There are 2 bars of gold-silver alloy; one piece has 2 parts

Official Solution:

There are two bars made of gold-silver alloy. The first bar contains 2 parts of gold and 3 parts of silver by weight, while the second contains 3 parts of gold and 7 parts of silver by weight. If both bars are melted and combined to form a single 8-kilogram bar with a gold-to-silver ratio of 5:11 by weight, what was the weight of the first bar?

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


Let's assume that the weight of the first bar is \(x\) kilograms and the weight of the second bar is \(8-x\) kilograms. To determine the value of \(x\), we must construct an equation that we can solve for \(x\). One possible method is to equate the weights of gold in the first and second bars to the weight of gold in the final mixture.

The weight of gold in the first bar is \(\frac{2}{2+3}*x\) kilograms, and the weight of gold in the second bar is \(\frac{3}{3+7}*(8-x)\) kilograms. By adding these two quantities and setting the result equal to the weight of gold in the final bar, which is \(\frac{5}{5+11} * 8 = 2.5\) kilograms, we get the equation \(\frac{2}{5}*x + \frac{3}{10}*(8-x)=2.5\).

To simplify the equation, we multiply both sides by 10, giving us \(4x + 3*(8-x)=25\). Solving for \(x\), which represents the weight of the first bar, we find that \(x=1\) kilogram.


Answer: A

I cannot give a generic solution to the mixture problems. These may vary and hence must be assessed on individual merit (some require systematic work, others can be solved by considering the LCMs).

Here is how you can solve the given problem
Let X and Y be the weights of 1st and 2nd bar respectively.
Thus, X+Y=8 ..................(1)

To get a second relation, recognize that 2/5 of first and 3/10 of second bar are gold. Similar 3/5 and 7/10 respectively for silver.
Use this relation
(2/5)X+(3/10)Y
-------------------
(3/5)X+(7/10)Y


The above ratio is equal to 5:11

Two equations and 2 variables. Solve to get X=1.

best,
parsifal

First bar has 40% gold and second bar has 30% gold by weight.

The resulting 8 kg bar has 31.25% of gold by weight.


So by obviously we know that maximum percent of second bar would go to make the final bar because 31.25 % is close to 30%. Working with options we get 1 kg of bar one as the answer. (we can use interpolation)


Hope this helps if time is short.

I solve mixture problems like simultaneous equations.
Look first at the amount of gold. In the final amount you have 8 kg, and since the ratio is 5:11, this means there must be 5/16 amount gold, or 5/2kg of gold in the end result.
So let x be the amount of the first gold bar and y the amount of y gold bar added to create an equation of gold amounts:
2/5X + 3/10Y = 5/2
You also know however that:
X + Y = 8
Since you want to know the value of X, rearrange to:
Y = 8 - X
Now substitute into the first equation to solve for X = 1

Responding to a pm:

Think of what the formula is:

w1/w2 = (C2 - Cavg)/(Cavg - C1)

First bar: C1 = 32/80, w1 = weight of first bar
Second bar: C2 = 24/80, w2 = weight of second bar
Cavg = 25/80

Simply plug these values in the formula.

w1/w2 = (C2 - Cavg)/(Cavg - C1) = (24/80 - 25/80)/(25/80 - 32/80) = 1/7
w1:w2 = 1:7

You don't need to worry about anything else.

When using the scale method, we flip the ratio because we calculate (Cavg - C1) first and (C2 - Cavg) later. This is opposite to the way it is in the formula so we flip the ratios.

Above, when I made the scale, I put the second bar first and the first bar later. The reason was that it is more intuitive that way on the number line since C2 = 24/80 is smaller than C1 = 32/80. Since I was finding Cavg - C2 first and C1 - Cavg later, I didn't need to flip the ratios.

My advice would be to simply identify one element as Element1, another as Element2 and figure out C1, w1 and C2, w2 for the 2 of them and simply plug in the formula. There will be no confusion in that case.

For anybody who is used to solving mixture probs using the cross method



Hence, the ratio of the two gives 1/7 i.e 1 part of 1st bar and 7 parts of second.
Please notice that the question can be easily twisted to ask for the weight of first bar for minimum possible integer weight of final bar (Figure 8kg not provided) in which case this method would certainly help.

can some one solve it using cross method ?

A portion of the 85% solution of chemicals was replaced with an equal amount of 20% solution of chemicals. As a result, 40% solution of chemicals resulted. What part of the original solution was replaced?

When you remove a portion of 85% solution and replace it with 20% solution, you are basically just mixing 85% and 20% solution to get 40% solution.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (85 - 40)/(40 - 20) = 45/20 = 9/4

So 20% solution : 85% solution = 9:4

So out of a total 13 lts, 9 lts is 20% solution and 4 lts is 85% solution. This means 9/13 of the original solution was replaced by the 20% solution.

We are given that the first bar of gold-silver alloy has a ratio of gold to silver of 2x : 3x (thus it has a weight of 5x) and the second bar has a ratio of gold to silver of 3y : 7y (thus it has a weight of 10y). Both bars are melted into a new bar that has a final ratio of gold to silver of 5 : 11 and a weight of 8 kg. We can create the following two equations:

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

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

22x + 33y = 15x + 35y

7x = 2y

3.5x = y

We also know that 5x + 10y = 8. Since y = 3.5x, that gives us:

5x + 10(3.5x) = 8

40x = 8

x = 8/40 = 1/5

Since x = 1/5, the total weight of the first bar is 5(1/5) = 1 kg.

Answer: A

1. (weight of first bar*proportion of gold) +(weight of second bar*proportion of gold) / (weight of first bar*proportion of silver) +(weight of second bar*proportion of silver)= 5/11
2. x*(2/5) + y*(3/10) / x*(3/5) + y*(7/10) =5/11
3. So x/y =1/7 and we have x+y=8
x=1 kg or weight of the first bar is 1 kg

Hi All,

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:

GMAT assassins aren't born, they're made,
Rich

Hello stevie1111, can this question be solved by using the alligation method on the overall ratio of gold to silver, or do we need to use ratios for specifically one of the substances, either gold or silver, when performing the calculation?

I see above that you have solved this question using the alligation method with ratios of just gold in each bar (1,2, and 3), but was wondering if we can use the ratios 2/3, 3/7, and 5/11 (ratios of gold to silver in each bar) to solve the problem as well. I keep doing the calculation and getting the ratio of bar 1's weight to bar 2's weight as 6/49, which is very close to 7/49 (which is equivalent to 1/7), so I am wondering if I'm making a calculation mistake. I have solved this question via the algebraic (non-alligation) method as well, which makes sense, but I'm trying to understand the allegation method more clearly. Please assist.