Math 10

PTR
Simple interest (I) = 100

Where, P = Principal

R = Rate of interest per annum

T = Time (in year)

If the principal does not remain the same for the entire period, due to the addition of
interest after a certain period of time, then the money is said to be lent at compound
interest. If the interest is compounded annually, the interest is added to the principal after
the interval of one year.

P = Principal

T = Time (in year)

R = Rate percent per annum

PTR
Interest in one year = 100

= P×1× R
100

PR
= 100

Amount in one year = P + I

= PR
= P + 100
∴ Principal in the second year =
P 1 + R 
100 

P 1 + R 
100 

P 1 + R  × T × R
100 
Interest in the second year =
= 100
Amount in the second year =
P 1 + R  R PR 1 + R 
100  = 100 100 

100

PRP 1 + R  P1 + R 
100 100  + 100 

= P 1 + R  ×P 1 + R 
100  100 

Oasis School Mathematics-10 45

= P 1 + R 2
100



Similarly, Amount in the third year = P  1 + R 3
 100 

 R  4
 100 
Amount in the fourth year = P 1 +

 R  T
 100 
∴ Amount in Tth year = P 1 +

 R  T
 100 
∴ Compound Amount (C.A.) = P 1 +
Compound Interest (C.I.) =
= CA – P

P  1 + R T –P
 100 

= P 1 + R T – 1
100 



∴ Compound Interest (C.I.) = P 1 + R T – 1
If the interest is compounded half yearly 100  

Rate = R% p.a.

= R % per half year
2

Time = T years

= 2T half years

In this case, Compound Amount (C.A.) = P 1 + R 2T
200



Compound Interest (C.I.) = P 1 + R 2T – 1
200  

Similarly, i f the interest is compounded 'n' times a year,

Then, Compound Amount (C.A.) = P  1 + R nT
 100n



Compound Interest (C.I.) = P 1 + R nT – 1
 100n 



46 Oasis School Mathematics-10

Note:

I f R1, R2 and R3 be the rate percent per annum for the first, second and third year
respectively, then compound amount for three years is given by

C.A. = P  1 + R1   1 + R2   1 + R3 
 100   100   100 

• Compound interest compounded yearly means compounded annually.

• Compound interest compounded half yearly means compounded semi-annually.

• In T years and M months, Compound Amount (C.A.) = P 1 + R 1 + MR
100 1200

Worked Out Examples


Example: 1

Without using the formula, find the amount and compound interest on Rs. 1,200 for 3
years at the rate of 8% p.a. when the interest is compounded annually.
Solution:

Here, Principal (P) = Rs. 1200

Rate (R) = 8% p.a.

Using, I PTR
= 100

Interest in the first year = Rs. 1200 ×1× 8
100

= Rs. 96

∴ Principal in the second year = Rs. 1,200 + Rs. 96 = Rs. 1,296

Interest in the second year = Rs. 1296 ×1× 8
100

= Rs. 103.68

Principal in the 3rd year = Rs. 1,296 + Rs. 103.68 = Rs. 1,399.68

Interest in the 3rd year = 1399.68 × 1 × 8100
100

= Rs. 111.97

Amount in the 3rd year = Rs. 1,399.68 + Rs. 111.97
= Rs. 1,511.65
∴ Compound interest = A – P
= Rs. 1,511.65 – Rs. 1,200

= Rs. 311.65

Oasis School Mathematics-10 47

Example : 2
Find the amount and compound interest of Rs. 3000 for 2 years at the rate of 8% p.a.
(i) if the interest is compounded annually.
(ii) if the interest is compounded semi-annually.

Solution:

(i) Here, Principal (P) = Rs. 3,000
Time (T) = 2 years

Rate (R) = 8% p.a.

We have, R T
100
Annual c ompound amount (C.A.) = P  1 + 
 

. = Rs. 3,000  1 + 8 2
 100 

= Rs. 3,000 1 + 2 2


25 

 27 2


= Rs. 3,000  25 

729
= Rs. 3,000 × 625

= Rs. 3499.20

Thus, Compound Amount (C.A.) = Rs. 3499.20

Also, Compound Interest = C.A. – P

= Rs. (3,499.20 – 3,000) = Rs. 499.20

(ii) Here, Principal (P) = Rs. 3,000

Time (T) = 2 years

Rate (R) = 8% p.a.

We have, amount (C.A.) = P1 + R 2T
Semi annual compound

200 

= Rs. 3,000 1 + 8 4
200 

= Rs. 3,000 1 + 1 4


25 

 26 4
= Rs. 3,000  25 

456976
= Rs. 3,000 × 390625

48 Oasis School Mathematics-10

= Rs. 3509.58

Thus, Compound Amount (C.A.) = Rs.3509.58

Also Compound Interest (C.I.) = C.A. – P

= Rs. (3509.58 – 3,000)

= Rs. 509.58
Example: 3

Calculate the compound interest of Rs. 700 lent for 3 years at 6% p.a. for the first

year, 6 1 % p.a. for the second year and 7% p.a. for the third year, interest being
2
compounded yearly.

Solution:

Here, Principal (P) = Rs. 700

Rate for the first year (R1) = 6% p.a.

Rate for the second year (R2) = 6 1 % p.a. = 13 % p.a.
2 2

Rate for the third year (R3) = 7% p.a.

We have,

Compound Amount (C.A.) = P 1 + R1  1 + R2  1 + R3 
100  200  100 

= Rs. 700 1 + 6  1 + 13  1 + 7 
100  200  100 

= Rs. 700  106  213   107 
 100   200   100 

= Rs. 855.55

∴ Compound Interest (C.I.) = C.A. – P

= Rs. 855.55 – Rs. 700

= Rs. 155.55

Example: 4

Find the difference between simple interest and the annual compound interest on
Rs. 12,000 for 2 years at the rate of 15% p.a.
Solution:

Here, Principal (P) = Rs. 12000
Time (T) = 2 years
Rate (R) = 15% p.a.

We have,

Oasis School Mathematics-10 49

Simple interest (S.I.) = PTR
100
=
12000 × 2 ×15
100

= Rs. 3600

Again,

Compound interest (C.I.) = P 1 + R T – 1
100 



= 12,000  1 + 15 2 – 1
 100  

= 12,000  115 2 – 1
 100  

= 12,000  529 – 1
 400

129
= 12,000 × 400

= 30 × 129
= Rs. 3,870

∴ C.I. – S.I. = Rs. (3,870 – 3,600)

= Rs. 270

Example: 5

A compound interest on a sum of money in 2 years at 8% per annum will be Rs. 320
more than the simple interest. Find the sum.

Solution:

Here, Principal (P) = ?

Time (T) = 2 years

Rate (R) = 8% p.a.

We have, PTR
Simple Interest (S.I.) =
100
= P×2×8

= 100
4P

25

50 Oasis School Mathematics-10

Compound Interest (C.I.) = P 1 + R T – 1
100 



= P  1 + 8 2 – 1
 100 



 108 2 – 1


 100 
= P

= P  729 – 1
 625

= 104 = 104P
P 625 625

From the given condition,

C.I. – S.I. = Rs. 320

or, 104P – 4P = 320
625 25

or, 104P – 100P = 320
or, 625 320
4P
625 =

or, 4P = Rs. 320 × 625

or, P = Rs. 320 × 625
or, 4

P = Rs. 50,000

∴ The sum is Rs. 50,000.

Example 6
In what time will Rs. 16,000 invested at 10% p.a. compounded semi-annually amounts
to Rs. 18,522?

Solution:

Here, Principal (P) = Rs. 16,000

Rate (R) = 10% p.a.

Compound amount (C.A.) = Rs. 18,522

We have, C.A. = P 1 + R 2T
200



Oasis School Mathematics-10 51

or, ( )= 10 2T
200
18522 16000 1+
or,
18522 ( )=210 2T
16000 200

or, 18522  21 2T
16000 =  20 

or, 9261  21 2T
8000 =

 20 

or,  21 3  21 2T
or,
 
 20  =  20 

2T = 3

or, T = 3 years
2

∴ Time period = 1 1 years.
Example: 7 2

The compound amount of a sum of money in 2 years and 3 years are Rs. 5,832 and
Rs. 6,298.56 respectively. Find the rate of interest compounded yearly and the sum.

Solution:

Here, Compound amount in 2 years = Rs. 5,832

Compound amount in 3 years = Rs. 6,298.56

We have, Compound amount = PP  1 + R T
 100 
When,
T = 2 yrs.

5,832 = P 1 + R 2 ................... (i)
100



When, T = 3yrs

6,298.56 = P  1 + R 3 ................... (ii)
 100 

Dividing equation (ii) by (i),

6298.56 P  1 + R 3
5832 =  100 

 R 2
P  1 + 100



or, 629856 = P 1 + R 
583200 100 

52 Oasis School Mathematics-10

1.08 = P 1 + R 
or, 100 
or,
From equation (i) P 1 + R  = 0.08
100 
or,
R = 8%
or, 5,832 =
or, P  1 + 8 2
or,  100 
∴
Hence, 5,832 = P  108 2
 100



5,832 = P  27 2
 25 

P = 5832×(25)2
(27)2

P = Rs. 5,000
P = Rs. 5,000
Rate (R) = 8 % p.a. and the sum is Rs. 5,000.

Example: 8

Compound interest on a certain sum of money for 2 years at 5% p.a. compounded
annually is Rs. 820. Find the simple interest of the same sum at the same rate and for
the same period of time.
Solution:

Here, Time (T) = 2 years

Rate (R) = 5% p.a.

Compound interest (C.I.) = Rs. 820

We have, C.I = P 1 + R T – 1
100  

820 = P 1 + 5 2 – 1
100  

or, 820 = P  105 2 – 1
 100 



or, 820 = P  21 2 – 1
20  

Oasis School Mathematics-10 53

or, 820 = P  441 – 1
 400

41
or, 820 = P × 400

Again, or, P = 820 × 400
41

or, P = Rs. 8000

Principal (P) = Rs. 8000

Time (T) = 2yrs.

Rate (R) = 5% P.a.

Simple interest (I) = ?
PTR
We have, I = 100

= 8000 × 2 × 5 = Rs. 800
Hence, simple interest is Rs. 800. 100

Example: 9

Rs. 16,000 invested at 10% p.a. compounded annually amounts to Rs. 19,360 in a
certain period of time. What would be the amount of the same sum at the same rate
and for the same period of time if the interest is being compounded semi-annually?

