Great explanation

I solve most of the work related problems with below approach,

Machine B eats 40 chocolates per day => Machine A eats (3/4)*40 = 30 chocolates per day.

Machine A eats x chocolates in 2 days (days = lots) => 60 chocolates in 2 days. (x=60)

Machine A and Machine B working simultaneously will reduce efficiency of each by 20% => Machine A eats 24 chocolates/day and Machine B eats 32 chocolates/day.

There are 3x chocolates => 180 chocolates.

Machine A and B together eat 56 chocolates per day (32+24) => in 3 days they eat 168 chocolates.

If in three days they eat 168 chocolates, the remaining 12 chocolates, they eat in 0.21 days.

So total number of days to complete 180 chocolates is 3.21 days.

In this approach, choosing appropriate numbers to solve the problem is most important as it eases the calculation. I chose Machine B efficiency as 40 chocolates/day because of 3 reasons:
1) 40 is multiple of 10.
2) 20% of 40 is 8 (an integer)
3) (3/4) * 40 is also a multiple of 10 which is 30, so 20% of 30 is also an integer which is 6

To clarify the intent of the question stem, I've replaced how long with how many hours:



The constant rate at which machines A works is $\frac{3}{4}$ of the constant rate at which machine B works.
Let B's rate = 4 lots per day, implying that A's rate = 3 lots per day.

Under normal conditions, machine A would complete x lots in 2 days.
Since x = the number of lots produced by A working at a rate of 3 lots per day for 2 days, we get:
x = (rate)(time) = 3*2 = 6 lots

Normal time for A and B to produce 6 lots $= \frac{work}{combined-rate-for-A-and-B} = \frac{6}{3+4} = \frac{6}{7}$ days

Running the machines simultaneously reduces the work rate of each machine by 20%.
How many hours will it take the machines to complete 3x lots, working simultaneously?

To produce 3x lots -- in other words, 3 times the amount of work -- A and B would normally require 3 times the number of days above:
$3*\frac{6}{7} = \frac{18}{7}$ days

Rate and time have a RECIPROCAL RELATIONSHIP.
When the combined rate for A and B is reduced by 20%, the resulting rate is 4/5 their normal rate.
As a result, A and B will require 5/4 of their normal time:
$\frac{5}{4} * \frac{18}{7}$ days

To convert from days to hours, we multiply by 24 (since there are 24 hours in each day):
$\frac{5}{4} * \frac{18}{7} * 24 = 77.14$ hours

This was how I solved it.

Work rate of B = Y

Work rate of A = 3/4Y

If A completes X work in 2 days, rate per day = X/2

Therefore 3/4Y = X/2

Y = 2X/3

Substituting Y in rates for A and B gives
A = X/2
B = 2X/3
Effficiency of machine reduce by 20% when they work simultaneously so
New work rate for A = 80/100 x X/2 = 2X/5
New work rate for B = 80/100 x 2X/3 = 8X/15
Combined work rate = 2X/5 + 8X/15 = 14X/15
So 14X/15 = 3X/T

I discovered that, it was because I reduced it to the lowest term was why I got the answer wrong.
Can someone please tell me why I cant reduce to lowest term.

Your approach is sound.
Just a bit more work will yield the correct answer.



For A and B to produce 3x units at the combined rate above, the time $= \frac{work}{rate} = 3x ÷ \frac{14x}{15} = \frac{45}{14}$ days
Since there are 24 hours in a day, the required number of hours $= \frac{45}{14} * 24 = 77.14$ hours

THANKS

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