class 10 compound interest

CI 2-12 4:53 PM

Tuesday, June 30, 2020

formula and definition from the class work
Interest is the
principle
Rate
units of time=

compound amount yearly (

CI=

AT (half yearly)=

if time is given in the years and month

if different rate

1a) find the compound interest without using formula
P=10000
Rate= 10 %
Time=2 years
for first years ,

Amount for first year
P+ SI=10000+1000=11000
Principal for the second year= Amount of first year=11000

=1000+1100=2100

1920

a) 2 ) Find the compound interest on Rs 3000 at 10 % for 2 years
Principal= 3000

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Principal= 3000
Rate=10%
Time=2 years
We have,

6300

2) B Find the compound interest on Rs 4000 at 10 % for 3 years

3) Samudra deposits Rs 850 at NBL at the half yearly rate of 8% for 1 year .how much interest
will he get ?
Solution
Principal (P)= 850
Time(T)= 1 years
Rate( R) =8%
CI half yearly=?

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4)Jharans got Rs 1881 interest of certain sum for 2 years
at 9% compounded yearly . Find the sum solution
CI=1881
T=2
R=9

5) In what time the compound amount of Rs 60000 at
10% rate is Rs 79860 ?
Solution
P= 60000
CA=798T0
R=10

T=3
6) Aman lent Rs 5000 in a bank at the rate of 12% p.a for

three years
I) HOW MUCH SIMPLE INTEREST WILL BE RECEIVE

AFTER 3 YEARS ?
II) ) HOW much CI will he receive after 3 years ?

Solution ,
p=5000
R=12%
T=3

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7)Hari borrowed Rs 130000 from krisha at the rate of 21% per
annum . At the end of 1 year and 6 month
i) how much simple interest will he has to pay ?
Ii) how much compound interest will he has to pay ?
Solution ,
P=130000
R=21%
T= 1 year 6 month =

8) Find the difference between the compound interest and simple
interest on Rs 7500 in 2 years 10% per annum
P=
R=
T=

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Now
9) The difference between the compound interest and the simple interest on a sum

of money for two years at interest rate 20% per annum is Rs 400 . Find sum
Solution,
R=20%
T=2
CI-SI=400

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10 mohan deposit Rs 5000 at 8% Pa compound interest in a bank . Find the difference
between amount compounded yearly and half yearly in two years.
Solution ,
p= 5000
R=8
T=2
We have,

=849.2928

1N1o)wtdhifefesruemnceo=f simple interest and compound interest after 2 year is Rs 202.5 and
the interest rate is 5% per annum. Find the principal
Solution ,
Time(T) =2 Years
Rate ( R) = 5
CI+SI= 202.50

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Now
CI+SI= 202.5
0.1025+0.1P=202.5
0.2025P=202.5
P=1000

12)the simple interest of 2 years is Rs 91 less than its compound interest . If the rate of
interest in a year is 15% . Find the sum
SOLUTION ,

Time(T) =2 Years
Rate ( R) = 15
CI+SI= 202.50

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Now
CI-SI= 91
0.3225P -0.3P=91
0.0225P=

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CI 13-18 4:59 PM

Tuesday, June 30, 2020

13) Ram borrowed a sum of money at the rate of 5% simple interest for 2 years and he
lent in compound interest at the same duration of time . In this transaction if he gained Rs
30 find the sum he borrowed .
Solution,

Time(T) =2 Years
Rate ( R) = 5
CI+SI=30

Now
CI-SI= 30
0.1025-0.1P=30
0.0025P=30
P=

14) Ram borrowed Rs 4800 from sita at the rate of 10% . At the end of one year
i) How much simple interest will ram have to pay ?
ii) How much compound interest semiannually will Ram have to pay ?
Solution
P=4800
R=10%
T=1year

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=5292
15) The compound interest on a certain sum for 2 year at 10% per annum is Rs 420 .what

would be the simple interest on the same st the same rate for the same time ?
Solution ,
T=2
R=10%
CI=

420
P=

16) According to the system of compound interest , a sum of money in 2 years
amounts to Rs 7260 and in 3 years it amounts to Rs 7986. find the rate of
interest and sum
Solution ,

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Solution ,
CA for 2 years=7260
CA for 3 years=7986
We have ,

R= 10

FROM EQUATION 1

16 b)
Solution ,
CA for 2 years=6050
CA for 3 years=6655

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CA for 3 years=6655
We have ,

R= 10

FROM EQUATION 1

17) In how many years will Rs 8000 mounts to Rs 13824 at 20% interest compounded annually?
Solution ,
P=
CA=
R=

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18) At what rate the compound amount on Rs 576 will be
Rs 625 in 2 years ?
P=
CA=
R=

R= 4.1

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CI18-24 5:01 PM

Tuesday, June 30, 2020

19) A man deposits a sum of moment at the annual rate of interest
10% and gets the compound interest of Rs 1986 after 3
years .What was the deposited principle ? What is the simple
interest of the sum for 3 years .

