Question

a farmer plough 7/12 part to his field in first day, 3/12 part in second day and the rest in third day. what part of the field is ploughed in the third day​

Answer 1

Answer:1/6

Step-by-step explanation:

1-(7/12)-3/12=2/12=1/6

Answer 2

Answer:

The farmer ploughs $\frac{1}{6} th$ part of field on third day

Step-by-step explanation:

Given that,

A farmer ploughs  $\frac{7}{12} th$ part of field on first day and

He ploughs $\frac{3}{12} th$ part of field on second day and the rest is done on  third day.We are required to find what part of field is ploughed on the third day.

We shall write the given data in symbolic form

$t_{1}=\frac{7}{12} \\t_{2}=\frac{5}{12}$

We know that sum of work done on all the 3 days is equal to one.Therefore we have

$t_{1}+t_{2}+t_{3}=1\\= > \frac{7}{12} +\frac{3}{12} +t_{3}=1\\= > t_{3}=1-\frac{10}{12} =\frac{2}{12} =\frac{1}{6}$

Therefore,the farmer ploughs $\frac{1}{6} th$ of field on third day.

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