CAT 2024 Slot 3 · Quant · Question 11

With annual compounding at rate r, the value after n years is 4000{\left( 1+r \right)}^{n}.

The ratio of the 3-year value to the 5-year value is \frac{{\left( 1+r \right)}^{3}}{{\left( 1+r \right)}^{5}}=\frac{1}{{\left( 1+r \right)}^{2}}.

Setting this equal to \frac{25}{36} gives {\left( 1+r \right)}^{2}=\frac{36}{25}.

Taking the positive root, 1+r=\frac{6}{5}, so r=0.2, a rate of 20% per annum.

The value must exceed ₹20000, so 4000{\left( \frac{6}{5}\right)}^{n}>20000.

Dividing by 4000, the condition is {\left( \frac{6}{5}\right)}^{n}>5.

Computing the powers, {\left( 1.2 \right)}^{8}\approx 4.300, which falls short.

Next, {\left( 1.2 \right)}^{9}\approx 5.160, which clears 5.

So 9 years are needed, and the value then is about ₹20640.

Shortcut. The two-year gap carries the whole rate: \frac{36}{25} is {\left( \frac{6}{5}\right)}^{2}, so r=20\% falls out at sight. After that, keep multiplying by 1.2 from 4000 — 4800, 5760, 6912, 8294, 9953, 11944, 14332, 17199, 20639 — and count to the first value past 20000.