Question

A certain sum was invested by a person on si at 5% per annum. After 6 months, he again made an investment of the same amount on si at 6% p.A. After a certain period of time, the amount received from both the investments are equal which is rs.4600. Find the sum he has invested in each investment?

Answer 1

Answer:

Adam borrowed some money at the rate of 6% p.a. for the first two .... We need to know the S.I., principal and time to find the rate.

Answer 2

The person invested Rs 4000 in each investment.

Solution:

Given, a certain sum was invested by a person on SI at 5% per annum.  

After 6 months, he again made an investment of the same amount on si at 6% p.A.  

After a certain period of time, the amount received from both the investments are equal which is rs.4600.  

We have to Find the sum he has invested in each investment?

Let the sum of amount invested be s.

We know that, simple interest $=\frac{\mathrm{p} \times \mathrm{r} \times \mathrm{t}}{100}$

Where, p is sum of amount invested, r is rate percentage, t is time period of investment

Now, S.I at 5% for 6 months $=\frac{s \times 5 \times 6 \text { months }}{100}=\frac{s \times 5 \times \frac{1}{2} \text { years }}{100}=\frac{s \times \frac{1}{2}}{20}=\frac{s}{20} \times \frac{1}{2}=\frac{s}{40}$

Let, the time period after 6 months be h.

Then S.I of 1st investment for time period h  $=\frac{s \times 5 \times h}{100}=\frac{s \times h}{20}$

And S.I of 2nd investment at 6% rate $=\frac{s \times 6 \times h}{100}$

Now, according to given information,

1st investment + S.I of 1st investment for 6 months + S.I of 1st investment for time period h = 4600

$\begin{array}{l}{S+\frac{s}{40}+\frac{s \times h}{20}=4600} \\\\ {S\left(1+\frac{1}{40}+\frac{2 \times h}{2 \times 20}\right)=4600} \\\\ {S\left(1+\frac{(1+2 h)}{40}\right)=4600 \rightarrow(1)}\end{array}$

And, 2nd investment + S.I for time period h = 4600

$\begin{array}{l}{S+\frac{s \times 6 \times h}{100}=4600} \\\\ {S\left(1+\frac{3 h}{50}\right)=4600 \rightarrow(2)}\end{array}$

Now, equate (1) and (2)

$\begin{array}{l}{S\left(1+\frac{1+2 h)}{40}\right)=S\left(1+\frac{3 h}{50}\right)} \\\\ {1+\frac{1+2 h}{40}=1+\frac{3 h}{50}} \\\\ {\frac{1+2 h}{40}=\frac{3 h}{50}}\end{array}$

50(1 + 2h) = 40(3h)

5(1 + 2h) = 4 x 3h

5 + 10h = 12h

12h – 10h = 5

2h = 5

h = 2.5

now, substitute h value in (2)

$\begin{array}{l}{S\left(1+\frac{3 \times 2.5}{50}\right)=4600} \\\\ {S\left(1+\frac{3}{20}\right)=4600} \\\\ {S \times \frac{23}{20}=4600} \\\\ {S=4600 \times \frac{20}{23}} \\\\ {S=200 \times 20=4000}\end{array}$

Hence, he invested Rs 4000 in each investment.