Chapter 13: SIMPLE INTEREST


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Note: The day on which money is deposited isn't counted but date on which money is withdrawn is counted.

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

The 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))^2 - 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)^n = 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

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