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(B) Let the distance between the house and the school be $x 	ext{ km}$.Initial speed $v_1 = 2.5 	ext{ km/h} = \frac{5}{2} 	ext{ km/h}$.New speed $v

Vedclass · Accepted Answer

$\frac{7}{4}$

Vedclass · Answer

(B) Let the distance between the house and the school be $x 	ext{ km}$.Initial speed $v_1 = 2.5 	ext{ km/h} = \frac{5}{2} 	ext{ km/h}$.New speed $v_2 = 2.5 + 1 = 3.5 	ext{ km/h} = \frac{7}{2} 	ext{ km/h}$.Time taken at speed $v_1$ is $t_1 = \frac{x}{v_1} = \frac{x}{5/2} = \frac{2x}{5} 	ext{ hours}$.Time taken at speed $v_2$ is $t_2 = \frac{x}{v_2} = \frac{x}{7/2} = \frac{2x}{7} 	ext{ hours}$.The difference in time is $6 	ext{ min late} - (-6 	ext{ min early}) = 12 	ext{ min} = \frac{12}{60} 	ext{ hours} = \frac{1}{5} 	ext{ hours}$.So,$t_1 - t_2 = \frac{1}{5} \Rightarrow \frac{2x}{5} - \frac{2x}{7} = \frac{1}{5}$.Multiplying by $35$,we get $14x - 10x = 7$.$4x = 7 \Rightarrow x = \frac{7}{4} 	ext{ km}$.

$A$ student walks from his house at a speed of $2 \frac{1}{2} \text{ km/h}$ and reaches his school $6 \text{ min}$ late. The next day,he increases his speed by $1 \text{ km/h}$ and reaches $6 \text{ min}$ before school time. How far is the school from his house? (in $\text{km}$)

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