# INTEREST WORD PROBLEMS WORKSHEET

## About "Interest word problems worksheet"

Interest word problems worksheet :

Worksheet on interest word problems is much useful to the students who would like to practice problems on simple interest.

## Interest word problems worksheet - Questions

1.  Alex invested \$500 and received \$650 after three years. What had been the interest rate ?

2.  Arthur invests his inheritance of \$24000 in two different accounts which pays 6% and 5% annual interest. After one year, he received \$1340 in interest. How much did he invest in each account ?

3.    Part of \$5000 was invested at 5% and the other part at 6%. The 6% investment yielded \$135 more in profit than the other investment. How much money was invested at each rate ?

4.   Josh invested some money at 6% annual interest and three times as much at 8%. The total interest after one year was \$660. How much did he invest at each rate ?

## Interest word problems worksheet - Answers

Question 1 :

Alex invested \$500 and received \$650 after three years. What had been the interest rate ?

Step 1 :

Interest earned  =  A - P

Interest earned  =  650 - 500  =  150

And also,

P  =  500

T  =  3 years

Step 2 :

The interest formula is

I  =  PRT

Plug the values of I, P and T in the interest formula.

150  =  500 x R x 3

150  =  1500 x R

Divide both sides by 1500

150 / 1500  =  R

0.1  =  R

To convert decimal into percentage, multiply by 100.

(0.1 x 100) %  =  R

10 %  =  R

Hence, the interest rate is 10%.

Question 2 :

Arthur invests his inheritance of \$24000 in two different accounts which pays 6% and 5% annual interest. After one year, he received \$1340 in interest. How much did he invest in each account ?

Step 1 :

Let "x" be the money invested at 6% rate.

Then, the money invested at 5% rate  =  24000 - x

Step 2 :

The interest formula is

I  =  PRT

Step 3 :

Interest earned at 6% rate is

I  =  (x)(0.06)(1)

I  =  0.06x

Step 4 :

Interest earned at 5% rate is

I  =  (24000 - x)(0.05)(1)

I  =  (24000 - x)(0.05)

I  =  1200 - 0.05x

Step 5 :

Adding the results of step 3 and step 4, we get

0.06x + 1200 - 0.05x  =  Total interest

0.01x + 1200  =  Total interest

In the question, the total interest is given \$1340.

Then, we have

0.01x + 1200  =  \$1340

Step 6 :

Solve for "x"

0.01x + 1200  =  \$1340

Subtract 1200 from both sides

0.01x  =  \$140

Divide both sides by 0.01

x  =  140 / 0.01

x  =  14000

Then,

24000 - x  =  24000 - 14000

24000 - x  =  10000

Hence, the money invested at 6% rate is \$14000 and 5% rate is \$10000.

Question 3 :

Part of \$5000 was invested at 5% and the other part at 6%. The 6% investment yielded \$135 more in profit than the other investment. How much money was invested at each rate ?

Step 1 :

Let "x" be the money invested at 6% rate.

Then, the money invested at 6% rate  =  5000 - x

Step 2 :

The interest formula is

I  =  PRT

Step 3 :

Interest earned at 5% rate is

I  =  (x)(0.05)(1)

I  =  0.05x

Step 4 :

Interest earned at 6% rate is

I  =  (5000 - x)(0.06)(1)

I  =  (5000 - x)(0.06)

I  =  300 - 0.06x

Step 5 :

From, the given information, the difference between the interests earned at 6% and 5% is \$135.

(300 - 0.06x) - (0.05x)  =  135

300 - 0.06x - 0.05x  =  135

300 - 0.11x  =  135

300 - 135  =  0.11x

165  =  0.11x

Divide both sides by 0.11

165/0.11  =  x

1500  =  x

Then,

5000 - x  =  5000 - 1500

5000 - x  =  3500

Hence, the money invested at 5% rate is \$1500 and 6% rate is \$3500.

Question 4 :

Josh invested some money at 6% annual interest and three times as much at 8%. The total interest after one year was \$660. How much did he invest at each rate ?

Step 1 :

Let "x" be the money invested at 6% rate.

Then, the money invested at 8% rate  =  3x

Step 2 :

The interest formula is

I  =  PRT

Step 3 :

Interest earned at 6% rate is

I  =  (x)(0.06)(1)

I  =  0.06x

Step 4 :

Interest earned at 8% rate is

I  =  (3x)(0.08)(1)

I  =  0.24x

Step 5 :

Adding the results of step 4 and step 5, we get

0.06x + 0.24x  =  Total interest

0.3x  =  Total interest

In the question, the total interest is given \$660.

Then, we have

0.3x  =  \$660

Step 6 :

Solve for "x"

0.3x  =  \$660

Divide both sides by 0.3.

x  =  660 / 0.3

x  =  2200

Then,

3x  =  3(2200)

3x  =  6600

Hence, the money invested at 6% rate is \$2200 and 8% rate is \$6600.

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