**Principal: ** The money which we deposit in or the lower from the bank or the money learned called the principal.

**Rate of interest: ** The interest paid on $ 100 for one year is called the rate per cent per year or rate per cent per annum.

**Time: ** The period of time for which the money is lent or invested.

**Interest: ** Additional money paid by the borrowed to the lender for using the money is called interest.

**Simple Interest: **If the interest is calculated uniformly on the original principal throughout the lone period, it is called simple interest.

**Amount: ** The total money paid back to the lender is called the amount.

Calculate Simple Interest

If P denotes the principal ($), R denotes the rate (percentage p.a.) and T denotes time (years), then:-

**S.I = (P × R × T)/100**

**
R = (S.I × 100)/(P × T)
**

**P = (S.I × 100)/(R × T)
**

**T = (S.I × 100)/(P × R)**

If the denotes the amount, then **A = P + S.I**

**Note:**

● When we calculated the time period between two dates, we do not could the day on which money is deposited but we count the day on which money is retuned.

● Time is always taken according to the per cent rat.

● For converting time in days into years, divide th number of days by 365 (for ordering or lap year.)

● For converting time in month into years, divide th number of month by 12 (for ordering or lap year.)

Calculate Simple Interest

*Find the simple interest on:*

**(a)*** $ 900 for 3 years 4 months at 5% per annum. Find the amount also.*

**Solution:**

P = $ 900,

R = 5% p.a.

T = 3 years 4 months = 40/12 years = 10/3 years

Therefore, S.I = (P × R × T)/100 = (900 × 5 × 10)/(100 × 3) = $ 150

Amount = P + S.I = $ 900 + $ 150 = $ 1050

**(b)*** $ 1000 for 6 months at 4% per annum. Find the amount also.*

**Solution:**

P = $ 1000,

R = 4% p.a.

T = 6 months = 6/12 years

S.I = (P × R × T)/100 = (1000 × 4 × 1)/(100 × 2) = $ 20

Therefore, A = P + I = $( 1000 + 20) = $ 1020

**(c) ***$ 5000 for 146 days at 15¹/₂% per annum. *

**Solution:**

P = $ 5000,
R = 151/2% p.a.
T = 146 days

S.I = ( 5000 × 31 × 146)/(100 × 2 × 365)

= $ 10 × 31 = $ 310

**(d)*** $ 1200 from 9ᵗʰ April to 21ˢᵗ June at 10% per annum. *

**Solution:**

P = $ 1200,
R = 10% p.a.
T = 9th April to 21st June

= 73 days [April = 21, May = 31, Jun = 21, 73 days]

= 73/365 years

S.I = (1200 × 10 × 73)/(100 × 365) = $ 24

Calculate Simple Interest

**1.** *In how much time dose $ 500 invested at the rate of 8% p.a. simple interest amounts to $ 580.*

**Solution:**

Here P = $ 500, R = 8% p.a A = $ 580

Therefore S.I = A - P = $ (580 - 500) = $ 80

Therefore T = (100 × S.I)/(P × R) = (100 × 80)/(500 × 3) = 2 years

**2.** * In how many years will a sum of $ 400 yield an interest of $ 132 at 11% per annum?*

**Solution:**

P = $ 400,
R = 11%
S.I = $ 132

T = (100 × S.I)/(P × R) = (132 × 100)/(400 × 11) = 3 years

Calculate Simple Interest

**3.** * In how many years will a sum double itself at 8 % per annum? *

**Solution:**

Let Principal = P, then, Amount = 2P

So , S.I. = A - P = 2P – P = P

T = (100 × S.I)/(P × R) = ( 100 × P)/(P × 8) = 25/2 = 121/2 years

Calculate Simple Interest

**4.** * In how many years will simple interest on certain sum of money at 6 1/4% Per annum be 5/8 of itself?*

**Solution:**

Let P = $ x, then S.I = $ 5/8 x

Rate = 6 1/4% = 25/4 %

Therefore T = ( 100 × S.I)/(P × R) = ( 100 × 5/8)/(x × 25/4) x = ( 100 × 5 × x × 4)/(x × 8 × 25)

T = 10 years

**1.** * Find at what rate of interest per annum will $ 600 amount to $ 708 in 3 years.*

**Solution:**

P= $ 600 , A = $ 708 Time = 3 years

Therefore S.I. = $ 708 - $ 600 = $ Rs. 108

Now, R = ( 100 × S.I)/(P × R) = (100 × 108)/(600 × 3) = 6% p.a.

