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apples and oranges ratio

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by vishubn » Thu Oct 02, 2008 7:42 pm
At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. 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. How many oranges must Mary put back so that the average
price of the pieces of fruit that she keeps is ¢ 52?
A. 1
B. 2
C. 3
D. 4
E. 5
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by Azntycoon » Thu Oct 02, 2008 8:04 pm
Apples = A
Orange = R

i) We are told that:
(0.4A + 0.6R)/10 = 0.56

ii) Solving it further, we get: (0.4A + 0.6R) = 5.6

iii) Picking numbers, we know that A + R = 10. Start with picking numbers for R. 10 won't work because it will be (0.4(0) + 0.6(10) = 6. So you keep working your way, and you will find that the optimal combination is A = 2 and R = 8.

iv) Set up another equation to solve for the "new" R. We know that A is fixed at 2, so let the new total "x" = 2 + R

v) Therefore: 0.52x = (0.4*2) + (0.6)(x-2)

vi) x from step above equals 5. The answer is therefore C because 8 - 5 = 3.
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by stop@800 » Thu Oct 02, 2008 8:13 pm
avg = 56
total = 560

A no of apples
B no of oranges

(40A + 60B)/(A+B) = 56


40a+60b = 56a+56b
4b = 16a
b=4a

so
apples = 2
oranges = 8


for new avg

(40*2 + 60(8-x))/(10-x) = 52
80 + 480 - 60x = 520 - 52x

560 - 520 = 8x
40 = 8x
x = 5

Ans 5
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by vishubn » Thu Oct 02, 2008 8:24 pm
yaaa

thanksss stop@800!! u nailed it for sure

vishu
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by stubbornp » Thu Oct 02, 2008 9:47 pm
stop@800 wrote:avg = 56
total = 560

A no of apples
B no of oranges

(40A + 60B)/(A+B) = 56


40a+60b = 56a+56b
4b = 16a
b=4a

so
apples = 2
oranges = 8


for new avg

(40*2 + 60(8-x))/(10-x) = 52
80 + 480 - 60x = 520 - 52x

560 - 520 = 8x
40 = 8x
x = 5

Ans 5

with 2 apples and 8 oranges--80+480/10=56

with 4 apples and 6 oranges-160+360/10=52.....

answer should be 2.....
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by stop@800 » Thu Oct 02, 2008 9:53 pm
stubbornp wrote:
stop@800 wrote:avg = 56
total = 560

A no of apples
B no of oranges

(40A + 60B)/(A+B) = 56


40a+60b = 56a+56b
4b = 16a
b=4a

so
apples = 2
oranges = 8


for new avg

(40*2 + 60(8-x))/(10-x) = 52
80 + 480 - 60x = 520 - 52x

560 - 520 = 8x
40 = 8x
x = 5

Ans 5

with 2 apples and 8 oranges--80+480/10=56

with 4 apples and 6 oranges-160+360/10=52.....

answer should be 2.....
The Q says

How many oranges must Mary put back so that the average
price of the pieces of fruit that she keeps is ¢ 52?


What you are doing is replacing apples with oranges
the Q is only asking to remove apples.

Hope this helps!!
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by stubbornp » Fri Oct 03, 2008 12:30 am
stop@800 wrote:
stubbornp wrote:
stop@800 wrote:avg = 56
total = 560

A no of apples
B no of oranges

(40A + 60B)/(A+B) = 56


40a+60b = 56a+56b
4b = 16a
b=4a

so
apples = 2
oranges = 8


for new avg

(40*2 + 60(8-x))/(10-x) = 52
80 + 480 - 60x = 520 - 52x

560 - 520 = 8x
40 = 8x
x = 5

Ans 5

with 2 apples and 8 oranges--80+480/10=56

with 4 apples and 6 oranges-160+360/10=52.....

answer should be 2.....
The Q says

How many oranges must Mary put back so that the average
price of the pieces of fruit that she keeps is ¢ 52?


What you are doing is replacing apples with oranges
the Q is only asking to remove apples.

Hope this helps!!

tnx buddy...thats a big mistake i did...hope ll not do the same on exam day... :o
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by stop@800 » Fri Oct 03, 2008 1:01 am
stubbornp wrote:
stop@800 wrote:
stubbornp wrote:
stop@800 wrote:avg = 56
total = 560

A no of apples
B no of oranges

(40A + 60B)/(A+B) = 56


40a+60b = 56a+56b
4b = 16a
b=4a

so
apples = 2
oranges = 8


for new avg

(40*2 + 60(8-x))/(10-x) = 52
80 + 480 - 60x = 520 - 52x

560 - 520 = 8x
40 = 8x
x = 5

Ans 5

with 2 apples and 8 oranges--80+480/10=56

with 4 apples and 6 oranges-160+360/10=52.....

answer should be 2.....
The Q says

How many oranges must Mary put back so that the average
price of the pieces of fruit that she keeps is ¢ 52?


What you are doing is replacing apples with oranges
the Q is only asking to remove apples.

Hope this helps!!

tnx buddy...thats a big mistake i did...hope ll not do the same on exam day... :o
All the best!!
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by Azntycoon » Fri Oct 03, 2008 3:55 am
stop@800 wrote:avg = 56
total = 560

A no of apples
B no of oranges

(40A + 60B)/(A+B) = 56


40a+60b = 56a+56b
4b = 16a
b=4a

so
apples = 2
oranges = 8


for new avg

(40*2 + 60(8-x))/(10-x) = 52
80 + 480 - 60x = 520 - 52x

560 - 520 = 8x
40 = 8x
x = 5

Ans 5
thanks. I inadvertently mix-ed up the numbers in the last step. The answer is indeed 5.
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by nitin86 » Fri Oct 03, 2008 10:05 am
I think, we can solve it as

560 - n* ( 60) = 52* (10 - n)
where n is the number of oranges removed.

1. Total cost of 10 (apple + orange) = 56 * 10 = 560
2. Let say n oranges are removed, so the cost decreased by n * 60
3. Now, if n oranges are removed, then

Total fruits = 10 - n

4. Total cost, if the average is 52, is 52 * (10-n)
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by stop@800 » Fri Oct 03, 2008 2:23 pm
nitin86 wrote:I think, we can solve it as

560 - n* ( 60) = 52* (10 - n)
where n is the number of oranges removed.

1. Total cost of 10 (apple + orange) = 56 * 10 = 560
2. Let say n oranges are removed, so the cost decreased by n * 60
3. Now, if n oranges are removed, then

Total fruits = 10 - n

4. Total cost, if the average is 52, is 52 * (10-n)
yes we can
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