Time and Distance Practice Test 3 with solutions

Time taken by B = $\frac{100}{\mathit{x}}$ second
Time taken by A = $\frac{100}{\mathit{x}}$ second
Distance covered by A = $\frac{100}{\mathit{x}}$ × y
= $100\left(\frac{\mathit{y}}{\mathit{x}}\right)\mathit{m}$
Circumference = $2\left[100+\left(100\frac{\mathit{y}}{\mathit{x}}\right)\right]$
= $200\left(1+\frac{\mathit{y}}{\mathit{x}}\right)$ ... (i)
Distance covered by B = $100\frac{\left(\frac{\mathit{y}}{\mathit{x}}\right)}{\mathit{x}}+60$
=
= $\mathit{x}\left[200+200\left(\frac{\mathit{y}}{\mathit{x}}\right)-60-100\frac{\mathit{y}}{\mathit{x}}\right]$
$100\frac{{\mathit{y}}^2}{\mathit{x}}$ = 200x + 200y - 60x - 100y
$100\frac{{\mathit{y}}^2}{\mathit{x}}+60\mathit{y}$ = 140x + 100y
$100\frac{{\mathit{y}}^2}{\mathit{x}}$ = 140x + 40y
$10{\mathit{y}}^2$ = $14{\mathit{x}}^2+4\mathit{y}\mathit{x}$
$5{\mathit{y}}^2$ = $7{\mathit{x}}^2+2\mathit{x}\mathit{y}$
$5{\left(\frac{\mathit{y}}{\mathit{x}}\right)}^2$ = $7+2\left(\frac{\mathit{y}}{\mathit{x}}\right)$
Let $\frac{\mathit{y}}{\mathit{x}}$ = t
$5{\mathit{t}}^2$ - 2t - 7 = 0
5t 2- 7t + 5t - 7 = 0
(5t + 7) (t + 1) = 0
t(5t - 7) + 1 (5t - 7)
t = -1 not possible
t = $\frac{7}{5}$
Substituting value of t in equation (i),
$200\left(1+\frac{7}{5}\right)$ = $200\left(\frac{12}{5}\right)$ = 480 m