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Ratios/Rates Problem

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by sunnyjohn » Tue Aug 16, 2011 6:59 pm
1:2:3:5

Time x :
First machine can sew : 1 shirt
Second machine can sew: 2 shirts
third ---------------: 3 shirts
fourth ---------------: 5 shirts

Its obvious that 4th machine is fastest.
Given:
The fastest can sew a shirt in 2 hours.
So:
Total time to sew 5 shirts = 2*5 = 10 hours.
This means-
Time 10 hours :
First machine can sew : 1 shirt ( 10 hours for 1 shirt)
Second machine can sew: 2 shirts ( 5 hours for 1 shirt)
third ---------------: 3 shirts (3 1/3 hours for 1 shirt)
fourth ---------------: 5 shirts ( 2 hours for 1 shirt)

Ideally - if all other 3 machines work together, how much they can complete in 1 hour

1/10 + 1/5 + 1/3.33 ==> 0.1 + 0.2 + 0.3 = 0.6

total shirts required = 3
0.6 * 5 = 3
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by gmatboost » Wed Aug 17, 2011 2:24 pm
I don't understand this statement: "We can use the rate of the fastest machine to determine the unknown multiplier: 5x = 0.5 ==> x = 1/10"
On ratio questions, people often like to let the unknown quantities be equal to the know ratio times x

Example: white marbles to red marbles are in ratio 2:3, then let the amounts be 2x and 3x

Here, it is the rates that are given in the ratio, 1:2:3:5
So, we could say: Let the rates be x, 2x, 3x, and 5x
x is a multiplier because it is what we multiply the ratio values (1, 2, 3, 5) by to get the actual rate

So, the fastest has a rate of 0.5 as you said, so 0.5 = 5x, and x = 0.1
Now we can multiply the other ratio values (1, 2, 3) to get the rates for the other machines (.1, .2, .3)

The combined rate of the others is then 0.6, and we use the equation 0.6 * T = 3 to solve.
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