A boy goes from his home to school at a speed | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the distance between home and school be $S 	ext{ km}$.Time taken to go from home to school is $t_1 = \frac{S}{x} 	ext{ hours}$.Time taken to

Vedclass · Accepted Answer

$\frac{2xy}{x + y} 	ext{ km/hr}$

Vedclass · Answer

(C) Let the distance between home and school be $S 	ext{ km}$.Time taken to go from home to school is $t_1 = \frac{S}{x} 	ext{ hours}$.Time taken to return from school to home is $t_2 = \frac{S}{y} 	ext{ hours}$.Average speed is defined as the total distance divided by the total time taken.Total distance $= S + S = 2S 	ext{ km}$.Total time $= \frac{S}{x} + \frac{S}{y} = S \left( \frac{1}{x} + \frac{1}{y} ight) = S \left( \frac{x + y}{xy} ight) 	ext{ hours}$.Average speed $= \frac{2S}{S \left( \frac{x + y}{xy} ight)} = \frac{2xy}{x + y} 	ext{ km/hr}$.

$A$ boy goes from his home to school at a speed of $x \text{ km/hr}$ and returns at a speed of $y \text{ km/hr}$. What is the average speed of the boy?

Similar Questions

If $a, b, c$ are in $H.P.$,then the value of $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$ is

If $x = \sum_{n=0}^{\infty} a^n$,$y = \sum_{n=0}^{\infty} b^n$,and $z = \sum_{n=0}^{\infty} c^n$,where $a, b, c$ are in arithmetic progression and $|a| < 1, |b| < 1, |c| < 1$,then $x, y, z$ are in which progression?

If $(y - x)$,$2(y - a)$,and $(y - z)$ are in $H.P.$,then $(x - a)$,$(y - a)$,and $(z - a)$ are in

Find the Harmonic Mean $(H.M.)$ of the sequence $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots, \frac{1}{17}$.

If $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ are in $H.P.$,then $a, b, c$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module