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m03: Q30

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m03: Q30 [#permalink] New post 14 Jan 2009, 12:58
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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

[Reveal] Spoiler: OA
A

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Re: m03: Q30 [#permalink] New post 14 Jan 2009, 19:56
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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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Re: m03: Q30 [#permalink] New post 16 Jan 2009, 08:12
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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.

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Re: m03: Q30 [#permalink] New post 16 Jan 2009, 15:37
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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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Re: m03: Q30 [#permalink] New post 20 Jan 2009, 01:11
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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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Re: m03: Q30 [#permalink] New post 16 Feb 2010, 09:27
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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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Re: m03: Q30 [#permalink] New post 16 Jul 2010, 05:06
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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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Re: m03: Q30 [#permalink] New post 16 Jul 2010, 06:15
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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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Re: m03: Q30 [#permalink] New post 16 Jul 2010, 06:22
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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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Re: m03: Q30 [#permalink] New post 16 Jul 2010, 18:44
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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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Re: m03: Q30 [#permalink] New post 18 Jul 2010, 09:58
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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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Re: m03: Q30 [#permalink] New post 18 Jul 2010, 10:33
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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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Re: m03: Q30 [#permalink] New post 19 Jul 2010, 21:27
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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.html

In 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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Re: m03: Q30 [#permalink] New post 19 Jul 2010, 21:50
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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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Re: m03: Q30 [#permalink] New post 28 Jul 2010, 00:06
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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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Re: m03: Q30 [#permalink] New post 28 Jul 2010, 12:01
Why do you need to take ratio of
Gold / Total
or
Silver / Total

Rather than using the ratio of gold/silver:
ie. (2x + 3y) / (3x + 7y) = 5/11
or someting like that?
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Re: m03: Q30 [#permalink] New post 23 Jul 2011, 11:08
5g + 11s = 8
Therefore,
2.5 kgs of gold
5.5 kgs of silver
Solving for gold, there is 1 kg of gold in the first bar
Solving for silver, there is 1.65 kg of silver in the first bar
The total weight of the first bar is 2.65 kg.
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Re: m03: Q30 [#permalink] New post 01 Jun 2012, 23:55
This is how I did it but there seems to be a hole in my logic:

1st Bar: 2 : 3
2nd Bar: 3 : 7

Final Bar: 5 : 11 (given 8kg total)

So for the final bar, the multiplier x = 5x + 11x = 8 and x = 1/2.
So for the Final Bar the qty is 2.5 : 5.5

Now using this multiplier for the 1st and 2nd bar and comparing it to the 3rd bar:

1st Bar: 1 : 1.5
2nd Bar: 1.5 : 3.5

Thus 1 + 1.5 = 2.5 (OK) and

1.5 + 3.5 = 5 but, according to the original ratio this should be 5.5.

Where am I going wrong?
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Re: m03: Q30 [#permalink] New post 24 Jul 2012, 04:40
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use allegation:

golda 2/5 goldb 3/10


5/16(mixture)



5/16-3/10=1/80 2/5-5/16=7/80

1:7mixture ratio
ie.===1kg out of 8 for GoldA
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Re: m03: Q30 [#permalink] New post 24 Jul 2013, 06:43
HI Karishma/Bunuel,

Could you please explain this problem with weighted avgs?

I did the problem using the weighted average method like below

w1/w2= [A2- A avrg]/[A avrg- A1]

A1- 2/5
A2- 3/10
A avrg- 5/16

i got w1/w2= 1/7

I conclude weight of first mixture as 1. But is this the right approach? Can you please confirm?
Re: m03: Q30   [#permalink] 24 Jul 2013, 06:43
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