M12-18

Question

A metal bar weighing 20 kilograms, consisting of tin and silver, has lost 2 kilograms of its weight when placed in water. If 10 kilograms of tin lose 1.375 kilograms in water and 5 kilograms of silver lose 0.375 kilograms in water, what is the proportion of tin to silver in the bar?

A. \(\frac{1}{4}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{5}\)
E. \(\frac{2}{3}\)

Show Answer

Official Answer

E

A.  \frac{1}{4}  
B.  \frac{2}{5}  
C.  \frac{1}{2}  
D.  \frac{3}{5}  
E.  \frac{2}{3}

Bunuel · Accepted Answer

Official Solution:

A metal bar weighing 20 kilograms, consisting of tin and silver, has lost 2 kilograms of its weight when placed in water. If 10 kilograms of tin lose 1.375 kilograms in water and 5 kilograms of silver lose 0.375 kilograms in water, what is the proportion of tin to silver in the bar?

A. \(\frac{1}{4}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{5}\)
E. \(\frac{2}{3}\)

Show Answer

Official Answer

E

A.  \frac{1}{4}  
B.  \frac{2}{5}  
C.  \frac{1}{2}  
D.  \frac{3}{5}  
E.  \frac{2}{3}

Let  t  be the amount of tin in the bar (in kilograms) and  s  be the amount of silver in the bar (in kilograms). Since 10 kilograms of tin lose 1.375 kilograms in water,  t  kilograms of tin will lose  \frac{t}{10}*1.375=0.1375*t  kilograms in water. 
 
Since 5 kilograms of silver lose 0.375 kilograms in water,  s  kilograms of silver will lose  \frac{s}{5}*0.375= 0.075*s  kilograms in water. 
 
The total weight lost by both tin and silver is  0.1375*t + 0.075*s = 2  kilograms.
 
Since  t  is equal to  20 - s , we can substitute this into the equation to get  0.1375 *(20 - s) + 0.075 * s = 2 . Solving for  s , we find that  s = 12 . Hence,  t = 20 - s = 20 - 12 = 8 .
 
Finally, the ratio of tin to silver in the bar is calculated as  \frac{t}{s} = \frac{8}{12} = \frac{2}{3} .

Answer: E

szaszgt · Answer

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.

shasadou · Answer

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  choose E  and move forward.

naumyuk · Answer

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)

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

spence11 · Answer

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.

Attachments

Allegation.PNG [ 4.08 KiB | Viewed 35497 times ]

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.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Josvit · Answer

1 kg of tin loses 0.1375kg (1.375/10) in water
1 kg of silver loses 0.075kg (0.375/5) in water

In the BAR:
If X is the quantity of Tin, 0.1375 X will be the weight lost.
If Y is the quantity of Silver, 0.075 Y will be the weight lost.

0.1375 X + 0.075 Y = 2 .... (I)
and X + Y = 20 ....(II)

Given that we need to calculate X/Y, if we multiply by 10 the first equation, we have:

1.375X + 0.75 Y = 20 ... (1)

X + Y = 20 ....(2)

Then, when we subtract (2) from (1), it results:

0.375 X - 0.25Y = 0
0.375 x = 0.25 Y
x/y= 2/3

The8 · Answer

One elegant solution which I learned from GMAT Ninja videos is to use number line and plot (Basically this is weighted mean approach)
0.750-------1.00-----------------1.375
Silver loss total tin loss

then use proportion of difference = 0.375/(0.25+0.375) and from there everything will be easy to calculate - one important notice is to have losses converted into per 10kg
let me know if it helps

sindhureddyalla · Answer

I like the solution - it’s helpful.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=83864 --- Harsha GMAT-Club-Forum-7wadnh2d.png

Swanoff · Answer

I like the solution - it’s helpful. Easy to grasp

06learner · Answer

I like the solution - it’s helpful.

repellatquas · Answer

I like the solution - it’s helpful. Here's an alternative: 
The sum of N&D in ratios should be a multiple of 20 as in -
A) 1/4 = 1x+4x = 5x (multiple of 20)
B) 2/5 = 7x (not a multiple of 20)
C) 1/2 = 3x (not a multiple of 20)
D) 3/5 = 8x (not a multiple of 20)
E) 2/3 = 5x (multiple of 20)

Basis this, we canelimiate B,C,D and left with A and E

Now, option A is 1/4 or 4/16 (as 4+16 = 20, or 5x=20 and x=4)
If 10 kgs of tin loses 1.375, 4 kgs will lose 0.55
Similarly, 16 kgs of silver loses 1.2 kgs 
But 0.55+1.2 = 1.75 (not equal to 2, so ELIMINATE)

Answer is E. You can verify using the same method above but we don't need to do so as all others have been elimiated.

M12-18

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