Solution:

Here, Principal (P) = Rs. 16,000

Rate (R) = 10% p.a.

Compound amount (C.A.) = Rs. 19360

Time (T) = ? R T
We have, amount (C.A.) = PP 1 100 
Compound +

or, 19360 = 16000  1 + 10 T
 100 

or,  11 T
19360 = 16000  10 

or, 19360  11 T


16000 =  10 

or, 121  11 T
100 =  10 

54 Oasis School Mathematics-10

or,  11 2  11 T

 
 10  =  10 

∴ T = 2 years.

Again, if the interest is being compounded semi-annually,

We have, P 1 + R 2T
Compound Amount (C.A.) = 200



= Rs. 16,000 1 + 10 2×2
200 

 21 4
= Rs. 16,000  20 

 21 4


= Rs. 16,000  20 

194481
= Rs. 16,000 × 160000

= Rs. 19,448.10
Example: 10

The compound interest on a certain sum of money for 2 years is Rs. 330 and the
simple interest on the same sum for 3 years at the same rate is Rs. 480. Find the rate
percent and the sum.

Solution:

Here,

S.I. = Rs. 480, T = 3 yrs.

We have, PTR
S.I. = 100

3PR
or, 480 = 100

PR
or, 160 = 100
or, PR = 16000 .................. (i)

Again, C.I. = Rs. 330, T = 2 yrs.

We have, C.I = P 1 + R T – 1
100  

or, 330 = P 1 + R 2 – 1
100 



Oasis School Mathematics-10 55

or, 330 = P 1 + 2R + R² – 1
100 10000

or, 330 = PR 2 + R 
100 10000 

or, 330 = 1600P0R 2 + R  [from (i)]
100 10000 

8R
or, 330 = 320 + 5

8R
or, 330 – 320 = 5

8R
or, 10 = 5

or, 50 = 8R

or, R = 50 % = 25 %= 6 1 %
8 4 4

From (i) P = 16000
R

= 16000 × 4 = Rs. 2,560
Example: 11 25

A man took a loan of Rs. 4,50,000 from the bank at the rate of 10% p.a. compounded
annually. After 1 year he paid Rs. 1,50,000 then bank reduced the interest rate to 8% p.a.
compounded semi-annually. At the end of second year he again paid an installment
of Rs. 1,50,000. Find how much he has to pay at the end of third year to clear the loan.

Solution:

For the first year,

Principal (P) = Rs. 4,50,000

Rate (R) = 10% p.a.

Time (T) = 1 year

We have, compound amount (C.A.) = P (1+ R )T
100

= 4,50,000 (1+ 10 )1
100

= 4,50,000 ( 110 )= 4,95,000.
100

Since, he paid an installment of Rs. 1,50,000.

For second year,

Principal (P) = Rs. 4,95,000 – Rs. 1,50,000

= Rs. 3,45,000,

56 Oasis School Mathematics-10

Rate (R) = 8% p.a.

Time (T) = 1 year

Since the interest is compound semi-annually,

We have, compound amount (C.A.) = P (1+ R )2T
200

= 3,45,000 (1+ )8 2×1

200

= 3,45,000 ( 208 )2
200

= 3,73,152

For third year,

Principal (P) = Rs. 3,73,152 – Rs. 1,50,000 = Rs. 2,23,152

Rate (R) = 8% p.a.

Time (T) = 1 year

We have, compound amount (C.A.) = P (1+ R )2T
= 200
2,23,152 (1+ 8 )2×1
200

= 2,23,152 ( 208 )2
200

= 2,41,361.20

Hence, he has to pay Rs. 241361.20 at the end of third year to clear the loan.

Example: 12

A man deposited Rs. 1,00,000 in a bank at the interest rate of 10% p.a. compounded
annually. After 2 years he withdrew Rs. 11,000 and after 3 years bank reduced the
interest rate to 6% p.a. compounded semi-annually. Find the total interest received by
him at the end of 4 years.
Solution:

For the two year,

Principal (P) = Rs. 1,00,000

Rate (R) = 10% p.a.

Time (T) = 2 year

We have, compound amount (C.A.) = P (1+ R )T
100

= 1,00,000 (1+ 10 )2
100

= 1,00,000 ( 110 )
100

= Rs. 1,21,000.

Hence, the interest in the first year = Rs. (1,21,000 – 1,00,000)

= Rs. 21,000

Since, he withdrew Rs. 11,000 from the bank.

Oasis School Mathematics-10 57

For third year,
Principal (P) = Rs. (1,21,000 – 11,000) = Rs. 1,10,000

Rate (R) = 10% p.a.

Time (T) = 1 year

We have, compound amount (C.A.) = P (1+ R )T
100

= 1,10,000 (1+ 10 )1
100

= 1,10,000 × 1.1. = Rs. 1,21,000.

Interest in third year = Rs. (1,21,000 – 1,10,000) = Rs. 11,000.

Again,

For fourth year

Principal (P) = Rs. 1,21,000

Rate (R) = 6% p.a.

Time (T) = 1 year

Since the interest is compounded half yearly.

We have, compound amount (C.A.) = P (1+ R )2T
200
)6 2×1
= 1,21,000 (1+
200

= 1,21,000 ( 206 )2 = Rs. 128368.90
200

Interest on 4th year = Rs. (128368.90 – 1,21,000) = Rs. 7368.90.

Hence, total interest received in 4 years = Rs. 21,000 + Rs. 11,000 + Rs. 7368.90 = Rs. 39368.90.


Exercise 3.1

1. (a) According to the annual compound interest, write the relation among compound
amount (C.A.), rate percent (R), time in year (T) and principal (P).

(b) According to the annual compound interest, write the relation among compound
interest (C.I.), rate percent (R), time in year (T) and principal (P).

(c) According to the semi annual compound interest, write the relation among
compound interest (C.I.), rate percent per annum (R), time in year (T) and
principal (P).

(d) According to the semi annual compound interest, write the relation among compound
amount (A), rate percent per annum (R), time in year (T) and principal (P).

(e) If the interest is compounded yearly, till what period of time are the simple interest
and compound interest equal?

(f) If the interest is compounded half yearly, till what time are the simple interest and
compound interest are equal?

(g) If the rate of interest for years consecutive 3 is R1, R2 and R3 respectively, write the
relation among C.A., T, P, R1, R2 and R3.

58 Oasis School Mathematics-10

(h) If the interest per rupee per year is 6 paisa, what is the rate of interest?
(i) If the interest per rupee per month is 1 paisa, find the rate of interest.

2. Without using the formula, find the compound amount and the compound interest
compounded yearly in the following.

(a) Rs. 2500 for 2 years at the rate of 10% p.a.

(b) Rs. 350 for 2 years at the rate of 12% p.a.

(c) Rs. 800 for 3 years at the rate of 5% p.a.

3. Using the formula, find the compound amount and the compound interest compounded

yearly of the following:

(a) Rs. 1,200 for 2 years at the rate of 8% p.a.

(b) Rs. 3,000 for 3 years at the rate of 10% p.a.

(c) Rs. 5,000 for 3 years at the rate of 15% p.a.

4. Find the compound amount and compound interest of the following if the interest is
compounded half-yearly.

(a) Rs. 1,000 for 1 year 6 months at the rate of 6% p.a.

(b) Rs. 2,250 for 2 years at the rate of 10% p.a.

(c) Rs. 5,000 for 18 months at the rate of 8% p.a.

5. (a) Ram borrowed Rs. 4,800 from Shyam at the rate of 10% per annum. At the
end of 2 years,

(i) how much simple interest will Ram have to pay?

(ii) how much compound interest will Ram have to pay?

(b) A borrowed Rs. 24,000 from B for 1 1 years at the rate of 10% p.a. Calculate,
2
(i) simple interest that A has to pay to B.

(ii) compound interest that A has to pay to B if the interest is compounded
half-yearly.

6. (a) Find the difference between the compound interest and simple interest of Rs. 2,500
for 3 years at the rate of 10% p.a.

(b) A borrowed Rs. 18,000 from bank at the rate of 15% p.a. simple interest. If he lent the
same amount of money to B at the same rate of compound interest compounded
annually, how much would he gain after 2 years?

7. (a) Find the difference between compound interest payable annually and
semi annually for an amount of Rs. 14,000 at the rate of 12% per annum for 2 years.

(b) Ajit borrowed Rs. 10,000 from a bank at the rate of 6% per annum compounded
annually. Immediately he lent it to Sujit at the same rate of compound interest
payable semi-annually. How much did Ajit gain in 2 years ?

8. Which scheme do you prefer out of the given two rates with the given principal and time?

(a) Rs. 12,000 for 2 years, C.I. at the rate of 10% p.a. or S.I. at the rate of 12% p.a.

(b) Rs. 5,000 for 2 years, C.I. compounded annually at the rate of 12% p.a. or C.I.
compounded semi-annually at the rate of 10% p.a.

Oasis School Mathematics-10 59

(c) A bank has fixed the rate of interest 10% p.a. semi-annually compound interest in
account M and 12% per annum annually compound interest in account N. If you are
going to deposit Rs. 40,000 for 2 years, in which account will you deposit and why?

9. (a) What sum invested for 3 years amounts to Rs. 9261 at 5% p.a. compound interest?

(b) What sum of money produces an interest of Rs. 524.95 in 2 years at the rate of 12%
per annum compounded semi-annually?

10. (a) Find the compound interest of Rs. 6,000 for 2 years if the rate of interest for the first
year is 5% p.a. and for the second year 10% p.a.

(b) A man borrowed Rs. 4,000 for 3 years at compound interest. How much should
he pay at the end of 3 years if the rate for the first 2 years is 8% p.a. and that of the
third year is 10% p.a.?

11. (a) In how many years will Rs. 7500 amount to Rs. 9408 at the rate of 12% p.a.
compounded annually?

(b) A sum of Rs. 25,000 invested at 8% p.a. compounded semi-annually amounts to
Rs. 28,121. 60. Calculate the time period.

(c) In how many years will the compound interest payable yearly on Rs. 10,000 at 10%
per year be Rs. 3,310?

(d) A man took a loan of Rs. 46,875. If the rate of compound interest is 4 paisa per
rupee per year, in how many years will the compound interest become Rs. 5,853 ?

12. (a) A man invested Rs. 3,000 in a finance company and received Rs. 3,993 after 3 years.
Find the rate of interest, if the interest is payable annually.

(b) At what rate of compound interest payable semi-annually on Rs. 20,000 is Rs.
2,050 in 1 year ?

