A and B run a 1 , km race. If A gives B a sta | vedclass

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(A) Let $x$ and $y$ be the time (in $sec$) taken by $A$ and $B$ to run $1 \, km$ $(1000 \, m)$,respectively.Case $1$: $A$ gives $B$ a start of $50 \,

Vedclass · Accepted Answer

$100$

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(A) Let $x$ and $y$ be the time (in $sec$) taken by $A$ and $B$ to run $1 \, km$ $(1000 \, m)$,respectively.Case $1$: $A$ gives $B$ a start of $50 \, m$. $B$ runs $950 \, m$ in time $y$,while $A$ runs $1000 \, m$ in time $x$. Since $A$ wins by $14 \, sec$,$A$ takes $14 \, sec$ less than $B$ to cover the distance.Time taken by $B$ to run $950 \, m = \frac{950}{1000} y = 0.95 y$.So,$0.95 y - x = 14$ $....(1)$Case $2$: $A$ gives $B$ a start of $22 \, sec$. $A$ runs for $(y - 22) \, sec$ and $B$ runs for $y \, sec$. $B$ wins by $20 \, m$,meaning $B$ covers $1000 \, m$ while $A$ covers $980 \, m$.Speed of $A = \frac{1000}{x} \, m/sec$.Distance covered by $A$ in $(y - 22) \, sec = \frac{1000}{x} (y - 22) = 980$.$1000(y - 22) = 980x \Rightarrow 100y - 2200 = 98x \Rightarrow 50y - 49x = 1100$ $....(2)$From $(1)$,$x = 0.95y - 14$. Substitute into $(2)$:$50y - 49(0.95y - 14) = 1100$$50y - 46.55y + 686 = 1100$$3.45y = 414 \Rightarrow y = 120 \, sec$.Substituting $y = 120$ in $(1)$:$x = 0.95(120) - 14 = 114 - 14 = 100 \, sec$.

$A$ and $B$ run a $1 \, km$ race. If $A$ gives $B$ a start of $50 \, m$,$A$ wins by $14 \, sec$. If $A$ gives $B$ a start of $22 \, sec$,$B$ wins by $20 \, m$. The time taken by $A$ to run a $1 \, km$ is (in $sec$):

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