Math, asked by Arinkishore, 1 year ago

ten men,working for 6 days of 10 hours each

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Answered by realsujaykumar
68
1 day working of 10 men = (5/21)×(1/6)
1 day working of 1 man =(5/21) ×(1/6)×(1/10)
8 day working of 1 men =(5/21)×(1/6)×(1/10)×8
= 2/63
remaining work = 1-(5/21) = 16/21
hence, number of men required to complete the work in 8 days = (16/21)÷(2/63)
=(16×63)/(2×21) = 8×3 = 24 men
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I hope this helps.

Answered by tabbymykitten
3

one day working of 10 man is 5 by 21 into 1 by 6 one day working of one man is equal to 5 by 21 into 1 by 6 into 1 by 10 a day working of one man is equal to 5 by 21 into 1 by 6 into 1 by 10 in 28 which is equal to 2 by 63 remaining work 1 minus 5 by 21 which equals to 16 by 21 hence number of men required to complete work in 8 days is equal to 16 by 21 / 2 by 6316 in 263 by 2 into 21 is equal to 8 into 3 is equals to 24 men

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