(c) Compound interest on Rs. 10,000 for 2 years compounded annually is Rs. 1,025.
Find the interest per rupee per year.

13. (a) The difference between the compound interest and simple interest on a sum of
money for 2 years at the rate 10% p.a. is Rs. 240. Find the sum.

(b) The difference between the compound and simple interest on a certain sum at 15%
p.a. for 3 years is Rs. 567. Find the sum.

(c) A man borrowed a sum of money for 2 years at 5% p.a. simple interest, and he
immediately lent the sum at compound interest at the same rate for the same
period of time. In the transaction, if he gained Rs. 11, find the sum borrowed.

14. (a) The difference between the annual compound interest and the semi annually
compound interest on a sum of money for 2 years at the rate of 20% p. a. is Rs. 482.
Find the sum.

(b) If the compound interest on a sum of money compounded semi annually in one
year at 10% per annum is Rs. 40 more than the compound interest on the same
sum compounded annually in the same time and at the same rate, find the sum.

15. (a) According to the system of compound interest, a sum of money in 2 years amounts to
Rs. 7,260 and in 3 years it amounts to Rs. 7,986. Find the rate of interest and the sum.

60 Oasis School Mathematics-10

(b) A certain sum of money amounts to Rs. 3,380 in 2 years and Rs. 3,515.20 in 3 years
at compound interest. Find the rate of interest and the sum.

(c) The semi annual compound amount of a sum of money in 1 year is Rs. 400 and in
2 years is Rs. 441. Find the principal and the rate of interest.

16. (a) The compound interest of a certain sum for 2 years at 10% p.a is Rs. 420. What
would be the simple interest on the same sum at the same rate of interest for the
same period of time?

(b) A man borrowed a certain sum of money at the rate of 12% p.a. simple interest. If he
has paid Rs. 510 as S.I. at the end of a year, how much extra money should he pay if the
interest is compounded semi annually at the same rate and for the same period of time?

(c) A man borrowed a certain sum of money at the rate of 10% p.a. for 2 years and

paid Rs. 2100 as compound interest compounded yearly. How much extra money

should he pay if the interest is compounded half yearly?

17. (a) Simple interest of a certain sum is 1 of the principal at the rate of 5% for a certain
10

period of time. Obtain the ratio of C.I. and S.I of the same sum in the same time

and at the same rate of interest.

(b) Simple interest on a certain sum of money is 1 of the principal at the rate of 10%
5

p.a. for a certain period of time, find the time. What part of the sum is the compound

interest for the same period of time? Also obtain the ratio of S.I. and C.I. for the same
period.

18. (a) What will be the compound interest on Rs. 5,000 for 2 years at the rate of 20%
compounded per annum ? At what time, will the same interest be produced by the
same sum at the same rate of simple interest ?

(b) What will be the compound interest on Rs. 2,000 for 2 years at the rate of 10%
per annum? At what rate of simple interest will the same sum produce the same
interest in the same period of time?

19. (a) Mohan lends altogether Rs. 6,600 to Ram and Shyam for 2 years. Ram agrees to
pay a simple interest at 15% p.a. and Shyam agrees to pay a compound interest at
the same rate. If Ram paid Rs. 112.50 more than Shyam in interest, find how much
he lent to each ?

(b) Divide Rs. 4,500 into two parts such that the sum of the simple interest on the first
at the rate of 10% p. a. for 2 years and the compound interest on the second at the
same rate for the same period of time is Rs. 925.

20. (a) The compound interest on a certain sum of money for 2 years is Rs. 137.50 and the
simple interest on the same sum for 3 years at the same rate is Rs. 198. Find the rate
percent and the sum.

(b) Compound interest on a sum of money for 2 years compounded annually is Rs.
1680. Simple interest on the same sum for the same period and at the same rate is
Rs. 1600. Find the sum and the rate of interest.

21. (a) A man has invested a certain sum to get compound interest. Compound interests for
the first and second year are Rs. 200 and Rs. 210 respectively. Find the rate percent
and the sum.

Oasis School Mathematics-10 61

(b) The compound interest on a certain sum of money for the 1 year and 2 years are
Rs. 500 and Rs. 1,050 respectively. Find the sum and the rate of interest.

22. (a) A man took a loan of Rs. 2,50,000 at the rate of 8% p.a. compounded annually. At
the end of first year he paid first installment of Rs. 1,00,000. At the same time bank
reduced the interest rate to 6% p.a. compounded semi-annually. Again, at the end
of second year he again paid second installment of Rs. 1,00,000. Find what amount
should he pay at the end of third year to clear the loan.

(b) Saleem took a loan of Rs. 3,00,000 from the bank at the rate of 10% p.a. compounded
annuallly. At the end of first year he paid the first installment of Rs. 1,00,000.
Again, at the end of second year he paid the second installment of Rs. 1,00,000.
What should be paid in the end of third year to clear the loan if the bank changed
the interest rate to 8% p.a. compounded semi-annually

23. (a) Aadhya deposited Rs. 20,000 in a bank at the rate of 10% C.I. compounded
annually. At the end of second year she withdrew Rs. 5000 and at the end fourth
year she deposited Rs. 15000. Find the total interest received by him in 5 years if
the interest is compounded half yearly in the last year.

(b) Mingma deposited Rs. 50,000 in a bank at the rate of 6% p.a. C.I. compounded half
yearly. Again, in the beginning of the second year bank chaned the interest rate
to 10% p.a. compounded annually. If he deposited Rs. 25,000 in the beginning of
third year, find the total interest received by him in 3 years.

Answers 

1. Consult your teacher 2. (a) Rs. 3025, Rs. 525 (b) Rs. 439.04, Rs. 89.04

(c) Rs. 926.10,Rs. 126.10 3. (a) Rs. 1399.68, Rs. 199.68 (b) Rs. 3993,Rs. 993

(c) Rs. 7604.37, Rs. 2604.37 4. (a) Rs. 1092.73, Rs. 92.73 (b) Rs. 2734.89, Rs. 484.89

(c) Rs. 5624.32, Rs. 624.32
5. (a) (i) Rs. 960, (ii) Rs. 1008 (b) (i) Rs. 3600, (ii) Rs. 3783 6. (a) Rs. 77.5 (b) Rs. 405
7. (a) Rs. 113.08 (b) Rs. 19.09 8. (a) Second scheme (b) First scheme (c) Account N

9. (a) Rs. 8000 (b) Rs. 2000 10. (a) Rs. 930 (b) Rs. 5132.16

11. (a) 2 years (b) 1 1 years (c) 3 years (d) 3 years 12. (a) 10% (b) 10%
2

(c) 5 paisa 13. (a) Rs. 24000 (b) Rs. 8000 (c) Rs. 4400

14. (a) 20,000 (b) Rs. 16,000 15. (a) 10%, Rs. 6,000 (b) 4%, Rs. 3125 (c) 10%, Rs. 362.81
21
16. (a) Rs. 400 (b) Rs. 15.30 (c) Rs. 55.0625 17. (a) 4R1a:4m0:(Rb)s.23y6e0a0r,sS,h1y0a0 m, 2: 0R:s2.13,000
18. (a) Rs. 2200, 2.2 years (b) Rs. 420, 10.5% 19. (a) 1

(b) First: Rs. 2,000, Second Rs. 2,500 20. (a) 8 3 %, Rs. 792 (b) Rs. 8,000, 10%

21. (a) 5%, Rs. 4,000 (b) 10%, Rs. 5,000 22. (a) Rs. 85246.50 (b) Rs. 165484.80

23. (a) Rs. 12150.78 (b) Rs. 16684.45

Project Work

Get into a small group of 4/5 students. As a group, visit different financial institutions
(Banks, cooperatives, etc.) of your locality. Find out the financial institution with a
better scheme of taking a loan or depositing money.

62 Oasis School Mathematics-10

Unit Population Growth
and Depreciation
4

4.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.

Sunayana kept Rs. 1200 in a bank for 2 years at the rate of 4% p.a. How much will he
get after 2 years if the interest is compounded annually?

The present population of a village is 1200. If it is growing at the rate of 4% p.a., what
will be the population of the village after 2 years?

What are the similarities in these two questions? Can we calculate the population of a
village after 2 years by using the same formula of compound amount?

The first question is about the growth of money whereas the second one concerns
growth of population.

4.2 Population Growth

While calculating the compound amount and the compound interest, there is growth
of money at a certain rate after some time. There may be similar type of situation in the
growth in population also. Therefore, using the similar type of formulae, we can obtain
the population after a certain period of time.

If, P0 = Initial population
PT = Population after T years
R = Rate of population growth per annum



T = Time period

then, Population after T years R T
100
PT = P0 1 +  ....................... (i)


But the population of a country is also affected by the migration and the death of people.

∴ Actual population after T years

PT = P0 1 + R T – D + Min – Mout....................... (ii)
100



Where, D = Number of dead people

Min = Number of in-migrants

Mout = Number of out-migrants

If there is no migration and only deaths then,

PT = P0 1 + R T – D....................... (iii)
100 

Oasis School Mathematics-10 63

If there is no death, only migration then,

PT = P0  1 + R T + Min – Mout ....................... (iv)
 100



If there is no death and no out migrants, only in-migrants then,
T
PT = P0  1 + R  + Min ....................... (v)
 100

If the rate of population growth is different in different years, population after T years

PT = P0 1 + R1  1 + R2  1 + R3  ....................... (vi)
100  100  100 

If the population decreases with the rate R% p.a.,

then, population after T years T

PT = P0  1 − R ....................... (vii)
 100

Note: • If both the birth rate and death rate are given, then R = birth rate – death rate.
• If the rate of out-migrants and in-migrants are given, R = Rate of Min– Rate of Mout

Remember ! P0 PT

Population after T years
Present population
Present population

Population before T years



Worked Out Examples

Example: 1

The population of a village was 5000. Within a year, 2% people migrated in and 1% people
died due to different causes. What would be the population of the village after one year?

Solution: Alternative method

Here, P opulation of a village = 5000 Here,

Number of migrants entered = 2% of 5,000 Population of the village
2
(P0) = 5,000
= 100 × 5,000 = 100 Growth rate (R)

Number of dead people = 1% of 5,000 = (2-1)% = 1%

Time (T) = 1 year

1 We have, PT = P0 (1+1R00)T
= 100 × 5,000 = 50
= 5,000 (1+1010)1
∴ N umber of people after 1 year = (5,000 + 100 – 50)
= 5,000 × 1.01
= 5,050
= 5,050

64 Oasis School Mathematics-10

Example: 2

Two years ago, the population of a village was 12000. It is increasing at the rate of 5%
every year. If 150 people entered and 175 people left the village, calculate the present
population of the village.

