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COMMERCIAL ARITHMETIC
2.1 SIMPLE INTEREST
hen an investor lends money to a borrower, the borrower must pay back the money
originally borrowed and also the fee charged for the use of the money. This fee charged
is called as interest and the capital originally lent by the investor is called the principal.
From the investor’s point of view, interest is the income from invested capital. The sum
of the principal and the interest due is called the amount or accumulated value. Any interest
transaction can be described by the rate of interest, which is the ratio of the interest earned in one
time unit to the principal.
We shall use the following notations:
N timeperiod
R rateof
A amount or accumulated
I simple
P principal
interest
valueofP
interest
The simple interest I on principal P for N years at annual rate R is given by
PNR
I
and the amount A is given by
PNR NR
A P I P P
The factor 1 100
^ NR
in Error! Reference source not found. is called an accumulation factor at
Unit
W
simple interest, and the process of calculating A from P using Error! Reference source not found.
is called accumulation at simple interest. From Error! Reference source not found. , we have
1
1 100 1 100
A NR
P A
NR
(^)
as the present or discounted value at rate R of A due in N years. The factor
1
1 100
NR
in
Error! Reference source not found. is called a discount factor at simple interest, and the process of
calculating P from A using Error! Reference source not found. is called discounting at simple
interest.
The time N must be in years. If the time is given in months, then
number of months N
If the time is given in days, we may calculate exact simple interest or simple interest, on the basis of
a 365-day year (leap year or not), that is,
number of days N
Solved Problems
- Find the simple interest and the maturity amount on Rs. 20,000 for 6 years at 10 % per annum.
Solution:
Given P Rs .20,000, N 6 years , R 10% p a..
Then the simple interest
PNR
I
= Rs. 12,000.
Hence the maturity amount A P I Rs. 20,000 Rs. 12,
Rs. 32,
- Find the simple interest and the maturity amount on Rs. 16,000 for 2 years at 17.5 % per annum.
Solution:
Given (^) P Rs .16,000, (^) N 2 years , R 17.5% p a..
years years.
Then the simple interest 100
PNR
I =
= Rs. 9,360.
Hence the maturity amount A P I Rs. 18,000 Rs. 9,
Rs. 27,.
- Find the simple interest and the maturity amount on Rs. 35,000 for 146 days at 13% per annum.
Solution:
Given P Rs .35,000, R 13% p a. .and N 146 days
years.
Then the simple interest 100
PNR
I =
= Rs. 1,820.
Hence the maturity amount A P I Rs. 35,000 Rs. 1,
Rs. 36,..
- Find the simple interest and the maturity amount on Rs. 72,000 for 2.5 years at 14.5% per annum.
Solution:
Given P Rs .72,000, N 2.5 years , R 14.5% p a..
Then the simple interest 100
PNR
I =
= Rs. 26,100.
Hence the maturity amount A P I Rs. 72,000 Rs. 26,
Rs. 98,.
- Find the simple interest and the maturity amount on Rs. 1,80,000 for 4 years and 6 months at
1 3
12 % per annum.
Solution:
Given P Rs .1,80,000,
1 3
R 12 % p a. .=
p a
And N 4 years and 6 months
years and years
years years
Then the simple interest 100
PNR
I =
= Rs. 99,900.
Hence the maturity amount A P I Rs. 1,80,000 Rs .99,
Rs. 2,79,^.
- Find the simple interest and the maturity amount on Rs. 60,000 for 216 days at 1 2
8 % per annum.
Solution:
Given P Rs .60,000, 1 2
R 8 % p a. .=
p a
and N 216 days
years.
Then the simple interest 100
PNR
I =
= Rs. 3,018.08.
Hence the maturity amount A P I Rs. 60,000 Rs .3,018.
Rs. 63,018..
- Find the simple interest and maturity amount on Rs. 1,00,000 for 292 days at 3% per month.
Solution:
Given P Rs .1,80,000, R 3% p m. =(3 12) % p a. .36 % p a..
and N 292 days
days
Then the simple interest 100
PNR
I =
= Rs. 28,800.
and N = (31-11) days in March + 30 days in April + 31 days in May
- 28 days in June. (excluding 29
th June )
= 109 days =
. (Since, 2004 is a leap year)
Then the simple interest 100
PNR
I =
= Rs. 558.40.
