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SVP  Joined: 04 May 2006
Posts: 1563
Schools: CBS, Kellogg
A 20 kg metal bar made of alloy of tin and silver lost 2 kg  [#permalink]

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4
29 00:00

Difficulty:   95% (hard)

Question Stats: 57% (02:58) correct 43% (03:05) wrong based on 257 sessions

### HideShow timer Statistics 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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Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9433
Location: Pune, India
Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg  [#permalink]

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2
6
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/
_________________
Karishma
Veritas Prep GMAT Instructor

Manager  Joined: 11 Apr 2008
Posts: 128
Schools: Kellogg(A), Wharton(W), Columbia(D)

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7
1
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 ???
##### General Discussion
Intern  Joined: 26 Jul 2007
Posts: 26

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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. Senior Manager  Joined: 19 Apr 2008 Posts: 280 Re: M12-18 [#permalink] ### Show Tags 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? Manager  Joined: 11 Apr 2008 Posts: 128 Schools: Kellogg(A), Wharton(W), Columbia(D) Re: M12-18 [#permalink] ### Show Tags 1 1 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. CEO  Joined: 17 May 2007 Posts: 2803 Re: M12-18 [#permalink] ### Show Tags 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 ? Manager  Joined: 11 Apr 2008 Posts: 128 Schools: Kellogg(A), Wharton(W), Columbia(D) Re: M12-18 [#permalink] ### Show Tags 2 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. SVP  Joined: 04 May 2006 Posts: 1563 Schools: CBS, Kellogg Re: M12-18 [#permalink] ### Show Tags 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 _________________ Intern  Joined: 27 Jun 2013 Posts: 3 Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 3 1 [*] 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! Attachments duong cheo.PNG [ 3.32 KiB | Viewed 8249 times ] Senior Manager  Joined: 10 Jul 2013 Posts: 306 Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 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 _________________ Asif vai..... SVP  Joined: 06 Sep 2013 Posts: 1649 Concentration: Finance Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 1 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 Manager  B Status: Student Joined: 26 Aug 2013 Posts: 178 Location: France Concentration: Finance, General Management Schools: EMLYON FT'16 GMAT 1: 650 Q47 V32 GPA: 3.44 Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 1 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 _________________ Think outside the box Current Student B Joined: 23 May 2013 Posts: 186 Location: United States Concentration: Technology, Healthcare GMAT 1: 760 Q49 V45 GPA: 3.5 Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 6 1 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 VP  P Joined: 07 Dec 2014 Posts: 1206 A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 1 .10(T+S)=.1375T+.075S T/S=2/3 Veritas Prep GMAT Instructor D Joined: 16 Oct 2010 Posts: 9433 Location: Pune, India Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 2 2 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. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Veritas Prep GMAT Instructor D Joined: 16 Oct 2010 Posts: 9433 Location: Pune, India A 20 kg metal bar made of tin and silver lost 2 kg of its weight in th [#permalink] ### Show Tags 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/ _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Current Student B Joined: 17 Sep 2014 Posts: 148 Location: India Concentration: Operations, Strategy GMAT 1: 710 Q49 V38 GPA: 3.65 WE: Engineering (Manufacturing) Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 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. Intern  B Joined: 06 Oct 2013 Posts: 45 Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 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-->> $$x/10*(1.375)+y/5*(0.375)=2$$ $$x=20-y$$---->$$(20-y)/10*1.375+y/5*0.375=2$$ 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 EMPOWERgmat Instructor V Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14563 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg [#permalink] ### Show Tags 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** # Rich Cohen Co-Founder & GMAT Assassin Follow Special Offer: Save$75 + GMAT Club Tests Free
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www.empowergmat.com/ Re: A 20 kg metal bar made of alloy of tin and silver lost 2 kg   [#permalink] 14 Jan 2018, 12:40

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