Management Aptitude Test 2015 Paper

(c) Given A, = ₹ 944, T = 3 yr, ${\mathit{R}}_1$=R (let)
and ${\mathit{A}}_2$ = ₹ 980,
${\mathit{R}}_2$ = $\left(\mathit{R}+\frac{25}{100}\mathrm{of}\mathit{R}\right)$
SI is the difference of amount as the time period is
same i.e.
SI = ${\mathit{A}}_2-{\mathit{A}}_1$ = 980 - 944 = 36
As the rate is increased by 25%.
So, 25% of SI = 36
SI = $\frac{36\times 100}{100}$= 144
We know that, A = P + SI
⇒ 944 = P + 144
∴ P = 944 -144 = ₹ 800
To find the rate of interest,
SI = $\frac{\mathit{P}\times {\mathit{R}}_1\times \mathit{T}}{100}$
⇒ 144 = $\frac{800\times \mathit{R}\times 3}{100}\Rightarrow \frac{144}{8\times 3}$ = R
∴ R = 6%