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mixture problem

by buoyant » Sat Jan 17, 2015 8:03 pm
How to solve this problem using alligation method?

There are two bars of gold-silver alloy. The first bar has 2 parts of gold and 3 parts of silver, and the other has 3 parts of gold and 7 parts of silver. If both bars are melted into an 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

[spoiler]OA: A [/spoiler]
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by Brent@GMATPrepNow » Sat Jan 17, 2015 8:35 pm
buoyant wrote:How to solve this problem using alligation method?

There are two bars of gold-silver alloy. The first bar has 2 parts of gold and 3 parts of silver, and the other has 3 parts of gold and 7 parts of silver. If both bars are melted into an 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
Let x = weight of the first bar (our goal is to find the value of x)
8-x = weight of the second bar [since the TOTAL weight is 8kg.]

Let's just keep track of the weight of gold in each bar.
First bar: 2 parts of gold to 3 parts of silver
So, for every 5 parts, 2 parts are gold and 3 parts are silver
In other words this bar is 2/5 gold
So, the weight of the gold in this bar is (2/5)x

Second bar: 3 parts of gold to 7 parts of silver
So, for every 10 parts, 3 parts are gold and 7 parts are silver
In other words this bar is 3/10 gold
So, the weight of the gold in the 2nd bar is (3/10)(8-x)

Final bar (after melting: 5 parts of gold to 11 parts of silver
So, for every 16 parts, 5 parts are gold, and 11 parts are silver
In other words this final bar is 5/16 gold
If the weight of the entire bar is 8 kg, the weight of the gold in this bar is (5/16)(8)

Now, we'll write an equation using the fact that:
(weight of gold in 1st bar) + (weight of gold in 2nd bar) + (weight of gold in final bar)
So, (2/5)x + (3/10)(8-x) = (5/16)(8)
Simplify to get: 2x/5 + 12/5 - 3x/10 = 5/2
Eliminate fractions by multiplying both sides by 10 to get: 4x + 24 - 3x = 25
Simplify: 24 + x = 25
x = 1

Answer = A
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Here's a similar question to practice with: https://www.beatthegmat.com/mixture-rati ... 89861.html

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by [email protected] » Sat Jan 17, 2015 10:20 pm
Hi buoyant,

This question is essentially just a wordy weighted average question.

We're given the composition of 2 types of alloy bar:

Bar A: 2 parts gold, 3 parts silver = 2/5 gold = 40% gold
Bar B: 3 parts gold, 7 parts silver = 3/10 gold = 30% gold

We're going to "mix" these two bar "types" and end up with a mixture that is 5/16 gold

We know that the total mixture will be 8kg. We're asked how many of the kg are Bar A.

A = # of kg of Bar A
B = # of kg of Bar B

Now we can set up the weighted average formula:

(.4A + .3B) / (A + B) = 5/16

And cross-multiply and simplify....

6.4A + 4.8B = 5A + 5B
1.4A = 0.2B

Multiply by 5 to get rid of the decimals...

7A = B

A/B = 1/7

This tells us that the ratio of A to B is 1:7. Since the total weight is 8kg, we must have 1 kg of Bar A and 7 kg of Bar B.

Final Answer: A

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by GMATGuruNY » Sat Jan 17, 2015 11:33 pm
buoyant wrote:How to solve this problem using alligation method?

There are two bars of gold-silver alloy. The first bar has 2 parts of gold and 3 parts of silver, and the other has 3 parts of gold and 7 parts of silver. If both bars are melted into an 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
Let F = the first bar, S = the second bar, and M = the mixture of the two bars.
Alligation can be performed only with percentages or fractions.

Step 1: Convert the ratios to FRACTIONS.
F:
Since gold:silver = 2:3, and 2+3=5, gold/total = 2/5.
S:
Since gold:silver = 3:7, and 3+7=10, gold/total = 3/10.
M:
Since gold:silver = 5:11, and 5+11=16, gold/total = 5/16.

Step 2: Put the fractions over a COMMON DENOMINATOR.
F = 2/5 = 32/80.
S = 3/10 = 24/80.
M = 5/16 = 25/80.

Step 3: Plot the 3 numerators on a number line, with the numerators for F and S on the ends and the numerator for M in the middle.
F 32-----------25-----------24 S

Step 4: Calculate the distances between the numerators.
F 32-----7-----25-----1-----24 S

Step 5: Determine the ratio in the mixture.
The required ratio of F to S is equal to the RECIPROCAL of the distances in red.
F:S = 1:7.

Implication:
The 8kg MIXTURE of F and S -- in other words, the resulting 8kg bar -- is composed of 1kg of F and 7kg of S.

The correct answer is A.

More practice with alligation:
https://www.beatthegmat.com/ratios-fract ... 15365.html
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by buoyant » Sun Jan 18, 2015 12:39 pm
what i don't understand is that we took the fraction of GOLD in the alligation method, but ended up with the ratios of two MIXTURES.
does that mean we are talking about the % or fraction of gold in total amount of mixture and therefore, the ratio we got represents the parts of each bar in the final 8kg bar?
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by GMATGuruNY » Sun Jan 18, 2015 1:34 pm
buoyant wrote:what i don't understand is that we took the fraction of GOLD in the alligation method, but ended up with the ratios of two MIXTURES.
does that mean we are talking about the % or fraction of gold in total amount of mixture and therefore, the ratio we got represents the parts of each bar in the final 8kg bar?
The problem above requires that we determine following:
In what proportion should a bar that is 2/5 gold be combined with a bar that is 3/10 gold to yield a bar that is 5/16 gold?

Alligation determines the required ratio:
For every 1kg of the bar that is 2/5 gold, there must be 7kg of the bar that is 3/10 gold.
The resulting bar will be 5/16 gold.

Examples:
If the resulting bar is to be 8kg, 1kg of the bar that is 2/5 gold must be combined with 7kg of the bar that is 3/10 gold.
If the resulting bar is to be 16kg, 2kg of the bar that is 2/5 gold must be combined with 14kg of the bar that is 3/10 gold.
If the resulting bar is to be 24 kg, 3kg of the bar that is 2/5 gold must be combined with 21kg of the bar that is 3/10 gold.

In the problem above, the weight of the resulting bar is 8kg.
Thus, 1kg of the bar that is 2/5 gold must be combined with 7kg of the bar that is 3/10 gold.
The resulting 8kg bar will be 5/16 gold.
Proof:
Amount of gold in the first bar = (2/5)(1) = 2/5 kilograms.
Amount of gold in the second bar = (3/10)(7) = 21/10 kilograms.
Total amount of gold = (2/5) + (21/10) = 25/10 = 5/2 kilograms.
Fraction of gold in the resulting 8kg bar = (5/2)/8 = 5/16.
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by subhakimi » Mon Jan 19, 2015 9:49 am
Ratio are 2:3 and 3:7
so amount 2x/3x for the first bar (first bar is 5x)
and for second bar 3y/7y (second bar is 10y)

The ratio is 5:11 and notice the summation is 8
So we can say that 8 kg are divided into 2.5 and 5.5

2x + 3y=2.5 .............1
3x + 7y=5.5 ............ 2

Multiply Equation 1 by 3 ->
6x + 9y =7.5
Multiply Equation 2 by 2 ->
6x + 14y=11

So 5y=3.5 and y=0.7 so 10y=7
And 5x=8-7=1

Alternate Method
2:3=40% Gold
3:7=30% gold

5:11=31.25%

Applying alligation->
So the ratio is (31.25-30): (40-31.25)
1:7
Since the summation is already 8 we can say the answer is 1 kg.
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