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Sugar - Bit on the Difficult side

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Sugar - Bit on the Difficult side

by harsh.champ » Thu Feb 18, 2010 11:42 pm
A salesman mixes three varieties of sugar costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?

(A) 1 kg
(B) 2 kgs
(C) 3 kgs
(D) 5 kgs
(E) 6 kgs


The OA is D.
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by ajith » Fri Feb 19, 2010 12:04 am
harsh.champ wrote:A salesman mixes three varieties of sugar costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?

(A) 1 kg
(B) 2 kgs
(C) 3 kgs
(D) 5 kgs
(E) 6 kgs


The OA is D.
20% increased price = 30 Rs

Cost = 30/1.2 = 25 Rs

2 Kgs of 30 Rs/Kg is there in the mixture

x kgs of 24Rs/Kg and y Kgs of 22 Rs/Kg sugar are there

now 30*2 + x*24+y*22/(x+y+2) = 25
30*2 + x*24 + y*22 = 50 +25 x+ 25 y

x+3y = 10

Which has infinite solutions

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by shashank.ism » Sun Feb 21, 2010 3:26 pm
harsh.champ wrote:A salesman mixes three varieties of sugar costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?
(A) 1 kg
(B) 2 kgs
(C) 3 kgs
(D) 5 kgs
(E) 6 kgs The OA is D.
salesman mixes three varieties of sugar costing Rs.20/kg, Rs.24/kg and Rs.30/kg
profit = 20% = 0.2 x 30 = Rs. 6 /kg
So CP = 30 -6 = Rs. 24 /kg
let x and y be the amount of 1st and 2nd added
--> 24 = 20 x +24 y + 2 (30)
--> 20 x +24y + 36 = 0
--> 5x + 6y+9 =0

I am not reaching upto the answer.....Harsh is the question correct...
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by ashforgmat » Mon Feb 22, 2010 1:05 am
Info is incomplete. this cannot be solved. We need to find out how much of the total mixture he made and probably one more equation to solve this.
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