Problem 1 :
Alex invested $500 and received $650 after three years. What had been the interest rate ?
Problem 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 ?
Problem 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 ?
Problem 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 ?
Problem 1 :
Alex invested $500 and received $650 after three years. What had been the interest rate ?
Answer :
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
So, the interest rate is 10%.
Problem 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 ?
Answer :
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
So, the money invested at 6% rate is $14000 and 5% rate is $10000.
Problem 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 ?
Answer :
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
So, the money invested at 5% rate is $1500 and 6% rate is $3500.
Problem 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 ?
Answer :
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
So, the money invested at 6% rate is $2200 and 8% rate is $6600.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 14, 25 11:50 PM
May 11, 25 12:34 AM
May 10, 25 09:56 AM