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A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink]
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14 May 2008, 01:09
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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
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Re: M1218 [#permalink]
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14 May 2008, 05:00
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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: M1218 [#permalink]
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14 May 2008, 08:04
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+0.7501 => 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: M1218 [#permalink]
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14 May 2008, 22:45
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: M1218 [#permalink]
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15 May 2008, 00:05
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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: M1218 [#permalink]
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15 May 2008, 04:48
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: M1218 [#permalink]
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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/AlligationPls let me know if this explanation is satisfactory. In the menawhile I will look for a better link. Thanks.



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Re: M1218 [#permalink]
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15 May 2008, 19:42
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/AlligationPls 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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09 Aug 2013, 21:32
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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 ... gebrutes/
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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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10 Aug 2013, 06:11
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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09 Jan 2014, 05:43
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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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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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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,
\(2.75(x) + 1.5(1x) = 2\) \(2.75x + 1.5  1.5x = 2\) \(1.25x = .5\) \(x = (1/2)/(5/4) = 2/5.\)
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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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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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 ... gebrutes/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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A 20 kg metal bar made of tin and silver lost 2 kg of its weight in th [#permalink]
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29 May 2016, 22:36
Responding to a pm: A 20 kg metal bar made of tin and silver lost 2 kg of its weight in the water. If 10 kg of tin loses 1.375 kg in the water and 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 The bar lost certain percentage of its weight. We don't know how much tin was lost and how much silver was lost but in all 2 kg was lost with is 10% of its overall weight. Tin loses 1.375 kg in 10 kg so 13.75% of its weight when it is put in water. Silver loses .375 kg in 5 kg so .375/5 * 100 = 7.5% of its weight in water. Now, we just need to use weighted averages: Wt/Ws = (7.5  10)/(10  13.75) = 2.5/3.75 = 2/3 Answer (E) For more on weighted averages and mixtures, check: http://www.veritasprep.com/blog/2011/03 ... averages/http://www.veritasprep.com/blog/2011/04 ... mixtures/
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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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07 Jun 2016, 02:13
Say in 20kg, let 'x'kg be tin and 'y'kg be silver. Now, find the losses of tin and silver for 20kg. So, tin looses 2.75 and silver looses 1.5 resp for 20 kgs.
So, 2.75x + 1.5y = 2(x+y). Solving, we get x/y=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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16 Jun 2016, 10:33
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 Solution: let Tin is x kg and Silver is y kg in 20 kg metal bar>\(x+y=20\) for 10 kg of Tin it loses 1.375 kg , then for x kg it loses \(x/10*(1.375) kg\) for 5 kg of Silver it loses 0.375 kg , then for y kg it loses \(y/5*(0.375) kg\) now both of these weights should be equal to 2 kg>> \(x/10*(1.375)+y/5*(0.375)=2\) \(x=20y\)>\((20y)/10*1.375+y/5*0.375=2\) Solve for y, then \(y=7.5/0.625\). \(x=20(7.5/0.625)>x=5/0.625\) \(x/y=5/0.625/7.5/0.625\)>\(x/y=5/7.5>x/y=2/3.\) Ans 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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14 Jan 2018, 11:40
Hi All, This question can be solved by TESTing THE ANSWERS. Since we're told that the total weight of the metal bar is 20 kg, and the additional information is in 10 kg and 5 kg 'increments', I want to do a quick hypothetical calculation of a 20 kg bar that is exactly 10 kg tin and 10 kg silver.... 10 kg tin = loses 1.375 kg of water 10 kg silver = loses 2(.375) = .750 kg of water This 'mix' of metals (1:1) will lose... 1.375 + .75 = 2.125 kg of water.... This is pretty close to the 2 kg that it's supposed to be, but it's a little too much. Thus, we need a little more silver and a little less tin. Looking at the answer choices, the closest ratio to 1:1 is Answer E (2:3)  so let's TEST that Answer... Answer E: 2:3 Tin:Silver in a 2:3 ratio gives us.... 8 kg of tin and 12 kg of silver.... 8 kg of tin = (8/10)(1.375) = (8/10)(11/8) = 88/80 = 11/10 kg of water lost 12 kg of silver = (12/10)(.750) = (12/10)(3/4) = 36/40 = 9/10 kg of water lost 11/10 + 9/10 = 20/10 = 2 kg... This is an exact match for what we were told, so this MUST be the answer. Final Answer: GMAT assassins aren't born, they're made, Rich
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