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Question Stats:
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46% (01:21) wrong based on 1 sessions
There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 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 Source: GMAT Club Tests - hardest GMAT questions
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I think there is an easier way to do this. Using the algebra is great but is time consuming and leaves a lot of room for mistakes. Convert the ratios into percentages. 2/5 = .4 and 3/10 = .3 5/16 = .31 Then using the famous mixture method in one of the prior postings.. mixture-problems-made-easy-49897.html.4 .3 \ / .31 / \ .01 .09 gives the ratio for bar 1 to bar 2 is 1:9 bar 1 to total is 1:10 1/10 = x/8 x =.8 ~ 1 therefore A
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vksunder wrote: There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 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 w1= weight of Ist bar w2= weight of IInd bar w1 +w2 = 8 w2 = 8 - w1 I bar: g1/s1 = 2/3 gold weight = w1(2/5) = 2w1/5 silver weight = w1(3/5) = 3w1/5 II bar: g2/s2 = 3/7 gold weight = w2(3/10) = (8-w1) (3/10) = 3(8-w1)/10 silver weight = w2(7/10) = (8-w1) (7/10) = 7(8-w1)/10 Togather: g/s = 5/11 [(2w1/5) + 3(8-w1)/10]/[(3w1/5) + 7(8-w1)/10] = 5/11 (4w1 + 24 - 3w1) / (6w1 + 56 -7w1) = 5/11 44w1 + 11*24 - 33w1 = 30w1 + 280 -35 w1 16w1 = 280 - 264 w1 = 1 A.
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vksunder wrote: There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 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 Here is how I solve it: 1)First bar: 2x+3x=a 2)Second bar: 3y+7y=b 3)Combined bar: 5n+11n=8; n=0.5; So, we have 2.5kg gold and 5.5kg silver in total. 4)total gold: 2x+3y=2.5 5)total silver: 3x+7y=5.5 From 4&5 we find that x=0.2 kg => Substituting in equation 1 yields a=1
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There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 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 weight of bar1 be x kgs. So the weight of bar2 will be ( 8 - x) kgs The first bar is 2/5 gold, the second 3/10 gold 2/5 (x) + 3/10 ( 8 - x) 2x/5 + 24/10 - 3x/10 4x/10 + 24/10 - 3x/10 x/10 + 24/10 ( This is the amount of gold in the mixture) given: weight of the mixture is 8 kgs and the ratio of gold to total weight is 5/16 8 [ x/10 + 24/10 ] = 5/16 x/10 + 24/10 = 5/16 times 8 x/10 + 24/10 = 5/2 x/10 + 24/10 = 25/10 x + 24 = 25 x = 1 So the answer is A. www.graduatetutor.com
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x - wieght of the first alloy y - weight of the socond 0,4*x+0,3*y=2,5 x+y=8 Hence x=1
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I think the fastest way to answer the question is to look at the composition of the two bars. The first bar contains 40% (2/5) of gold, while the second bar contains 30% (3/10) of gold. They are to be melted into a 8kg bar, which contains slightly more than 30% of gold (5/16).
Given that the gold content of the final bar is very close to the gold content of the second bar, it has to consist to an overwhelming majority of the second bar (i.e. close to 8kg). This means that the weight of the first bar has to be very low, hence choice A.
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There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 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
We can solve this by the rule of Alligation
1) Gold content in bar 1 = 2/2+3 = 2/5
2) Gold content in bar 2 = 3/3+7 = 3/10
3) Gold content in mixture of bar 1 and bar 2 = 5/16
4) Now by rule of Alligation we have:
2/5 3/10
5/16
So ratio of mixing is (5/16-3/10)/ (2/5 - 5/16) = 1/7
Since total weight given is 8 kg so 1 Kg is weight of bar 1 and 7 kg is weight of bar 2.
This is known as rule of alligation which can be used to solve questions on mixtures.
Using this technique we can solve the problem is less than 2 minutes.
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GMAT TIGER wrote: vksunder wrote: There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 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 w1= weight of Ist bar w2= weight of IInd bar w1 +w2 = 8 w2 = 8 - w1 I bar: g1/s1 = 2/3 gold weight = w1(2/5) = 2w1/5 silver weight = w1(3/5) = 3w1/5 II bar: g2/s2 = 3/7 gold weight = w2(3/10) = (8-w1) (3/10) = 3(8-w1)/10 silver weight = w2(7/10) = (8-w1) (7/10) = 7(8-w1)/10 Togather: g/s = 5/11 [(2w1/5) + 3(8-w1)/10]/[(3w1/5) + 7(8-w1)/10] = 5/11 (4w1 + 24 - 3w1) / (6w1 + 56 -7w1) = 5/11 44w1 + 11*24 - 33w1 = 30w1 + 280 -35 w1 16w1 = 280 - 264 w1 = 1 A. I am confused. I hope these kind questions are not in reag GMAT!
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...
given: weight of the mixture is 8 kgs and the ratio of gold to total weight is 5/16
8 [ x/10 + 24/10 ] = 5/16
x/10 + 24/10 = 5/16 times 8
x/10 + 24/10 = 5/2
x/10 + 24/10 = 25/10
x + 24 = 25
x = 1
Sorry, I am a bit confused.. why do we multiply 5/16 by 8 rather than divide by 8?
