A road 7/4 km. long was prepared in 4 days. On the first day 2/5of the road, on 2nd day 1/7 of the road and on the 3rd day 5/21 of the road was repaired. What length of the road was repaired on the 4th day ?
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Step-by-step explanation:
First of all take the lcm of
4, 5, 7 and 21
i.e. 420
Then equalise (MAKE THE DENOMINATORS OF ALL THE FRACTIONS EQUAL) all the fractions related to it i.e.
7/4 = 735/420 (TOTAL WORK)
2/5 = 168/420
1/7 = 60 /420
5/21=100/420
Remaining work = 735 - 168 - 60 - 100
= 407
Therefore on the last day of work, the work done is
= 407/420
Hope it helped you ...............
ARINDOMDEBBARMA:
my answer was also 407/420 arrived but according to book result is 23/60
Sorry my mistake....
I'll just correct it for you
I'm unable to edit the question, can you please delete this question. I will answer the same question you have asked.
actually I'm unable to find delete option
ok no problem
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