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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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 20 Oct 2015, 03:19
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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20 Oct 2015, 04:15
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Bunuel wrote: At a certain fruit 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 fruit 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
Kudos for a correct solution. Apple (A) = 40 cemts Orange (B) = 60 cents Average of 10 fruits = 56 cents u.e. Total Price of 10 Fruits = 56*10 = 560 cents i.e. 40A + 60 B = 560 and A + B = 10 i.e. 40A + 60 (10-A) = 560 i.e. -20A + 600 = 560 i.e. A = 2 i.e. B = 8 Average to be brought at = 52 cents let oranges to be put back = C i.e. Total Fruits = 10-C where total Oranges left = 8-C i.e. Total Price of New lot of fruits = 52*(10-C) i.e. 40*2 + 60*(8-C) = 52*(10-C) i.e. 80 + 480 - 60C = 520 - 52C i.e. 8 C = 40 i.e. C = 5 Answer: Option E
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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20 Oct 2015, 04:31
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Ans E
Sum of Apples and Oranges is 10
A + O = 10
Apple cost 40 cent and Orange cost 60 cent and there mean is 56 , therefore
(40*A + 60*O)/10 = 56 , which gives us
4A + 6O = 56
Solving the above 2 equation we get , Apples = 2 , Oranges = 8.
Now as she has remove some oranges ( x ) such that the new mean of the remaining fruits become 52
note that the number of Apples remain the same , therefore new equation will become
( 2*40 + (8-x)60 ) / 10 -x = 52
As the remaining fruits will be 10 - x
On solving for x we will get our ans 5
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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20 Oct 2015, 05:39
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Let number of Apples = A number of oranges = B A+B=10 --- 1 .56 =(.4A + .6 B)/10 => 56 = 4A + 6B ----2 Solving 1 and 2, we get A= 2 B= 8 Let the number of oranges put back = C 52*(10-c) = 40*2 + 60(8-C) => C= 5 Answer E Alternatively , we can use Scale method Since average is .56 , distance of the average from price of apple =.56-.4=.16 and distance of the average from price of orange = .6 - .56 = .04 Ratio of distances is .16:.04 = 4:1 Therefore , number of apples and oranges will be in inverse proportion =1:4 Number of apples = 2 Number of oranges = 8 After removing a certain number of oranges C , the new average will be .52 Ratio of distances is .08:.12 = 2:3 Number of oranges will be 3 Therefore Mary must put 5 oranges back . Answer E
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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20 Oct 2015, 06:34
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Step 1: We start with the first average to find the total price of the apples and oranges.
Using the formula : Sum = Average * Number of terms
Sum = 56 * 10 = 560
Step 2: Set up the new average based on the first one.
x : number of oranges we need to take away.
We will have the previous sum (i.e. 560) minus the price of all oranges we will take away (i.e. 60x) and the number of elements will be 10 minus x.
So :
(560-60x) / (10-x) = 52
560 - 60x = (10-x) * 52
560 - 520 = -52x + 60x
40 = 8x
x=5
The answer is E.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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20 Oct 2015, 21:37
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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20 Oct 2015, 22:25
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Bunuel wrote: At a certain fruit 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 fruit 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
Kudos for a correct solution. here price of apple = 40 cents and price of orange = 60 cents average price = 56 cents so, the apples and oranges are in ratio apple : orange = 60 - 56 : 56 - 40 => apple : orange = 4 : 16 => apple : orange = 1 : 4 total no =10, therefore apples = 2 nos. and oranges = 8 nos. now we have to make average price = 52 centsso, the apples and oranges will be in ratio apple : orange = 60 - 52 : 52 - 40 => apple : orange = 8 : 12 => apple : orange = 2 : 3 now we have apples = 2 nos. from above, therefore oranges must be 3 in number oranges to be put back = 8 - 3 = 5 nosAnswer choice Ekudos if you like the explanation
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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22 Mar 2016, 03:34
v12345 wrote: Bunuel wrote: At a certain fruit 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 fruit 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
Kudos for a correct solution. here price of apple = 40 cents and price of orange = 60 cents average price = 56 cents so, the apples and oranges are in ratio apple : orange = 60 - 56 : 56 - 40 => apple : orange = 4 : 16 => apple : orange = 1 : 4 total no =10, therefore apples = 2 nos. and oranges = 8 nos. now we have to make average price = 52 centsso, the apples and oranges will be in ratio apple : orange = 60 - 52 : 52 - 40 => apple : orange = 8 : 12 => apple : orange = 2 : 3 now we have apples = 2 nos. from above, therefore oranges must be 3 in number oranges to be put back = 8 - 3 = 5 nosAnswer choice Ekudos if you like the explanation Hi! I used the exact same way as you did in solving this question, but when reaching the last part 2:3 I interpereted it as 3/5 oranges and 2/5 apples? Seeing as we're still closer to 60 than 40. And then we should have 6 oranges and 4 apples? Please explain!