Solution:

Here, Population before 2 years (P0) = 12,000

Growth rate (R) = 5%

Min = 150
Mout = 175

T = 2 years

We have,  R  T
 100 
PT = P0 1 + + Min – Mout

= 12000 1 + 5 2 + 150 – 175
100



= 12000 1 + 1 2 – 25


20 

= 12000  21 2 – 25
 20 

= 12000 × 441 – 25
400

= 30 × 441 – 25

= 13230 – 25 = 13205

∴ Present population = 13,205

Example: 3

The population of a town increases every year by 10%. At the end of 2 years, the total

population of the town was 30,000. If 5,800 people were added by migration, what

was the population of the town at the beginning? Alternative method
Solution: Here,
We have,

( )PT = P01 + R T + Min
100
Rate (R) = 10% per year.
R )T
Time (T) = 2 years or, 30,000 = P0(1 + 100 + 5800

Population after 2 yrs. = 30,000 or, 30,000–5800 = P0 (111000)2

or, 24,200 = P0 ( 11 )2
10
Migrated population = 5,800
or, 24,200 = P0 × 121
PT = 30,000 – 5,800 100
= 24,200.
or, P0 = 24,200×100
121

or, P0 = 20,000

Oasis School Mathematics-10 65

We have, PT = P 1 + R 2
24,200 = 100 
or, 24,200 =
P0  1 + 10 2
 100 

P0  11 2
 10



or, 121
24200 = P0 × 100
or,
or, P0 = 24,200 ×100

121

P0 = 200 × 100

or, P0 = 20,000

∴ Population at the beginning = 20,000

Example: 4

Two years ago the population of a town was 40,000. The population of the town at
present is 44,100. Find the population growth rate.
Solution:

Here, Time (T) = 2 years


Population before 2 years (P) = 40,000

Present population (PT) = 44,100

Growth rate (R) = ? T

We have, PT = P0 1 + R 
100 

44,100 = 40,000P 1 + R 2
100 

or, 44100 = P 1 + R 2
40000 100 

or, 441 = P 1 + R 2
or, 100 
400
2 2
 21 = P 1 + R
 100 
 20  

or, 21 = P 1 + R 2
20 100 

or, P 1 + R  = 21
100  20

66 Oasis School Mathematics-10

or, P 1 + R  = 21 – 1
or, P 1 + 100  20
R 
100  = 1
20

or, R = 5%

∴ Population growth rate = 5%

Example: 5
In how many years will the population of a town be 26,901 from 24,400 at a growth
rate of 5% p.a.?

Solution:

Here,

P0 = 24,400, PT = 26,901 R = 5% p.a. T=?
We have,

PT = P0  1 − R T
 100 

26,901 = 24,400  1 + 5 T
 100



26901  105 T


or, 24400 =  100 

441  21 T
or, 400 =  20 

 21 2  21 T
or,  20  =  20 

∴ Time (T) = 2 years

Exercise 4.2

1. (a) If the initial population (Po) is increasing at the rate of R% per annum for T years,
write the relation of population after T years (PT) with Po, R and T.

(b) Write the relation among the initial population (Po), population after T years (PT)
and time T years if the population is decreasing at the rate of R% per annum.

(c) If the population (Po) of a certain village is increasing at the rate of R% per annum
and D be the number of dead people in an interval of T years, write the relation
among PT, Po, R, T and D.

Oasis School Mathematics-10 67

(d) If Po be the original population, PT the population after T years R the population
growth rate, Min the number of people who are migrated in and Mout the number
of people who migrated out, write the relation among Po, PT, T, R, Min and Mout.

(e) If R1 be the birth rate and R2 the death rate of the population, then what is the
population growth rate.

2. Find the present population in the following cases.

(a) Population before 2 years = 3,600, growth rate = 5% p.a.

(b) Population before 3 years = 2,50,000, growth rate = 2% p.a.

3. Find the population after the following period of time.

(a) Present population = 5,000, growth rate = 10% p.a., time = 2 years.

(b) Present population = 1,20,000, growth rate = 5% p.a., time = 3 years.

4. Find the population before the following period of time.

(a) Present population = 5,408, growth rate = 4% p.a., time = 2 years.

(b) Present population = 33,075, growth rate = 5% p.a., time = 2 years.

(c) Present population = 10,648, growth rate = 10% p.a., time = 3 years.

5. (a) The population of a village is 20,400. If the yearly growth rate of the population
is 2%, what will be the population after one year?

(b) The present population of a village is 30,000. If it is increased at the rate of 10%
p.a., what will be the population after 2 years?

(c) 2 years ago, the population of a town was 1,20,000. If the rate of growth of
population is 3% p. a., what is the present population of the town?

(d) One year ago, the population of Kusunde village was 10,000. If the population
growth rate is 2%, what will be the population of the village after 1 year?

6. (a) The present population of a town is 2,74,000. If the annual birth and death rate
are 8% and 3% respectively, what will be the population of the town after 2 years?

(b) The present population of Birtamod Bazaar is 45,000. If the annual birth rate is 1%,
and 1% people migrated from different places, find the population of Birtamod after
2 years.

7. (a) The present population of a country is 2,58,519. What was its population one
year ago if it increased by 2% annually?

(b) The population of a village is 17,640. If the population growth rate is 5% p.a., find
its population 2 years ago.

(c) The present population of a town is 33,800. If the birth rate is 6% p.a. and death
rate is 2% p.a., what was the population of the town 2 years ago?

8. (a) The population of a village 2 years ago was 4,000. If the annual growth rate of
population in the last 2 years was 5% and 10% respectively, find the present
population.

68 Oasis School Mathematics-10

(b) The population of a town 3 years ago was 45,000. In the last 3 years the annual growth
rate of the population was 4%, 5% and 10% respectively, find the present population of
the town.

9. (a) The present population of a village is 2,205. 2 years ago, the population was 2000.
Find the annual growth rate.

(b) One year ago the population of a town was 10,000. After one year the population
will be 12,100. Find the growth rate of the population.

(c) The population of a village was 10,000 one year ago. The population at present is
9,780. Find the decreasing rate of the population.

10. (a) In how many years does the population of a village increases from 2500 to 2525
at the rate of 1% per annum?

(b) The present population of a town is 50,000. If the annual growth rate is 4%, after
how many years will the population be 54,080?

(c) The present population of a town is 66,550. If the annual growth rate is 10%, how
many years ago was the population 50,000?

11. (a) The number of curd clotting bacteria increases at the rate of 5% per hour. At 5
P.M., if the number of bacteria is 2 × 108, find its number at 7 P.M. of the day.

(b) The number of bacteria in an infected person increases every day at the rate of
10%, if the number of bacteria in a day is 3.2 × 107, find its number after 2 days.

12. (a) The population of a town increases every year by 10%. At the end of two years
the total population of the town was 30,000. If 5,800 people were added by
migration, what was the population of the town at the beginning?

(b) Population of Chhaimale village increases every year by 5%. At the end of 2 years,
the total population of the village is 3,835. If 134 people migrated to other places
during this interval, what was the population of the village at the beginning?

13. (a) The population of a village at a certain instant of time is 12,000. Population
growth rate is 5%. If 1,400 people migrated in from different villages at the end
of 1 year, find the population of the village after 3 years.

(b) In the beginning of 2070 B.S., the population of Butwal was around 1,00,000.
The growth rate of population is 2%. At the beginning of 2071 B.S., 8,000 people
migrated here from different places. Find the population of the town at the
beginning of 2073 B.S.

(c) Three years ago, the population of Barpak village was 5,000. The population
growth rate was 1%.

(i) What was the population of the village 2 years ago if 950 people migrated
here from different places?

(ii) What is the present population?

Oasis School Mathematics-10 69

Answers
1. Consult your teacher 2. (a) 3,969 (b) 26,5,302 3. (a) 6,050 (b) 1,32,300
4. (a) 5,000 (b) 30,000 (c) 8,000 5. (a) 20,808 (b) 36,300 (c) 1,27,308 (d) 10,404
6. (a) 302085 (b) 46,818 7. (a) 25,3,450 (b) 16,000 (c) 31,250
8. (a) 4,620 (b) 54,054 9. (a) R = 5% (b) 10% (c) 2.2%
10. (a) 1 year (b) 2 years (c) 3 years 11. (a) 2.205 × 108 (b) 3.872 × 107 12. (a) 20,000 (b) 3600
13. (a) 15,435 (b) 1,14,444 (c) (i) 6,000 (ii) 6,121 (around)

Project Work

Visit the office of your ward or village. Take information of the population of your
village according to the 2068 B.S. Census. Assuming different growth rates like
1%, 1.5%, 2%, etc., estimate the present population.

4.3 Depreciation

Let's discuss these questions in your class.

A mobile set is bought for Rs. 15,000. What will be the price of the mobile after 2 years?

• More than Rs. 15,000
• Rs. 15,000

• Less than Rs. 15,000
A man bought a motorcycle for Rs. 2,00,000.
What will be its value after 2 years?

• More than Rs. 2,00,000 • Rs. 2,00,000 • Less than Rs. 2,00,000

The value of fixed assets like machines, equipment decreases gradually after being used
for a certain period of time. The decline in the original value of fixed assets is called
depreciation. There are many causes of depreciation like wear and tear, efflux of time,
obsolescence, fall in market price, etc. The depreciation of fixed assets in unit time is the
rate of depreciation. The residual value of the fixed assets to which it is reduced with time
is known as scrap value.

Depreciation may be simple or compound. In simple depreciation, the reduction in
the value of the fixed asset is the same in each year. But in compound depreciation,
reduction in the value of the assets is not the same. It is calculated from the depreciated
value of the asset at the end of the previous year.

If, V0 = Original value of the asset
R = Yearly rate of depreciation

T = Time period

VT = Value after T years

70 Oasis School Mathematics-10

then VT = V0 1 − R T
100 

Amount of depreciation = V0 – VT

If the rate of depreciation is different in different years, then,

VT = V0 (1– R1 ) (1– R2 ) (1– R3 )...
100 100 100

Where, R1 = rate of depreciation in first year
R2 = rate of depreciation in second year
R3 = rate of depreciation in third year, and so on

Worked Out Examples

Example: 1

The original value of a machine is Rs. 60,000. Find its value after 2 years if the annual
rate of depreciation is 4%.

