Question

$8$ men and $12$ boys can finish a piece of work in $10$ days while $6$ men and $8$ boys can finish in it $14$ days. Find the time taken by $1$ man alone and that by one boy alone to finish the work.

Answer 1

Let the time taken by man be $xV days

time taken by boys be$y$ days

work done by 1 man in one day$=1/x$

work done by 1 boy in one day$=1/y$

$8/x+12/y=1/10$............eq 1

$6/x+8/y=1/14$..........eq2

Let u=1/x & v=1/y

$8u+12y=1/10.$........eq3

$6u+8v=1/14$...........eq4

Multiplying eq3 by 2 & eq4 by 3

$16u+24y=2/10$.........eq5

$18u+24v=3/14$.........eq6

subtract eq 5 & 6

$-2u=-3/14+2/10$

$-2u=(-3\times 10+2\times 14)/140$

$-2u=(-30+28)/140$

$-2u=-2/140$

$u=1/140$

put this value in eq 5

$16u+24y=2/10$

$16\times 1/140+24y=1/5$

$4/35+24y=1/5$

$24y=1/5-4/35$

$24y=(1\times 7-4)/35$

$24y=(7-4)/35$

$24y=3/35$

$y=3/35\times 24$

$y=1/280$

$u=1/x$

$1/140=1/x$

$x=140$

$v=1/y$

$1/280=1/y$

$y=280$

A $man$ complete the work in 140 days

A $boy$ can complete the work in 280 days