## Introduction

Simple interest is calculated as
P * R * T / 100
Where P = Principal or the amount borrowed
R = Rate of interest
T = years
Note: The day on which money is deposited isn’t counted but date on which money is withdrawn is counted.

• Relation Among Principal, Time, Rate Percent of Interest Per Annum and Total Interest :

• Amount = Principal + Total interest

• A = P + Ptr100

• Time = Total interestInterest on the principal in one year. Thus, if we have the total interest as  300 and the interest per year is  50, then we can say that the number of years is 300/50 = 6 years.

COMPOUND INTEREST

• Let principal = P, time = n years and rate = r% per annum and let A be the total amount at the end of n years, then A = P[1+r100]n

• When compound interest is reckoned half-yearly. If the annual rate is r% per annum and is to be calculated for n years. Then in this case, rate = (r/2)% half-yearly and time = (2n) half-years then A = P[1+r/2100]2n

• When compound interest is reckoned quarterly. In this case, rate = (r/4)% quarterly and time = (4n) quarter years. \ As before, then A = P[1+r/4100]4n

• The difference between the compound interest and the simple interest over two years is given by Pr21002

### Solved Question Papers

Q.
The SI on a sum of money is 25% of the principal, and the rate per annum is equal to the number of years. Find the rate percent.

1.4.5
2.6
3.5
4.8

Ans.3

Explanation :
Let principal = x, time = t years
Then interest = x/4, rate = t%
Now, using the SI formula, we get
Interest = (Principal × Rate × Time)/100
so x/4 = (x × t × t)/100
so t2 = 25
and we get t = 5%

Q.
The rate of interest for first 3 years is 6% per annum, for the next 4 years, 7 per cent per annum and for the period beyond 7 years, 7.5 percentages per annum. If a man lent out  1200 for 11 years, find the total interest earned by him?

1.1002
2.912
3.864
4.948

Ans.2

Explanation :
Whenever it is not mentioned whether we have to assume SI or CI we should assume SI.
For any amount, interest for the 1st three years @ 6% SI will be equal to 6 × 3 = 18%
Again, interest for next 4 years will be equal to 7 × 4 = 28%.
And interest for next 4 years (till 11 years) –7.5 × 4 = 30%
So, total interest = 18 + 28 + 30 = 76%
So, total interest earned by him = 76% of the amount = (76*1200 / 100) = 912

Q.
A sum of money doubles itself in 12 years. Find the rate percentage per annum

1.12.5%
2.8.33%
3.10%
4.7.51%

Ans.2

8.33%

Explanation :
Let principal = x, then interest = x, time = 12 years.
Using the formula, Rate = (Interest × 100)/Principal × Time
= (x × 100)/(x × 12) = 8.33%

Q.
A certain sum of money amounts to  704 in 2 years and  800 in 5 years. Find the principal.

1.580
2.600
3.660
4.640

Ans.4

Explanation :
Let the principal be  x and rate = r%.
Then, difference in between the interest of 5 years and of 2 years equals to
 800 –  704 =  96
So, interest for 3 years =  96
Hence, interest/year =  96/3 =  32
So, interest for 2 years Æ 2 ×  32 =  64
So, the principal =  704 –  64 =  640

Q.
A sum of money was invested at SI at a certain rate for 3 years. Had it been invested at a 4% higher rate, it would have fetched  480 more. Find the principal.

1.4000
2.4400
3.5000
4.3500

Ans.1

Explanation :
Let the rate be y% and principal be  x and the time be 3 years.
Then according to the question = (x(y + 4) × 3)/100 – (xy × 3)/100 = 480
so xy + 4x – xy = 160 × 100
so x = (160 × 100)/4 =  4000

Q.
A certain sum of money trebles itself in 8 years. In how many years it will be five times?

1.22
2.16
3.20
4.24

Ans.2

Explanation :
It trebles itself in 8 years, which makes interest equal to 200% of principal.
So, 200% is added in 8 years.
Hence, 400%, which makes the whole amount equal to five times of the principal, which will be added in
16 years.

Q.
If CI is charged on a certain sum for 2 years at 10% the amount becomes 605. Find the principal?

1.550
2450
3.480
4.500

Ans.4

Explanation :
Using the formula, amount = Principal (1 + rate/100)time
605 = p(1 + 10/100)2 = p(11/10)2
p = 605(100/121) =  500

Q.
If the difference between the CI and SI on a certain sum of money is  72 at 12 per cent per annum for 2 years, then find the amount.

1.6000
2.5000
3.5500
4.6500

Ans.2

Explanation :
Simple interest and compound interest for the first year on any amount is the same.
Difference in the second year’s interest is due to the fact that compound interest is calculated over the first
year’s interest also.
Hence, we can say that  72 = Interest on first year’s interest Æ 12% on first year’s interest =  72.
Hence, first year’s interest =  600 which should be 12% of the original capital. Hence, original capital =
5000 (this whole process can be done mentally).

Q.
The population of Jhumri Tilaiya increases by 10% in the first year, it increases by 20% in the second year and due to mass exodus, it decreases by 5% in the third year. What will be its population after 3 years, if today it is 10,000?

1.11,540
2.13,860
3.12,860
4.12,540

Ans.4

Explanation :
Population at the end of 1 year will be Æ 10,000 + 10% of 10,000 = 11,000
At the end of second year it will be 11,000 + 20% of 11,000 = 13,200
At the end of third year it will be 13,200-5% of 13,200 = 12,540.

