I believe that the following reflects the intent of the problem:
Mo2men wrote: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.
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