Solution: Here,

Original value (V0) = Rs. 60,000

Rate of depreciation (R) = 4% p.a.

Time period (T) = Rs. 2 years

Value after T years (VT) = ?

We have, VT = V0  1 − R T
 100



= 60, 000  1 – 4 2
=  100 
=
60, 000  96 2
 100



60, 000  24 2
 25



= Rs. 55,296

∴ Value of the machine after 2 years = Rs. 55,296

Example 2

The present value of a machine is Rs. 17496. If the annual rate of depreciation is 10%,
what was the value of the machine 3 years before?
Solution: Here,

Oasis School Mathematics-10 71

Original value (Vo) = ?
Rate of depreciation (R) = 10%

Time period (T) = 3 years

We have, Value after T years (VT) = Rs. 17496

VT = V0  1 − R T
 100 

17,496 = V0  1 – 10 3
 100



or, 17,496 = V0  90 3
 100



or, 17,496 = V0  9 3
 10 

or, 17,496 = 729Vo
1000

or, V0 = Rs. 24,000

∴ Original value of the machine = Rs. 24,000

Example: 3

The original value of a television set was Rs. 16,000. Find the annual rate of
depreciation if its value after 3 years is Rs. 13,718.

Solution: Here,

Original value (V0) = Rs. 16,000
Time (T) = 3 years

Annual rate (R) = ?

Value after T years (VT) = Rs. 13,718

We have, VT = V0 1 − R T
100 

or, 13718 = 16000  1 – R 3
 100 
3
or, 11630701081 60=0 0  1 – R
 100 


or, 860805091 60=0 0  1 – R 3
 100 

72 Oasis School Mathematics-10

or,  19 3  1 – R 3
 20  100
1 60=0 0 


or, 121096 0=0 0 11–– R 3
100 

o1r6,0 00  1 – R 3 1 – 19
 100  = 20

o1r6,0 00  1 – R 3 1
 20

100  =

∴ R = 5%

Hence, the rate of depreciation = 5%

Example: 4
A motorcycle is bought for Rs. 1,80,000. If the rate of compound depreciation is 8%
p.a., when will its value be Rs. 1,52,352 ?

Solution: Here,

Original value (Vo) = Rs. 1,80,000
8% p.a.
Rate (R) = Rs. 1,52,352
?
Value after T years (VT) =

Time (T) =

We have, VT = V0  1 − R T
 100 

or, 1,52,352 = 1, 80, 000  1 − 8 T
or,  100
or, 
or, 
or,
∴ 1, 52, 352  92 T
1,80,000 =

 100 

4761  23 T
5625 =

 25 

529  23 T
625 =

 25 

 23 2  23 T

 
 25  =  25 

T = 2 years

Example: 5

A man bought a taxi for Rs. 4,50,000. Using the taxi, he earned Rs. 50,000 in 2 years. If
he sold the taxi after 2 years at the rate of 10% p.a. compound depreciation, calculate
his gain or loss in 2 years.
Solution:

Oasis School Mathematics-10 73

Here, Original value (V0) = Rs. 4,50,000

Time (T) = 2 years

Rate (R) = 10% p.a.

Value after T years (VT) = ?

We have, VT = V0  1 − R T
or,  100 

PT = 4, 50, 000  1 – 10 2
 100 

= 4, 50, 000  90 2
 100 

= 4, 50, 000 × 9 × 9
10 10

= Rs. 3,64,500

Amount earned by using taxi = Rs. 50,000

Now, value after depreciation + amount earned by taxi = Rs.(3,64,500 + 50,000)

= Rs. 4,14,500

Since the original value is greater than this amount,

loss = Rs. 4,50,000 – Rs. 4,14,500

= Rs. 35,500

Exercise 4.3

1. (a) If Vo be the original value of an asset, VT the value of the asset after T years, R the
rate of depreciation, write the relation among Vo, VT, R and T.

(b) If Vo be the value of a piece of land, R be the growth rate of the value, T the time,
then write the relation of the value after T years VT with Vo, R and T.

(c) If Vo be the original value and VT the value of the asset after T years. Then, what
is the amount of depreciation?

(d) If R1 and R2 be the rate of depreciation for two consecutive years and Vo the
original value, write the relation of VT with Vo, R1 and R2.

2. (a) A machine costing Rs. 10,000 is depreciated 5% every year, what will be its value
next year?

(b) A mobile set costing Rs. 15,000 is depreciated 10% every year, what will be its
value next year?

(c) The value of a piece of land is Rs. 15,00,000. Its value is increasing at the rate of
20%, what will be its value next year?

74 Oasis School Mathematics-10

3. Find the value after the given period of time with given rate in each of the following cases.

(a) Original value = Rs. 15,000, annual rate of depreciation = 8%, time = 2 years.

(b) Original value = Rs. 36,000, annual rate of depreciation = 10%, time = 3 years.

4. Find the original value in each of the following cases.

(a) Value after 2 years = Rs. 10,000, annual rate of depreciation = 10%

(b) Value after 3 years = Rs. 1,20,000, annual rate of depreciation = 5%

5. (a) Present value of a machine is Rs. 2,50,000. Its value depreciated by 12% per annum,
then

(i) what will be its value next year?

(ii) what was its value one year before?

(b) A man bought a second hand computer at Rs. 24,000 in 2060 B.S. If the annual
rate of depreciation is 20%,

(i) what was its value in 2058 B.S.?

(ii) what was its value in 2062 B.S.?

(c) The present price of a scooter is Rs. 95,000. If it is depreciated at 6% per year,
what will be the price of the scooter after 2 years?

(d) A television set costing Rs. 11,600 is depreciated at the rate of 15% per annum.
What will be the cost of the television set after 2 years ?

6. (a) A man bought 250 shares of Rs. 100 each. The value of the shares got devaluated
by 5% every year due to political instability. Find his profit or loss at the end of 2
years.

(b) Sangita bought 150 secondary shares of a bank at the rate of Rs. 500 per share.
The value of share depreciated by 3% for 2 years. Find her profit or loss.

7. (a) A company bought 5 ropanis of land at Rs. 2,80,00,000. Due to the political
instability, its value got devaluated by 5% every year. Find the price of the land
after 2 years.

(b) A man sold a piece of land at Rs. 19,44,000 after the devaluation of the price of
land by 10% every year. Find at what price he had bought the land 2 years before.

8. (a) The number of WBC in the infected blood was 14 × 109 per litre. After using
antibiotics its number decreased by 10% every day. Find the number of WBC per
liter after 3 days.

(b) After using medicine the number of bacteria decreased by 5% per hour. If its
number was 2.5×105 per ml at 11:00 A.M., what was its number at 9:00 A.M.?

9. (a) A machine costing Rs. 50,000 is sold at Rs. 40,500 after 2 years. Find the annual
rate of depreciation.

(b) A man bought a motorcycle at Rs. 1,50,000. After 2 years, he sold it at Rs. 1,16,160.
Find the annual rate of depreciation.

10. (a) In how many years is the value of an article depreciated from Rs. 18000 to
Rs. 14580 at the rate of 10% p.a.?

Oasis School Mathematics-10 75

(b) In how many years is the value of an article depreciated from Rs. 2,00,000 to
Rs. 1,70,000 at the rate 15% p.a.?

(c) A radio costing Rs. 1200 depreciates 10% every year. After how many years will
its value be reduced by Rs. 228?

11. (a) A microbus was bought for Rs. 15,00,000. It was used for 2 years and the net
profit during that time was Rs. 4,50,000. After 2 years it was sold at the rate of
compound depreciation of 10% p.a. Find the profit or loss.

(b) Bhai Raja bought a tractor for Rs. 8,50,000. He earned Rs. 4,80,000 from the tractor.
If he sold the tractor after 3 years at compound depreciation of 5%, calculate his
profit or loss.

Answers

1. Consult your teacher 2. (a) Rs. 9,500 (b) Rs. 13,500 (c) Rs. 18,00,000

3. (a) Rs. 12,696 (b) Rs. 26,244 4. (a) Rs. 12345.68 (b) Rs. 1,39,962.09

5. (a) (i) Rs. 2,20,000, (ii) Rs. 2,84,090.90 (b) (i) Rs. 37,500, (ii) Rs.15,360, (c) Rs. 83,942, d. Rs. 8,381

6. (a) Loss Rs. 2,437.50 (b) Loss Rs. 4,432.50 7. (a) Rs. 2,52,70,000 (b) Rs. 24,00,000

8. (a) 10.206 × 109 (b) 2.77 × 105 9. (a) 10% (b) 12% 10. (a) 2 years (b) 1 year (c) 2 years

11. (a) Rs. 1,65,000 profit, (b) Profit 3,58,768.50

Project Work

Visit different showrooms and garages. Ask the resale value of the vehicles, their
original value, time period and calculate the rate of depreciation.

76 Oasis School Mathematics-10

Miscellaneous Exercise

Tax and Money Exchange

1. If a tourist paid Rs. 7480 for a carved window made of wood with a discount of 15% and
including 10% VAT, find

(i) the marked price (ii) the amount of VAT [Ans: Rs. 8,000, Rs. 680]

2. A foreigner paid Rs. 5,610 for a carved window made of wood with a discount of 15%,
including 10% VAT. How much does he get back while leaving Nepal? [Ans: Rs. 510]

3. Which scheme do you prefer on the M.P. Rs. 5,000?

(i) Discount of 10% and no VAT levied.

(ii) Discount of 20% and 10% VAT levied. [Ans: Second one]

4. Supplier 'A' sold some construction materials costing Rs. 5,40,000 to another supplier 'B'
without VAT. If the VAT is 13%, find

(i) How much VAT the supplier 'B' paid?

(ii) What is the net cost price for supplier 'B' if he paid Rs. 5,000 local tax and Rs. 6000
transportation charge?

(iii) What should be his net selling price including VAT if he made a profit of 10%.

