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There are 2 bars of gold-silver alloy; one piece has 2 parts [#permalink]
18 Jun 2007, 18:15

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Difficulty:

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Question Stats:

63% (03:27) correct
37% (02:43) wrong based on 257 sessions

There are 2 bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver and another has 3 parts of gold to 7 parts of silver. If both bars are melted into 8 kg bar with the final gold to silver ratio of 5:11. What was the weight of the first bar?

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

Can somebody please explain me how to solve these kind of questions.. I'm having a very tough time with mixture problems. Thanks

There are 2 bars of gold-silver alloy ;one pice has 2 parts of gold to 3 parts of silver and another has 3 parts of gold to 7 parts of silver. If both bars are melted into 8 kg bar with the final gold to silver ratio of 5:11. What was the weight of the first bar?

Let x be the amount of original solution.
8-x would be the amount added.

äh how to make 3silve+7silver=11 silver?
maybe it should be 10..
If so i have still the problem that I dont know what for "parts"
are they the same weight? Or the same volume? (gold has nearby the double weight!)
If, if , if:
2) 3kg

Re: PS: Melted Gold-Silver [#permalink]
20 Mar 2010, 23:02

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This post received KUDOS

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. _________________

Focus on any one of the two elements say Gold. First bar has 2/5 gold. Second bar has 3/10 gold and mixture has 5/16 gold. Make the fractions comparable for easy calculation i.e. give them the same denominator. LCM of 5, 10 and 16 is 80. First bar has 32/80 gold. Second bar has 24/80 gold and mixture has 25/80 gold.

Attachment:

Ques3.jpg [ 5.9 KiB | Viewed 9242 times ]

So first bar weight:second bar weight should be in the ratio 1:7. Out of 8 total kgs, first bar must have been 1 kg. _________________

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

Re: PS: Melted Gold-Silver [#permalink]
01 Aug 2011, 08:47

Shortest way to solve this Problem is to analyze it for about 30 secs and you have an answer in the 31st.

Here we go-:

We have been given the Gold (G) to Silver (S) ratio for the first alloy which is 2:3.

We can formulate the question as 2G+3S= What?

And the question stem says that, when we have merged both first and the second piece of the alloys, the ratio for the merged piece is 5:11.

The second equation can be written as 5G+11S=8

Now, our job is to apply our mind a little

If 5G+3S=8 From, the first equation, it can be inferred that, in the second equation G is 2.5 times and S is approx. 3.7 times of what they are in the first equation 2*2.5= 5 and 3*3.7= 11 approx.

Since, 2.5 + 3.7= 6.2 approx.; which is an addition to the existing first alloy

And 8 - 6.2= 1.8, which could have been the weight for the first alloy. Nearest figure in the answer choices available to us is 1 and that's our answer.

gold = x silver 8-x 2/5*x+3/10*(8-x)=5/10*8 x=1 Ans.

Hi Baten Can you explain the purpose of taking the 3/10 into consideration? I guess you have counted the gold of the 1st bar and gold of first bard. In that case that (8-x) will not be applicable to 3/10 ratio? Correct me if I misunderstood.

Focus on any one of the two elements say Gold. First bar has 2/5 gold. Second bar has 3/10 gold and mixture has 5/16 gold. Make the fractions comparable for easy calculation i.e. give them the same denominator. LCM of 5, 10 and 16 is 80. First bar has 32/80 gold. Second bar has 24/80 gold and mixture has 25/80 gold.

Attachment:

Ques3.jpg

So first bar weight:second bar weight should be in the ratio 1:7. Out of 8 total kgs, first bar must have been 1 kg.

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

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. _________________

Re: There are 2 bars of gold-silver alloy; one piece has 2 parts [#permalink]
10 Apr 2012, 03:29

2

This post received KUDOS

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

Attachment:

IMG.JPG [ 11.34 KiB | Viewed 5929 times ]

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.

Re: There are 2 bars of gold-silver alloy; one piece has 2 parts [#permalink]
24 Jul 2012, 22:05

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?

Focus on any one of the two elements say Gold. First bar has 2/5 gold. Second bar has 3/10 gold and mixture has 5/16 gold. Make the fractions comparable for easy calculation i.e. give them the same denominator. LCM of 5, 10 and 16 is 80. First bar has 32/80 gold. Second bar has 24/80 gold and mixture has 25/80 gold.

Attachment:

Ques3.jpg

So first bar weight:second bar weight should be in the ratio 1:7. Out of 8 total kgs, first bar must have been 1 kg.

By far the fastest method to deal with these questions. Kudos! _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: There are 2 bars of gold-silver alloy; one piece has 2 parts [#permalink]
24 Jul 2012, 22:45

Expert's post

smartmanav wrote:

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.