Introduction
First, we know the basic things related to "simple interest" chapter.
Principal:-
It is the money that a person takes in the form of a loan from any person or bank or other financial institutions. This is denoted by P.
Rate:-
Each loan is given at a fixed rate, usually expressed in percentages. This is denoted by r.
Time:-
The money taken as a loan is payable after a certain period. This period is called time. This is denoted by t.
Interest:-
The borrower pays some amount in addition to the principal amount after the expiry of the period. This extra amount is called interest. This is denoted by I.
Amount:-
This is the entire amount that is to be returned by the borrower. Obviously, this would include both principal and interest. This is denoted by A.
Read Compound Interest on click here
Some Formula:-
1. \(A=P+I\) . 2. \(I=\frac{Prt}{100}\) .
3. \(P=\frac{100×I}{rt}\) . 4. \(r=\frac{100×I}{Pt}\) .
5. \(t=\frac{100×I}{Pr}\) . 6. \(P=\frac{100×A}{100+rt}\) .
Some Example
1. A sum of money becomes \(\frac{7}{6}\) of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is:
a). \(5\frac{5}{9}\%\)
b). \(18\%\)
c). \(6\frac{5}{9}\%\)
d). \(25\%\)
Ans: -
Here I=\((\frac{7}{6}-1)\) P.=\(\frac{1}{6}\)P.
So, r = \(\frac{100×I}{Pt}=\frac{100×\frac{P}{6}}{P×3}=\frac{100}{18}=\frac{50}{9}=5\frac{5}{9}\%\).
2. In how many years the simple interest of the principal amount 400 at \(5\%\) per annum will be Rs. 100?
Ans: - Time = \(\frac{100×100}{400×5}=5\) years.
Note:-
1. Simple interest has a proportional relationship with P, t, and r. When only P is changed by n or by \(x\%\) then S. I. also changes in the same ratio.
For example, if P is doubled, then S.I. will also be double. If it is increased by \(40\%\) then S.I. will also increase by \(40\%\) provided r and t are constant.
2. When P is increased by m times, r is increased by m times, and t is increased by q times then S.I. is increased by \( (m×n×q)\). Similar to the case of decrease.
Trick:- 1
A sum of money n multiplies in t years at the rate of simple interest.
So rate of interest = \(\frac{(n-1)×100}{t}\)
Example:-
1. A sum of money becomes nine times in 12 years. What is the simple interest rate per annum?
Ans:- rate of interest = \(\frac{(9-1)×100}{12}\)
= \(\frac{8×100}{12}\) = \(\frac{200}{3}\%\)
Trick:- 2
A sum becomes n times at the rate of \(r\%\) at simple interest then time
t = \(\frac{[(n-1)×100]}{r}\).
Example:-
1. A sum of money becomes 16 times at the rate of \(\frac{25}{3}\%\) per annum. For how long was that amount lent at simple interest?
Ans:- time = \(\frac{[(16-1)×100]}{\frac{25}{3}}\) = 180years.
Trick:- 3
# If the sum is n times in t years then it will be m times in \(\frac{(m-1)}{(n-1)}t\) years. As the rate of interest is constant.
# If the sum is n times in \(r\%\) rate of interest then it will be m times in \(\frac{(m-1)}{(n-1)}r \%\) rate of interest. As time is constant.
Example:-
1. A sum becomes 5 times in 3 years. In how many years it will be 10 times?
Ans:- time = \(\frac{(10-1)}{(5-1)}3\) = \(\frac{27}{4}\) years.
2. If a sum becomes 4 times with \(6.25\%\) per year simple interest, then find the rate of simple interest for which the sum will be 6 times at the same time.
and:- rate = \(\frac{ (6-1)}{(4-1)}× 6.25 \) = \(\frac{ 125}{12}\% \).
Trick:-4
The sum of money was given for a certain period of time t at the rate of \(r\%\) per annum. If it is given at the rate of \(v\%\) per annum then interest will be x rupees more or less,
Here the principle is = \(\frac{100x}{t(difference \ of \ r \ and \ v)}\).
