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# CAT 2021 Quant Question [Slot 1] with Solution 12

Question
Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to
1. 929.48
2. 934.65
3. 931.72
4. 926.84
Option: 3
Solution:
Let the principal amount be P and the interest rate be r.
Then $P{\left( {1 + r} \right)^2} - P\left( {1 + r} \right) = 806.25$ -(1)
$P{\left( {1 + r} \right)^3} - P{\left( {1 + r} \right)^2} = 866.72$ -(2)
Dividing (2) by (1), we get:
$\frac{{\left( {P{{\left( {1 + r} \right)}^3} - P{{\left( {1 + r} \right)}^2}} \right)}}{{P{{\left( {1 + r} \right)}^2} - P\left( {1 + r} \right)}} = \frac{{866.72}}{{806.25}}$
$\frac{{\left( {{{\left( {1 + r} \right)}^2} - 1 - r} \right)}}{{1 + r - 1}} = 1.075$
$\frac{{{r^2} + r}}{r} = 1.075$
r=0.075 or 7.5%
$\frac{{\left( {Interest\;accrued\;in\;4th\;yr} \right)}}{{Interest\;accrued\;in\;3rd\;yr}} = \frac{X}{{866.72}}$
$\frac{{\left( {P{{\left( {1 + r} \right)}^4} - P{{\left( {1 + r} \right)}^3}} \right)}}{{P{{\left( {1 + r} \right)}^3} - P{{\left( {1 + r} \right)}^2}}} = \frac{X}{{866.72}}$
Dividing numerator and denominator by $P{\left( {1 + r} \right)^2}$
$\frac{{{r^2} + 2r + 1 - 1 - r}}{{1 + r - 1}} = \frac{X}{{866.72}}$
$r + 1 = \frac{X}{{866.72}}$
$X = 1.075 \times \;866.72 = 931.72$
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## CAT 2021 Quant questions with Solutions

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