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midpoint / word problem

by smallsorrow » Thu Nov 06, 2008 1:31 am
At a certain food stand, the price of each apple is $0.40 and the price of each orange is
$0.60. 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 $0.56. How many oranges must Mary
put back so that the average price of the pieces of fruit that she keeps is $0.52?
A. 1
B. 2
C. 3
D. 4
E. 5
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by raunekk » Thu Nov 06, 2008 3:10 am
imo:E
(5)

A+O= 10...(1)

0.40A + 0.60B/ 10 = 0.56...(2)

solving (1) and (2)

O= 8 and A = 2

We can form an equation to get mean as 0.52

let x be the number of oranges 2b removed

0.40 (2) + 0.60(8-x) / 10 -x = 0.52

solving

x=5

thus E

but this took me more around 3 mins,,,

wats d OA??
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Re: midpoint / word problem

by sudhir3127 » Thu Nov 06, 2008 3:25 am
smallsorrow wrote:At a certain food stand, the price of each apple is $0.40 and the price of each orange is
$0.60. 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 $0.56. How many oranges must Mary
put back so that the average price of the pieces of fruit that she keeps is $0.52?
A. 1
B. 2
C. 3
D. 4
E. 5

even i go with E.5
As raunekk did ... i dont think so there is any faster way of doing it .. its that we need to solve them fast... i just did as u did .. i took me 2 mints .. so i think its more of practice .. but still wud be interested if theres any shorter way ...
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by jnellaz » Thu Nov 06, 2008 7:03 am
At a certain food stand, the price of each apple is $0.40 and the price of each orange is
$0.60. 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 $0.56. How many oranges must Mary
put back so that the average price of the pieces of fruit that she keeps is $0.52?
A. 1
B. 2
C. 3
D. 4
E. 5

I did it this way - which may be a bit quicker -

We know that the average price of friut (for 10 pieces) is $0.56 - the total spent should be 10 X 0.56 = $5.60.

Now they are asking what would bring the average price of the pieces of fruit down to $0.52. So working backwards - we use the answer choices. Lets start with C. If we put back 3. We are left with 7 pieces of fruit that need to average to $.52. BUT when we multiply 7 by $.52 we gat $3.64. There is no combination of apples and oranges that could come up to that amount because the units position is not zero. Remember that apples and oranges are both .40 cents and .60 cents. If you try all the answer choices, the only answer choice that could have a zero in the units position is E)5.

Hope this helps!
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by cramya » Thu Nov 06, 2008 5:24 pm
Let x be the apples , y be the oranges

.40x+.60y = 5.6

4x+6y=56 (1)

x+y=10 (2)

From 1) and 2) y=8 x=2

Let k be the oranges that are present after Mary puts back a certain number of oranges (which we need to find)

Given .60k + .40 * 2 / k+2 = .52

Solving for k; k= 3

k=3 means Mary put back 5 oranges since she had 8 oranges originally!!

E)
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by logitech » Thu Nov 06, 2008 5:31 pm
cramya wrote:Let x be the apples , y be the oranges

.40x+.60y = 5.6

4x+6y=56 (1)

x+y=10 (2)

From 1) and 2) y=8 x=2

Let k be the oranges that are present after Mary puts back a certain number of oranges (which we need to find)

Given .60k + .40 * 2 / k+2 = .52

Solving for k; k= 3

k=3 means Mary put back 5 oranges since she had 8 oranges originally!!

E)
Too many ZERO's again :lol:

Just use 40, 60 and 56 to make the calculations easy :idea:
LGTCH
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Re: midpoint / word problem

by logitech » Thu Nov 06, 2008 5:41 pm
smallsorrow wrote:At a certain food stand, the price of each apple is $0.40 and the price of each orange is
$0.60. 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 $0.56. How many oranges must Mary
put back so that the average price of the pieces of fruit that she keeps is $0.52?
A. 1
B. 2
C. 3
D. 4
E. 5
For the ease of calculation, I will use x100 of the numbers

So Apple : 40, Orange :60

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

Lets call this X

Apple's contribution to average will not change, so we don't need to worry about apple. It is ORANGE TIME! :lol:

The difference between 10 items and 10-x items can be formulated as:

60X = (56*10) - 52*(10-X)

X=5


Solved in 20 seconds! :twisted:
LGTCH
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Re: midpoint / word problem

by mental » Sat Nov 08, 2008 12:15 am
Apple's contribution to average will not change, so we don't need to worry about apple. It is ORANGE TIME!

