COMPOUND INTEREST PROBLEMS
Problem 1 :
$800 is invested in C.I where the rate of interest is 20% per year. If interest is compounded half yearly, what will be the accumulated value and C.I after 2 years ?
Solution :
The formula to find accumulated value in C.I is
A = P(1 + r/n)nt
A -----> Accumulated value (final value )
P -----> Principal (initial value of an investment)
r -----> Annual interest rate (in decimal)
n -----> Number of times interest compounded per year
t -----> Time (in years)
C.I -----> Amount of interest
Here,
P = 800
r = 20% = 0.2
n = 2
t = 2
Then, the accumulated value is
A = 800(1 + 0.1)⁴
A = 800(1.1)⁴
A = 800 ⋅ 1.4641
A = $1171.28
Compound interest is
C.I = A - P
C.I = 1171.28 - 800
C.I = $371.28
Problem 2 :
A sum of money placed at compound interest doubles itself in 3 years. In how many years will it amount to four times itself ?
Solution :
Let P be the amount invested initially.
From the given information,
P becomes 2P in 3 years
Because the investment is in C.I, the principal in the 4th year will be 2P.
And 2P becomes 4P (it doubles itself) in the next 3 years.
Therefore, at the end of 6 years accumulated value will be 4P.
So, the amount deposited will amount to 4 times itself in 6 years.
Problem 3 :
The compound interest and simple interest on a certain sum for 2 years is $ 1230 and $ 1200 respectively. The rate of interest is same for both compound interest and simple interest and it is compounded annually. What is the principle ?
Solution :
Simple interest for two years is 1200 and interest for one year is 600
So, C.I for 1st year is 600 and for 2nd year is 630.
(Since it is compounded annually, S.I and C.I for 1st year would be same)
When we compare the C.I for 1st year and 2nd year, it is clear that the interest earned in 2nd year is 30 more than the first year.
Because, interest 600 earned in 1st year earned this 30 in 2nd year.
It can be considered as simple interest for one year.
That is, principal = 600, interest = 30
I = Prt / 100
30 = (600 ⋅ r ⋅ 1) / 100
30 = 6r
5% = r
In the given problem, simple interest earned in two years is 1200.
I = Prt / 100
1200 = (P ⋅ 5 ⋅ 2) / 100
1200 = P / 10
Multiply each side by 10.
12000 = P
So, the principal is $ 12,000.
Problem 4 :
Mr. David borrowed $ 15,000 at 12% per year compounded annually. He repaid $ 7000 at the end of 1st year. What amount should he pay at the end of second year to completely discharge the load ?
Solution :
The formula to find accumulated value in C.I is
A = P(1 + r/n)nt
A -----> Accumulated value (final value )
P -----> Principal (initial value of an investment)
r -----> Annual interest rate (in decimal)
n -----> Number of times interest compounded per year
t -----> Time (in years)
C.I -----> Amount of interest
To find the accumulated value for the first year,
Substitute P = 15000, r = 0.12, n = 1 and t = 1 in the above formula.
A = 15000(1 + 0.12/1)1x1
A = 15000(1 + 0.12)1
A = 15000(1.12)
A = 16800
Amount paid at the end of 1st year is 7000.
Balance to be repaid :
= 16800 - 7000
= 9800
This 9800 is going to be the principal for the 2nd year.
Now we need the accumulated value for the principal 9800 in one year. (That is, at the end of 2nd year)
A = 9800(1 + 0.12/1)1x1
A = 9800(1 + 0.12)1
A = 9800 ⋅ 1.12
A = 10976
So, to completely discharge the loan, at the end of 2nd year, Mr. David has to pay $ 10,976.
Problem 5 :
There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate ?
Solution :
Let the principal in simple interest be $ 100.
Since there is 60% increase, simple interest = 60
We already know the formula for S.I.
That is,
I = Prt/100
Here, I = 60, P = 100, t = 6
60 = (100 ⋅ r ⋅ 6) / 100
60 = 6r
Divide each side by 6.
10% = r
Because the rate of interest is same for both S.I and C.I, we can use the rate of interest 10% in C.I.
To know compound interest for 3 years,
Substitute P = 12000, r = 0.1, n = 1 and t = 3 in the formula of C.I.
A = 12000(1 + 0.1)3
A = 12000(1.331)
A = 15972
C.I = A - P
C.I = 15972 - 12000
C.I = 3972
So, the compound interest after 3 years at the same rate of interest is $ 3972.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
If you have any feedback about our math content, please mail us :
v4formath@gmail.com
We always appreciate your feedback.
You can also visit the following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Trigonometry word problems
Markup and markdown word problems
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Word problems on sets and venn diagrams
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Recent Articles
-
Solving Problems with Patterns
Jul 25, 21 12:02 PM
Solving Problems with Patterns
-
Example Problems on Formula by Induction
Jul 25, 21 10:54 AM
Example Problems on Formula by Induction
-
Area and Volume of Similar Figures Example Problems
Jul 25, 21 02:02 AM
Area and Volume of Similar Figures Example Problems
