multiple of 7. 100 - 45 = 55 => not a multiple of 7 so exclude 100 - 21 = 79 => not a multiple of 7 so exclude 100 -9 = 91 => a multiple of 7 so keep 100 - 15 = 85 => not a multiple of 7 so exclude 100 - 12 = 88 => not a multiple of 7 so exclude Hence, answer choice 9 is correct. Senior Manager Status: mba here i come! Joined: 07 Aug 2011 Posts: 271 Location: Pakistan Concentration: Strategy, Marketing GMAT 1: 680 Q46 V37 GMAT 2: Q V Followers: 13 Kudos [?]: 459 [2] , given: 48 Re: Oranges sold at different rates [#permalink] 29 Feb 2012, 10:03 2 This post received KUDOS \frac{35}{3}x+\frac{85}{7}(100-x)=1210 solve and you'll get x = 9 _________________ press +1 Kudos to appreciate posts Senior Manager Joined: 10 Nov 2010 Posts: 270 Location: India Concentration: Strategy, Operations GMAT 1: 520 Q42 V19 GMAT 2: 540 Q44 V21 WE: Information Technology (Computer Software) Followers: 4 Kudos [?]: 19 [0], given: 22 Re: Oranges sold at different rates [#permalink] 15 Mar 2012, 21:18 \frac{35}{3}x+\frac{85}{7}(100-x)=1210 Can you pls explain the Right hand side of equation. Won't it be 12.10 _________________ The proof of understanding is the ability to explain it. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 3114 Location: Pune, India Followers: 573 Kudos [?]: 2020 [2] , given: 92 Re: Oranges sold at different rates [#permalink] 15 Mar 2012, 22:45 2 This post received KUDOS GMATD11 wrote: \frac{35}{3}x+\frac{85}{7}(100-x)=1210 Can you pls explain the Right hand side of equation. Won't it be 12.10 The equation equates the total selling price. He gets$12.10 i.e. 1210 cents.
If he sold x oranges for 35/3 cents and (100-x) for 85/7 cents, this is a total of
(35/3) * x + (85/7) * (100-x) cents. You equate cents to cents.

Also, you can use the weighted average formula here:

w1/w2 = (85/7 - 121/10)/(121/10 - 35/3) = 9/91

Total 9+91 is 100. So he sells 9 oranges at 35 for 3 and 91 oranges at 85 for 7.
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Re: Oranges sold at different rates [#permalink]  17 Aug 2012, 00:47
GMATD11 wrote:
\frac{35}{3}x+\frac{85}{7}(100-x)=1210

Can you pls explain the Right hand side of equation.
Won't it be 12.10

Hi,

Because we are taking all the values in cents so we have converted $12.10 into cents which is 1210 cents. as 1 cent =0.01 Dollar Hope this helps. Try and fail but never fail to try. Re: Oranges sold at different rates [#permalink] 17 Aug 2012, 00:47 Similar topics Replies Last post Similar Topics: A woman will win$3.2 if in 5 tosses of a coin she gets 2 09 Oct 2003, 21:34
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