THREE PAINTERS COMPLETED 2/7 , 1/5 AND 3/10 A WORK EACH ON THREE CONSECUTIVE DAYS. FIND OUT WHICH PAINTERS DID THE MOST WORK IN THE THREE DAYS.
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Step-by-step explanation:
Third painter did most work
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Answer:
worker C did the most work in three days.
given,
work done by three workers each day:
- A = 2/7
- B = 1/5
- C = 3/10
to find,
which worker did the most work in three days.
solution,
to find the worker who did the maximum work we must compare the total work done by each worker.
to compare their works, the denominator must be made equal. so, let's find the LCM of the denominators.
LCM of 7, 5, and 10:
so,
total work done by A= 2 x 10 / ( 7 x 10 ) = 20/70
total work done by B= 1 x 14 / ( 5 x 14 ) = 14/70
total work done by C= 3 x 7 / ( 10 x 7 ) = 21/70
on comparing the work done by A, B, and C, it can be seen that the work done by C is maximum.
therefore, worker C did the most work in three days.
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