Investment Word Problems

Investment is the commitment of money or capital to purchase financial instruments or other assets in order to gain profitable returns in form of interest, income, or appreciation of the value of the instrument. Investment is involved in many areas of the economy, such as business management and finance no matter for households, firms, or governments. Investment word problems typically involve Simple interest calculations. In this article we will see few Investment word problems . (Source: Wikipedia)
Investment Word Problems Examples:

Example 1:

Rajul put `$` 4000 into an investment at 5% annual interest for 2 years. How much interest will he get at the end of two years?

Solution:

Given,

Principal = P= `$` 4000

Interest rate = R = 5%

=` 5/100` = 0.05

Time = T= 2 years

By formula,

Simple interest = P`xx` R`xx` T

I= 4000 `xx ` 0.05 `xx` 2

= 400

Hence Rajul will get the interest `$` 400.

Example 2:

Mr. David invested `$` 56,000 part at 5% and the rest at 7% annual interest.  If his annual incomes from both investments are `$` 3500, find the amount invested at each rate.

Solution:

Given,

Investment = `$` 56000,

Interest rates = 5% and 7%,

Interest earned = `$` 3500,

Time = 1 year.

We can derive the given statement as,

The amount invested at 5% + the amount invested at 7% = 56,000

Let x = amount invested at 5% and 56000-x = amount invested at 7%.

And,

The interest earned at 5% + the interest earned at 7% = 3500

By formula,

I= PRT

(x `xx` 0.05 `xx` 1 )+ ((56000-x) `xx` 0.07 `xx` 1) = 3500

0.05x+3920-0.07x= 3500

Combine like terms,

0.05x-0.07x= 3500 -3920

-0.02x=-420

x= `420/0.02`

=21000

x = amount invested at 5% = `$` 21000

56000-x = amount invested at 7%= 56000-21000 = `$` 35000
Investment Word Problems Examples Continued:

Example 3:

Mr. John wishes to get a return of 10% on a total of two investments. If he has `$` 8,000 invested at 8%, how much additional money should be invested at 12%?

Solution:

Let, x= amount invested at 12%

0.08(8000) + .12x = 0.10(8000 + x)

640 + .12x = 800 + .10x

Combine like terms,

0.12x-0.10x = 800-640

0.02x = 160

x = `160/0.02`

=8000

Hence John should invest `$` 8000 at 12% to get return of 10%.