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# John bought a total of 12 Mangoes and Oranges. Each Mango co

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Director
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John bought a total of 12 Mangoes and Oranges. Each Mango co [#permalink]

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31 Jul 2013, 08:04
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Question Stats:

73% (02:46) correct 27% (02:50) wrong based on 173 sessions

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John bought a total of 12 Mangoes and Oranges. Each Mango costs 80 cents and each orange costs 60 cents. If the average price of the 12 mangoes and oranges that John originally purchased was 65 cents, then how many oranges needs to return to raise the average price of his purchase to 72 cents?

a) 4
b) 5
c) 6
d) 7
e) 8
[Reveal] Spoiler: OA

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Re: John bought a total of 12 Mangoes and Oranges. Each Mango co [#permalink]

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31 Jul 2013, 08:09
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Using alligation (refer to my Mixture post in my signature) we get that $$\frac{M}{O}=\frac{1}{3}$$, so we have 3 Mangoes and 9 Oranges.

The number x of Oranges he has to return is:

$$\frac{3*0.8+(9-x)*0.6}{12-x}=0.72$$, solve for x and obtain $$x=7$$.

Hope it's clear, for the method I've used refer here tips-and-tricks-mixtures-151906.html
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Re: John bought a total of 12 Mangoes and Oranges. Each Mango co [#permalink]

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31 Jul 2013, 09:55
Let number of mangoes be x, number of oranges be 12-x

0.80x +(12-x)0.60/12 = 0.65
solving for x, we get x = 3 --> Mangoes 3, Oranges 9

Now, number of oranges to be returned be y

0.80*3 + (9-y)*0.60/12-y = 0.72

solving for y, y = 7
Ans: D

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Re: John bought a total of 12 Mangoes and Oranges. Each Mango co [#permalink]

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24 Sep 2013, 10:16
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M 80 cents O=60cents Average= 65 then number of M/number of O=(65-60)/(80-65)=1/3

Since there are 12 fruits in total then we have 1x+3x=12 > x=3 M=1*3=3 ; O=3*3=9
**
M 80 cents O=60cents Average= 72 then number of M/number of O=(72-60)/(80-72)=3/2

note that the number of M hasnt changed, but the number of O has changed from 9 to 2
so, we need to remove 9-2=7 yummy oranges
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John bought a total of 12 Mangoes and Oranges. Each Mango co [#permalink]

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02 Jul 2014, 17:32
fozzzy wrote:
John bought a total of 12 Mangoes and Oranges. Each Mango costs 80 cents and each orange costs 60 cents. If the average price of the 12 mangoes and oranges that John originally purchased was 65 cents, then how many oranges needs to return to raise the average price of his purchase to 72 cents?

a) 4
b) 5
c) 6
d) 7
e) 8

There is no need to even calculate this. A 50-50 split between mangoes and oranges give a price of 70 cents. Since 72 cents is a little bit above the arithmetic mean, we should also pick an answer a little bit above the arithmetic mean.

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Re: John bought a total of 12 Mangoes and Oranges. Each Mango co [#permalink]

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20 Jun 2017, 07:30
fozzzy wrote:
John bought a total of 12 Mangoes and Oranges. Each Mango costs 80 cents and each orange costs 60 cents. If the average price of the 12 mangoes and oranges that John originally purchased was 65 cents, then how many oranges needs to return to raise the average price of his purchase to 72 cents?

a) 4
b) 5
c) 6
d) 7
e) 8

We can let the number of mangos = m and number of oranges = r. We can create two equations:

m + r = 12

m = 12 - r

and

65 = (80m + 60r)/12

780 = 80m + 60r

78 = 8m + 6r

39 = 4m + 3r

Since m = (12 - r), we have:

39 = 4(12 - r) + 3r

39 = 48 - 4r + 3r

9 = r

Since r = 9, m = 3.

Let’s let x = the number of oranges to return. We can create the following average equation:

72 = [80*3 + 60*(9-x)]/(12 - x)

72(12 - x) = 240 + 540 - 60x

864 - 72x = 780 - 60x

84 = 12x

7 = x

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Re: John bought a total of 12 Mangoes and Oranges. Each Mango co [#permalink]

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29 Oct 2017, 12:02
fozzzy wrote:
John bought a total of 12 Mangoes and Oranges. Each Mango costs 80 cents and each orange costs 60 cents. If the average price of the 12 mangoes and oranges that John originally purchased was 65 cents, then how many oranges needs to return to raise the average price of his purchase to 72 cents?

a) 4
b) 5
c) 6
d) 7
e) 8

let x=number of oranges needed to return
72¢=(12*65¢-x*60¢)/(12-x)
x=7 oranges
D

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Re: John bought a total of 12 Mangoes and Oranges. Each Mango co   [#permalink] 29 Oct 2017, 12:02
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