Test Index

Time Value of Money

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Question : 36 of 99

Marks: +1, -0

A sum of 44,000 is divided into 3 parts such that the corresponding interest earned after 2 years, 3 years and 6 years may be equal at the rate of simple interest are

6 \%

8 \%

6 \%

p.a. respectively. Then the smallest part of the sum will be :

$\text{Rs. }4,000$
$\text{Rs. }8,000$
$\text{Rs. }10,000$
$\text{₹}12,000$

Solution:

Concept:

Equate the simple interest from three parts to find their ratio using

SI = \frac{P \times R \times T}{100}

.

Explanation:

Let the three parts be

P_1

,

P_2

, and

P_3

.

Given

SI_1 = SI_2 = SI_3 = k

(say).

SI_1 = \frac{P_1 \times 6 \times 2}{100} = 0.12P_1 = k \implies P_1 = \frac{k}{0.12}

.

SI_2 = \frac{P_2 \times 8 \times 3}{100} = 0.24P_2 = k \implies P_2 = \frac{k}{0.24}

.

SI_3 = \frac{P_3 \times 6 \times 6}{100} = 0.36P_3 = k \implies P_3 = \frac{k}{0.36}

.

So

P_1 : P_2 : P_3 = \frac{1}{0.12} : \frac{1}{0.24} : \frac{1}{0.36}

.

Convert to fractions:

0.12 = \frac{12}{100} = \frac{3}{25}

,

0.24 = \frac{6}{25}

,

0.36 = \frac{9}{25}

.

Thus

P_1 : P_2 : P_3 = \frac{25}{3} : \frac{25}{6} : \frac{25}{9} = \frac{1}{3} : \frac{1}{6} : \frac{1}{9}

.

Multiply by 18:

P_1 : P_2 : P_3 = 6 : 3 : 2

.

Total sum = ₹44,000. Sum of ratios =

6+3+2 = 11

.

P_1 = \frac{6}{11} \times 44000 = ₹24,000

,

P_2 = \frac{3}{11} \times 44000 = ₹12,000

,

P_3 = \frac{2}{11} \times 44000 = ₹8,000

.

Smallest part is ₹8,000.

Answer:

₹8,000 (Option B).

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