Question

three painters completed 2/5 , 1/5 and 3/10 of a work each of three consecutive days . find out which painter did the most work in the three days .

Answer 1

Answer:

Step-by-step explanation:

Three painters completed in consequence days = 2/5+1/5+3/10

=4+2+3/10 = 9/10

Total work =9/10

Therefore

Worker a 9/10-2/5=9-4/10=5/10

Worker b 9/10-1/5=9-2/10=7/10

Worker c 9/10-3/10=6/10

So here worker B complete 7/10th of painting work

Answer 2

Step-by-step explanation:

Let first painter completed the work = 2/7

second painter completed the work = 1/5

third painter completed the work = 3/10

Therefore, LCM of 7,5 and 10 is 70

2/7 = 2×10/7×10 = 20/70

1/5 = 1×14/5×14 = 14/70

3/10 = 3×7/10×7 = 21/70

Therefore,21/70,20/70,14/70

Hence, the third painter did the most work in the three days.