Example:-
1. A sum of money lent at 4% per annum for 3 years. If that amount had been paid at the rate of 5.5% per annum, Rs. 1800 would have received more interest. What was that amount?
Ans:- The principle is
= \(\frac{(1800×100)}{[(5.5-4)×3]}\)= 40000.
Trick:-5
If the sum will be A in t years and it will be B in s years, then sum = A - \(\frac{ (B-A)}{ (s-t) }t\).
Example:-
If some money will be 500 in 3 years and 700 in 5 years. Find the money.
Ans:- principle = 500 - \(\frac{(700-500)}{(5-3)}3\)
= 500 - \((\frac{200}{2})×3\).
= 500 - 300 = 200.
Trick:-6
The simple interest on an amount is that \(\frac{m}{n}\) of the principal. If the rate of interest per annum and the time are the same. Then time = rate per annum =
\(\sqrt{\frac{100m}{n}}\) .
Example:-
If the simple interest on a sum is \(\frac{4}{9}\) of the principal and the rate of interest and the time are the same, then what is the annual rate of interest?
Ans:- the rate of interest= \(\sqrt{\frac{4×100}{9}}\) =\(\sqrt{\frac{400}{9}}\)
= \(\frac{20}{3 }\%\).
Trick:-7
The rate on a sum of money P is r% for the first t years, s% for the next f years, and c% for the next b years and the total interest is x then the principal = \(\frac{100x}{(rt+fs+cb)}\).
Example:-
The rate on a principal is 3% for the first 2 years, 5% for the next 3 years, and 6% for the next 2 years, and the total interest at the end of seven years is Rs.660. If so, what is that principle?
Ans:- principal = \(\frac{(660×100)}{[(3×2)+(5×3)+(6×2)] }\) = 2000.
Trick:-8
A sum P is divided into two parts in such a way that the interest on the first part at the rate of r% for t years is the same on the second part at the rate of s% for c years. In this case, we get,
\(\frac{(1st \ part)}{(second \ part)}=\frac{sc}{(rt)}\).
Example:-
Rs.3600 Rs. was divided into two parts in such a way that interest on the first part at 6% rate in 2 years, the second part at 8% rate in 3 years yields equal interest. What is the share of the principal given the rate of 8%?
Ans:- \(\frac{(1st \ part)}{(second \ part)} = \frac{(8×3)}{(6×2)}= \frac{2}{1}\).
So, 2nd part = \(\frac{1}{2+1}×3600\) = \(\frac{3600}{3}\) = 1200.
Trick:-9
A sum of money out to P is lent at the rate of r% per annum and the rest at the rate of s% per annum. If the total interest received after t year is m. then the loan amount, which was given at the rate of s%
= \([m - \frac{prt}{100}]×[\frac{100}{t×(difference \ between \ s \ and \ t)}]\).
And the loan amount, which was given at the rate of r%
= \([m - \frac{pst}{100}]×[\frac{100}{t×(difference \ between \ s \ and \ t)}]\).
Note:- negative sign neglect.
Example:-
If one part of 6000 rupees was lent at the rate of 4% per annum and the rest at the rate of 7% per annum. If after 3 years Rs. 900 interest comes, then how much Rs. lent at the rate of 4%.
Ans:- loan amount = \([900-\frac{(6000×7×3)}{100}]×[\frac{100}{(7-4)×3}]\)
= -4000
= 4000 (neglect negative sign).
Trick:- 10
If m amount given in t instalment with rate of interest \(r\%\) per annum, then the amount of each instalment is
= \(\frac{m×100}{(100×t)+[(t-1)+(t-2)+....+(t-t)]×r}\).
Example: -
The amount 3052 payable in 4 installments when the interest rate is 6% per annum, then find the instalment.
Ans: - The instalment
= \(\frac{3052×100}{(100×4)+[(4-1)+(4-2)+(4-3)+(4-4)]×6}\)
= \(\frac{3052×100}{400+[3+2+1]×6}\)
= \(\frac{305200}{436}\)
= 700.