### Problems

Q. Find the simple interest on Rs. 3000 at rate of interest 6 1/4% for 73 days

A. 73 days = 73/365 = 1/5 yrs.

SI = (3000 * 25/4 * 1/5) / 100

Q. An amount loaned at interest rate 13.5% per annum becomes Rs. 2502.50 after 4 years. Find the sum.

A. P + SI = P + P * 13.5 * 4 / 100 (Adding P on both sides)

2502.5 = P ( 1 + 54/100)

2502.5 = P (1.54)

P = 2502.5 / 1.54 = 1625

Q. A borrowed money at interest rate 6% for first two years, 9% for next three years and 14% for period beyond 5 years. If he pays a total interest of Rs. 11400 after 9 yrs how much did he borrow?

A. 11400 = (P *6*2/100) + (P*9*3/100) + (P*14*4/100)

11400 = (12+27+56)P / 100

11400 * 100 / 95 = P

P = 12000

Q. A certain sum of money amounts to Rs 1008 in 2 years and Rs. 1164 in 3.5 yrs.Find sum and RI

A. SI for 1.5 yrs = 1164 – 1008 = 156

SI for 1 yrs = 156 * 2 / 3 = 104 and two years = 208.

Principal = 1008 – 208 = 800

Use simple interest formula to get RI.

Q. A sum of 1500 is lent in 2 parts where one is at 8% and second is at 6%. If the total annual income is Rs. 106, find money lent at each rate.

A. (x*1*8/100) + ((1500-x) * 1 * 6 / 100 ) = 106

8x/100 + (9000 – 6x)/100 = 106

8x +9000 – 6x = 10600

2x = 1600

x=800

### Compound Interest

When P = principal, n = years and R = rate of interest compounded annually

Amount = P (1+R/100)^n

When P = principal, R = rate of interest compounded half yearly the
Amount = P ( 1 + (R/2) / 100)^2n

When P = principal, R = rate of interest compounded quarterly the
Amount = P ( 1 + (R/4) / 100)^4n

When P = principal, R = rate of interest compounded annually but the time is in fraction then like 3 2/5 yrs
Amount = P ( 1 + R / 100)^3 * (1 + (2R/5)/100)

When P = principal, R = rate of interest compounded annually but is different for first year R1, second year R2 and third year R3 then
Amount = P ( 1 + R1 / 100) * ( 1 + R2 / 100) * ( 1 + R3 / 100)

Present worth of Rs. x due n years hence is
Present worth = x / (1 + R/100)^n

### Solved Problems

Q. Find CI on Rs.7500, compounded annually at RI of 4% for 2 years.

A. Amount = 7500 * (1 + 4/100)^2

Then amount – 7500 gives CI.

Q. CI is compounded half yearly, principal = Rs. 10000 in rate 4% for 2 years.

A.  Amount = 10000( 1+ 2/100)^4

Amount = 10824.32

CI = Amount – principal = 10824.32 – 10000 = 824.32

Q. Difference between SI and CI accrued on an amount of Rs. 18000 in 2 years is Rs.405. What is the RI.

A. P( 1+ (R/100) )^n – P*R*T/100 = 405

{18000 ( 1 + (R/100))^– 18000}  – (18000 * R * 2 / 100) = 405

Solving this you can get R.

Q. Divide 1301 between A and B such that the amount of A after 7 years is equal to amount in B after 9 years. Interest is compounded at 4%.

A. Let the amount be ‘x’ and 1301 – x.

x(1+4/100)^7 = (1301-x)(1+4/100)^9

Solving this we can get ‘x’.

Q. A sum of money amounts to Rs.6690 after 3 years and Rs. 10035 after 6 years on CI. Find the sum.

A. P(1+R/100)^3 = 6690 ; P(1+R/100)^6 = 10035

Dividing first eqn by second eqn we get (1+R/100)^3 = 10035/6690 = 3/2

Substituting this value in first equation we get P = 6690 * 2 / 3 = 4460.

Q. A sum doubles itself in 9 years how many will it take to become 8 times.

A. P(1+R/100)^9 = 2P

(1+R/100)^9 = 2

Now finding P(1+R/100)^n = 8P

we need to get (1 + R/100)^= 8 but 8 = 2^3

(1+R/100)^n = (1 + R/100)^ 9 ^3

we know that A^b^c = A^b*c

so (1+R/100)^n = (1 + R/100)^ 9*3 = (1 + R/100)^27

n = 27

### CAT Problems

Q.A sum of money compounded annually becomes Rs.625 in two years and Rs.675 in three years. The rate of interest per annum is

1. 7
2. 8
3. 6
4. 5

Ans.b

### Quiz

Score more than 80% marks and move ahead else stay back and read again!

Q1: Find the simple interest on Rs. 300 at rate of interest 10% for 730 days
1.60
2.75
3.80
4.90

ANS.1

Q2: An amount loaned at interest rate 13.5% per annum becomes Rs. 2502.50 after 4 years. Find the sum.
1.1625
2.1650
3.1700
4.1800

ANS.1

Q3: A borrowed money at interest rate 6% for first two years, 9% for next three years and 14% for period beyond 5 years. If he pays a total interest of Rs. 11400 after 9 yrs how much did he borrow?
1.12000
2.15000
3.18000
4.21000

ANS.1

Q4: A certain sum of money amounts to Rs 1008 in 2 years and Rs. 1164 in 3.5 yrs.Find the sum of money.
1.800
2.900
3.1000
4.1100

ANS.1

Q5: When P = principal, R = rate of interest compounded annually but is different for first year R1, second year R2 and third year R3 then Amount =
1.P ( 1 + R1 / 100) * ( 1 + R2 / 100) * ( 1 + R3 / 100)
2.P ( 1 + R1 / 3*100)
3.P ( 1 + R / 100)^3
4.P ( 1 + 3*R / 100)

ANS.1