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Orange

Expert replies
by MBA.Aspirant » Sun Jun 12, 2011 10:03 am
By selling 45 oranges for Rs 160, a women loses 20%. How many oranges
should she sell for Rs 112 to gain 20% on the whole ?
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Source: — Problem Solving |

by Frankenstein » Sun Jun 12, 2011 10:13 am
Hi,
When she sells for Rs160, loss is 20%.
If C is her cost price then C(1-0.2) = 160 => C=200.
So, cost of each orange is 200/45 = 40/9
Now, she sells some more oranges(n) for 112 to make profit of 20% on the whole.
So, total selling price is 160+112 = 272.
Cost price of (45+n) oranges = 272/1.2 = 680/3
So, 45+n = (680/3)/(40/9) = 51.
So, n =6
So she should sell 6 oranges for Rs 112.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by Surev » Sun Jun 12, 2011 10:48 am
S.P = 160; Loss% = 20; So C.P(of 45 Oranges) = 200/-
C.P of 1 orange = 200/45
Suppose she sells n oranges
C.P of n oranges = (200/45)*n
Profit % = 20; S.P of n oranges = 112
So 1.2 * (200/45)*n = 112
n = 21
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by finites » Sun Jun 12, 2011 10:49 am
45 - 160 - 20%loss ie 160 is 80% of the toatl value x...80/100x = 160, x = 200..
200 is buying price of 45 mangoes. and selling price is 160..
each mango will cost 200/45 ie 40/9

let 45 + n be equal to z. ie total mangoes and total selling price is 272.
objective is to make 20% profit on the whole by selling n mangoes at 112.

buying price of z mangoes + 20/100 of buying price of z mangoes (ie 20%profit) = total selling price
ie z * 40/9 + 20/100 * z *40/9 = 160 + 112

z * 40/9(1.2) = 272
z = 272*9*10/(40 * 12)
z = 51.

z = 45 + n
51 = 45 +n
n = 6..
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by finites » Sun Jun 12, 2011 10:50 am
In the above .forget mangoes..take oranges !
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