# 8 men and 12 boys can finish a piece of work in 10 days while 6 men and boys can finish it in 14 days find the time taken by one man alone and that by one boy alone to finish the work

Let m = the men's rate of work in jobs/man/day
Let b = the boys' rate of work in jobs/boy/day
Given:
8 men and 12 boys finish in 10 days
6 men and 8 boys finish in 14 days
For these two jobs, we can write
m*8*10 + b*12*10 = 1[job]
m*6*14 + b*8*14 = 1[job]
80m + 120b = 1
84m + 112b = 1
m + 3b/2 = 1/80 -> m = 1/80 - 3b/2
3m + 4b = 1/28 -> 3(1/80 - 3b/2) + 4b = 1/28
Solve for b:
3/80 - 9b/2 + 4b = 1/28
b/2 = 3/80 - 1/28 = 21/560 - 20/560 = 1/560
b = 1/280 [jobs/boy/day]
If one boy is working alone, the rate is 1 job per 280 days, so it would take 280 days to finish
The men's rate = m = 1/80 - (3/2)(1/280) = 7/560 - 3/560 = 4/560 = 1/140
Thus a man working alone would need 140 days to complete the job

• -13

Let the part of the job done by 1 men per day be x

Let the part of the job done by 1 boy per day be y

So,

Multiply equation 1 by 3 and equation 2 by 4 to get,

From equation 3 and 4,

So the time taken by the boy to complete the job alone will be 280 days and for the men will be 140.

• 61
Let m = the men's rate of work in jobs/man/day
Let b = the boys' rate of work in jobs/boy/day
Given:
8 men and 12 boys finish in 10 days
6 men and 8 boys finish in 14 days
For these two jobs, we can write
m*8*10 + b*12*10 = 1[job]
m*6*14 + b*8*14 = 1[job]
80m + 120b = 1

84m + 112b = 1
m + 3b/2 = 1/80 -> m = 1/80 - 3b/2
3m + 4b = 1/28 -> 3(1/80 - 3b/2) + 4b = 1/28
Solve for b:
3/80 - 9b/2 + 4b = 1/28
b/2 = 3/80 - 1/28 = 21/560 - 20/560 = 1/560
b = 1/280 [jobs/boy/day]
If one boy is working alone, the rate is 1 job per 280 days, so it would take 280 days to finish
The men's rate = m = 1/80 - (3/2)(1/280) = 7/560 - 3/560 = 4/560 = 1/140
Thus a man working alone would need 140 days to complete the job
• 7
140
• -17
One man will take 9 days to complete
One boy will take 5.1 days to complete

• -30
mine is also 140
• -18
1 man will  take 140 days
• -8
man takes 140 days and boy takes 280 days
• -12
Time taken for work done by 1 man= 1/x days,  Time taken for work done by 1 boy=1/y days

8/x + 12/y= 1/10,  multiply by 3=>24/x+36/y=3/10--(1)
6/x + 8/y = 1/14, multiply by 4=>24/x+32/y=4/14--(2)

(1)-(2)
24/x-24/x+36/y-32/y=3/10-4/14
4/y=42-40/140
4/y=1/70
y=280--(3)

substitiute (3) in (1)
24/x+ 36/280=3/10
x=140

• 18
I/I40 Da dei lossu Answer Maja VA
• 2
Let m = the men's rate of work in jobs/man/day
Let b = the boys' rate of work in jobs/boy/day
Given:
8 men and 12 boys finish in 10 days
6 men and 8 boys finish in 14 days
For these two jobs, we can write
m*8*10 + b*12*10 = 1[job]
m*6*14 + b*8*14 = 1[job]
80m + 120b = 1

84m + 112b = 1
m + 3b/2 = 1/80 -> m = 1/80 - 3b/2
3m + 4b = 1/28 -> 3(1/80 - 3b/2) + 4b = 1/28
Solve for b:
3/80 - 9b/2 + 4b = 1/28
b/2 = 3/80 - 1/28 = 21/560 - 20/560 = 1/560
b = 1/280 [jobs/boy/day]
If one boy is working alone, the rate is 1 job per 280 days, so it would take 280 days to finish
The men's rate = m = 1/80 - (3/2)(1/280) = 7/560 - 3/560 = 4/560 = 1/140
Thus a man working alone would need 140 days to complete the job
• 0
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