Simple interest is the money paid on an original amount that has been either borrowed or invested. Or interest is the money paid by borrower to lender for using his money for specified time.
TERMINOLOGIES:-
Interest = money paid by borrower to lender.
Principle = the original/actual sum borrowed.
Time = some specified time for which money is borrowed on interest
Rate of interest (r) = rate at which principal is given and it is calculated on principal
Amount (A) = it is the sum of principal and interest
Interest = (Principal × time ×rate of interest)/100
Interest = (p×n×r)/100
Let’s understand above formula with example
Q.1. Person A gave 50,000 Rs. To person B for one year at the rate of interest 10% per annum. What will be the amount received by person A at the end of the year?
Solution:
Here, P = 50,000
n = 1 year
r = 10%
Interest = (p×n×r)/100
= (50,000×1×10)/100
= 5000 Rs.
Amount received = principal + interest
= 50000 + 5000
= 55000 Rs.
SOLVED PROBLEMS ON SIMPLE INTEREST
Q.1. A certain sum of money amounts to 1560 Rs. In 3 years at 2% per annum find the amount of money borrowed?
Solution:
Here, P=?
A = 1560 rs
n = 2 years
r = 2%
Here A = P + (P×n×r)/100
1560 = P (1+(2×2)/100)
1560 = P ((100+4)/100)
(1560×100)/104 = P
P = 1500 Rs.
1500 Rs. Was borrowed.
Q.2. A certain sum of money amounts to 3924 Rs. In 3 years and 4140 in 5 years find the rate of interest?
Solution:
Here A_{1} = 3924 n_{1} = 3 r =? And P is same
A_{1} = 4140 n_{2} = 5
Method 1
A_{1} = P + (Pn_1 r)/100 A_{1 }= P + (Pn_2 r)/100
3924 = P + 3Pr/100 4140 = P + 5Pr/100
P = 3924 - 3Pr/100 …(1) P = 4140 - 5Pr/100 …(2)
From 1 and 2
3924 - 3Pr/100 = 4140 - 5Pr/100
2Pr/100 = 4140 - 3924
2Pr/100 = 216
Pr = 10800
P = 3924 - (10800×3)/100
P = 3924 – 324
P = 3600
P = 3924 - 3Pr/100
3600 = 3924 - (3×3600×r)/100
(3×3600×r)/100 =3924 – 3600
r = (324×100)/(3600×3)
= 3%
Above method is time taking the simpler method with less time taking is,
Method 2
A_{1} = P + (SI)_{1}
A_{2} = P + (SI)_{2}
A_{2} – A_{1} = (SI)_{2} - (SI)_{1}
4140 – 3924 = (SI)_{2} - (SI)_{1}
216 = (SI)_{2} - (SI)_{1}
Here, 216 -> SI for (n_{2} - n_{1}) ‘2’ years
By Unitary Method
2 → 216
3 → ?
? = interest for 3 years = (216×3)/2
= 324
P = A – simple interest
= 3924 – 324
= 3600
S.I. = Pnr/100
324 = (3600×3×r)/100
r = 3%
Q.3. A sum of Rs. 800 amounts to Rs. 920 in 3 years at simple interest if the interest rate is increased by 3% it would amount to how much?
Solution:
Here, P = 800
n = 3 years
A = 920
SI = 920 – 800
= 120
(SI)_{1} = Pnr/100
= (800×3×r)/100
= 120
r = 5%
Now, new rate = r + 3%
= 5% + 3%
= 8%
(SI)_{2} = (800×8×3)/100
= 192
New amount = P + (SI)_{2}
= 800 + 192
= 992 Rs.
Q.4. A sum of money becomes 3 times in 5 years at simple interest. In how many years, will the same sum becomes 6 times at the same rate of interest?
Solution:
Here first,
A = 3P
n = 5 years
r =?
A = P + Pnr/100
3P = P + (P×5×r)/100
3 – 1 = 5r/100
r = (100×2)/5
= 40%
Now, secondly A = 6P, r = 40% and n =?
