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OG mix problem and alligation technique

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OG mix problem and alligation technique

by Mo2men » Mon Oct 10, 2016 2:00 am
At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

I'm confused about use the alligation technique in this problem. How can I create common denominator for the apple and orange??
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by GMATGuruNY » Mon Oct 10, 2016 2:07 am
I post two approaches here:
https://www.beatthegmat.com/a-certain-fo ... 16515.html
The second post discusses how to use alligation.
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by Mo2men » Mon Oct 10, 2016 2:38 am
GMATGuruNY wrote:I post two approaches here:
https://www.beatthegmat.com/a-certain-fo ... 16515.html
The second post discusses how to use alligation.
Thanks for your reply.

I have read the link attached above. You always create common denominator. However, This is not clear for me how the denominatr is the same. I copied first step from the other post.

Step 1: Plot the 3 averages on a number line, with the average apple cost (40) and the average orange cost (60) on the ends and the average cost of all the fruit (56) in the middle.
A(40)--------------------56-------O(60)

I hope you clear my confusion.

Thanks
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by GMATGuruNY » Mon Oct 10, 2016 3:22 am
Mo2men wrote:
GMATGuruNY wrote:I post two approaches here:
https://www.beatthegmat.com/a-certain-fo ... 16515.html
The second post discusses how to use alligation.
Thanks for your reply.

I have read the link attached above. You always create common denominator. However, This is not clear for me how the denominatr is the same. I copied first step from the other post.

Step 1: Plot the 3 averages on a number line, with the average apple cost (40) and the average orange cost (60) on the ends and the average cost of all the fruit (56) in the middle.
A(40)--------------------56-------O(60)

I hope you clear my confusion.

Thanks
Alligation does not require that the values in the number line be over a common denominator.
In some of my solutions -- for example, my second post at https://www.beatthegmat.com/alligation-t283497.html -- I put the fractions over a common denominator to make the math easier.
But this step is not required.
Here, alligation can be performed directly with the given averages.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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