Question

If a person increases his speed by 1 km/h, he reaches the
office in 3th/4
of the time he normally takes and if he
decreases his speed by 1 km/h, he reaches his office 2 hours
late. Find the distance that the person travels to his office.​

Answer 1

$Let \\ i)normal \: speed \: of \: a \: person = x \:km/h$

$Time = t \: hours$

$Distance \: travelled = d \: km$

$\blue { d = xt \: km} \: ---(1)$

/* If he increases the speed by 1 km/h then he reaches the office (3/4)th of normal time */

$Increased \:speed = (x+1) \: km/h$

$time (t_{2}) = \frac{d}{(x+1)}$

$\implies \frac{3t}{4} = \frac{d}{(x+1)}$

$\implies \frac{3t}{4} = \frac{xt}{(x+1)}\: [From \:(1)]$

$\implies \frac{3}{4} = \frac{x}{(x+1)}$

$\implies 3(x+1) = 4x$

$\implies 3x + 3 = 4x$

$\implies 3 = 4x - 3x$

$\implies x = 3 \: ---(2)$

/* If he decreases the speed by 1 km/h then he reaches the office 2 hours late of normal time */

$Decreased \: speed = (x-1) \: km/h$

$\implies t + 2 = \frac{d}{(x-1)}$

$\implies t + 2 = \frac{xt}{(x-1)} \: [From \:(1)]$

$\implies t + 2 = \frac{3t}{(3-1)} \: [From \: (2) ]$

$\implies t + 2 = \frac{3t}{2}$

$\implies 2(t+2) = 3t$

$\implies 2t + 4 = 3t$

$\implies 4 = 3t - 2t$

$\implies t = 4 \: ---(3)$

/* Substitute x and t values in equation (1) ,we get*/

$\implies d = 3 \times 4 = 12 \: km$

Therefore.,

$\red { Distance \: that \: the \: person }$

$\red { travels \: to \: the \: office } \green {= 12 \: km}$

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