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by [email protected] » Wed Jun 08, 2011 6:29 am
A lump of two metals weighing 18 gms. is worth Rs. 87 but if their weights be interchanged, it would be worth Rs. 78.60. If the price of one metal be Rs. 6.70 per gm, find the weight of the other metal in the mixture.

A] 12
B] 20
C] 30
D] 8
E] 10


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by Frankenstein » Wed Jun 08, 2011 6:45 am
Hi,
Let x be the weight of metal costing 6.7/gm and 'p' be the cost/gm of the other metal
So, (6.7)x+p(18-x) = 87
p(x)+(6.7)(18-x) = 78.6
Solving we get p = 2.5 and x=10.
So, 18-x = 8

Hence, D
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by cans » Wed Jun 08, 2011 7:03 am
A,B two metals
weight= a,b
price=p,q per gram
p=6.7
ap+bq=87
bp+aq=78.6
a+b=18
to find b. a=18-b
q=(78.6 - 6.7b)/(18-b)
(18-b)6.7 + b((78.6 - 6.7b)/(18-b))=87
solving, b=8
IMO D
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by sunilramu » Wed Jun 08, 2011 7:13 am
Plugging in results.

Eliminate B and C since they are greater than 18.

plugging in 8.

8(x) + 10 (6.7) = 87
=> x = 2.5
validate if, 8(6.7) + 10(2.5) = 78.6 and answer is yes.

I chose 8 because 8 * 6.7 would give me a unit digit of 6 in 78.6 and the other number would be 10 which would cancel out decimal on the other side. It was a lucky guess.

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