6P = P + (P×40×n)/100
5 = (40×n)/100
n = (100×5)/40
n = 12.5 years
Q.5. When the article is sold at 20% discount, the selling price is 24 Rs. What will be selling price when the discount is 30%?
Solution:
As we know, Discount is always given on marked price,
S.P. = (100 – D1) % of M.P.
Here, D_{1} = 20%
And (SP)_{1} = 24 Rs.
24 = 20/100× M.P.
M.P. = 30 Rs.
(SP)_{1} =? And D_{2} = 30%
(SP)_{2} = (100 - D_{2})% M.P.
= 70/100 × 30
(SP)_{2 }= 21 Rs.
Q.6. A man invested 1000 Rs. On a simple interest at a certain rate and 1500 Rs. At 2% higher rate. The total rate of interest in three years is 390 Rs. What is the rate of interest for 1000 Rs.?
Solution:
Here, P = 1000 Rs.
R_{1} = r %
R2 = (r + 2)%
By the condition given
(Pnr1)/100 + (Pnr2)/100 = 390
(100×r×3)/100 + (1500(r+2)×3)/100 = 390
30r + 45r + 90 = 390
75 × r = 300
r = 4%
Therefore, rate of interest for 1000 Rs. = 4%
In simple interest (R = 10%) In compound interest (R = 10%)
Table – 1 Table – 2
Principal Interest Amount Principal Interest Amount
Year1 1000 100 1100 year1 1000 100 1100
Year2 1000 100 1200 year2 1000 100 1210
Year3 1000 100 1300 year3 1210 121 1330
Total interest = 300 Total interest = 331
So you can see in the above tables that interest earned by compound interest is greater than interest earned by simple interest and reason behind that is in case of simple interest the principle was same while in case of compound interest the principal amount was increasing as interest earned was being reinvested.
Terminologies used in the compound interest is same as that of simple interest only one term is added in compound interest which is,
Compounding frequency:
Compounding frequency is the number of times the interest earned is added to principal in a year.
Compounding frequency could be yearly, half yearly, quarterly etc.
Formula of the compound interest when compounded annually,
A = P ( 1+ r/100 )^{n}
Where P = Principal
R = rate of the interest
n = time in years
When compounded half yearly
A = P (1+ r/200 )^{2n}
When compounded quarterly,
A = P (1+ r/400 )^{4n}
In general,
A = P ( 1+ r/(k*100))^{kn}
Where, r = rate of interest
n = time
k = compounding frequency
P = principal
SOLVED PROBLEMS:
Q.1. Find the amount after 3 years if the sum is 2000 at rate of interest 5% annually when compounded yearly?
Solution:
Here, P = 2000 r = 5% n = 3
When compounded yearly
A = P (1+ r/100 )n
= 2000 (1+ 5/100)3
= 2000 (1+ 1/20)3
= 2000 (21/20)3
= 2000 (1.05)3
= 2000 × 1.157
= 2315.25 ≈ 2315
Q.2. What will be the amount after 3 years when 5000 Rs. Is given on compound interest of 10% per annum compounded annually?
also find,
1. Interest earned in 1st year
2. Interest earned in 2nd year
3. Interest earned in 3rd year
Solution:
Here, P = 5000
r = 10%
A =?
A = P (1+ r/100 )n
= 5000 (1+ 10/100)3
= 5000 (1.1)3
= 5000 (1.331)
A = 6655 Rs.
Therefore amount after 3 years will be 6655 Rs.
let I_{1} be Interest earned in 1st year
Let A_{1} = amount after one year
A_{1} = P (1+ r/100 )n
= 5000 (1+ 10/100 )1
= 5000(1.1)
= 5500
Interest = A_{1}– P
= 5500 – 5000
I_{1} = 500
let I_{2} be Interest earned in 2nd year
Let A_{2} = amount after 2nd year
A_{2} = P (1+ r/100 )n
A_{2} = 5000 (1+ 10/100 )2
= 5000 (1.1)2
= 6050
Interest earned in 2nd year = A_{2} – A_{1}
= 6050 – 5500
I_{2} = 550 Rs.
let I_{3} be Interest earned in 3rd year
Let A_{3} = amount after 3rd year
As we have already calculated
A_{3} = 6655 Rs.
