Question

Two cranes worked for $15$ h to unload the barge. One crane began operating $7$ h later than the other. It is known that the first crane alone can unload the barge $5$ h faster than the second crane. How many hours does it take each crane alone to unload the barge?

Answer 1

Let the first crane take $x-5$ hours

let the second crane take $x$ hours

Second crane can do $\frac{1}{x}$ part in $1$ hour

second crane can do $\frac{7}{x}$ part in $7$ hours.

Remaining work $1-\frac{7}{x}=\frac{x-7}{x}$

They both do the remaining work together in 8 hours. i.e. $\frac{1}{x}+\frac{1}{x-5}$

$\frac{x-5+x}{x^2-5x}=\frac{2x-5}{x^2-5x}$ part in $1$ hour.

$\frac{\frac{2x-5}{(x-5)x}}{\frac{x-7}{x}}=8$

$\frac{2x-5}{x^2-12x+35}=8$

$2x-5={8x}^2-96x+280$

${8x}^2-98x+285=0$

${8x}^2-60x-38x+285=0$

$4x(2x-15)-19(2x-15)=0$

$x=\frac{19}{4},\frac{15}{2}$ $\to$ first crane take $2.5$ hours

second crane take $7.5$ hours.