GRE Statistics
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Author:  Ignacio Pastor [ Thu Jul 19, 2012 4:14 pm ]
Post subject:  GRE Statistics

Could you please
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?

Author:  Gennadiy [ Fri Jul 20, 2012 2:11 pm ]
Post subject:  Re: average problem

What did you try to do? What went wrong?

Author:  Ignacio Pastor [ Fri Jul 20, 2012 7:54 pm ]
Post subject:  Re: average problem

I got stuck when I arrived at the question how many oranges should their be in the first place.

Author:  Gennadiy [ Sat Jul 21, 2012 5:45 pm ]
Post subject:  Re: average problem

I got stuck when I arrived at the question how many oranges should their be in the first place.
If there is some unknown quantity – write it down as a variable. Don't be afraid to use many variables if needed. Some of them might get a fixed value after.

So let's denote the number of oranges by o and the number of apples by a. What do we know from the question statement?

1. o + a = 10

What else do we have in the statement? The average price! Write down the equation for the average price of a fruit and you'll have the system of equations to solve.

Author:  Ignacio Pastor [ Sat Jul 21, 2012 7:20 pm ]
Post subject:  Re: average problem

okay so this is the following equation im getting



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

Author:  Gennadiy [ Sun Jul 22, 2012 3:04 am ]
Post subject:  Re: average problem

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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