SIMPLE INTEREST - II

It is the sum which is paid by the borrower to the lender for using the money for a specific time period. The money borrowed is called the Principal. The rate at which the interest is calculated on the principal is called Rate of Interest. The time for which the money is borrowed is Time and the total sum of principal and interest is called the Amount.
Simple Interest

If P = Principal, R = Rate per cent per annum T = Number of years, SI = Simple Interest and

A = Amount. Then,

Also, Sum = (A1R2 - A2R1) / (R2-R1)
Sum = (A1T2 - A2T1) / (T2-T1)

NOTE :-The interest is calculated on the original principal ie, the principal to calculate the interest remains constant throughout the time period. The interest earned on the principal is not taken into account for the purpose of calculating interest for later years.

EXERCISE

1. A person borrows Rs. 5000 for 2 years at 4 % P.A simple interest. He immediately lends it to another person at 6.25 % P.A for 2 years. Find his gain in the transaction per year.

A) 112.50 B) 125 C) 150 D) 167.50

2. What is the present worth of Rs. 132 due in 2 years at 5 % simple interest per annum?

A) 112 B) 118.80 C) 120 D) 122

3. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest increased by 4%, what amount will Rs. 800 become in 3 years?

A) 1020.80 B)1025 C) 1052 D)data inadequate

4. Arun Jaitly borrowed Rs. 830 from Sushma Swaraj at 12% P.A S.I. for 3 years. He then added some more money to the borrowed sum and lent it to Baba Ramdev for the same period at 14% P.A rate of interest. If Arun Jaitly gains Rs. 93.90 in the whole transaction, how much money did he add from his side?

A)35 B)55 C)80 D)105

5. Salman Khan invested an amount of Rs. 12000 at the rate of 10% P.A Simple interest and another amount at the rate of 20% P.A simple interest. The total interest earned at the end of one year on the total amount invested become 14% P.A. Find the total amount invested.

A)20000 B)22000 C)24000 D)25000

6. Mohan deposits Rs. 1000 in a bank, which gives simple interest at 2% per annum. He withdrew Rs. 600 at end of the first year. How much money will he have in the bank after 3 years?

A) 432.80 B) 436.80 C) 480.20 D) 448.40

7. Mr. Thomas invested an amount of Rs. 13900 divided into two different schemes A and B at the simple interest rate of 14% p.a and 11% p.a respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, What was the amount invested in sheme B?

A)6400 B)6500 C)7200 D)7500

8. The effective annual rate of interest corresponding to a nominal rate of 8% per annum payable half yearly is ?

A)8% B) 8.01% C) 8.13% D) 8.16%

9. A certain amount is lent at X % P.A simple interest for two years. Instead, if the amount was lent at 2X% P.A simple interest for ‘Y’ more years, then the interest would have been five times the earlier interest, what is the value of Y?

A) 2 years B) 3 years C) 4 years D) 5 years

10. A money lender, Rahul Gandhi ,lends a part of money at 10% per annum and the rest at 15% per annum. His annual income is 1900. However , if he had interchanged the rate of interest on the two sums , he would had earned RS 200 more . The amount lent will fetch what 15 % ?

A)6000 B) 4000 C) 10000 D) 4400




1. A Gain in 2 years = [5000x(25/4)x2/ 100] – [5000x4x2/100]
Gain for 2 years = 625-400 = 225
Gain per year = 225/2 = 112.5

2. C
A=132, R=5%, T=2 years
A= P[1+RT/100]
132=P[1 + 10/100]
132= P[11/10]
Solving we get P=120

3.C
I=156
R= 100x156/800x3 = 6.5%
New R = 6.5+4 = 10.5%, P=800
A= P[1+RT/100]
A= 800[1+ 10.5x3/100]
A= 800[(1000+315)/1000]
A= 800x1315/1000 = 1052

4.D
Gain = [(830+X)x14x3/100] – [830x12x3/100]
93.90 = 42X/100 + 830x(42-36)/100
9390 = 42X + 4980
42X= 4410
X=105

5.A
I1 = 12000x10x1/100 = 1200
I2 = P2 x20x 1/100 = P2 /5
Total Interest = 1200 + P2 /5 = (6000+P2 )/5
ATQ,
(6000+P2 )/5 = (12000+P2)x14x1/100
Solving P2 = 8000
Sum = P1+P2 = 20000

6. B
Interest for 1 st year = 1000x2x1/100 =20.
Principle from 2nd year onwards as 400 debited 1020-600 = 420.
A= 420[1+2x2/100]= 420(26/25)= Rs. 436.80

7. D
A=X then, B= 13900-X
3508 = (X*14*2)/100 + (13900-X)*11*2/100
3508 = 28X/100 + 13900x22/100 – 22X/100
3508 = 6X/100 + 3058
Solving we get,
X= 7500.

8. D
100x(1/2)x8/100 = 4
104x(1/2)x8/100= 4.16
For 1 year 4+4.16 = 8.16

9.B
I1 = P*X*2/100 = PX/50
Acc to question
I2 = 5I1
I2 = P*2X*(2+Y)/100 = 5PX/50
(4+2Y)/100 = 5/50
4+2Y= 10
2Y= 6
Y=3years

10. A
Let amount lent @10% be X and @ 15% be Y
Now, acc to question:-
10X/100 + 15Y/100 = 1900
10Y/100 + 15X/100 = 2100
ð 10X + 15Y = 190000
ð 15Y + 10X = 210000
Solving we get Y= 6000
Hence, amount lent @15% is 6000.