Hence the maturity amount A P I Rs. 50,000 Rs. 558.
Rs. 50,558.^.
He withdraws Rs. 50,558.40 from his account.
- An amount given on simple interest of 13% per year becomes Rs.22,240 after 3 years. What was
the principal amount?
Solution:
Given A Rs .22, 240, N 3 years , R 13% p a..
Since,
1 100
A
P
NR
= A
1
1 100
NR
We have, P = 22,
1 13 3 1 100
= Rs. 16,000.
- An amount given on simple interest of 1% per month becomes Rs. 4,00,000 after 5 years. What
was the principal amount?
Solution:
Given A Rs. 4,00,000, N 5 years and R 1% p m. . (1 12)% p a. .12% p a..
Since,
1 100
A
P
NR
= A
1
1 100
NR
We have, P = 4,00,
1 12 5 1 100
= Rs. 2,50,000.
- An amount given on simple interest of 2% per month becomes Rs. 52,400 after 73 days. What
was the principal amount?
Solution:
Given A Rs. 52 , 400 ; R 2 % p. m .( 2 12 )% p. a . 24 % p. a.
and N 73 days years 365
Since,
1 100
A
P
NR
= A
1
NR
We have, P = 52,
1
= Rs. 50,000.
- An amount given on simple interest of 16.5% per year becomes Rs.23,662.50 after 3 years and 6
months. What was the principal amount?
Solution:
Given A Rs. 23 , 662. 50 ; R 16. 5 % p. a.
and N 3 yearsand 6 months years and years 12
years 3. 5 years 2
Since,
1 100
A
P
NR
= A
1
NR
We have, P = (23,662.50)
1
= Rs. 15,000.
- In how many years do an amount becomes 4 times itself at 12% per annum simple interest?
Solution:
If P is the principal, then given that A = 4P ; R 12 % p. a.
- The interest on a certain sum of money at the end of 4
6 years amounts to 8
of the sum itself.
What rate percent was charged?
Solution:
Let P be the principal and R be the rate of interest per annum.
Given N = 4
6 years and I = P 8
Then the simple interest 100
PNR
I and hence PN
I
R
Hence,
P
R
P
= 6% p.a.
- If the rate of interest is reduced from % 2
4 to % 4
3 per year, find the decrease in a half yearly
interest on Rs. 1,600.
Solution:
Given P = Rs. 1,600 and N = 2
year.
When R = R 1 = % 2
p.a., we have the simple interest
1 1
PNR
I
= Rs. 36, and
When R = R 2 = % 4
p.a., we have the simple interest
2 2
PNR
I
= Rs. 30.
Hence the decrease in a half yearly interest = Rs. 36 – Rs. 30 = Rs. 6.
- A man borrows Rs. 50,000 on simple interest at the rate of 6% per annum. After 3 years he
pays back Rs. 17,700, how much he has to pay after 5 years to clear the loan?
Solution:
Given P Rs. 50,000; R = 6%.
Amount repaid after 3 years = Rs. 17,
Therefore, the corresponding principal =
= Rs. 15,000.
theremaining 2 years
Principalamountleftfor = Rs. 50,000 – Rs. 15,
= Rs. 35,
Then the simple interest 100
PNR
I
= 100
= Rs. 10,500.
after 5 years
Amount toberepaid
Rs. 35 , 000 Rs. 10 , 500
Rs. 45 , 500 .
- Divide Rs. 9000 into two parts such that simple interest on one at 5% for 4 years may be
double to that on other at 3% for 5 years.
Solution:
Let one part of Rs.9000 be x.
Then the remaining part is Rs. 9000 – x.
Let I 1 be the interest on Part I and I 2 be the interest on Part II
i.e., with the given data, we have
Part I Part II
P 1 =Rs. x ; N 1 = 4 years P 2 = Rs. (9000 – x ); N 2 = 5 years
R 1 =5% p.a. ; I 1 = 2 I 2 R 2 = 3% p.a.
1 1 1 1 2
PN R
I I
x I ------(1)
2 2 2 2
PN R
I
2
x I
Substituting (2) in (1) we get,
Solution:
Let P be the sum deposited in the bank at the beginning of each year.
Given P = Rs.7,500 and R = 5% p.a. for each year.
For the first year deposit, the duration is 4 years.
For the second year deposit, the duration is 3 years.
