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A 20 kg metal bar made of alloy of tin and silver lost 2 kg

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A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

A. 1/4
B. 2/5
C. 1/2
D. 3/5
E. 2/3
[Reveal] Spoiler: OA

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New post 14 May 2008, 06:00
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sondenso wrote:
A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

1/4
2/5
1/2
3/5
2/3


10 kg tin loses 1.375 kg=> 20 kg loses 2.75 kg
5 kg sliver loses 0.375 kg=> 20 kg loses 1.5 kg
Actual loss is 2 kg.
Apply the alligation rule=>

2.75 1.5
\ /
2
/ \
0.5 0.75


ratio of tin/silver => 0.5/0.75 => 1/2/3/4 => 3/8 ???
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Re: M12-18 [#permalink]

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New post 14 May 2008, 09:04
Hello All,

Solution is 2/3.
How i did it :
Let x be ratio we want.
If r(s) is rate of loss for silver and r(t) that of tin, T total weigh of tin and S total weigh of silver
We have 20=T+S (equation $) and 2=r(t)*T+r(s)*S
We also know r(s)=0.375/5 and r(t)=1.375/10
So 2 = r(t)*T+r(s)*S => 20= 1.375*T+0.750*S (equation £) and we know (equation $) => 1+x = 20/S
So (equation $) => x=1.375*x+0.750-1 => 0.375*x = 0.25 => x= 2/3

A little long but that's how i solved it!
I'm sure there is a more simple way to do.
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Re: M12-18 [#permalink]

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New post 14 May 2008, 23:45
anirudhoswal wrote:
sondenso wrote:
A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

1/4
2/5
1/2
3/5
2/3


10 kg tin loses 1.375 kg=> 20 kg loses 2.75 kg
5 kg sliver loses 0.375 kg=> 20 kg loses 1.5 kg
Actual loss is 2 kg.
Apply the alligation rule=>

2.75 1.5
\ /
2
/ \
0.5 0.75


ratio of tin/silver => 0.5/0.75 => 1/2/3/4 => 3/8 ???


your rule is correct , calculation is wrong . I m interested in knowing this rule , can you elaborate , when can we apply this rule?
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Re: M12-18 [#permalink]

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New post 15 May 2008, 01:05
rpmodi wrote:
anirudhoswal wrote:
sondenso wrote:
A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

1/4
2/5
1/2
3/5
2/3


10 kg tin loses 1.375 kg=> 20 kg loses 2.75 kg
5 kg sliver loses 0.375 kg=> 20 kg loses 1.5 kg
Actual loss is 2 kg.
Apply the alligation rule=>

2.75 1.5
\ /
2
/ \
0.5 0.75


ratio of tin/silver => 0.5/0.75 => 1/2/3/4 => 3/8 ???


your rule is correct , calculation is wrong . I m interested in knowing this rule , can you elaborate , when can we apply this rule?


Oh Crap !! obviously.. (1/2)/(3/4) = 4/6 = 2/3.

What the hell will i do on the GMAT!!

Neways, the rule I used is called "alligations". It is useful for quickly calculating ratios of individual components in MIXTURES.
Please bear with me... I will post a more detailed expl asap.

thnx.
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Re: M12-18 [#permalink]

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New post 15 May 2008, 05:48
This alligations rule may help me with word translations problems for ratios and mixtures. Do you happen to have a detailed description or a link that describes it properly ?
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Re: M12-18 [#permalink]

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New post 15 May 2008, 05:57
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I have been looking for a good explanation. Unfortunately, I have found none.
There is a brief mention of the method on wikipedia -
http://en.wikipedia.org/wiki/Alligation

Pls let me know if this explanation is satisfactory.
In the menawhile I will look for a better link.

Thanks.
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Re: M12-18 [#permalink]

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New post 15 May 2008, 20:42
anirudhoswal wrote:
I have been looking for a good explanation. Unfortunately, I have found none.
There is a brief mention of the method on wikipedia -
http://en.wikipedia.org/wiki/Alligation

Pls let me know if this explanation is satisfactory.
In the menawhile I will look for a better link.

Thanks.


Many thanks anirudhoswal. The link is good, but it is somewhat difficult for me to understand. Do you have the reference that is the same way as you apply to in previous post? Thanks!

OA is E
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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[*]
sondenso wrote:
A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

1/4
2/5
1/2
3/5
2/3


interesting question.

This is my solution for the problem:

10 kg tin loses 1.375 kg => lost 13.75% its weight
5 kg Silver loses 0.375 kg => lost 7.5% of its weight

Initially, 20 kg of alloy loses 2kg => 10% of its weight

Call X is the weight of Tin the alloy
Y is the weight of Silver in the alloy

We have


=> X/Y = 2.5%/ 3.75% = 2/3

Very nice question!
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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Campanella1989 wrote:
[*]
sondenso wrote:
A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

1/4
2/5
1/2
3/5
2/3


interesting question.

This is my solution for the problem:

10 kg tin loses 1.375 kg => lost 13.75% its weight
5 kg Silver loses 0.375 kg => lost 7.5% of its weight

Initially, 20 kg of alloy loses 2kg => 10% of its weight

Call X is the weight of Tin the alloy
Y is the weight of Silver in the alloy

We have


=> X/Y = 2.5%/ 3.75% = 2/3

Very nice question!


