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There are 2 bars of goldsilver alloy; one piece has 2 parts
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18 Jun 2007, 18:15
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There are 2 bars of goldsilver 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
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Re: PS: Melted GoldSilver
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13 Mar 2011, 11:04
You can easily use the scale method here. The scale method is explained here: toughds105651.html#p828579Focus 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.
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X is weight of first bar and (8x) is weight of second bar.
(2/5)x + (3/10)(8x) = (5/16)*8
=>x = 1




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First bar is x kg, then second bar is (8x)kg
First bar:
Amt of gold = 2x/5kg
Amt of silver = 3x/5 kg
Second bar:
Amt of gold = 3(8x)/10 kg
Amt of silver = 7(8x)/10 kg
Total amt of gold = 2x/5 + 3(8x)/10 = (4x+243x)/10 = (x+24)/10 kg
Total amt of silver = 3x/5 + 7(8x)/10 = (6x+567x)/10 = (56x)/10 kg
Ratio of gold/silver = 5/11 = (x+24)/10 * 10/(56x)
5/11 = (x+24)/(56x)
5(56x) = (x+24)(11)
x = 1kg



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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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10 Apr 2012, 03:29
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.



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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



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Re: PS: Melted GoldSilver
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31 Mar 2010, 09:03
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



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Re: PS: Melted GoldSilver
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20 Mar 2010, 23:02
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.



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Re: PS: Melted GoldSilver
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09 Apr 2012, 20:56
VeritasPrepKarishma wrote: You can easily use the scale method here. The scale method is explained here: toughds105651.html#p828579Focus 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.
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Re: PS: Melted GoldSilver
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25 Apr 2011, 16:24
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



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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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03 Mar 2014, 23:56
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  (2524)  (3225) = 1:7 = 1/8 * 8 = 1kg = Answer = A



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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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02 Aug 2014, 13:30
Hi,
Lets do it by a different approach:
Lets take two Bars of GoldSilver 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.



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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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27 Mar 2017, 10:57
trahul4 wrote: There are 2 bars of goldsilver 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 goldsilver 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
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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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27 Feb 2018, 09:58
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 crossmultiply 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
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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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05 Mar 2020, 13:07
Mansoor50 wrote: my answer is switching bars!...HALP!!
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
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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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16 Mar 2020, 04:28
VeritasKarishma wrote: You can easily use the scale method here. The scale method is explained here: http://gmatclub.com/forum/toughds105651.html#p828579Focus 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.
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Re: PS: Melted GoldSilver
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13 Mar 2011, 03:25
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 * (0.40.3) = 2.5 2.4 = 0.1 => W1 = 1 So the answer is A.
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Re: PS: Melted GoldSilver
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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=8From, 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.



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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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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?



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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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24 Jul 2012, 22:45
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/
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Re: There are 2 bars of goldsilver alloy; one piece has 2 parts
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