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A kilogram of nut mixture contains X% chestnuts and Y% walnu

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by varun289 » Sat Apr 13, 2013 10:44 am
A kilogram of nut mixture contains X% chestnuts and Y% walnuts and sells for $7.00/kg. If the ratio of chestnuts is increased by 50% so that the new mixture is sold for $8.00/kg, what is the price of a kg of walnuts?

A. $1.00
B. $2.50
C. $5.00
D. $7.50
E. $10.00
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by Anju@Gurome » Sat Apr 13, 2013 7:07 pm
varun289 wrote:A kilogram of nut mixture contains X% chestnuts and Y% walnuts and sells for $7.00/kg. If the ratio of chestnuts is increased by 50% so that the new mixture is sold for $8.00/kg, what is the price of a kg of walnuts?
Let us assume, price of a kg of chestnut is C and that of walnuts is W.

Initial ratio of chestnuts and walnuts = X/Y = n
Let us assume that we have mixed n kg of chestnuts and (1 - n) kg of walnuts to get a total of 1 kg mixture.
So, total price = nC + (1 - n)W = 7 ........................ (A)

New ratio = x/y + 50% 0f x/y = 3x/2y = 3n/2 = 1.5n
Let us assume, we have mixed 1.5n kg of chestnuts and (1 - 1.5n) kg of walnuts to get a total of 1 kg mixture.
So, total price = 1.5nC + (1 - 1.5n)W = 8 ---> 3nC + (2 - 3n)W = 16 ........................ (B)

Multiplying (A) with 3 and subtracting (B) from it, 3*(1 - n)W - (2 - 3n)W = 21 - 16
--> 3W - 3nW - 2W + 3nW = 5
--> W = 5

The correct answer is C.
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by CSASHISHPANDAY » Sun Apr 14, 2013 1:35 am
Weight of chestnut = x% of 1 kg
Price of chestnut = A
Weight of walnut = y% of 1 kg = (1-x)% of 1 kg
Price of walnut = B

XA + (1-X) B = 7 ……………………….i
(1.5X) A + (1 - 1.5X) B = 8 …………….ii

From 1 and 2:
0.5XA – 0.5XB = 1
XA – XB = 2
XA = 2 + XB…………………………… iii

Substitute the value of XA on Equation i:
XA + (1-X) B = 7
2 + XB + B - XB = 7
B = 7 - 2
B = 5.00

Therefore, it is C.
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by veenu08 » Thu Apr 18, 2013 2:11 am
Can anyone please suggest where I am going wrong-

aX + bY = 700

3aX + bY = 3200
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