The difference between 10 items and 10-x items can be formulated
great approach Logitech
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by Acetylcholine » Sun Feb 08, 2009 5:42 am
Hi all,

I just started preparing for the GMAT and the apples/oranges problem is the first one I wanted to tackle ... and failed miserably. In search of help I came across this thread. I can follow as far as this:

A+O= 10...(1)

0.40A + 0.60B/ 10 = 0.56...(2)


as posted by raunekk, but how do you then go about resolving both equations so that you get O= 8 and A = 2? Could someone please post a step-by-step explanation on how to resolve (1) and (2)?

Thanks - I know I have a long way to go to beating the GMAT!

Cheers!
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by sureshbala » Sun Feb 08, 2009 6:46 am
Folks, if you aregood at the average concepts, this can be answered much faster than what we discussed above.

First let me get out of those decimals. I would take the price of an apple to be 40 and orange as 60. Given that the average of the 10 fruits purchased initially is 56.

For every orange fruit given back since it's price is 4 more than the present average obviously for certain number of oranges given back we want the average to come down to 52 from 56 that is we are looking to reduce the average by 4.

Let's say this happens for x oranges. The relative reduction in the sum is 4x. This reduction of 4x in the sum upon the present 10-x fruits should give us a reduction of 4 in the average.

Hence 4x/(10-x) = 4. Hence x=5.

Definitely this is much quicker...Get into this relative calculation
Last edited by sureshbala on Sun Feb 08, 2009 6:56 am, edited 1 time in total.
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by sureshbala » Sun Feb 08, 2009 6:54 am
Folks, those of you who could not follow the above relative calculation look at the following.

First of all, I don't understand why most of you are calculating the initial number of apples and oranges. Definitely we can avoid this calculation.

First let me get out of those decimals. I would take the price of an apple to be 40 and orange as 60. Given that the average of the 10 fruits purchased initially is 56. So total money spent is 560.

Let's say we are giving back x oranges so that the average of the remaining fruits comes down to 52.

Since we gave back x oranges, the money that we spend in this case is 560-60x.......(1)

Also since the average of these remaining 10-x fruits is 52, the money spent by us should be 52(10-x).....(2)

Obviously (1) and (2) are same...so 560-60x = 52(10-x)

Thus x = 5. Hence 5 oranges must be given back.
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by Acetylcholine » Sun Feb 08, 2009 10:23 am
Cheers sureshbala, much appreciated. I understand that in order to solve this problem, calculating the number of apples of oranges is not required. However, for the sake of math practice, I'd love understand how to calculate the original number of apples and oranges. Could you or someone else provide a step-by-step guide on how to go about this after putting together the two equations


A+O= 10...(1)

0.40A + 0.60B/ 10 = 0.56...(2)

?

thanks, Thomas
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by gmatfailed » Wed Feb 24, 2010 10:13 am
Acetylcholine wrote:Cheers sureshbala, much appreciated. I understand that in order to solve this problem, calculating the number of apples of oranges is not required. However, for the sake of math practice, I'd love understand how to calculate the original number of apples and oranges. Could you or someone else provide a step-by-step guide on how to go about this after putting together the two equations


A+O= 10...(1)

0.40A + 0.60B/ 10 = 0.56...(2)

?

thanks, Thomas
hey here is my explanation that has lot of math but its easy

Step 1. Let the no of organges be x....then the no of apple will be 10-x

Its given that average of fruits is 0.56 cents which means each of the fruit costs .56 cents...and we have 10 fruits...so the total cost would be $5.6

lets calculate how many apples and organges we initially have

.40(10-x)+ .60x= 5.6
x=8 which is the no of oranges....no of apples is 2

Now the question reads as or states that how many oranges we can drop so our new average is 0.52

x(.60)+2(.40)= (x+2).52
x=3
only if we have 3 organges and 2 apples our average be .52
but initially we had 8 oranges...so if he drop 5 only can we bring the average to .52
This isn't complicated..all you have to do is understand the concept....first try to solve mentally or just prepare a flow chart that shows what steps are required to answer and what is given....with bit of practice you can tackle such questions easily....the other folks here have solved this question intuitively which is extremely good. Remember to ace on the gmat you need to make sure you know how to tackle problems from multiple approaches.
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by mcdesty » Wed Mar 06, 2013 9:52 am
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by mcdesty » Wed Mar 06, 2013 9:53 am
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