Time Value of Money Chapter 4 Exercise 4A CA foundation maths solutions
Exercise 4E chapter 4 Time Value of Money C A Foundation Maths solution for some problems are given.
First you should the textbook lesson Time value of Money very well.
You should observe and practice all example problems and solutions which are given in the textbook.
You must observe the given below solutions and will try them in your own method.
Solutions exercise 4A Chapter 4 CA foundation maths ( Mathematics of Finance)
CA Foundation maths solutions
Chapter 4
Time Value of Money
Mathematics of Finance
Exercise 4A
Problem
1. S.I. on Rs. 3,500 for 3 years at 12% per annum is Rs.
Solution.
P = Rs. 3,500, R = 12%, N = 3
S.I. = PNR/100
= 3,500 (12) (3)/100 = 1,260
Therefore, S.I. = Rs. !.260
Problem
2. P = Rs.5,000, R = 15%, T = 4 1/2 using I = PRT/100, I will be Rs.
Solution.
P = Rs.5,000, R = 15, T = 4 1/2
I = PRT/100
= 5000 (15) (4.5) /100 = 3375
Therefore, I = Rs. 3.375
Problem
3.If P = Rs. 5000, T = 1, I = Rs. 300, R will be
Solution.
P = Rs. 5000, T = 1, I = Rs. 300
I = PRT/ 100
300 = 5000 (R) (1) /100
R = 300/50 = 6
Therefore, R = 6%
Problem
4. If P = Rs. 4,500, A = Rs. 7,200 then Simple Interest. i.e. I will be Rs.
Solution:
P = Rs.4,500, A = Rs. 7,200
A = P + I
7200 = 4500 + I
I = = 7200 – 4500 = 2700
Therefore, I = Rs. 2,700
Problem
5. P = Rs. 12,000, A = 16,500, T = 2 1/2 years, Rate percent per annum Simple interest will be
Solution
P = Rs. 12,000, A = 16,500, T = 2 1/2 years
I = A – P
= 16,500 – 12,000 = 4,500
I = Rs.4,500
I = PRT/100
4,500 = 12000 (R) (5)/ (2)100
R = 4,500/60(5) = 15
Therefore, R = 15%
Problem
6. P = Rs. 10,000, I = Rs. 2,500, R = 12 1/2%, S.I. The number of years T will be
Solution:
P = Rs. 10,000, I = Rs. 2,500, R = 12 1/2%
I = PRT/100
2,500 = 10,000 (12.5) (T) /100
T = 25,000 (100)/ 125000 = 10/5 = 2
Therefore, T = 2 years
Problem
7. P = Rs. 8,500, A = 10,200, R = 12 1/2, S.I. T will be
Solution:
P = Rs. 8,500, A = 10,200, R = 12 1/2
I = A – P
= 10200 – 8500 = 1700
I = Rs. 1,700
I = PRT/100
1,700 = 8,500 (12.5) (T)/ 100
T = 1700 (100)/8500 (12.5) = 1.6
Therefore, T = 1.6 years or 1 year 7 months (option)
Problem
8. The sum required to earn a monthly interest of Rs. 1200 at 18% per annum S.I. is Rs.
Solution:
R = 18%, I = 1200 per month, T = 1/12
I = PRT/100
1200 = P (18) (1)/100 (12)
P = 1200 (1200)/18 = 80,000
Therefore, P = Rs. 80,000
Problem
9.The sum of money amount to Rs. 6,200 in 2 years and Rs. 7, 400 in 3 years. The principle and rate of interest are
Solution:
A = Rs. 6,200, when T = 2 years
A = Rs. 7,400, when T = 3 years
A = P (1 + RT/100)
A = P (1 + 2R/100)
6,200 = P [(100 + 2R)/100] …… (1)
7,400 = P [(100 + 3R)/100] …… (2)
Divide (2) by (1), we get
74/62 = (100 + 3R)/ (100 + 2R)
7400 + 148 R = 6200 + 186 R
186 R – 148 R = 7400 – 6200
38 R = 1200
R = 1200/ 38 = 31.57
Therefore, R = 31.57%
Substitute value of R in (1), we get
6,200 = P [(100 + 2 (31.57)/100)]
P = Rs. 3800
Therefore, P = Rs. 3,800, R = 31.57%
Problem
10. A sum of money doubles itself in 10 years. The number of years it would triple itself is
Solution:
A sum of money doubles itself in 10 years
Let p = 100, A = 200 in 10 years
I = 200 – 100 = 100
I = PRT/100
100 = 100 (R)(10)/100
R = 10
R = 10%
I = A – P = 300 – 100 = 200
I = PRT/100
200 = 100 (10) (T)/ 100
T = 20 years
Problem
11. If P = Rs.1000, R = 5% p.a. n = 4, what is amount and C.I. is
Solution:
P = Rs. 1000, R = 5% p.a., n = 4
A = P (1 + R/100) ^n
A = 1000 (1 + 5/100) ^4
A = 1000 (121550625)
A = 1215.50
I = A – P = 1215.50 – 1000 = 215.50
Therefore, A = Rs. 1,215.50 and I = Rs. 215.50
Problem
12. Rs. 100 will become after 20 years at 5% p.a. compound interest amount of
Solution:
P = 100, n = 20, R = 5%
A = P (1 + R/100) ^n
A = 100 (1 + 5/100) ^20
A = 265.33
Therefore A = 265.50
Problem
!3. The effective rate of interest corresponding to a normal rate 3% p.a. payable half yearly is
Solution:
Interest is payable half yearly, so for a year interest is 2 times
Therefore n = 2
Effective rate of interest E = [(1 + i/n) ^n – 1] (100)
E = [(1 + 0.03/2) ^2 – 1] (100)
E = 3.0225
Therefore, E = 3.0225% p.a.
Problem
CA foundation maths solutions
Note: You should observe the solution and will try them in your own way.