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Re: Oranges sold at different rates [#permalink]
26 Jul 2010, 08:58

13

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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

Re: Oranges sold at different rates [#permalink]
15 Mar 2012, 21:45

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Expert's post

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. _________________

Re: A woman sold 100 oranges at $12.10, some at the rate of 3 [#permalink]
29 Sep 2013, 07:23

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Re: A woman sold 100 oranges at $12.10, some at the rate of 3 [#permalink]
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 _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: A woman sold 100 oranges at $12.10, some at the rate of 3 [#permalink]
02 May 2015, 00:35

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