Mixture Problem

[Mo2men]

A 20 kg metal bar made of allow of tin and silver lost 2 kg of its weight in the water. If 10 kg of tin loses 1.375 kg in 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

OA:E

Source: GMATClub

[Jay@ManhattanReview]

A 20 kg metal bar made of allow of tin and silver lost 2 kg of its weight in the water. 10 kg of tin loses 1.375 kg in water; 5 kg of silver loses .375 kg. What is the ration of tin to silver in the bar?

a) 1/4
b) 2/5
c) 1/2
d) 3/5
e) 2/3

OA:E

Source: GMATClub

Hi Mo2men,

The question is not types correctly. Many errors.

We have

A 20 kg metal bar made of alloy of tin and silver lost 2 kg of its weight in the water. If 10 kg of tin loses 1.375 kg in water, and 5 kg of silver loses 0.375 kg, what is the ratio of tin to silver in the bar?

Given that the 20 kg metal bar alloy loses 2 kg.

10 kg of tin loses 1.375 kg => 20 kg of tin loses 2*1.375 = 2.75 kg

5 kg of silver loses 0.375 kg => 20 kg of silver loses 4*0.375 = 1.5 kg

Now we have to find out the ratio of tin to silver.

Say the ratio of tin to silver = 1 : x

=> (2.75*1 + 1.5*x)/(1 + x) = 2

=> 2.75 + 1.5x = 2 + 2x

=> 0.75 = 0.50x

=> x = 0.75 / 0.50 = 3/2

=> Ratio of tin to silver = x : 1 = [spoiler]2 : 3[/spoiler].

The correct answer: E

Hope this helps!

Relevant book: Manhattan Review GMAT Word Problems Guide

-Jay
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[GMATGuruNY]

I believe that the following reflects the intent of the problem:

A 20 kg metal bar made of an alloy of tin and silver lost 2 kg of its weight in water. If every 10 kg of tin loses 1.375 kg of water, while every 5 kg of silver loses 0.375 kg of water, 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 following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Put the loss rates over a COMMON DENOMINATOR
Tin:
Water loss per kilogram = (1.375)/(10) = (11/8)/10 = 11/80.
Silver:
Water loss per kilogram = (0.375)/5 = (3/8)/5 = 3/40 = 6/80.
Metal bar:
Water loss per kilogram = 2/20 = 8/80.

Step 2: Plot the 3 numerators on a number line, with the numerators for T and S on the ends and the numerator for the metal bar in the middle.
T 11-----------8-----------6 S

Step 3: Calculate the distances between the numerators.
T 11-----3-----8-----2-----6 S

Step 4: Determine the ratio in the mixture.
The ratio of T to S is equal to the RECIPROCAL of the distances in red.
T:S = 2:3.

The correct answer is E.

[Mo2men]

Thanks Jay.
Edited the problem. My apology.

[[email protected]]

Hi Mo2men,

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: E

GMAT assassins aren't born, they're made,
Rich