Ratios

[beater]

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 kg
b. 3 kg
c. 5 kg
d. 6 kg
e. 7 kg

[piyush_nitt]

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 kg
b. 3 kg
c. 5 kg
d. 6 kg
e. 7 kg

Weight of Gold = 5/16* 8 = 2.5 KG
weight of Silver = 8-2.5 = 5.5 KG

Also,
In first Bar 2 parts of Gold and in second 3 parts of gold , therefore
5parts = 2.5Kg
hence 1 Part of Gold = .5Kg
similarly 1 Part of Silver = .55KG

Weight of 1st Bar = 2*.5 +3*.55 = 2.65 KG

IMO B ???

[Ian Stewart]

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 kg
b. 3 kg
c. 5 kg
d. 6 kg
e. 7 kg

There are a few ways to do this problem. It is a weighted average problem, so you could use a weighted average approach here. We have:

first bar: 2/5 gold
second bar: 3/10 gold
mixture: 5/16 gold

We should get a common denominator:

first: 32/80 gold
second: 24/80 gold
mix: 25/80 gold

It's now clear we need a lot more of the second bar than the first to get a mix with such a small amount of gold; indeed they should be mixed in a 7 to 1 ratio (to see this quickly, I'm using weighted average principles: the ratio of the distances to the combined average is the reverse of the ratio of the amounts of the two components of the mixture; that is, the ratio of the two parts is (32 -25) to (25 - 24), or 7 to 1). So the answer should be A: we want 1 kg of the first bar, 7 kg of the second.

That explanation may not make much sense if you haven't worked much with weighted averages. You could also do this algebraically:

say we have x kg of bar 1. We then have 8-x kg of bar two. The first bar is 2/5 gold, the second 3/10 gold, so when we mix them, we will have this much gold in the resulting mixture:

(2/5) * x + (3/10)*(8 - x) = (x/10) + 2.4

We know that the resulting bar weighs 8 kg and the ratio of gold to the total weight is 5/16:

[(x/10) + 2.4]/8 = 5/16
x/5 + 4.8 = 5
x = 1

Or you could likely backsolve this one, though I haven't tried it to see how fast it is.

[beater]

Fantastic explanation. Thanks Ian!

[dendude]

The weight of G in the mixture
= (5/16)*8 = 2.5

The weight of S in the mixture
= (11/16)*8 = 5.5

Lets,
Wt of 1st bar X
Wt of 2nd bar Y

Wt of G is
= (2/5)*X + (3/10)*Y from the 1st and 2nd bars
= 2.5 (which we have already established above)

Similarly Wt of S is
= (3/5)*X + (7/10)*Y
= 5.5

We now have two equations in X & Y for two unknowns

Solving these two equations (either plug-in or solve simultaneously)
X = 1 & Y = 7

Hence A