In a race of 1 , km,A gives B a start of 100 | vedclass

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Question

(A) Let $A$ take $x$ seconds to cover $1000 \, m$ and $B$ take $y$ seconds to cover $1000 \, m$.In the first case,$A$ runs $1000 \, m$ in $x$ seconds,

Vedclass · Accepted Answer

$\frac{500}{29}$

Vedclass · Answer

(A) Let $A$ take $x$ seconds to cover $1000 \, m$ and $B$ take $y$ seconds to cover $1000 \, m$.In the first case,$A$ runs $1000 \, m$ in $x$ seconds,while $B$ runs $900 \, m$ in $(x + 20)$ seconds.So,the speed of $B$ is $v_B = \frac{900}{x + 20}$.In the second case,$A$ runs $950 \, m$ in $(y - 25)$ seconds,while $B$ runs $1000 \, m$ in $y$ seconds.So,the speed of $B$ is $v_B = \frac{1000}{y}$.Equating the speeds: $\frac{900}{x + 20} = \frac{1000}{y} \Rightarrow y = \frac{10}{9}(x + 20) = \frac{10x + 200}{9} \dots (i)$.Also,$A$'s speed is $v_A = \frac{1000}{x}$. In the second case,$A$ covers $950 \, m$ in $(y - 25)$ seconds: $\frac{950}{y - 25} = \frac{1000}{x} \Rightarrow \frac{19}{y - 25} = \frac{20}{x} \Rightarrow 19x = 20y - 500 \dots (ii)$.Substituting $(i)$ into $(ii)$: $19x = 20(\frac{10x + 200}{9}) - 500$.$171x = 200x + 4000 - 4500$.$29x = 500 \Rightarrow x = \frac{500}{29} \, s$.

In a race of $1 \, km$,$A$ gives $B$ a start of $100 \, m$ and still wins by $20 \, s$. But if $A$ gives $B$ a start of $25 \, s$,$B$ wins by $50 \, m$. The time taken by $A$ to run $1 \, km$ is (in $s$):

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