[Ans: (i) Rs. 70,200 (ii) Rs. 6,21,200 (iii) Rs. 7,72,151.60]

5. After allowing 15% discount on the marked price of Rs. 60,000 and then levying VAT, a
shopkeeper sold an article for Rs. 57,630. What is the rate of value added tax? [Ans: 13%]

6. The marked price of a laptop is Rs. 75,000. After allowing certain percent of discount and

including 15% VAT, the laptop is sold at Rs. 73,312.50. Calculate the discount percent

and VAT amount. [Ans: 15%, Rs. 9562.50]

7. A dealer sold a machine for Rs. 50,000 and adding 13% VAT to the retailer. Find the

selling price. The retailer spent Rs. 2700 on transportation and Rs. 4,500 on local tax and

sold it to the customer at a profit of 20%. At the rate of 13%, how much does a customer

have to pay as VAT? [Ans: Rs. 9750]

8. A dealer sold a laptop for Rs. 60,000 adding 13% VAT to the retailer. If the retailer sold at

10% loss, spent Rs. 3,980 on transportation and Rs. 10,000 in local tax then supplied it to

a customer, how much VAT amount did the customer pay for the laptop at the current

VAT rate. [Ans: Rs. 9,750]

9. Given, $ 1 = Rs. 108.36 and 1 Euro = Rs. 113.87

(i) Find how much Euros can be bought for $ 800.

(ii) If the Nepali rupees is devalueated by 2%, find how many Euro can be bought for
$800 after giving 3% commission to the bank.[ Ans: (i) 721.29 Euro (ii) 744.08 Euro]

10. Ramesh exchanged Rs. 2,50,000 into Euro for his Europe tour at the rate of 1
Euro = Rs. 113. He spent 2000 Euros on his tour. During the time of his tour the Nepali
currency is devaluated by 2%, find how much Nepali currency is left with him after the tour.

[Ans: Rs. 24,480]

Oasis School Mathematics-10 77

11. The buying rate and selling rate of 1 US Dollar are Rs. 99 and Rs. 100 respectively. A

man bought some US Dollars for Rs. 40,000 and sold them the next day. If the Nepalese

currency is devaluated by 10%, find his profit percentage. [Ans: 8.9%]

12. Some electronic goods costing $ 25,000 were imported from America through India. A
2% shipment charge was paid in India. These goods were imported to Nepal from India
by paying 150% customs duty. If $1 = Rs. 104 then,

(i) find the cost of goods in Nepali market.
(ii) what is the selling price to make a profit of 10%?
(iii) what is the net selling price including 13% VAT?

[Ans:(i) 66,3000, (ii) Rs. 72,9300, (iii) 82, 41,090]

13. 5,000 Nepali caps were bought for Rs. 400 per piece in Nepal and sold it in the UK after
paying a 5% export tax. If a 150% profit was made, find the selling price of a cap in the UK.

[Given £1 = Rs. 150] [Ans: £ 7 ]

Compound Interest

1. What is the difference between the compound interest and simple interest on Rs. 5,120

for 3 years at 12.5% per annum? [Ans: Rs. 250]

2. A certain sum invested at 4% per annum compounded semi-annually amounts to

Rs. 7,803 at the end of one year. Find the sum. [Ans: Rs. 7500]

3. The compound interest of a certain sum of money for 2 years at the rate of 5% p.a. is

Rs. 410. Find the principal. [Ans: Rs. 4000]

4. Rs. 16,000 invested at 10% p.a. compounded semi-annually amounts to Rs. 18,522. Find
1
the time period of the investment. [Ans: 1 2 yrs.]

5. A man invested Rs. 60,000 at the rate of 1 paisa per rupee per month. In how many years

will he get an interest of Rs. 5,264? [Ans: 2 yrs.]

6. At what rate of compound interest per annum interest on Rs. 1,25,000 be Rs. 91,000 in 3

years ? [Ans: 20%]

7. At what rate of compound interest will the sum of money become 125 times the original
64
sum in 3 years? [Ans: 20%]

8. The compound interest on a sum of money in 2 years at 8% per annum is Rs. 360 more

than the simple interest. Find the sum. [Rs. 56, 250]

9. The simple interest on a sum of money in 2 years is Rs. 90 less than its compound

interest. If the rate of interest is 15%, find the sum. [Ans: Rs. 4000]

10. According to the system of yearly compound interest, a sum of money amounts in 2

years and 3 years for Rs. 12,100 and Rs. 13,310 respectively. Find the sum and the rate of

interest. [Ans: 10%, Rs. 10,000]

78 Oasis School Mathematics-10

11. A sum of money amounts to Rs. 19,360 in 2 years and Rs. 23,425.60 in 4 years at the rate

of compound interest compounded annually. Find the rate of compound interest and

the sum. [Ans: 10%, Rs. 16,000]

12. 'A' invested Rs. 25000 for 3 years at the rate of 12% p.a. simple interest and 'B' invested the
same amount for the same time at the rate of 10% annual compound interest. Calculate

(i) interest received by 'A'. (ii) interest received by 'B'.

(iii) without altering the time and the rate of interest, how much more or less money
should 'A' invest for equal interest? [Ans: Rs. 9,000, Rs. 8,275, Rs. 2,013.89 less]

13. What will be the simple interest on Rs. 20,000 for 2.1 years at the rate of 10% per annum ?

At what time does the same sum produce the same interest at the same rate of compound

interest compounded annually ? [Ans: Rs. 4,200, 2 yrs.]

14. Divide 21,000 into two parts such that the annual compound amount of the first part for

3 years is equal to the compound amount of second part for 4 years at the rate of 4% p.a.

compounded annually. [Ans: Rs.10,000 and Rs. 11,000]

15. A man borrowed Rs. 30,000 for 2 years at 10% per annum compounded annually. He

paid only half of the principal at the end of 2 years. He paid the remaining principal and

interest at the same rate at the end of the following two years. How much did he pay at

to clear the debt? [Ans: Rs. 25773]

16. The semi-annual compound interest on a sum of money in 1 years and 2 years are
Rs. 2,100 and Rs. 3,310 respectively. Find the rate of interest and the principal.

[Ans: Rs. 10,000 and 20%]

17. The compound interest of sum of money in 3 years and 6 years is Rs. 33,100 and Rs.
77,156.10 respectively. Find the rate of interest and the sum. [Ans: 10%, Rs. 10,000]

Population Growth

1. The population of a village was 7,200. If 5% of the population migrated and 2% died

due to different causes within a year, what would be the population of the village after

a year? [Ans: 6696]

2. The present population of a town is 8,484. If the annual birth rate and death rate are 2%

and 1% respectively, find the population of the town before 1 year. [Ans: 8400]

3. The present population of a village is 1,800. One year ago, the population of the town

was 2,000. Find the rate at which the population is decreasing. [Ans: 10%]

4. In how many years will the population of a town be 2,09,475 from 1,90,000 at the growth

rate of 5% per annum? [Ans: 2 yrs.]

5. The population of Beni Bazaar increases every year by 2.5%. At the end of 2 years, the

total population of the town was 24,895. If 320 people migrated to the other village,

what was the population of the town at the beginning? [Ans: 24,000]

Oasis School Mathematics-10 79

6. The growth of curd clotting bacteria is 5% per hour. If the number of bacteria at 5 pm is

9.261×1011, what was the number of bacteria at 2 pm? [Ans: 8×1011]

7. The present price of the land is Rs. 10,40,000 per aana. If the annual price increases at the

rate of 10%, find the price per ropani after 2 years. [Given, 1 ropani = 16 aana]
[Ans: 2,01,34,400]

8. 3 years ago the population of a town is 3,75,000 and the annual growth rate is 2%. At the

end of the second year 1m 480 people were migrated in and 2,750 people died. Find the

present population of the town. [Ans: 3,96,658]

9. The population of a town in 2072 B.S. and 2073 B.S. was 2,20,500 and 2,31,525 respectively.

What was the population of the village in 2070 B.S. if the growth rate is the same?

[Ans: 2,00,000]

10. 40,000 students were registered in a university. The number of students increased at
tshtuedreantetsobfe2c10omeevser4y6,y3e0a5r..
Find the time period at the end of which the total number of
[Ans: 3 years]

Depreciation

1. If the cost is depreciated at the rate of 10% p.a., the cost of an article becomes Rs. 92,583

after 3 years. Find the original price of the article. [Ans: 1,27,000]

2. A man bought a computer for Rs. 44,100 and after using it for 2 years, he sold it for
[41261%]
Rs. 40,000. Find the rate of compound depreciation.

3. The original value of a car is Rs. 4,50,000. If its value depreciates by 1 the value at the
beginning of each year, when will its value be reduced to 2,88,000? 5 [Ans: 2 yrs.]

4. A truck was bought at Rs. 15,50,000. It earned Rs. 1,40,000 in 3 years. If it is sold after 3

years at 20% depreciation per year, find the profit or loss. [Ans: 6,16,400 loss]

5. A man bought a taxi for Rs. 8,00,000. He earned Rs. 2,50,000 in 3 years from this and sold
it at a price which was depreciated by 10% p.a. What is his profit or loss?

[Ans: Profit Rs. 33200]

6. The present price of a machine is Rs. 5,00,000. It is depreciated at the rate of 5%, 10% and
5% in three consecutive years. Find the price of the machine after 3 years.

[Ans: Rs. 4,06,125]

7. A building worth Rs. 50,00,000 is constructed on a plot of land worth Rs. 60,00,000. If the
price of the house depreciates at the rate of 10% p.a. and the price of land increases at
the rate of 10% p.a., find the combined price of the house and land after 3 years.

[Ans: Rs. 1, 16, 31,000]

8. A factory was bought for Rs. 3,00,000 some years ago, and now its value is Rs. 2,43,000.
If the value of the factory is depreciated at 10% p.a., when was the factory bought?

[Ans: 2 years ago]

80 Oasis School Mathematics-10

Attempt all the questions: Full Marks: 25

Group – A [4 × 1 = 4]

1. (a) Selling price of an article is Rs. 2,700. How much VAT should be paid according to
the current VAT rate of Nepal?

(b) If $ 1 = Rs. 108.35, convert Rs. 7,000 into $.

2. (a) If P0 represents the present population, Pn is the population after 'n' years, R be the
growth rate,

(b) Express Pn in terms of P0, n and R.

How much value is depreciated on a computer worth Rs. 20,000 at the rate 10% in
1 year?

Group – B [4 × 2 = 8]

3. (a) If $1 = Rs.108.35 and 1 Euro = Rs. 113, convert $ 60 into Euro.

(b) The S.P. of an article, including 13% VAT is Rs. 1,695. Find the selling price excluding
VAT.

4. (a) The population of a village is 5,000. If the annual growth rate of population is 10%,
find the population after 2 years.

(b) A mobile set was bought for Rs. 36,000. After 2 years, it was sold at Rs. 29,160. Find
the rate of depreciation.