Some Important Example: -
1. Rs. 500 was invested at 12% per annum simple interest and a certain sum of money invested at 10% per annum simple interest. If the total interest on both the sum after 4 years is Rs. 480, the latter sum of money is :
(a) Rs. 450 (b) Rs. 750 (c) Rs. 600 (d) Rs. 550
2. A money lender finds that due to fall in the annual rate of interest from 8% to \(7\frac{3}{4}\%\), his yearly income diminishes by Rs. 61.50. His capital is
(a) Rs. 22400. (b) Rs. 23800
(c) Rs. 24600. (d) Rs.26000
3. A lends Rs. 2500 to B and a certain sum to C at the same time at 7% annual simple interest. If after 4 years A received interest of Rs. 1120 from B and C, the sum lent to C is
(a) Rs. 700. (b) Rs. 6500
(c) Rs. 40000. (d) Rs. 1500
4. A certain sum of money amounts to Rs. 756 in 2 years and to Rs. 873 in \(3\frac{1}{2}\) years at a certain rate of simple interest. The rate of interest per annum is
(a) 10%. (b) 11%. (c) 12%. (d) 18%
5. What sum will amount to Rs. 7000 in 5 years at \(3 \frac{1}{3}\%\) simple interest?
(a) Rs. 6300 (b) Rs. 6500
(c) Rs. 6000. (d) Rs. 5000
6. A sum of money becomes \(\frac{41}{40}\) itself in \(\frac{1}{4}\) year at a certain rate of simple interest. The rate of interest per annum is
(a) 10%. (b) 1% (c) 2.5%. (d) 5%.
7. The simple interest on a certain sum for 8 months at 4% per annum is Rs. 129 less than the simple interest on the same sum for 15 months at 5% per annum. The sum is:
(a) Rs. 2,580. (b) Rs. 2400.
(c) Rs. 2529. (d) Rs. 3600
8. A man loses Rs. 55.50 yearly when the annual rate of interest falls from 11.5% to 10%. His capital (in rupees) is
(a) 3700. (b) 7400. (c) 8325. (d) 1110
9. A sum of Rs. 1000 is lent out partly at 6% and the remaining at 10% per annum. If the yearly income on the average is 9.2%. the both parts respectively are:
(a) Rs. 400, Rs. 600. (b) Rs. 450, Rs. 550. (c) Rs. 200, Rs. 800 (d) Rs. 550, Rs. 450
10. A man took a loan from a bank at the rate of 12% per annum at simple interest. After 3 year he had to pay Rs. 5400 as interest only for that period. The principal amount borrowed by him was:
(a) Rs. 2000 (b) Rs. 10,000
(c) Rs. 20,000. (d) Rs. 15,000
11. A sum of money at simple interest amounts to 1,012 in \(2\frac{1}{2}\) year and to Rs. 1067.20 in 4 years. The rate of interest per annum is :
(a) 2.5% (b) 3%. (c) 4% (d) 5%
12. A sum of money lent out at simple
interest amounts to Rs. 720 after 2 years and to Rs 1020 after 5 years. The sum is
(a) Rs. 520. (b) Rs. 600
(c) Rs 700. (d) Rs. 710
13. The sum of money, that will give Rs. 1 as interest per day at the rate of 5% per annum simple interest is:
(a) Rs. 3650. (b) Rs. 36500
(c) Rs. 730. (d) Rs. 7300
14. Mohan lent some amount of money at 9% simple interest and an equal amount of money at 10% simple interest each for two years. If his total interest was Rs. 760, what amount was lent in each case ?
(a) Rs. 1700. (b) Rs. 1800
(c) Rs. 1900. (d) Rs. 2000
15. Simple interest on a certain sum at a certain annual rate of interest is \(\frac{16}{25}\) of the sum. If the number representing rate percent and time in years be equal, then the rate of interest is :
(a) 8%. (b) \(11\frac{1}{2}\%\). (c) \(12\frac{1}{2} \%\) (d) \(12\frac{1}{4}\%\).
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