A certain food stand

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alex.gellatly: Master | Next Rank: 500 Posts; Posts: 435; Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members; Thanked: 48 times

by alex.gellatly » Sat Jul 28, 2012 11:23 pm

At a certain food 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 food stand, and the average (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?
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Last edited by alex.gellatly on Tue Aug 14, 2012 7:19 pm, edited 1 time in total.

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Birottam Dutta: Master | Next Rank: 500 Posts; Posts: 342; Joined: Wed Jul 08, 2009 8:50 amFollowed by: 19 members; Thanked: 214 times; GMAT Score:740

by Birottam Dutta » Sat Jul 28, 2012 11:37 pm

If mary takes x apples and (10-x) oranges then,

x(40) + (10-x)60 = 560 or x=2.

So, there are 2 apples and 8 oranges.

To make the average 53 now, mary is keeping oranges.

If she keeps 1, then she has 9 fruits and has spent a total of (560-60) = 500 cents. So, average = 500/9.

In this way, for 2 fruits, average = 440/8
In this way, for 3 fruits, average = 380/7
In this way, for 4 fruits, average = 320/6

So, after keeping 4 fruits, 320/6 is approx 53.

So, D is the answer.

Hope this is correct.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Jul 29, 2012 3:16 am

At a certain food 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 food 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 believe that the resulting average price is supposed to be 52 cents.

The original total cost of the 10 pieces of fruit = 10*56 = 560.
According to the answers, after 1, 2, 3, 4, or 5 pieces are removed -- so that 9, 8, 7, 6, or 5 pieces remain -- the average cost decreases to 52.
Since the prices are each a multiple of 10, the new total cost after the oranges are removed must also be a multiple of 10.
Only answer choice E will yield a new total cost that is a multiple of 10:
5*52 = 260.

The correct answer is E.

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ksc1940: Senior | Next Rank: 100 Posts; Posts: 68; Joined: Wed Oct 19, 2011 2:27 pmFollowed by: 2 members; Thanked: 2 times

by ksc1940 » Sun Jul 29, 2012 4:54 pm

Once we figure out the original number of apples and oranges, is there a quicker way to solve rather than plugging in for each combination? This invariably will take way longer than 2 minutes to solve.

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Sun Jul 29, 2012 5:15 pm

ksc1940 wrote:Once we figure out the original number of apples and oranges, is there a quicker way to solve rather than plugging in for each combination? This invariably will take way longer than 2 minutes to solve.

This is how i would do it using the balancing method.
Originally oranges/apples = (average price - price of apples)/(price of oranges-average price)
= (56-40)/(60-56) = 16/4 = 4:1
Since there are 10 fruit, we must have 8 oranges and 2 apples (4:1 ratio)

Lets assume that mary put back "x" oranges. Then we have 2 apples and 8-x oranges.
Then using the same method as above:

(8-x)/2 = (52-40)/(60-52) = 12/8 = 3/2
=>( 8-x)/2 = 3/2 => 8-x = 3
=> x = 8-3 = 5.

E is correct.

Alternatively, if you are not comfortable with the approach above, using good old algebra, and definition of averages:

Let original number of oranges = y, no. of apples = 10-y.
Then:
40*(10-y)+60*(y) = 56*10
400+20y = 560
=> y = 8.

Hence we have 8 oranges and 2 apples.
Let's say Mary removes "x" oranges, then oranges left =8-x, apples still =2
Then:
40*2 + 60*(8-x) = 52*(8-x+2)
=> 80 +480-60x = 520 - 52x
=> (60-52)x = 560-520 = 40
=> x= 40/8 = 5

E is correct.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Jul 29, 2012 6:50 pm

alex.gellatly wrote:At a certain food 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 food stand, and the average (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?
1
2
3
4
5

This is a mixture problem.
An average of 40 and an average of 60 are being mixed to yield an average of 56 in the first case and of 52 in the second case.
To determine the ratio of apples to oranges in each case, use ALLIGATION.

Case 1: average cost = 56.
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)

Step 2: Calculate the distances between the averages.
A(40)----------16----------56---4----O(60)

Step 3: Determine the ratio of apple to oranges.
The ratio of A to O is the RECIPROCAL of the distances in red.
A : O = 4:16 = 1:4.

Since A : O = 1:4 = 2:8, we know that A=2 and O=8, for a total of 10 pieces of fruit.

Case 2: average cost = 52.
A(40)----------12----------52---8----O(60)
A : O = 8:12 = 2:3.
Since the number of apples isn't changing, A=2 (same as above) and new O=3.

Thus:
Old O - new O = 8-3 = 5.

The correct answer is E.

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ksc1940: Senior | Next Rank: 100 Posts; Posts: 68; Joined: Wed Oct 19, 2011 2:27 pmFollowed by: 2 members; Thanked: 2 times

by ksc1940 » Sun Jul 29, 2012 9:49 pm

eagleeye wrote:
ksc1940 wrote:Once we figure out the original number of apples and oranges, is there a quicker way to solve rather than plugging in for each combination? This invariably will take way longer than 2 minutes to solve.
This is how i would do it using the balancing method.
Originally oranges/apples = (average price - price of apples)/(price of oranges-average price)
= (56-40)/(60-56) = 16/4 = 4:1
Since there are 10 fruit, we must have 8 oranges and 2 apples (4:1 ratio)

Lets assume that mary put back "x" oranges. Then we have 2 apples and 8-x oranges.
Then using the same method as above:

(8-x)/2 = (52-40)/(60-52) = 12/8 = 3/2
=>( 8-x)/2 = 3/2 => 8-x = 3
=> x = 8-3 = 5.

E is correct.

Alternatively, if you are not comfortable with the approach above, using good old algebra, and definition of averages:

Let original number of oranges = y, no. of apples = 10-y.
Then:
40*(10-y)+60*(y) = 56*10
400+20y = 560
=> y = 8.

Hence we have 8 oranges and 2 apples.
Let's say Mary removes "x" oranges, then oranges left =8-x, apples still =2
Then:
40*2 + 60*(8-x) = 52*(8-x+2)
=> 80 +480-60x = 520 - 52x
=> (60-52)x = 560-520 = 40
=> x= 40/8 = 5

E is correct.

Thanks for this. In the very last part where did you get that 52?

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Tue Jul 31, 2012 8:49 am

ksc1940 wrote: Thanks for this. In the very last part where did you get that 52?

The original question said 53 which did not match any of the options. (As x was coming out to be a fraction). So I changed it to 52.

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tisrar02: Master | Next Rank: 500 Posts; Posts: 130; Joined: Fri Apr 20, 2012 8:13 amFollowed by: 4 members; Location: Toronto, Ontario; Thanked: 16 times; GMAT Score:650

by tisrar02 » Tue Jul 31, 2012 5:33 pm

Here's how I solved this problem:

10 fruits cost an average of 56 cents.
10 x 0.56= 5.60

5 fruits cost an average of 52 cents.

5 x 0.52= 2.60

We need to find the number of oranges that will satisfy 5.60- 2.60= 3.00

Let X represent the number of Oranges that must be taken away.
X(.60)=3.00
X=3.00/.60
X=5

Mary must take away 5 oranges to fulfill the requirements.

Hope this helps

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