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Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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14 00:00

Difficulty:   75% (hard)

Question Stats: 62% (02:51) correct 38% (02:47) wrong based on 73 sessions

### HideShow timer Statistics 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. $$\frac{2}{3}$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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1
2
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 $$t = 8$$. $$\frac{t}{s} = \frac{8}{12} = \frac{2}{3}$$.

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Manager  Joined: 06 Mar 2014
Posts: 228
Location: India
GMAT Date: 04-30-2015

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1
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 $$t = 8$$. $$\frac{t}{s} = \frac{8}{12} = \frac{2}{3}$$.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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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 $$t = 8$$. $$\frac{t}{s} = \frac{8}{12} = \frac{2}{3}$$.

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.
_________________
Manager  Joined: 06 Mar 2014
Posts: 228
Location: India
GMAT Date: 04-30-2015

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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 $$t = 8$$. $$\frac{t}{s} = \frac{8}{12} = \frac{2}{3}$$.

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.
Intern  Joined: 17 Jul 2015
Posts: 13
Schools: Booth '18 (II)

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16
4
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.

Originally posted by szaszgt on 16 Sep 2015, 12:36.
Last edited by szaszgt on 16 Sep 2015, 22:43, edited 1 time in total.
Manager  Joined: 06 Mar 2014
Posts: 228
Location: India
GMAT Date: 04-30-2015

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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.
Intern  Joined: 17 Jul 2015
Posts: 13
Schools: Booth '18 (II)

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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. Senior Manager  Joined: 12 Aug 2015
Posts: 283
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37 GMAT 2: 650 Q43 V36 GMAT 3: 600 Q47 V27 GPA: 3.3
WE: Management Consulting (Consulting)

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2
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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KUDO me plenty
Current Student Joined: 29 Apr 2015
Posts: 26
Location: Russian Federation
GMAT 1: 710 Q48 V38 GPA: 4

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2
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)
Senior Manager  Joined: 31 Mar 2016
Posts: 376
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34 GPA: 3.8
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I think this is a high-quality question and I agree with explanation.
Manager  S
Joined: 13 Dec 2013
Posts: 150
Location: United States (NY)
Schools: Cambridge"19 (A)
GMAT 1: 710 Q46 V41 GMAT 2: 720 Q48 V40 GPA: 4
WE: Consulting (Consulting)

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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 $$t = 8$$. $$\frac{t}{s} = \frac{8}{12} = \frac{2}{3}$$.

Where does t=20-s come from?
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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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 $$t = 8$$. $$\frac{t}{s} = \frac{8}{12} = \frac{2}{3}$$.

Where does t=20-s 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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Intern  B
Joined: 28 Apr 2016
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Hi Bunuel,

Are there any page for these kind of problems? I mean there is a page of statistics questions etc. Like this:)
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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dyg wrote:
Hi Bunuel,

Are there any page for these kind of problems? I mean there is a page of statistics questions etc. Like this:)

Check our questions bank: http://gmatclub.com/forum/viewforumtags.php

Also, you might find the following topic useful ALL YOU NEED FOR QUANT ! ! !.
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Joined: 06 Apr 2017
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Location: United States (OR)
Schools: Haas EWMBA '21
GMAT 1: 730 Q48 V44 GMAT 2: 730 Q49 V40 GPA: 3.98
WE: Corporate Finance (Health Care)

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2
1
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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Manager  S
Joined: 23 Sep 2016
Posts: 235

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

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rishabhmishra wrote:
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

You are making a mistake in the very first step. The question asks: what is the ratio of tin to silver in the bar? So, if the ration tin/silver = 2/5, then tin = 2/(2 + 5)*20 = 40/7 and silver = 5/(2 + 5)*20 = 100/7.
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GMAT 1: 570 Q43 V25 GMAT 2: 640 Q42 V36 WE: Underwriter (Insurance)

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szaszgt wrote:
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.

Mate, if I could, I would give u a thousand kudos! I wanted to apply allegation method for wtd avrg but didn't know how. Thanks a ton!
Intern  Joined: 09 Dec 2018
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the best way to solve these questions is by answer choice . option a and e will remain .other options eliminated. Re M12-18   [#permalink] 14 Jul 2019, 02:14
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# M12-18

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