Mensuration Practice Test 7

Time taken by B = $\frac{100}{\mathit{x}}$ sec
Time taken by A = $\frac{100}{\mathit{x}}$ sec.

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