GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 06 Jul 2020, 17:04 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # There are 2 bars of gold-silver alloy; one piece has 2 parts

Author Message
TAGS:

### Hide Tags

Senior Manager  Joined: 16 May 2007
Posts: 465
There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

3
2
69 00:00

Difficulty:   65% (hard)

Question Stats: 64% (02:40) correct 36% (02:55) wrong based on 692 sessions

### HideShow timer Statistics

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?

A. 1 kg
B. 3 kg
C. 5 kg
D. 6 kg
E. 7 kg
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India

### Show Tags

28
17
You can easily use the scale method here. The scale method is explained here:
tough-ds-105651.html#p828579

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 29614 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.
_________________
Karishma
Veritas Prep GMAT Instructor

Senior Manager  Joined: 18 Jul 2006
Posts: 325

### Show Tags

18
1
8
X is weight of first bar and (8-x) is weight of second bar.
(2/5)x + (3/10)(8-x) = (5/16)*8

=>x = 1
##### General Discussion
CEO  Joined: 07 Jul 2004
Posts: 2903
Location: Singapore

### Show Tags

13
5
First bar is x kg, then second bar is (8-x)kg

First bar:
Amt of gold = 2x/5kg
Amt of silver = 3x/5 kg

Second bar:
Amt of gold = 3(8-x)/10 kg
Amt of silver = 7(8-x)/10 kg

Total amt of gold = 2x/5 + 3(8-x)/10 = (4x+24-3x)/10 = (x+24)/10 kg
Total amt of silver = 3x/5 + 7(8-x)/10 = (6x+56-7x)/10 = (56-x)/10 kg

Ratio of gold/silver = 5/11 = (x+24)/10 * 10/(56-x)

5/11 = (x+24)/(56-x)
5(56-x) = (x+24)(11)
x = 1kg
Intern  Joined: 19 Aug 2011
Posts: 20
Concentration: Finance, Entrepreneurship
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

5
1
For anybody who is used to solving mixture probs using the cross method

Attachment: IMG.JPG [ 11.34 KiB | Viewed 26188 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.
Manager  Joined: 21 May 2007
Posts: 103

### Show Tags

3
1
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
Intern  Joined: 23 Feb 2010
Posts: 7

### Show Tags

3
Bar 1 is x kg
Bar 2 is y kg

First equation (Gold):
2/5 * x + 3/10 * y = 5/16 * 8

Second equation (Silver):
3/5 * x + 7/10 * y = 11/16 * 8

Two equations, solve for x
X=1kg
Manager  Joined: 18 Feb 2010
Posts: 119
Schools: ISB

### Show Tags

2
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.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India

### Show Tags

2
VeritasPrepKarishma wrote:
You can easily use the scale method here. The scale method is explained here:
tough-ds-105651.html#p828579

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

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.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  Joined: 30 Mar 2011
Posts: 12

### Show Tags

1
1
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
SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1706
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

1
stevie1111 wrote:
For anybody who is used to solving mixture probs using the cross method

Attachment:
IMG.JPG

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.

Did in the same way, with one addition

Took LCM of 5, 10 & 16 which is 80

Multiplied 80 with all three fractions to get:

32 |||||||||||||||| 24

|||||| 25 ||||||||||||||

(25-24) |||||||| (32-25)

= 1:7 = 1/8 * 8 = 1kg = Answer = A
Intern  Joined: 16 Jan 2012
Posts: 7
WE: Information Technology (Computer Software)
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

1
Hi,

Lets do it by a different approach:

Lets take two Bars of Gold-Silver alloy and name them as A and B.

Now

let the total alloy content in A : 2x+3x
let the total alloy content in B : 3y+7y

now total weight of the mixture is 8 kg. That means its A+B

so the equation will be: 2x+3x+3y+7y=8

5x+10y=8 …….. a)

Also its given that the ration of the constituent mixture is 5:11( Gold: Silver)

Now that equates as:

2x+3y 5
———— = --------
3x+7y 11

i.e., 7x=2y……b)

now solving equations a and b gives x= 1/5 and y= 7/10

So the total weight of Alloy A= 2(1/5)+3(1/5)= 1

Therefore, the total weight of Alloy A is 1 Kilograms.
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11042
Location: United States (CA)
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

1
trahul4 wrote:
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?

A. 1 kg
B. 3 kg
C. 5 kg
D. 6 kg
E. 7 kg

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 17027
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

1
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.

GMAT assassins aren't born, they're made,
Rich
_________________
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 17027
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

1
Mansoor50 wrote:

1st bar: x kg with 2/5 gold
2nd bar: y kg with 3/10 gold
melted: (x plus y) kg 5/16 gold

now......

2x/5 + 3y/10 = 5/16(x +y) --> LCM is 80 and simpllifying gives us

32x + 24y = 25x + 25y --> 32x - 25x = 25y - 24y --> 7x = 1y

why is the switch occuring where as it shd be 1x and 7y!

Hi Mansoor50,

You have correctly calculated the ratio of the two weights, but you still have think in terms of the weight of X vs the weight of Y.

You properly determined that 7(X) = 1(Y).... this means that "7 times the value of X" is equal to "1 times the value of Y"

For example, when X = 1.... Y = 7

The prompt tells us that the total weight of the mixed bar is 8km, which means we'll have 1kg of X and 7 kg of Y in the mixture.

GMAT assassins aren't born, they're made,
Rich
_________________
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

1
You can easily use the scale method here. The scale method is explained here:
http://gmatclub.com/forum/tough-ds-105651.html#p828579

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.

Quote:
Hi Karishma,

For the scale method can we use gold to silver ratios rather than gold to total and work this problem out?
I understand in this case gold to total is an easier fraction to work with but trying to understand possibilities in scale method

No, you cannot use gold/silver ratio.
Gold/Total is the gold concentration (C1 and C2) in bar1 and bar2. You are finding the gold concentration in the two bars combined using volume as the weight of the weighted average.

Cavg = (C1 * Volume of bar1 + C2 * Volume of bar2) / (Total volume of the two bars)

Cavg = (Amount of gold in bar1 + Amount of gold in bar2) / (Total volume of the two bars)

So you can work with either gold concentration or silver concentration but not with gold/silver ratio. It doesn't give you any concentration.
_________________
Karishma
Veritas Prep GMAT Instructor

Retired Moderator B
Joined: 16 Nov 2010
Posts: 1163
Location: United States (IN)
Concentration: Strategy, Technology

### Show Tags

5/16 * 8 = Final weight of gold

2/5 * W1 = G1

3/10 * W2 = G2

W1 + W2 = 8

G1 + G2 = 5/16 * 8

2/5 * W1 + 3/10 * W2 = 5/16 * 8

2/5 * W1 + 3/10 * (8 - w1) = 5/16 * 8

W1(2/5 - 3/10) = 5/2 - 24/10

W1 * (0.4-0.3) = 2.5 -2.4 = 0.1

=> W1 = 1

_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings
Intern  B
Joined: 21 May 2011
Posts: 1

### Show Tags

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. Intern  Joined: 18 Jun 2012
Posts: 29
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

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?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India
Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

### Show Tags

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.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (85 - 40)/(40 - 20) = 45/20 = 9/4
(This is just the formula representing the scale. Check here if it's unclear: http://www.veritasprep.com/blog/2011/03 ... -averages/)

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.

For more on this, check out the first example here: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
_________________
Karishma
Veritas Prep GMAT Instructor Re: There are 2 bars of gold-silver alloy; one piece has 2 parts   [#permalink] 24 Jul 2012, 22:45

Go to page    1   2    Next  [ 30 posts ]

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