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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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22 Mar 2016, 03:37
My bad, I see now that we were not supposed to replace them with any apples but just remove them from the total. Didn't read carefully enough 
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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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Updated on: 04 Sep 2016, 06:29
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Bunuel wrote: At a certain fruit 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 fruit 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
Kudos for a correct solution. Solution: We can let the number of apples = x and the number of oranges = y. Using these variables we can create the following two equations: 1) x + y = 10 Using the formula average = sum/quantity, we have: 2) (40x + 60y)/10 = 56 Let’s first simplify equation 2: 40x + 60y = 560 4x + 6y = 56 2x + 3y = 28 Isolating for y in equation one gives us: y = 10 – x. Since y = 10 – x, we can substitute 10 – x for y in the equation 2x + 3y = 28. This gives us: 2x + 3(10 – x) = 28 2x + 30 – 3x = 28 -x = -2 x = 2 Since x + y = 10, then y = 8. We thus know that Mary originally selected 2 apples and 8 oranges. We must determine the number of oranges that Mary must put back so that the average price of the pieces of fruit that she keeps is 52¢. We can let n = the number of oranges Mary must put back. Let’s use a weighted average equation to determine the value of n. [40(2) + 60(8-n)]/(10 – n) = 52 (80 + 480 - 60n)/(10 – n) = 52 560 – 60n = 520 – 52n 40 = 8n 5 = n Thus, Mary must put back 5 oranges so that the average cost of the fruit she has kept would be 52 cents. Answer: E
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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04 May 2016, 15:07
let a=#apples
r=#oranges
we know that .4a+.6r=(.56)(10)=5.6➡.4a=5.6-.6r
after mary returns x oranges, then
.4a+.6(r-x)=(.52)(10-x)
substituting for .4a,
5.6-.6r+.6r-.6x=5.2-.52x
.08x=.4
8x=40
x=5 oranges put back
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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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Updated on: 18 May 2016, 10:02
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A = number of apples
B = number of oranges
We have :
\(\frac{40A + 60 B}{10}\) = 56
and \(\frac{40A + 60 (B - x)}{10 - x}\) = 52
\(\frac{40A + 60 B - 60 x}{10 - x}\) = 52
\(\frac{560 - 60 x}{10 - x}\) = 52
560 - 60 x = 520 - 52 x
40 = 8 x
x = 5
Originally posted by Alex75PAris on 04 May 2016, 15:19.
Last edited by Alex75PAris on 18 May 2016, 10:02, edited 1 time in total.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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14 May 2016, 11:28
Using Averages and Allegations
40 60
56
4 16
A:O =1: 4==>A=2 ,O=8(as total no of apples+ oranges =10)
now we know there are 2 Apples and 8 Oranges
Using Statistics
when Mean =52
52-40=12
12*2(bcos 2 apples)
60-52=8
x= no. of remaining oranges
24-(8*x)=0
==>x=3
Hence 5 oranges must be removed.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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17 May 2016, 21:17
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i did this one with a scale method
Apples : Price : Oranges
40 cents : 56 average : 60 cents
= 16 cents from average : average : 4 cents from average
= 1 apple for every 4 oranges
= 2 apples and 8 oranges = 10 fruits = 2*.4 + 8*.6 = 5.60 / 10 = .56
40 cents : 52 average : 60 cents
= 12 cents from average : average : 8 cents from average
= 2 apples for every 3 oranges
8 oranges - 3 oranges = 5 oranges must be put back
Hope this helps!
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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18 Jun 2016, 01:43
Although I got this right, it took me 10 minutes to solve it. Can anyone help me with a strategy to tackle this type of question on GMAT? There is no way im going to solve this question on the exam.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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18 Jun 2016, 05:08
total amount = avg * number of items
old amount = 56 * 10 = 560
new amount will be 52 *10-x
10-x , you have same number of apples , only decrease in number of oranges
now new amount is equal to what
560 - 60 x = new amount
solve and x= 5
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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The current total price is divisible by 10, and so is the price of each orange. The adjusted total price therefore will have to be divisible by 10, and at the same time divisible by 52. The only possible value for the adjusted total price is 260. Hence 300 cents' worth of oranges, or 5 oranges, will have to be removed.
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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09 Oct 2016, 20:13
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Bunuel wrote: At a certain fruit 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 fruit 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
Kudos for a correct solution. \(\frac{Wa}{Wo} = \frac{(60-56)}{(56-40}) = \frac{1}{4}\) Apple = 2 & Orange = 8 New Avg \(\frac{Wa}{Wo} = \frac{(60-52)}{(52-40)} = \frac{2}{3}\) \(\frac{2}{Wo} = \frac{2}{3}\) Orange = 3
Hence, Oranges have to reduce from 8 to 3 i.e. 5
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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15 Oct 2016, 03:01
Not convinced mate... I am able to get to fractions 1/4 and 2/3.... but given the total number of items is 10, why would make the ratio of 1/4 to 2/8 and keep the ratio 2/3 as it is ? If someone can breakdown this using the same formula that would be helpful. colorblind wrote: Bunuel wrote: At a certain fruit 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 fruit 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
Kudos for a correct solution. \(\frac{Wa}{Wo} = \frac{(60-56)}{(56-40}) = \frac{1}{4}\) Apple = 2 & Orange = 8 New Avg \(\frac{Wa}{Wo} = \frac{(60-52)}{(52-40)} = \frac{2}{3}\) \(\frac{2}{Wo} = \frac{2}{3}\) Orange = 3
Hence, Oranges have to reduce from 8 to 3 i.e. 5 |
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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]
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Dev1212 wrote: Not convinced mate... I am able to get to fractions 1/4 and 2/3.... but given the total number of items is 10, why would make the ratio of 1/4 to 2/8 and keep the ratio 2/3 as it is ? If someone can breakdown this using the same formula that would be helpful. Bunuel wrote: At a certain fruit 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 fruit 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
Kudos for a correct solution. Please find the solution as attached. Answer: option E I hope this helps!!!
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