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If a person increases his speed by 1 km/hr he reaches his office in 3/4th of the time he normally takes and if he decreases his speed by 1 km/hr, he reaches his office 2 hrs late. What distance dose the person travels to his office?

(a) 10 kms  (b) 8 kms

(b) 12 kms  (d) 9 kms

Let $t$ be the normal time he takes and let $S$ be the distance.

$$$$\frac{S}{\frac{S}{t} + 1} = \frac{3t}{4} \\\implies \frac{S}{S+t} =\frac{3}{4} \\\implies S = 3t \label{670eq1}$$$$

$$$$\frac{S}{\frac{S}{t} -1} = t + 2 \\\implies \frac{St}{S-t} = t + 2 \label{670eq2}$$$$

From $\ref{670eq1}$ and $\ref{670eq2}$,

$$\frac{3t^2}{2t} = t+2 \\\implies t = 4\\\implies S = 12kms$$.
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