For the third year deposit, the duration is 2 years.
For the fourth year deposit, the duration is 1 year
Then,
1 2 3 N 4 R
P
NR
P
N R
P
NR
A P
A P 1
A 7500
A 7500 (4.5) Rs. 33,750.
- What will be the amount in 6 years at 7% simple interest per annum of a sum whose amount
in 2 years at the same 7% rate of interest is Rs. 59,287?
Solution:
Let P be the principal.
Case I:
Given N 1 = 6 years and R 1 = 7% p.a.
Let A 1 be the maturity amount, then
1 1 1
N R
A P.
Hence
A 1 P^1 P ^1.^42 .
Case II:
Given N 2 = 2 years, R 2 = 7% p.a. and A 2 = Rs. 59,287.
Hence 2 2 2 1 100
N R
A P
, implies 59,287 =
P^1 , which on simplifying gives
P ( 1. 14 )
. Hence
P Rs^.
Using P = Rs. 52,000 in (1) we get A 1 = P (1.42) = 52,000 (1.42)
= Rs .73,840.
- What sum of money lent at simple interest of 5.5% per annum for 4 years will produce same
amount as Rs. 5,500 lent at 4% per annum, for 3.5 years?
Let P be the principal.
Case I:
Given N 1 = 4 years and R 1 = 5.5% p.a.
Let A 1 be the maturity amount, then
1 1 1 1 100
N R
A P
Hence
A 1 P^1 P 1. 22 . -----------(1)
Case II:
Given N 2 = 3.5 years, R 2 = 4% p.a. and A 2 = Rs. 5,500.
Hence
2 2 2
N R
A P implies A 2 =
55001 = Rs. 6,270. ---(2)
Given that A 1 = A 2. Therefore, using (1) and (2), we have
A 1 = P (1.22)= 6270 = A 2
Simplifying, we get P =
- 22
= Rs. 5139.34.
Supplementary Problems
- Find the simple interest and maturity amount on
a) Rs. 5,000 for 4 yeas at 8% p.a.
b) Rs. 7,000 for 3 years at 6% p.a.
c) Rs. 4,500 for 6 years at 14.5% p.a.
A
Rs. 1400 in 5 years. Find the sum and rate of interest.
- What will be the amount in 6 years at 7% per annum simple interest of a sum whose amount in 2
years at the same 7% per annum rate of interest is Rs. 19,380.
- What sum of money lent at simple interest of 3.5 % per annum for 5 years will produce same
amount as Rs. 3,000 lent at 5% per annum, for 4.5 years
- A sum of Rs. 2,40,000 is lent in 2 parts, I part is lent at 6% for 4 years and II part is lent at 9%
per annum for 3 years. If the total income earned is Rs. 61,800 find the sums lent at each rate.
- In what time will the simple interest on any sum of money at 4
3 3 % per annum be 0.09375 of the
principal?
- A man lent equal sums of money at 2
1 5 % and 4% per annum respectively for a period of 3 years.
If he earned Rs. 72 more from the money lent out at 2
1 5 %. Find the sum of money lent at 4 %.
Answers: (1) (a) Rs.1,600; Rs.6,600 (1) (b) Rs.1,260; Rs.8,260 (1) (c) Rs.3,915; Rs.8, (1) (d) Rs1,773.33; Rs.5,273.33 (1) (e) Rs.26,730; Rs.48,730 (1) (f) Rs. 8,835; Rs.53,
(1) (g) Rs.70,125; Rs.2,05,125 (1) (h) Rs.1,73,250; Rs.2,28,250 (1) (i) Rs. 26,040; Rs.88,
(1) (j) Rs. 1008; Rs.19,008 (1) (k) Rs.10,304;Rs.1,02,304 (1) (l) Rs. 70 ; Rs.
(1) (m)Rs.120; Rs.12,120 (1) (n) Rs.2,842; Rs.44,842.
(2) (^) Rs. 6,000 (3) (^) Rs.54,000 (4) (^) Rs.20,
(5) (^) Rs.80,000 (6) (^) Rs.40,000 (7) (^) 40 years
(8) (^) 12.5 years (9) (^) 7% p.a. (10) (^) 5%
(11) (^) 25% (12) (^) Rs.100 per half year. (13) (^) Rs.175 per half year
(14) (^) Rs.1,00,000 (15) (^) Rs. 23,606.25 (16) (^) Rs. 32,142.86;
Rs. 17,857. (17) (^) Rs.67,741.93;
Rs. 32,258.