Solution, CI= 1986
Rate= 10%
Time=3 years
We have,

Again

SI=1800

20) Nikkita receive the interest t Rs 254.40 at the end of 2yers 12% annual
compound irate of interest . If the semiannual compound interest is
allowed ,how much more interest will she receive
Solution ,
CI yearly = 254.40
R=12
T=2
We have,

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254.40

More amount =
21) The simple interest on a certain sum of money for 2 years at 5% pa is Rs 320 . What

will be the compound interest on the same sum for the same time at the same rate,
the interest being calculated yearly
Solution
T= 2
R=5%
SI=320

22) Bimal deposits Rs 1500 sita finance company and Rs 15000 in Lumbini Finance
company at the rate of 12% per annum. Sita finance company pays interest

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company at the rate of 12% per annum. Sita finance company pays interest
compounded half yearly and Lumbini finance company pays interest compounded
yearly . Calculate the interest paid by both finance companies in 2 years ?
Solution,
P=1500
T=2
R=

22 b) Rate for B is
Rate of per month per rupee 1 paisa simple interest

23) Compound interest on a sum of money for 2 years compounded annually is Rs 8034. simple
interest on the same sum for the same period an at the same rate is Rs 7800 .find the sum
and rate of interest
SOLUTION ,
CI=8034
SI=7800

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again

24) The differences between the semi annual compound interested the annul compound
interest on a sum of money for 2 years at the rate of 20% per annum is Rs 482. find
Ftrhoems1um
Solution
R=20%
T=2
We have,

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Now difference = 0.4641P-0.44P
482 = 0.4641P-0.44P

set 3 Q
total time 5 years
Rate = 10 %
time for simple interest = 3 years
time for compound interest = 2 years
Final amount = 471900
last 2 years is calculated as compound interest way

471900=P
p= 39000
for simple interest
p=300000

population

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Pop3-9 4:53 PM

Tuesday, June 30, 2020

3)The present population of a town is 64000 and it increase at the
rate of 5% per annum What will be the population of the town
after two years?
SOLUTION ,
P=64000
T=2
R=5%

=

4) The population of a town in the year 2012 was 425000. find its population in the years 2014 if
the rate of annual increaser is 4% per year
Solution
P=42500( initial population i.,te 2012)
R= 4%
T= 2 ie (2014 -2012)

5) Two years ago the population of a village was 15000 and the rate of growth of
pollution is 4% . What is the present population of the village?
SOLUTION
P=1500
R=
T=

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6) The present population of nagarkot is 48400. the annual growth rate is
10% . What was the population of Nagarkot before 2 years?
SOLUTION,
P= ?
R=
T=

7) The population of a village was 7200. if 5% of the population was migrated in
and 2% died due to different causes within a year .what would be the
population of the village after a year?

P=7200
R= (5-2)%=3%
T= 1

8) The present population of Ghandruk village is 13310 and rate of growth is 10%
how many years ago the population was 10000?

P=10000

R= 10
T= ?

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1.331=

9) One years ago the population of a village was 10000.
the population at present is 10210? Find the
population growth rate ?
SOLUTION
P=
R=
T=

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Pop 9-16 5:04 PM

Tuesday, June 30, 2020

10) The population of Dhading Bansi increase 10% every year . If the present population is 60000.
Find its population after 2 years and before two years
Solution
For after 2 years

P=600

R= 10
T= 2

for before 2 years
P=?

T= 2 years
Rate= 10
we have ,

11) The population of village increase every year by 5%. At the end of two years ,the
total population of the village was 10000. if 1025 were migrated to other places
what was the population of the village in the beginning ?
Solution ,
P=

R= 5
T= 2

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12) The population of Kathmandu valley was 1000000 in the year 2057 and the
population growth rate was 4.5% . If 25000 men migrated here from other places in
the year 2058 BS what will be the total population in the year 2060 ?
Solution ,
Population in 2057=1000000
R=4.5%
Population in 2058(after 1year before migration) =

Population in 2058(after 1year before migration) =

Population in 2058(after migration) =1045000+25000
=1070000

Population in 2060 ( after 2 years) =

=

13) Two years ago , the population of a village was 5000 . During the first
year ,it increased by 7% but due to an epidemic , it decreased by 6% in the
following year . What is the population of the village now ?
Solution,
P=

R1=7 % , R2=-6
T=

14) the population of a village 2 years ago was 31250 . The population growth rate of the
town is 6% . One year ago 625 people migrated in the village . Find the present population
Solution
P=31250

R= 6
population before 1 year before migration

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population before 1 year before migration

population before 1 year after migration =33125+625
=

present population =

15) The growth of curd clotting bacteria is 5% per hour . If the number of bacteria in the 7 pm evening

is 1.012 , what wasthe number beforre 4 hours ? Find it

Solution

P=

1.012

R= 5
T= 4

160) T the present population of a city is 200000. if the population of the city
increased first year by 2% then 4% and 5% respectively in continued 3 years .
What will be the population city after 3 years ?
Solution ,
P=200000

T= 3 years

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