Group – C [2 × 4 = 8]

5. A man buys an article at 20% discount on the marked price. If he paid Rs. 416 as the VAT
at the rate of 13%, find the (i) S.P. (ii) S.P. with VAT (iii) M.P. (iv) Discount.

6. The C.I. on a sum of money for 2 years compounded annually is Rs. 8,034. Simple interest
on the same sum for the same period is Rs. 7,800. Find the sum and rate of interest.

Group – D [1 × 5 = 5]

7. A man bought a motor in the Indian market at Rs. 55,000 I.C.

(i) What is its value in Nepali rupees if Rs. 100 IC = Rs. 160 N.C.?
(ii) If the customs duty is 80%, find the customs duty that he has to pay?
(iii) What is its value in the Nepali market?
(iv) If he makes a profit of 10%, what should be its S.P. in the Nepali market?
(v) How much does a customer have to pay for it including 13% VAT?

Oasis School Mathematics-10 81

Mensuration

24Estimated Teaching Hours

Contents

• Area of triangle and polygon
• Surface area and volume of cylinder, sphere and hemisphere
• Surface area and volume of cone
• Surface area and volume of combined solid having at most two

solids
• Surface area and volume of triangular prism
• Surface area and volume of pyramid
• Surface area and volume of geometrical bodies

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following competencies:
• To find the area of different types of triangles
• To find the area of a polygon (quadrilateral, trapezium, rhombus, etc.)
• To find CSA, TSA and volume of a cylinder, sphere and hemisphere
• To find the curved surface area, total surface area and volume of a cone
• To find the curved surface area, total surface area and volume of a combined

solid having at most two solids
• To find the lateral surface area, total surface area and volume of a triangular

prism
• To find the lateral surface area, total surface area and volume of a square-based

pyramid or equilateral triangular based pyramid
• To find the LSA, TSA and volume of a combined solid of a prism and a pyramid
• To estimate the cost of construction of geometrical bodies like a water tank,

well, pillar, etc.

Materials Required

• Models of a triangular prism, cone, sphere, hemisphere, cylinder,
combined solid, pyramid.

82 Oasis School Mathematics-10

Unit

5 Plane Surface

5.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.
• Write the formula to find the area and perimeter of a rectangle.
• How many altitudes can be drawn in a triangle?
• In a rectangle, how many altitudes can be drawn?
• Are all altitudes of a triangle equal? If all the altitudes are equal, what type of

triangle is this?
• Does a median divide a triangle into two equal parts?
• Does a diagonal divide a parallelogram into two equal parts?
• Is every rectangle a parallelogram?

5.2 Area of Triangle A

(i) When the base and altitude are given

Consider ΔABC having sides AB, BC and CA respectively. AM is
the altitude of the triangle, which is drawn from the vertex A on the
base BC. Then

Area of ΔABC = 1 base × altitude = 1 BC × AM B MC
2 2

Activity

Draw a triangle ABC. Draw AD perpendicular to BC. Take E and F, the mid points of

AB and AC respectively. Join EF, cut I and II from the triangle and paste it as shown in

the figure. A

E I II F h I II h/2
III
IV

III IV B DC

B D C base (b)

base (b)

Figure (i) Figure (ii)

Now it is a rectangle whose two sides are (b) and h2 .
Area of rectangle = b
× h
2
1
Area of triangle = 2 b × h [Area of figure (i) = Area of figure (ii)]

Hence, area of triangle = 1 base × height.
2

Oasis School Mathematics-10 83

Area of an equilateral triangle

Take an equilateral triangle ABC, having the length of each side 'a'. Draw AD

perpendicular to BC. Since the perpendicular drawn from the vertex of a triangle bisects

the base,

BD = DC = a A
2

In right angled triangle ABD,

AD = AB2–BD2 aa

= a2–(a/2)2

= a2 – a2 B a2 D a2 C
4

= 3a2
4

= a3
2
1
Area of ∆ABC = 2 BC × AD

= 1 a × a3 = 3 a2
2 2 4

∴ Area of equilateral triangle = 3 a2
4

Area of an Isosceles Triangle

Let's take an isosceles triangle ABC, where AB = AC = a (equal side) and BC (base) = b.

From A draw AD perpendicular to BC.

Then, A

BD = DC = b [In an isosceles triangle, the perpendicular drawn from the
2
vertex bisects the base] a a

Using the Pythagoras' theory

AD = AB2–BD2 b/2 b/2
= a2–(b/2)2
B DC

b

= a2 – b2
4

= 4a2–b2 = 4a2–b2
4 2

Area of ∆ABC = 1 BC × AD
2

= 1 b × 4a2–b2
2 2

=b 4a2–b2
4
Area of an isosceles triangle = b 4a2–b2
∴ 4

Where, b = base, a = equal side.

84 Oasis School Mathematics-10

When all three sides of a triangle are given

Consider the triangle ABC, whose sides AB, BC and CA are denoted by c, a and b
respectively.

Let 's' be the semi-perimeter of the triangle, then A

its perimeter = a + b + c = 2s c hb
Bx Ma a – x
Now, draw AM ⊥ BC and suppose BM = x, then C

MC = a – x.

In the right-angled ΔAMB, using the Pythagoras' theorem,

AB2 = BM2 + AM2

c2 = x2 + h2

or, h2 = c2 – x2 ....................... (i)

Also, in the right angled ΔAMC,

AC2 = AM2 + MC2

b2 = h2 + (a – x)2

or, h2 = b2 – (a – x)2 ....................... (ii)

From equations (i) and (ii), we get

c2 – x2 = b2 – (a – x)2

or, c2 – x2 = b2 – (a2 – 2ax + x2)

or, c2 – x2 = b2 – a2 + 2ax – x2

or, c2 + a2 – b2 = 2ax

or, x = c² + a² – b²
2a

Now, substituting the value of x in equation (i), we get

h2 = c² –  c² + a² – b ² 2
 2a



=  c + c² + a² – b ²   c – c² + a² – b² 
 2a   2a 

=  2ac + c² + a ² – b ²  2ac – c² – a² + b² 
 2a  2a 

= 1 [(a + c)² – b²][ b² – (c² + a² – 2ac)]
4a²

= 1 [(a + c)² – b²][ b² – (c – a)²]
4a²

= 1 (a + c + b) (a + c – b) (b + c – a) (b – c + a)
4a²

Oasis School Mathematics-10 85

= 1 2s (2s – 2 b) (2s – 2a) (2s – 2c)
4a²

= 16s (s – b) (s – a) (s – c)
=
4 s(s – 4a² b) (s – c)
a² a) (s –

∴ h = 2 s(s – a) (s – b) (s – c)
a

As we know, area of a triangle = 1 base × height
2
1
So, area of ΔABC = 2 BC × AM

= 1 × a ×h
2
= 1 × a × 22 s(s – a) (s – b) (s – c)
2 2 aa

Area of triangle (A) = a s(s – a) (s – b) (s – c) , where s = semi-perimeter = a+b+c
For equilateral triangle: A2

A B = BC = CA, i.e., a = b = c aa

∴ s = a+b+c = a+a+a 3a B aC
2 2 =2

a+b+c
2
Area of ΔABC (A) = s(s – a) (s – b) (s – c) , where, s =

= 3a  3a – a  3a – a   3a – a 
2  2   2   2 

= 3a . a . a . a
2 2 2 2

= 1 a2 . 3
= 4

3 .a²
4

∴ A rea of equilateral triangle (A) = 3 .×a(²side)².
4

Do you know!
Area of a triangle (A) = s (s–a) (s–b) (s–c) is known as Heron's formula.
It is also called Hero's formula, named after the Hero of Alexandria.
Its proof can be found in his book 'Metrica'.
Metrica is a collection of mathematical knowledge that was available in the ancient
world.

86 Oasis School Mathematics-10

Remember !

Area of a triangle = 1 base × height
2
a + b + c
Area of a triangle = s(s – a) (s – b) (s – c) , Where s = 2

Area of an equilateral triangle =3 4a2

Area of an isosceles triangle = b 4a2–b2 , where b = base, a = length of equal sides
4

Area of a parallelogram = base × height

Area of a rhombus = 1 × d1 × d2
2
1
Area of a trapezium = 2 height × sum of parallel sides

Area of a regular hexagon = 3 3a²
2
Area of a square = 1 d2 (if diagonal is given)
2
1
Area of a quadrilateral = 2 diagonal × sum of perpendicular draw from opposite

vertex to the diagonal

= 1 d (h1 + h2)
2

Worked Out Examples

3 cm A
4 cm
Example: 1

Find the area of a triangle of sides 3 cm, 4 cm and 5 cm.
Solution:

Here, in ΔABC, let a = 5 cm, b = 4 cm, c = 3 cm then, B 5 cm C

s= a+b+c = 4 + 5 + 3 = 6 cm
2 2
s(s – a) (s – b) (s – c), where s = a+b+c
We have, area of ∆ABC = 2

= 6(6 – 5) (6 – 4) (6 – 3)

= 6 ×1× 2 × 3 cm2
= 36 cm2 = 6 cm2

Example: 2 3 cm A 6 cm D12 cm
Find the area of the quadrilateral given below. 5 cm C
Solution:

The quadrilateral ABCD is divided into two triangles. So, the B 13 cm
sum of the areas of the two triangles is the required area of the
given quadrilateral.

Oasis School Mathematics-10 87

For ΔABD, 3+5+6
2
s = = 7 cm

ΔABD = s(s – a) (s – b) (s – c) , where s = a+b+c
= 2
7(7 – 3) (7 – 5) (7 – 6)

= 7×4×2×1
= 56 cm2

= 7.48 cm2

For ΔBCD, 5 12 13
2
s = + + = 15 cm

ΔBCD = s(s – a) (s – b) (s – c), where s = a+b+c
2

= 15(15 – 5) (15 – 12) (15 – 13) cm2

= 15×10×3×2 cm2

= 900 cm2 = 30 cm2

∴ Area of quadrilateral ABCD = (7.48 + 30) cm² = 37.48 cm2 A D
Example: 3

Find the area of a parallelogram in which its two adjacent 5 cm 8 cm
sides are 6 cm and 5 cm and the diagonal is 8 cm.

