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ratio: Gold and silver

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ratio: Gold and silver

by vittalgmat » Wed Dec 31, 2008 9:16 pm
There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 8-kg bar with the final ratio of 5:11 (gold to silver), what was the weight of the first bar?

A. 1
B. 3
C. 5
D. 6
E. 7

[spoiler]OA: A[/spoiler]

I am looking for multiple ways to solve this problem.

Thanks in advance.
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by crossingfingers » Wed Dec 31, 2008 10:17 pm
here is one way:

Based on the ratios:

1 equation is: (2/5)x + (3/10)y = (5/16)*8

2nd equation is x+y=8

solving the 2 we can get the answer for x (which in this case is gold) = 1
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by vivek.kapoor83 » Thu Jan 01, 2009 1:36 am
another way , wht i tried was similar as above but was lenghty..So cross 's way is better

i did it like this

(2/5)x+(3/10)y =(5*8)/16
(3/5)x+(7/10)y = (11*8)/16

Then if u solve for X u ll get 1.
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by vittalgmat » Thu Jan 01, 2009 1:47 am
Thanks Cross,
I guess his way is faster. To summarize his method.

part1/whole1 + part2/whole2 = combined_part/combined_whole * weight.

The other eqn gives the relation between weights.
as in x +y = 8

Solve the sumultaneous eqn to get the answer
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by ronniecoleman » Fri Jan 02, 2009 4:03 am
There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 8-kg bar with the final ratio of 5:11 (gold to silver), what was the weight of the first bar?

A. 1
B. 3
C. 5
D. 6
E. 7


Let first weigh: x
second y

x + y = 8

3/5 x + 7/10 y / (2/5 x+ 3/10 y) = 11/5

solve !!
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by korkun » Fri Jan 02, 2009 7:54 am
add 5 to 11 = 16
2+3=5
3+7=10

80 is the least possible multiple for 16, 5 and 10.

so the weights of gold are: 32/80 and 24/80
and the last weight is 25/80 ----->(5/(5+11))

24 25 .........32

25-24=1
32-25=7

so the ratio must be 7/1 =7

and according to this ratio the solution is 1.
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