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On a long stretch of east-west road, $A$ and $B$ are two points such that $B$ is $350$ km west of $A$. One car starts from $A$ and another from $B$ at the same time. If they move towards each other, then they meet after $1$ hour. If they both move towards east, then they meet in $7$ hrs. The difference between their speeds, in km per hour, is

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50

Given that, $A$ and $B$ are two points such that $B$ is $350 \; \text{km}$ of $A.$

Let $x\text{’}$ and $y\text{’}$ be the speed $(\text{in km/hr})$ of cars starting from both $A$ and $B$ respectively.

Let us assume $\text{car 2}$ traveled $d\;\text{km}$ and meet $\text{car 1},$ after $1 \; \text{hour}.$

$\boxed{\text{Speed} = \frac{\text{Distance}}{\text{Time}}}$

So, $S_{\text{car 1}} = \frac{350-d}{1}$

$\Rightarrow \boxed{x = 350 – d} \quad \longrightarrow (1)$

And, $S_{\text{car 2}} = \frac{d}{1}$

$\Rightarrow \boxed {y = d}$

From the equation $(1),$ we get

$x = 350 – d$

$\Rightarrow x = 350 – y$

$\Rightarrow \boxed{x+y = 350} \quad \longrightarrow (2)$

Let us assume, when they move toward east, they meet at point $P$ and distance traveled by $\text{car 1}$ is $p\text{’} \; \text{km},$ and distance traveled by $\text{car 2}$ is $350+p\text{’} \; \text{km}$ in $7 \; \text{hours}.$

So, $S_{\text{car 1}} = \text{p}{7}$

$\Rightarrow x = \frac{p}{7}$

$\Rightarrow \boxed{p = 7x} \quad \longrightarrow (3)$

And, $S_{\text{car 2}} = \frac{350+p}{7}$

$\Rightarrow y = \frac{350+7x}{7} \quad [\because \text{From equation} (3)]$

$\Rightarrow 7y = 350 + 7x$

$\Rightarrow 7y – 7x = 350$

$\Rightarrow 7(y – x) = 350$

$\Rightarrow y – x = \frac{350}{7}$

$\Rightarrow \boxed{y – x = 50 \; \text{km/hr}}$

$\textbf{PS:}$ If they both move in east direction, then $B$ will overtake $A$ only if $y>x.$

$\textbf{Short Method: }$

Concept of relative speed ;

• When  two bodies moves in the same direction then the $\boxed{\text{Relative speed = Difference of speeds}}$
• When two bodies move in opposite direction, then the $\boxed { \text {Relative speed = Sum of speeds}}$

Let $x\text{’}$ and $y\text{’}$ be the speed ( in km/hr) of cars starting from both $A$ and $B$ respectively.

$\text{Relative speed} = (y – x) \; \text{km/hr} \quad [\because y > x]$

They travel $350 \; \text{km}$ in $7 \; \text{hours}$ with a relative speed of $(y – x) \; \text{km/hr}.$

So, $(y -x) = \frac{350}{7}$

$\Rightarrow \boxed{ y – x = 50 \; \text{km/hr}}$

Correct Answer $: 50$

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