Mahesh, Suresh and Naresh together can complete a work in 4 days. The wages paid to Mahesh, Suresh and Naresh for completing the work are *6300, 37350 and *8400 respectively. Find the number of days in which Naresh alone can complete the work. #maths
Answers
Answer:
Below is the solution
Step-by-step explanation:
According to the given problem,
(i) Let W denotes the whole given piece of work.
(ii) Mahesh, Suresh & Naresh can together complete the work W in 4 days.
(iii) The wages paid to them for completing the work W is 6300, 7350 & 8400 respectively.
(iv) Let M, S & N denote the amounts of time (in days) required by Mahesh, Suresh & Naresh respectively to complete the work W seperately. Hence,
(v) Mahesh, Suresh & Naresh seperately in 1 day can complete the amounts of work W/M, W/S & W/N respectively.
From (ii) & (v) we get,
4*(W/M + W/S + W/N) = W
or W/M + W/S + W/N = W/4 …. (1a)
From (iii) & (v) we get,
W/M : W/S : W/N = 6300 : 7350 : 8400 [wage is proportional to rate of work done per day]
or W/M : W/S : W/N = 6 : 7 : 8 …. (1b) [dividing RHS by HCF 1050]
From (1b) we get,
W/N = [8/(6 + 7 + 8)]*(W/M + W/S + W/N)
or W/N = (8/21)*(W/4) [from (1b)]
or 1/N = 2/21 or N = 21/2 = 10 (1/2) (days)
Therefore it is evident from above that
the number of days in which Naresh alone
can complete the whole work = 10 (1/2) days [Ans]
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Answer:
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Step-by-step explanation:
M = 6300
S = 7350
N = 8400
Take LCM of these 176400
Efficiency of M , S , N = 28 , 24 , 21
Together complete the work in 4 days
Total work = efficiency * time
= (28 + 24 + 21) * 4
= 292
Naresh alone to complete the work in
= 292/21 days
= 13.9 days
= 14 days
= 2 weeks