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 Post subject: GRE Statistics Posted: Thu Jul 19, 2012 4:14 pm Joined: Thu Jul 19, 2012 3:34 pm
Posts: 3
help me answer this question.

Mary went to a grocery and bought oranges and apples. One apple cost 40 cents and one orange cost 60 cents. She bought 10 fruits in total, and the average cost of a fruit was \$0.56 . How many oranges must she take out of her basket in order to lower the average price to \$0.52?

Top Post subject: Re: average problem Posted: Fri Jul 20, 2012 2:11 pm Joined: Sun May 30, 2010 2:23 am
Posts: 498
What did you try to do? What went wrong?

Top Post subject: Re: average problem Posted: Fri Jul 20, 2012 7:54 pm Joined: Thu Jul 19, 2012 3:34 pm
Posts: 3
I got stuck when I arrived at the question how many oranges should their be in the first place.

Top Post subject: Re: average problem Posted: Sat Jul 21, 2012 5:45 pm Joined: Sun May 30, 2010 2:23 am
Posts: 498

Top Post subject: Re: average problem Posted: Sat Jul 21, 2012 7:20 pm Joined: Thu Jul 19, 2012 3:34 pm
Posts: 3
okay so this is the following equation im getting

o+a=10
-.6(o)+.4(a)=5.6
.4(o)+.6(a)=4.4
or

a=10-0
.4(o)+6-.6(o)=5.6

and then I end up with o= -1.6/.4
o=-4
and that is impossible, could you please show me how you would solve it, step by step, thank you.

Top Post subject: Re: average problem Posted: Sun Jul 22, 2012 3:04 am Joined: Sun May 30, 2010 2:23 am
Posts: 498
The prices were mistakenly switched, when the problem was rewritten in the proper form (Now corrected). So the proper prices were: an orange cost 60 cents and an apple cost 40 cents. (In order to decrease the average we must take out the more expensive fruit).

Here is the solution to find the original number of oranges.
We already have the first equation a + o = 10 . Let's write down the second equation.

The average is the total sum divided by the number of fruits. One orange cost \$0.6, so Mary paid \$(0.6 × o) for the oranges. Similarly, she paid \$(0.4a) for the apples. Therefore the total sum is
0.6o + 0.4a
There are 10 fruits, so the average is (0.6o + 0.4a)/10 . We know that it is \$0.56 . So here is the second equation:
(0.6o + 0.4a)/10 = 0.56
0.6o + 0.4a = 5.6

So we have the system of two equations:
o + a = 10
0.6o + 0.4a = 5.6

Manipulate the second equation to make it easier (another way would be to plug in a = 10 – o from the first into the second):
0.6o + 0.4a = 5.6
6o + 4a = 56
3o + 2a = 28

Our system becomes
o + a = 10
3o + 2a = 28

2o + 2a = 20
3o + 2a = 28

Deduct the first from the second to get:
3o – 2o + 2a – 2a = 28 – 20
o = 8
a = 10 – 8 = 2

Now you have the original number of oranges. Try to find the new number of oranges (Remember the number of apples stays the same). So now you have only one variable, the number of oranges. Do the same steps as for constructing the second equation. When you tried it by yourself – read further.
-----------

Here is the second part of the solution:
She has 2 apples and this number won't change Let's denote the new number of oranges by x. So there will be (2 + x) fruits.
She will pay \$(0.4 × 2) for apples and \$(0.6x) for oranges, which makes \$(0.6x + 0.8) in total.
The average becomes \$(0.6x + 0.8)/(2 + x) . It must be \$0.52 , so here is the new equation:
(0.6x + 0.8)/(2 + x) = 0.52
0.6x + 0.8 = 0.52(2 + x)
0.6x + 0.8 = 1.04 + 0.52x
0.08x = 0.24
x = 3

The new number of oranges is 3, the original number is 8, so Mary must take out 8 – 3 = 5 oranges.
The final answer is 5.

Note, that the numbers are quite simple (only 10 fruits) and this problem can also be solved by plugging in the values on either step. (You plug in the number of apples and oranges that add up to the total and check if the average fits). But it's better to use equations. Once you get comfortable with constructing them, any problem is solvable.

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