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### Show Tags

26 Jul 2010, 07:39
7
4
00:00

Difficulty:

55% (hard)

Question Stats:

69% (02:53) correct 31% (03:08) wrong based on 397 sessions

### HideShow timer Statistics

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 ##### Most Helpful Expert Reply Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8803 Location: Pune, India Re: Oranges sold at different rates [#permalink] ### Show Tags 15 Mar 2012, 21:45 5 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]

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26 Jul 2010, 08:58
16
1
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.
##### General Discussion
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Re: Oranges sold at different rates  [#permalink]

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29 Feb 2012, 09:03
5
$$\frac{35}{3}x+\frac{85}{7}(100-x)=1210$$

solve and you'll get x = 9
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Re: Oranges sold at different rates  [#permalink]

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15 Mar 2012, 20:18
$$\frac{35}{3}x+\frac{85}{7}(100-x)=1210$$

Can you pls explain the Right hand side of equation.
Won't it be 12.10
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Re: Oranges sold at different rates  [#permalink]

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16 Aug 2012, 23:47
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. Manager Joined: 23 May 2013 Posts: 104 Re: A woman sold 100 oranges at$12.10, some at the rate of 3  [#permalink]

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01 Oct 2013, 04:31
for such question need a will power to take on some random complicated numbers. i derived the equation but goofed up with the numbers
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Re: A woman sold 100 oranges at $12.10, some at the rate of 3 [#permalink] ### Show Tags 02 May 2015, 08:14 simple question.... killer calculation.. WOW... rxs0005 wrote: 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

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Re: A woman sold 100 oranges at $12.10, some at the rate of 3 [#permalink] ### Show Tags 13 May 2018, 02:47 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: A woman sold 100 oranges at$12.10, some at the rate of 3 &nbs [#permalink] 13 May 2018, 02:47
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