(18) (^) Rs.15,853.65;
Rs.4,146.
(19) (^) 8%; Rs. 1,
(20) (^) Rs. 24,140 (21) (^) Rs.3127.65 (22) (^) Rs. 1,00,000;
Rs. 1,40, (23) (^) 2 years and 6 months (24) (^) Rs. 1,
2.2 COMPOUND INTEREST
t the end of each interest period if the interest earned is added to the principal and thereafter
earns interest, then the interest is said to be compounded. The sum of the original principal
and total interest is called the amount. Hence the difference between the amount and
original principal is called the compound interest. For example, let a person takes a loan of Rs.
10000 from a moneylender for two years at 20% per annum interest. At the end of the first year the
principal along with the interest will amount to
1
PNR
A P
1
i e A
1
i e A
At the end of the second year, by using A 1 (^) 12000 as the principal, it amounts to
1 2 1
A NR
A A
2
i e A
Hence the compounded interest is 14400 10000 4400.
If P is the principal, R is the rate of interest, N is the time duration, the
the amount at the end of the year I :
R
P
PR
P
PNR
A P
the amount at the end of the year II :
1
1 1
1 2 1
R
A
AR
A
ANR
A A
2
R R R
A P P
Therefore the amount A at the end of (^) Nth year is given by:
N
N
R
A A P
The rate interest is usually stated as an annual interest rate, referred to as the nominal rate of interest.
Solved Problems
- Calculate the amount and compound interest on:
(i) Rs. 1,000 at 10 % p.a. for 5 years.
(ii) Rs. 12,000 at 5% p.a. for 7.5 years.
We have P = Rs. 15,000, N = 2 years.
During the first year R = R 1 = 5% and during the second year R = R 2 = 6%.
If A is the maturity amount, then
R 1 R 2
A P.
Hence, A =
150001 ^15000 (^1.^05 )(^1.^06 )
Rs. 16 , 695.
Also the compound interest C.I. = A – P
= Rs. 16,695 – Rs. 15,
= Rs. 1,695.
- Divide Rs. 39,030 between A and B so that when their shares are lent out, the amount that A
receives in 2 years is the same as what B receives in 4 years. The interest is compounded
annually at the rate of 4% per annum.
Solution:
Let the share of A be Rs. x , and hence the share of B is Rs. ( 39030 - x )
Then from the given data, we have R = R 1 = R 2 = 4% and
Part I Part II
P 1 =Rs. x ; N 1 = 2 years P 2 = Rs. (39030 - x ); N 2 = 4 years
A 1 =
1
1 1
N R P (^)
A 2 =
2
2 2
N R P (^)
2
(^) x =
4
x
Given that A 1 = A 2. Therefore,
2
(^) x =
4
x
Simplifying the above expression, we get
x =
- 0816
= Rs. 20,280.
- Rs. 16,820 is divided between two persons Mr. X and Mr. Y, 27 and 25 years old respectively;
and their money is invested at 5% per annum compound interest in such a way that both receive
equal money at the age of 40 years. Find the share of each out of Rs. 16,820.
Solution:
Let the share of Mr. X be Rs. x , and hence the share of Mr.Y is Rs. (16820 - x )
Then from the given data, we have R = 5% and
Mr. X Mr. Y
P 1 =Rs. x ;
N 1 = (40-27) years.
= 13 years.
P 2 = Rs. (16820 - x );
N 2 = (40-25) years
= 15 years
A 1 =
1
1 1
N R P (^)
A 2 =
2
2 2
N R P (^)
13
(^) x =
15
x
Given that A 1 = A 2. Therefore,
13
(^) x =
15
x
Simplifying the above expression, we get
x =
- 1025
= Rs. 8,820.
- What principal invested today will amount to Rs.1,630.80 in 4 years at 13% per annum
compound interest?
Solution:
Given A = Rs. 1,630.80; N = 4 years and R = 13% p.a.
Since,
N R A P
1 , we have
4
P
Hence P = 4 ( 1. 13 )
= Rs. 1,000.
- At what rate percent per annum will a sum of Rs. 50,000 become Rs. 80,000 if the loan given for
3 years attracts compound interest?
Solution:
