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
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