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Re M1218 [#permalink]
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16 Sep 2014, 00:47
Official Solution: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. \(\frac{1}{4}\) B. \(\frac{2}{5}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\) \(t\) represents the amount of tin in the bar, \(s\) represents the amount of silver in the bar (in kg). The tin lost equals \(0.1375*t\) kg; the silver lost equals \(\frac{0.375}{5}*s = 0.075*s\) kg. Together, the tin and silver lost equals \(0.1375*t + 0.075*s = 2\) kg. Because \(t = 20  s\), we have \(2.75  0.1375*s + 0.075*s = 2\) or \(0.75 = 0.0625*s\). Thus, \(s = 12\) and \(t = 8\). \(\frac{t}{s} = \frac{8}{12} = \frac{2}{3}\). Answer: E
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Re: M1218 [#permalink]
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17 Oct 2014, 16:41
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Bunuel wrote: Official Solution:
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. \(\frac{1}{4}\) B. \(\frac{2}{5}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)
\(t\) represents the amount of tin in the bar, \(s\) represents the amount of silver in the bar (in kg). The tin lost equals \(0.1375*t\) kg; the silver lost equals \(\frac{0.375}{5}*s = 0.075*s\) kg. Together, the tin and silver lost equals \(0.1375*t + 0.075*s = 2\) kg. Because \(t = 20  s\), we have \(2.75  0.1375*s + 0.075*s = 2\) or \(0.75 = 0.0625*s\). Thus, \(s = 12\) and \(t = 8\). \(\frac{t}{s} = \frac{8}{12} = \frac{2}{3}\).
Answer: E Hi Bunuel, So we should expect Questions like above which involves quite a bit of calculations? Also could you please provide links to Word Problem Questions that would help in better understanding of their types and grasp the common patterns. Thank you.



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Re: M1218 [#permalink]
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18 Oct 2014, 02:16
earnit wrote: Bunuel wrote: Official Solution:
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. \(\frac{1}{4}\) B. \(\frac{2}{5}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)
\(t\) represents the amount of tin in the bar, \(s\) represents the amount of silver in the bar (in kg). The tin lost equals \(0.1375*t\) kg; the silver lost equals \(\frac{0.375}{5}*s = 0.075*s\) kg. Together, the tin and silver lost equals \(0.1375*t + 0.075*s = 2\) kg. Because \(t = 20  s\), we have \(2.75  0.1375*s + 0.075*s = 2\) or \(0.75 = 0.0625*s\). Thus, \(s = 12\) and \(t = 8\). \(\frac{t}{s} = \frac{8}{12} = \frac{2}{3}\).
Answer: E Hi Bunuel, So we should expect Questions like above which involves quite a bit of calculations? Also could you please provide links to Word Problem Questions that would help in better understanding of their types and grasp the common patterns. Thank you. Yes, you should expect such questions. As for the questions. Here is the links to Question Bank: viewforumtags.php You can find all types of questions there.
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Re: M1218 [#permalink]
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18 Oct 2014, 11:09
Bunuel wrote: earnit wrote: Bunuel wrote: Official Solution:
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. \(\frac{1}{4}\) B. \(\frac{2}{5}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)
\(t\) represents the amount of tin in the bar, \(s\) represents the amount of silver in the bar (in kg). The tin lost equals \(0.1375*t\) kg; the silver lost equals \(\frac{0.375}{5}*s = 0.075*s\) kg. Together, the tin and silver lost equals \(0.1375*t + 0.075*s = 2\) kg. Because \(t = 20  s\), we have \(2.75  0.1375*s + 0.075*s = 2\) or \(0.75 = 0.0625*s\). Thus, \(s = 12\) and \(t = 8\). \(\frac{t}{s} = \frac{8}{12} = \frac{2}{3}\).
Answer: E Hi Bunuel, So we should expect Questions like above which involves quite a bit of calculations? Also could you please provide links to Word Problem Questions that would help in better understanding of their types and grasp the common patterns. Thank you. Yes, you should expect such questions. As for the questions. Here is the links to Question Bank: viewforumtags.php You can find all types of questions there. Thank you for that link but i am already aware of it. Usually you paste links to certain common types, and those are exact threads and their links with each link pointing to a specific question and not some big pool of questions where one gets lost. I was hoping to find something on those lines, to get a hang of few common types/patterns like i found in mixtures/ Simple/Compound interest Qs. Anyway, maybe there aren't any in Word problems, so i guess i'll figure it out. Thank you so much.



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If the bar had been made of Tin, it would lose 2.75kg of its weight. (2*1.375) If the bar had been made of Silver, it would lose 1.5kg of its weight. (4*0.375)
The metal bar loses 2kg of its weight, so there's more Silver in it than Tin, because 2 is nearer to 1.5 than to 2.75. Usually, I approach these problems visually. Draw a line and see where the average lies between the two extremes. In this case, 2 is 0.5 away from 1.5 and 0.75 away from 2.75. The ratio of Tin to Silver will be 0.5/0.75, which is 2/3.
Last edited by szaszgt on 16 Sep 2015, 22:43, edited 1 time in total.



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Re: M1218 [#permalink]
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16 Sep 2015, 16:00
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szaszgt wrote: Another approach would be to think in terms of weighted averages.
If the bar had been made of Tin, it would lose 2.75kg of its weight. (2*1.375) If the bar had been made of Silver, it would lose 1.5kg of its weight. (4*0.375)
The metal bar loses 2kg of its weight, so there's more Silver in it than Tin, because 2 is nearer to 1.5 than to 2.75. Usually, I approach these problems visually. Draw a line and see where the average lies between the two extremes. In this case, 2 is 0.5 away from 1.5 and 0.75 away from 2.75. The ratio of Tin to Silver will be 0.5/0.75, which is 2/3. U just nailed it! This was "the" approach.



