SSC CHSL 5 Mar 2018 Shift 1 Paper

Let distance between school and house = $\mathit{d}$ km and ideal time to reach school = $\mathit{t}$ hours
When he walks at a speed of 5 km/hr, he reaches school 3 minutes late, ⇒ using distance = speed × time
⇒ $\mathit{d}=5\times \left(\mathit{t}+\frac{3}{60}\right)$---------------(i)
Similarly,$\mathit{d}=9\times \left(\mathit{t}-\frac{3}{60}\right)$--------------(ii)
Comparing, equations (i) and (ii), we get :
⇒ $5\left(\mathit{t}+\frac{3}{60}\right)=9\left(\mathit{t}-\frac{3}{60}\right)$
⇒ $5\mathit{t}+\frac{1}{4}=9\mathit{t}-\frac{9}{20}$
⇒ $9\mathit{t}-5\mathit{t}=\frac{1}{4}+\frac{9}{20}$
⇒ $4\mathit{t}=\frac{\left(5+9\right)}{20}=\frac{14}{20}$
⇒ $\mathit{t}=\frac{7}{10}\times \frac{1}{4}=\frac{7}{40}$
Substituting it in equation (i), ⇒ $\mathit{d}=5\left(\frac{7}{40}+\frac{3}{60}\right)$
= $5\left(\frac{\left(21+6\right)}{120}\right)=\frac{27}{24}=1.125$ km