ratio: Gold and silver

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vittalgmat: Legendary Member; Posts: 621; Joined: Wed Apr 09, 2008 7:13 pmFollowed by: 4 members; Thanked: 33 times

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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Source: Beat The GMAT — Problem Solving | Wed Dec 31, 2008 9:16 pm

crossingfingers: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Wed Apr 18, 2007 7:09 am; Thanked: 1 times

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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vivek.kapoor83: Legendary Member; Posts: 833; Joined: Mon Aug 04, 2008 1:56 am; Thanked: 13 times

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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vittalgmat: Legendary Member; Posts: 621; Joined: Wed Apr 09, 2008 7:13 pmFollowed by: 4 members; Thanked: 33 times

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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ronniecoleman: Legendary Member; Posts: 546; Joined: Sun Nov 16, 2008 11:00 pm; Location: New Delhi , India; Thanked: 13 times

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 !!

Admission champion, Hauz khaz
011-27565856

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korkun: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Thu Jan 01, 2009 12:04 pm; GMAT Score:740

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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