# 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.

CA foundation maths solutions

## Solutions exercise 4A Chapter 4 CA foundation maths

CA Foundation maths solutions

Chapter 4

Time Value of Money

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.