Math, asked by dole1234567890, 1 day ago

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

Answered by PyBoi
1

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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Answered by abinaya052009
0

Answer:

Mark me as brainliest

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

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