Calculate Simple Interest

**2.** * Simple interest on a certain sum is 36/25 of the sum. Find the Rate per cent and time if they are both numerically equal.*

**Solution:**

Let the Principal be $ X Then S.I. = 36/25 x

R = ? T = ?

Let Rate = R % per annum, then Time = R years.

So S.I. = (P × R × T)/100 → 36/25 x = (x × R × T)/100

--- ( 36 × 10 × x)/(25 × x) = R2

----- R2 = 36 × 4 ----- R = √(36 × 4) = 6 × 2

Therefore Rate = 12 % p.a. and T = 12 years

Calculate Simple Interest

**3.** * At what rate per cent per annum will $ 6000 produce $ 300 as S.I. in 1 years?*

**Solution:**

P= $ 600, T = 1 year S.I. = $ 300

Therefore R = ( S.I × 100)/(P × R) = ( 300 × 100)/(6000 × 1) = 5% p.a

**4.** * At what rate per cent per annum will a sum triple itself in 12 years ? *

**Solution:**

Let the sum be $ P, then Amount = $ 3P

S.I. = $ 3P – P = $ 2P, Time = 12 years

Now, R =( S.I × 100)/(P × R) = (100 × 2P)/(P × 12) = 50/3 = 16.6 %

Calculate Simple Interest

**1.** *What sum will yield $ 144 as S.I. in 21/2 years at 16% per annum? *

**Solution:**

Let P = $ x, S.I. = $ 144

Time = 21/2 years or 5/2 years, Rate = 16%

So, P = ( 100 × S.I)/(P × R) = ( 100 × 144)/(16 × 5/2) = ( 100 × 144 × 2)/(16 × 5) = $ 360

**2.** * A some amount to $ 2040 in 21/2 years at , P = ? *

**Solution:**

Let the principal = $ x

S.I. = $ (x × 11 × 5/2 × 1/100) = $ 11x/40

Amount = P + S.I. = x/1 + 11x/40 = (40x × 11x)/40 = 51x/40

But 51x/40 = 2040

51x = 2040 × 40 ---- x = (2040 × 40)/51 = $ 1600

Calculate Simple Interest

**3.** * A certain sum amounts to $ 6500in 2 years and to $ 8750 in 5 years at S.I. Find the sum and rate per cent per annum. *

**Solution:**

S.I. for 3 years = Amount after 5 years – Amount after 2 years

= $ 8750 – $ 6500 = 2250

S.I. for 1 years = Rs. 2250/3 = $ 750

Therefore S.I. for 2 years = $ 500× 2 = $ 1500

So, sum = Amount after 2 years – S.I.for 2 years

= $ 6500- 1500 = $ 5000

Now, P = Rs.5000, S.I. = $ 1500, Time = 3 years

R = ( 100 × S.I)/(P × T) = (100 × 1500)/(5000 × 2) = 15%

Therefore The sum is $ 5000 and the rate of interest is 15%

**4.** * Divide $ 6500 in to two parts , such that if one part is lent out at 9% per annum and other at 10% per annum, the total yearly income is $ 605*

**Solution:**

Let the first part be $ x. Second part = $ (6500 - x )

Now S.I. on $ X at 9% per annum for 1 year = $ (x × 9 × 1)/100 = 9x/100

S.I. on $ (6500 – x ) at 10% per annum

1 year = $ ((6500-x) × 10 × 1)/100 = $ ((6500 - x))/10

Total S.I = $ (9x/100+ (6500 - x)/10) = ((9x + 6500 - 10x)/100)

= $ ( 65000 - x)/100

But given that total S.I.= $ 605

So, (6500 - x)/100 =605 -----65000 - x = 60500

----- 65000 – 60500 = x ---- x = $ 4500

Now, second part = 6500 – x = 6500 – 4500 = $ 2000

Hence, first part = $ 2000 and second part = $ 4500

**5.** * When the rate of interest in a bank is increased from 9% to 10% per annum; A person deposits $ 500 more into his account. If the annual interest now Received by him is $ 150more then before, find his original deposit.*

**Solution:**

Let the original deposits be $ x

Then, S.I. on $ x for 1 year at (10 - 9 )% = 1% per annum + S.I. on $ 500

For I year at 10% per annum = $ 15

----- ( x × 1 × 1)/100 + ( 500 × 10 × 1)/100 = 150

----- x/(100 ) + 50 = 150 ---- x/(100 ) + 150 – 50 ----- x/(100 ) + 100

----- x = 100 × 100 = $
10,000

Therefore, the original deposit is $ 10,000.

Calculate Simple Interest

● **Simple Interest**

**Practice Test on Simple Interest**

● **Simple Interest - Worksheets**

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