Thanks for the great work!
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GMAT TIGER wrote: vksunder wrote: There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 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 w1= weight of Ist bar w2= weight of IInd bar w1 +w2 = 8 w2 = 8 - w1 I bar: g1/s1 = 2/3 gold weight = w1(2/5) = 2w1/5 silver weight = w1(3/5) = 3w1/5 II bar: g2/s2 = 3/7 gold weight = w2(3/10) = (8-w1) (3/10) = 3(8-w1)/10 silver weight = w2(7/10) = (8-w1) (7/10) = 7(8-w1)/10 Togather: g/s = 5/11 [(2w1/5) + 3(8-w1)/10]/[(3w1/5) + 7(8-w1)/10] = 5/11 (4w1 + 24 - 3w1) / (6w1 + 56 -7w1) = 5/11 44w1 + 11*24 - 33w1 = 30w1 + 280 -35 w1 16w1 = 280 - 264 w1 = 1 A. thanks very helpful
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The best approach is graduatetutor’s but it has a mistake stated by ongste, correcting that we have:
w1= weight of first bar in kg
w2= weight of second bar in Kg
Since we know that the total weight of both bars (w1 + w2) is 8 and we are asked for w1 then
w2=8-w1
Using the proportion for gold. It could either be used for silver, but here for gold:
2/5 w1 + 3/10 w2 = 8 (5/16)
2/5 w1 + 3/10(8-w1) = 5/2
2/5 w1 + 12/5 – 3/10 w1 = 5/2
2/5 w1 – 3/10 w1 = 5/2 – 12/5
1/10 w1 = 1/10
w1 = 1
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Thanks for the solutions...
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seofah wrote: Here is how I solve it:
1)First bar: 2x+3x=a
2)Second bar: 3y+7y=b
3)Combined bar: 5n+11n=8; n=0.5; So, we have 2.5kg gold and 5.5kg silver in total.
4)total gold: 2x+3y=2.5
5)total silver: 3x+7y=5.5
From 4&5 we find that x=0.2 kg => Substituting in equation 1 yields a=1 This is the easiest to understand and usually how I solve this kind of problems. Thank you so much!
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Anyone do this one in 2 minutes? I am thinking it took me more like 5.
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How about this way...
(1) Assume first bar weight as: 1 KG (i.e, First answer choice)
2:3 = 0.4:0.6
Since total weight is given as 8 KG, the Bar2 weight must be 7 kg.
3:7 for bar means, 3*(7/10):7*(7/10) =2.1:4.9
Now total: (0.4+2.1):(0.6+4.9) = 2.5:5.5=5/11 (Thats what given in the problem)
Bingo!!!! first choice is correct
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frankida wrote: I think there is an easier way to do this. Using the algebra is great but is time consuming and leaves a lot of room for mistakes. Convert the ratios into percentages. 2/5 = .4 and 3/10 = .3 5/16 = .31 Then using the famous mixture method in one of the prior postings.. mixture-problems-made-easy-49897.html.4 .3 \ / .31 / \ .01 .09 gives the ratio for bar 1 to bar 2 is 1:9 bar 1 to total is 1:10 1/10 = x/8 x =.8 ~ 1 therefore A All you guys are spinning around in circles setting up equations and stuff. I think I'm pretty good at math; however, I think the mnemonic makes these mixtures very easy. It should take you a minute tops. mixture-problems-made-easy-49897.htmlIn my opinion, doing well on the GMAT is about eliminating steps in your processes so that you have less chances to make mistakes. Setting up equations is basically give yourself more opportunities to make simple mistakes that tend to happen when you're doing the GMAT at the 70th minute.
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Let weight of bar1 be x kg and weight of bar2 will be y kg. The first bar is 2/5 gold and 3/5 silver. 2/5(x) + 3/5(x) The first bar is 3/10 gold and 7/10 silver. 3/10(y) + 7/10(y) Amount of Gold in 8-kg bar is 5/16(8) = 5/2 Amount of Silver in 8-kg bar is 11/16(8) = 11/2 Equating gold and silver ratios. 2/5(x) + 3/10(y) = 5/2 3/5(x)+ 7/10(y) = 11/2 Solver for x.(as we need tofind weight of first bar) x=1
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Given Data:
Also my 2 cents:
1st bar= 2x gold, 3x silver total=5x
2nd bar= 3y gold, 7y silver total=10y
Melted into 8 kg= 2.5 gold, 5.5 silver according to the given ratio.
Entegration of the given data and solution:
So,
2x+3y=2.5
3x+7y=5.5
Solve for x, and 5x=1 is the answer
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stanford2012 wrote: I think the fastest way to answer the question is to look at the composition of the two bars. The first bar contains 40% (2/5) of gold, while the second bar contains 30% (3/10) of gold. They are to be melted into a 8kg bar, which contains slightly more than 30% of gold (5/16).
Given that the gold content of the final bar is very close to the gold content of the second bar, it has to consist to an overwhelming majority of the second bar (i.e. close to 8kg). This means that the weight of the first bar has to be very low, hence choice A. Great, quick way to look at it. Thanks!
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