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25 Apr 2025, 05:24
Kudos
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Net inc in efficiency = inc of y% for 40% of the machines, where
Y = 1/2-1/3 divided by 1/3 =50%
Thus, ans = 50% x 40% =20%
Y = 1/2-1/3 divided by 1/3 =50%
Thus, ans = 50% x 40% =20%
20 May 2025, 07:54
Kudos
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carcass
I guess, no one has used the weighted averages method for this question and hence posting this solution here for everyones benefit.
We are asked to determine the percent increase in number of toys over the original number of toys.
Option 1
If all the machines were old, working at 1 toy/3 min... then we will get 20 toys...
If all the machines were new, working at 1 toy/2 min... we will have 30 toys...
The ratio of new machines is 3:2...
Old (3) New (2)
20 -------------------------------------------------------30
We can see that the weighted average is 24... which will be 20% more than the current output.
Option 2
We are given the rates of old and new machines, and asked to find the latest rate if the ratio between old and new machine is 3:2
If all the machines were old, working at 1 toy/3 min... then we will get 20 toys...
If all the machines were new, working at 1 toy/2 min... i.e. 1.5 toys in 3 mins.
The ratio of new machines is 3:2...
Old (3) New (2)
1 -------------------------------------------------------1.5
We can see that the weighted average is 1.2... which will be 20% more than the current output.
Even though, I have posted the above method, I don't know whether I would be able to recollect this approach under time pressure during exam... and might have gone ahead with the number plugging approach here (The Good Old Fashioned Way
- Taking 5 machines, and calculating the output of each machine as per old and new rates... and finding out the percent increase) 
09 Jul 2025, 21:58
Kudos
Bookmarks
There are 2 types of machines.
M1 = 1/3 and M2 = 1/2
Work done by each of these machines in 1 hour:
M1 = (1/3)*60 M2 = (1/2)*60
Lets now calculate the original rate of work/rate of work using all old machines in an hour:
Here we can assume x to be the total number of machines:
60x/3
Now, since 40% of the machines were replaced with new ones. we will have 0.6x old machines and 0.4x new machines. Using this we will calculate the current rate of work:
(60(0.6x))/3 + (60(0.4x))/3
= 144x/3
Now we calculate the change in %
(((144x/3) - (60x/3))/(60x/3))*100 = 20%
M1 = 1/3 and M2 = 1/2
Work done by each of these machines in 1 hour:
M1 = (1/3)*60 M2 = (1/2)*60
Lets now calculate the original rate of work/rate of work using all old machines in an hour:
Here we can assume x to be the total number of machines:
60x/3
Now, since 40% of the machines were replaced with new ones. we will have 0.6x old machines and 0.4x new machines. Using this we will calculate the current rate of work:
(60(0.6x))/3 + (60(0.4x))/3
= 144x/3
Now we calculate the change in %
(((144x/3) - (60x/3))/(60x/3))*100 = 20%