Rather than using the alligation diagram, you can simply use this formula to avoid confusion:

w1/w2 = (A2 - Aavg)/(Avg - A1)

Weight of Tin/Weight of Silver = (Silver's loss - Avg loss)/(Avg loss - Tin's loss)

X/Y = (7.5 - 10)/(10 - 13.75) = 2/3

To check out a detailed discussion on this formula, see:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... ge-brutes/
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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New post 10 Aug 2013, 07:11
sondenso wrote:
A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

A. 1/4
B. 2/5
C. 1/2
D. 3/5
E. 2/3


10 kg tin loses 1.375 kg=> 20 kg loses 2.75 kg
5 kg sliver loses 0.375 kg=> 20 kg loses 1.5 kg
Actual loss is 2 kg.
Apply the alligation rule=>

2.75 1.5
\ /
2
/ \
0.5 0.75
! !
1/2 3/4

Ratio of tin/silver = (1/2) / (3/4) = 2/3
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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New post 09 Jan 2014, 06:43
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Asifpirlo wrote:
sondenso wrote:
A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

A. 1/4
B. 2/5
C. 1/2
D. 3/5
E. 2/3


10 kg tin loses 1.375 kg=> 20 kg loses 2.75 kg
5 kg sliver loses 0.375 kg=> 20 kg loses 1.5 kg
Actual loss is 2 kg.
Apply the alligation rule=>

2.75 1.5
\ /
2
/ \
0.5 0.75
! !
1/2 3/4

Ratio of tin/silver = (1/2) / (3/4) = 2/3


Concept of differentials may also be aplied here

Total loss 2

Loss 20kg of tin is 2.75
Loss 20kg of silver is 1.5

Therefore 0.75T - 0.5S = 0
75T = 50S
T/S = 50/75=2/3

Hope it helps
Cheers!
J :)
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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New post 24 Jan 2014, 14:44
Well, I found a tricky way for this one.

You have 20kg as a total.

A. 1/4 ==> Total 1+4=5 So can divide 20
B. 2/5 ==> Total 2+5=7 So cannot divide 20
C. 1/2 ==> Total 1+2=3 So cannot divide 20
D. 3/5 ==> Total 3+5=8 So cannot divide 20
E. 2/3 ==> Total \(1+4=5\) So can divide 20

Therefore you have A and E.

A cannot be the answer since the ratio is to big.

But let's look at E:

\(2/3\) of 20 is 8 and 12.

For the tin the water lose is: one kilo= \(1.375/10=0.1375\) and for the silverthe water lose is: one kilo=\(0.375/5=0.075\)

\((8*0.1375) + (12*0.075) = 2\)

Answer is E
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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New post 21 Mar 2014, 06:44
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Instead of using a formula, just think about it this way:

20kg of tin loses 2.75 kg in water.
20kg of silver loses 1.5kg in water.

Some sort of combination of these metals loses 2 kg in water.

Thus,

\(2.75(x) + 1.5(1-x) = 2\)
\(2.75x + 1.5 - 1.5x = 2\)
\(1.25x = .5\)
\(x = (1/2)/(5/4) = 2/5.\)

Thus, the alloy is 2/5 tin and thus 3/5 silver, and therefore the ratio of tin to silver is 2/3.

Answer: E
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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.10(T+S)=.1375T+.075S
T/S=2/3
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]

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New post 10 Feb 2016, 23:33
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VeritasPrepKarishma wrote:
Campanella1989 wrote:
[*]
sondenso wrote:
A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in the water; 5 kg of silver loses 0.375 kg. What is the ratio of tin to silver in the bar?

1/4
2/5
1/2
3/5
2/3


interesting question.

This is my solution for the problem:

10 kg tin loses 1.375 kg => lost 13.75% its weight
5 kg Silver loses 0.375 kg => lost 7.5% of its weight

Initially, 20 kg of alloy loses 2kg => 10% of its weight

Call X is the weight of Tin the alloy
Y is the weight of Silver in the alloy

We have


=> X/Y = 2.5%/ 3.75% = 2/3

Very nice question!


Rather than using the alligation diagram, you can simply use this formula to avoid confusion:

w1/w2 = (A2 - Aavg)/(Avg - A1)

Weight of Tin/Weight of Silver = (Silver's loss - Avg loss)/(Avg loss - Tin's loss)

X/Y = (7.5 - 10)/(10 - 13.75) = 2/3

To check out a detailed discussion on this formula, see:
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... ge-brutes/


Responding to a pm:

Here is how you will find the values of A1 an A2.

We have an overall loss (average loss). The average loss is 2 kg when 20 kg alloy is immersed.
This is a loss of (2/20)*100 = 10%.
This is Aavg

The loss of tin is 1.375 kg for every 10 kg.
This means it loses (1.375/10)*100 = 13.75% of its weight in water.
This is A1.

The loss of silver is .375 kg for every 5 kg.
This means it loses (.375/5)* 100 = 7.5% of its weight in water.
This is A2.
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Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg   [#permalink] 10 Feb 2016, 23:33
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