Three men A, B and C working together can do a job in 6 hours less time then A alone, in 1 hour less time than B alone and in one half the time needed by C when working alone. Then A and B together can do the job in :
Three men A, B and C working together can do a job in 6 hours less time then A alone, in 1 hour less time than B alone and in one half the time needed by C when working alone. Then A and B together can do the job in :
[A]$latex \frac{2}{3}\ hour&s=1$
[B]$latex \frac{3}{4}\ hour&s=1$
[C]$latex \frac{3}{2}\ hour&s=1$
[D]$latex \frac{4}{3}\ hour&s=1$
$latex \mathbf{\frac{4}{3}\ hour}&s=1$
Let A, B and C together do the work in x hours.
∴ Time taken by A = (x + 6) hours
Time taken by B = (x + 1) hours
Time taken by C = 2x hours
$latex \therefore \frac{1}{x+6}+\frac{1}{x+1}+\frac{1}{2x} = \frac{1}{x}&s=1$
$latex => \frac{1}{x+6}+\frac{1}{x+1} = \frac{1}{x} – \frac{1}{2x} = \frac{1}{2x}&s=1$
$latex => \frac{1}{x+6}+\frac{1}{x+1} = \frac{1}{2x}&s=1$
$latex => \frac{1}{x+6} = \frac{1}{2x} – \frac{1}{x+1} = \frac{x+1-2x}{2x(x+1)}&s=1$
$latex => \frac{1}{x+6} = \frac{1-x}{2x^{2}+2x}&s=1$
$latex => 2x^{2}+2x = x + 6 – x^{2} – 6x$
$latex => 3x^{2}+7x-6 = 0$
$latex => 3x^{2}+9x-2x-6 = 0$
$latex => 3x(x+3)-2(x+3) = 0$
$latex => (3x-2)(x+3) = 0$
$latex => 3x – 2 = 0\ as \ x+3 \neq 0$
$latex => x = \frac{2}{3}&s=1$
∴ Time taken by A $latex = 6+\frac{2}{3} = \frac{18+2}{3} = \frac{20}{3}\ hours&s=1$
Time taken by B $latex = 1+\frac{2}{3} = \frac{5}{3}\ hours&s=1$
∴ (A + B)’s 1 day’s work $latex = \frac{3}{20} + \frac{3}{5} = \frac{3+12}{20} = \frac{15}{20} = \frac{3}{4}&s=1$
∴ Required Time $latex = \frac{4}{3}\ hours&s=1$
Hence option [D] is correct answer.
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venkat
December 13, 2018 at 10:51 amis there any short cut method…. i’m unable to apply any short cut for this…..