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Re: M1218 [#permalink]
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17 Sep 2015, 00:37
earnit wrote: szaszgt wrote: Another approach would be to think in terms of weighted averages.
If the bar had been made of Tin, it would lose 2.75kg of its weight. (2*1.375) If the bar had been made of Silver, it would lose 1.5kg of its weight. (4*0.375)
The metal bar loses 2kg of its weight, so there's more Silver in it than Tin, because 2 is nearer to 1.5 than to 2.75. Usually, I approach these problems visually. Draw a line and see where the average lies between the two extremes. In this case, 2 is 0.5 away from 1.5 and 0.75 away from 2.75. The ratio of Tin to Silver will be 0.5/0.75, which is 2/3. U just nailed it! This was "the" approach. Haha yea, same weighted average, but slightly different.



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30 sec solution: eliminate B C D because only A and E are factors of 20: 1+4 and 2+3. Hence potential proprotion is either 4:16 or 8:12. if we roughly test A: "10 kg of tin loses 1.375 kg" > max of 0.7 kg lost weight of tin from 5 kg "5 kg of silver loses 0.375 kg" > 15 kg of silver will lose max 1.2 kg which together with max of 0.7 kg lost weight of tin from 5 kg  numbers taken roughly because this is a 30 sec approach  does not add up to 2 kg, so[b] choose E and move forward.[/b]
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Bunuel wrote: 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. \(\frac{1}{4}\) B. \(\frac{2}{5}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\) I have used a bit different approach. This problem is the same as solution problems, though it is formulated in other way. Let's translate given problem to conventional solution problem: Let's denote taken weight of tin as X and taken weight of silver as Y. 1. Tin loses 1.375/10 = 0.1375 kg for 1 kg of its weight. 2. Silver loses 0.375/5 = 0.075 kg for 1 kg of its weight. 3. Metal bar loses 2/20 = 0.1 kg for 1 kg of its weight. 0.1375X + 0.075Y  = 0.1; X+Y 0.1375X + 0.075Y = 0.1 (X+Y) 0.0375X = 0.025Y X/Y = 0.025/0.0375 = 2/3 You are welcome)



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Re M1218 [#permalink]
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26 Aug 2016, 08:24
I think this is a highquality question and I agree with explanation.



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Re: M1218 [#permalink]
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04 Jan 2017, 13:44
Bunuel wrote: Official Solution:
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. \(\frac{1}{4}\) B. \(\frac{2}{5}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)
\(t\) represents the amount of tin in the bar, \(s\) represents the amount of silver in the bar (in kg). The tin lost equals \(0.1375*t\) kg; the silver lost equals \(\frac{0.375}{5}*s = 0.075*s\) kg. Together, the tin and silver lost equals \(0.1375*t + 0.075*s = 2\) kg. Because \(t = 20  s\), we have \(2.75  0.1375*s + 0.075*s = 2\) or \(0.75 = 0.0625*s\). Thus, \(s = 12\) and \(t = 8\). \(\frac{t}{s} = \frac{8}{12} = \frac{2}{3}\).
Answer: E Where does t=20s come from?



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Re: M1218 [#permalink]
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05 Jan 2017, 03:22
Cez005 wrote: Bunuel wrote: Official Solution:
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. \(\frac{1}{4}\) B. \(\frac{2}{5}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)
\(t\) represents the amount of tin in the bar, \(s\) represents the amount of silver in the bar (in kg). The tin lost equals \(0.1375*t\) kg; the silver lost equals \(\frac{0.375}{5}*s = 0.075*s\) kg. Together, the tin and silver lost equals \(0.1375*t + 0.075*s = 2\) kg. Because \(t = 20  s\), we have \(2.75  0.1375*s + 0.075*s = 2\) or \(0.75 = 0.0625*s\). Thus, \(s = 12\) and \(t = 8\). \(\frac{t}{s} = \frac{8}{12} = \frac{2}{3}\).
Answer: E Where does t=20s come from? \(t\) represents the amount of tin in the bar, \(s\) represents the amount of silver in the bar (in kg). We are told that the metal bar made of tin and silver weighs 20 kg, thus t + s = 20, which leads to t = 20  s. Hope it's clear.
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Re: M1218 [#permalink]
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26 Mar 2017, 11:30
Hi Bunuel,
Are there any page for these kind of problems? I mean there is a page of statistics questions etc. Like this:)



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This question, like almost all weighted average questions, can be made simpler with allegation. For every 10 kg of tin, 1.375 kg dissolves in water. For every 10 kg of silver, .750 kg dissolves in water. We need a weighted average of tin and silver that will dissolve 1 kg per 10 kg of combined metal. Notice that 1 kg dissolving per 10 kg of combined metal is equivalent to 2 kg dissolving per 20 kg of combined metal: ratio of tin: \(1  .750 = .250\) ratio of silver: \(1.375  1 = .375\) \(\frac{.250}{.375}=\frac{2}{3}\) I've attached a visual representation of allegation for those who are interested.
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Re: M1218 [#permalink]
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08 Mar 2018, 01:33
Bunuel i am getting answer B in this question from plugging value let me solve for you we know total is 20kg and if we take B as answer then tin= 20*2/5= 8 kg so silver is 12 kg now from the question 1kg of tin loss 0.1375kg and 1kg of silver lost 0.075kg weight in water so 8 kg will loss 8*0.1375= 1.1 kg 12*0.075=0.9 kg so total weight lost is 2kg which is same according to question PLEASE EXPLAIN



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