four men and six boys can finish a piece of work bin 5 days while 3 men and four boys can finish it in 7 days . find time by one man alone and one boy alone
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Panzer786
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Let 1 man alone can finish the work in X days and 1 boy alone can finish it in Y .
Then,
1 man's 1 days work = 1/X
And,
1 boy's 1 day's work = 1/Y
(4 men's 1 day's work )+(6 boys 1 day's work )= 1/5
=> 4/X + 6/Y = 1/5
=> 4U + 6V = 1/5 [ Where 1/X = U and 1/Y =V]
=> 4U + 6V = 1/5 --------(1)
Again,
( 3 men's 1 day's work ) + ( 4 boys 1 day's work ) = 1/7
=> 3/X + 4/Y = 1/7
=> 3U + 4V = 1/7 -------(2)
On multiplying (1) by 3 and ,(2) by 4 we get,
12U + 18V = 3/5 --------(3)
And,
12U + 16 V = 4/7 --------(4)
Subtracting (3) and (4) we get,
2V = ( 3/5 - 4/7)
2V = 1/35
V = 1/35 ×2
V = 1/70
1/Y = V
1/Y = 1/70
Y = 70 days
Putting V = 1/70 in equation (1) we get,
4U + 6V = 1/5
4U = ( 1/5 - 6V )
4U = ( 1/5 - 6/70 )
4U = ( 14 - 6 /70)
4U = ( 8/70)
U = 8/70 × 1/4
U = 1/35
1/X = U
1/X = 1/35
X = 35 days
Therefore,
One man alone can finish the work in 70 days and One boy alone can finish the work in 35 days.
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