Interest earned in 3rd year = A_{3} – A_{2}
= 6655 – 6050
= 605 Rs.
SHORTCUT METHOD (SLAB METHOD)
P 1st year 2nd year 3rd year
5000 5000 5000 5000
500 500 500
50 50
50
5
Amount after 3 years = 6655 Rs.
Interest earned in 1st year = 500 Rs.
Interest earned in 2nd year = 550 Rs.
Interest earned in 3rd year = 605 Rs.
Q.3. If P = 16000, r = 10%, n = 2 years compounded annually then find the amount and compound interest?
Solution:
A = P (1+r/100) n
= 16000 (1+10/100) 2
= 16000(1.1)2
A = 19360 Rs.
And C.I. = A – P
= 19360 – 16000
= 3360 Rs.
Shortcut method
r = 10% = 10/100 = 1/10
Now, 1/10
10 11
×10 ×11 (two times because n = 2)
100 (P) 121(A)
CI = A – P = 21
When P = 100 A = 121
P = 16000 A =?
? = 16000×121/100
= 19360 Rs.
Same way interest can be found.
Q.4. When R = 20%, time = 2 year, and difference between compound interest and simple interest is 30 Rs. Then find the principal, SI and CI.
Solution:
With normal method it will be lengthy as well as time taking. So by doing this sum by ratio method,
Shortcut method (Ratio method)
Here
r = 20% = 20/100 = 1/5
Here P = 5 and n = 2 years
For two years
P = 5×5 = 25
P For first Year Second Year
25 25 25
5 5
1
CI = 5+5+1 = 11 and SI = 5+5 = 10
CI – SI = 11 – 10
= 1
1 is equivalent to 30 Rs.
25 is equivalent to X Rs.
X = (25×30)/1
= 750 Rs.
Principal = 750 Rs.
By Unitary Method,
When CI is 11, P = 25
CI is X P = 750
X = (750×11)/25
CI = 330 Rs.
And SI = 330 – 30
= 300 Rs.
Q.6. The compound interest on a sum for 2 years is 832 Rs. And simple interest on the sum at the same rate for the same period is 800 Rs. What is the rate of interest?
Solution:
Given, CI = 832 Rs.
SI = 800 Rs. And n = 2 years
CI = P {(1+r/100) n – 1} SI = Prn/100
832= P {(1+r/100) 2 – 1} 800 = (P×r×2)/100
832 = P [ 1+ r^2/10000 + 2r/100 – 1] P = 40000/r ………(2)
832 = P [r^2/10000 + 2r/100 ] ……..(1)
From (1) and (2)
832 = 40000/r (r^2/10000 + 2r/100)
832 = (4r/1 + 800)
832 – 800 = 4r
4r = 32
r = 8%
Rate of interest = 8%
Q.7. How much more or less would Rs.20,000 fetch after two years, if it is put at 10% per annum compound interest payable half yearly than if it is put at 15% per annum simple interest payable yearly?
Solution:
Compound interest Simple interest
P = 20,000 P = 20,000
r = 10% r = 15%
n = 2 years, compounded half yearly n = 2 years
A = P (1 + r/(2×100)) 2n S.I = Prn/100
= 20000(1 + 10/200) 4 = (20000×2×15)/100
= 20000(1.05) 4 = 6000
= 20000 × 1.215 SI – CI = 6000 – 4310
= 24310 = 1690 Rs.
CI = A – P
= 24310 – 20000
= 4310
∴ Money put on Compound interest would fetch 1690 Rs. Less than Money put on SI.
Posted By:
Hardik Kaneriya
The great mentor Leaderboard
Time Left
14 Hours
31 Minutes
58 Seconds
Time Left
5 Days
14 Hours
31 Minutes
58 Seconds
Time Left
4 Days
14 Hours
31 Minutes
58 Seconds
Copyright @ 2018 | Star Universal Mentors LLP | All Rights Reserved.
COMMENTS