Solution: B 6 cm C

Let ABCD be a parallelogram with BC = 6 cm, AC = 8cm and AB = 5 cm. Since the
diagonal bisects a parallelogram into two equal triangles,

Area of parallelogram ABCD = 2 × ΔABC

So, for ΔABC s = a+b+c = 8+5+6 = 9.5 cm
2 2

Area of ∆ABC = s(s – a) (s – b) (s – c) , where s = a+b+c
2

= 9.5(9.5 – 8) (9.5 – 5) (9.5 – 6) cm2

= 9.5 ×1.5 × 4.5 × 3.5 cm2

= 224.43 cm2 = 14.98cm2

∴ Area of parallelogram ABCD = 2 × ∆ABC [Diagonal bisects the parallelogram]

= 2 × 14.98 cm² = 29.96 cm²

88 Oasis School Mathematics-10

Example: 4 AB

Find the area of a regular hexagonal field of side 20 cm. O
Solution: FC

Let ABCDEF be a regular hexagonal field of side 20 cm. As we E D

know, the diagonal divides it into 6 equilateral triangles. So, the

Area of ABCDEF = 6 × ΔAOB Alternative method

= 6× 3 .(asi²de)2 Area of regular hexagon
4
3 3a2
3 = 2
4 .(a2²0)2
= 6× cm2 = 1039.23 cm2 3 3(20)2
2
3 =
4
= 6× .×a4² 00 = 600 3 = 1039.23 cm2 = 3 3 × 200
Example: 5
= 1039.23cm2

The three sides of a triangle are in the ratio 5 : 12 : 13. If the perimeter is 60 cm, find
the area of the triangle.

Solution:

Let the three sides of the triangle be 5x, 12x and 13x respectively.

Perimeter of the triangle = 60 cm

∴ 5x + 12x + 13x = 60 cm

or, 30x = 60 cm

x = 2 cm

Now, 5x = 5 × 2 = 10 cm

12x = 12 × 2 = 24 cm and

13x = 13 × 2 = 26 cm

∴ Semi-perimeter (s) = a+b+c
2

= 10 + 24 + 26 = 30 cm.
2

∴ Area of triangle = s(s – a) (s – b) (s – c) , where, s = a+b+c
2

= 30(30 – 10) (30 – 24) (30 – 26) cm2

= 30 × 20 × 6 × 4 cm2

= 14400 cm2 = 120 cm2
Example: 6

The perimeter of a right-angled triangle is 24 cm and its area is 24 cm2. Find the sides
of the triangle.

Solution:
In the right-angled ∆ABC, let 'p' 'b' and 'h' be the perpendicular, base and hypotenuse

respectively.

Oasis School Mathematics-10 89

Area of ∆ABC = 1 base × height A
2 p
24 cm2 =
or, p×b = 1 ×b×p B h
2 b

48 cm2 ...................... (i)

and, p + b + h = 24 cm C

or, p + b = 24 – h ........................ (ii)

Again, in the right-angled ∆ABC, we have,

h2 = p2 + b2

or, h2 = (p + b)2 – 2pb

or, h2 = (24 – h)2 – 2 × 48

or, h2 = 576 – 48 h + h2 – 96

or, 48h = 576 – 96

or, 480
h = 48 = 10 cm.

So, from equation (ii), p + b= 24 – 10 = 14 cm.

or, p = 14 – b ...................... (iii)

From (i) and (iii), (14 – b) b = 48

or, 14b – b2 = 48

or, 0 = b2 – 14 b + 48

or, b2 – 8b – 6b + 48 = 0

or, b(b – 8) – 6((b – 8) = 0

or, (b – 8) (b – 6) = 0

Either b – 8 = 0 ⇒ b = 8 cm

or, b – 6 = 0 ⇒ b = 6 cm.

So, from equation (iii)

When, b = 8 cm, p = 14 – 8 = 6 cm

and b = 6 cm, p = 14 – 6 = 8 cm.

So, the required sides of the right-angled triangle are 8 cm, 6 cm and 10 cm or 6 cm, 8 cm
and 10 cm, respectively.

Example: 7

An umbrella is made by stitching 10 triangular pieces of cloths each measuring 15 cm,
41 cm and 28 cm. How much cloths is required to make umbrella? If the total cost to
prepare umbrella is Rs. 630, find the cost per cm2 clothes.

Solution:
Taking a triangular piece of cloth measuring 15cm, 41 cm and 28 cm
a = 15 cm, b = 41 cm, c= 28 cm

Then, S = a+b+c = 15cm + 41cm + 28 cm = 47 cm.
2 2

90 Oasis School Mathematics-10

We have,

Area of a triangle = s(s – a) (s – b) (s – c)

= 42 (42–15) (42–41) (42–28)

= 42 × 27 × 1 × 14

= 15876

= 126 cm2

Area of 10 triangular pieces (A) = 126 × 10cm2

= 1260 cm2

Total cost to make umbrella = Rs. 630.

We have,

Total cost = Area × Rate

630 = 1260 × R

R = 630 = Rs. 0.50 per cm2.
1260

Hence, the cost per cm2 cloths = Rs. 0.50 per cm2

Exercise 5.1

1. a) What is the area of an equilateral triangle whose side is 'a' unit?

b) What is the area of a triangle whose sides are 'a', 'b' and 'c' units?

c) Find the area of a right-angled triangle whose base and height are 'a' cm and 'b' cm
respectively.

d) Find the area of a parallelogram whose base and height are 'a' cm and 'b' cm
respectively.

e) Write the formula of the area of a quadrilateral whose one diagonal is 'd' cm and the
perpendiculars on the diagonals are 'a' cm and 'b' cm respectively.

f) Write the formula of the area of a rhombus if 'd1' and 'd2' are its diagonals.
g) If 'a' and 'b' are the lengths of the two parallel sides of a trapezium and if 'h' be its

height, write the formula to calculate the area of the trapezium.

h) Write the formula to calculate the area of a square whose diagonal is d cm.

i) Write the formula to calculate the area of an isosceles triangle whose base is 'b' and
the length of the equal sides is 'a'.

2. Find the area of the following triangles. A

(a) A (b) X (c)

8 cm 8 cm 7 cm
5 cm
8 cm

B 10 cm C BDC
Y 8 cm 12 cm
Z

Oasis School Mathematics-10 91

A P D

(d) (e) (f)
5 cm 12 cm
6 cm 5 cm 12 cm

Q 5 cm R
41 cm E 12 cm F
D B 4 cm C
P A
X
(g) (i)
(h)

10 cm 13 cm
12 cm

Q R Y 40 cm Z BD C
F
12 cm (c) E

3. Find the area of given figures: 8 cm

(a) P S (b) D C

4 cm 7cm 5 cm 6 cm
5 cm 6cm

Q 6 cm R 5 cmA 6 cm B H G

(d) A (e) P (f) P

4cm 9cm 5cm
4cm
B5cm B C
T T
4cm 8cm Q S Q S

D
9cm
QS = 10cm QS = 12cm
C
R
4. Find the area of following hexagons. R
A B
(a) B C
(b)

A DF C

F 8 cm E E 6 cm D

5. Find the area of the following:
12 cm
(a) A B (b) W 8 cm X (c)P 6 cm Q
5 cm
8 cm 10 cm

D 6 cm C Z T 12 cm Y S 4 cm R

6. (a) Find the area of an equilateral triangle whose each side is 15 2 cm.
(b)
(c) Find the area of a right-angled isosceles triangle whose hypotenuse is 10cm.

(d) Find the area of a triangle having three sides, 9 cm, 11 cm and 13 cm.

Find the area of a parallelogram with its two adjacent sides 13 cm and 14 cm and
corresponding diagonal 15 cm.

92 Oasis School Mathematics-10

(e) If the diagonals of a rhombus are 20 cm and 12 cm, find its area.
(f) Find the area of a regular hexagon having each side 6 cm.
7. (a) The area of an equilateral triangle is 36 3 m2. Find the length of its sides.
(b) The area of an equilateral triangle is 16 3 cm². Find its perimeter.
The perimeter of an equilateral triangle is 30 cm. Find its area.
(c)
(d) Find the area of an isosceles triangle whose perimeter is 36 cm and the length of
the base 10 cm.
(e) The perimeter and the length of the base of an isosceles triangle are 25 cm and
9 cm respectively. Find the area of the triangle.

(f) Length of two equal sides of an isosceles triangle is 10cm. If its area is 48cm2, find
(g) the length of its base.
8. (a)
(b) The length of the base of an isosceles ∆ is 12 cm, if its area if 48cm2, find the
(c)
9. (a) length of its equal sides.

(b) If the ratio of the sides of a triangle is13 : 14 : 15 and its perimeter is 42 cm, find
the area of the triangle.
10. (a)
The perimeter and the area of a triangular field are 100 m. and 20036 5 m². If one
(b) side of the field is 40 m, find the other two sides.

11. (a) The perimeter of a right angled triangle is 30 cm and its area is 30 cm². Find the

(b) sides of the triangle. 12cm
15 cm
Find the cost to prepare a Nepali national flag of given
25cm
measurement at the rate of Rs. 3.50 per cm2.
4 ft

Find the cost to prepare a Nepali national flag of given 5 ft 2 ft 20cm

measurement at the rate of Rs. 50 per square feet. 3 ft

A tent is made by stitching 10 triangular pieces of cloths of two different colors,

each piece measuring 61 ft., 61 ft and 22 ft. How much cloth is required for the

tent? If the cost of 1 ft.2 cloth is Rs. 5, find the total cost of the cloth.

A tent is made by stitching 10 triangular pieces of cloths of two different colors,
each piece measuring 65 m, 65 m and 32 m. If the cost of 1 m2 cloth is Rs. 5, find
the total cost of the cloth.

A tent is made by stitching 10 triangular pieces of clothes of two different colors,
each piece measuring 101 cm, 101 cm and 40 cm. If the total cost of clothes is
Rs. 79,200, find the cost of 1 cm2 of cloth.
A tent is made by stitching 20 triangular pieces of cloths having dimension 3 m,
4m and 5m. If the cost to prepare the tent is Rs. 9600. Find the cost per m2 cloths.

Answers

1. Consult your teacher 2. (a) 31.22 cm² (b) 17.32 cm² (c) 48 cm² (d) 12 cm² (e) 10.83 cm²
(f) 62.35 cm² (g) 60 cm² (h) 180 cm² (i) 60 cm² 3. (a) 19.84 cm² (b) 24 cm² (c) 24 cm²
(d) 23.79cm2 (e) 60.59cm2 (f) 54cm2 4. (a) 166.27 cm² (b) 93.53 cm²

5. (a) 72 cm² (b) 100 cm² (c) 15 cm² 6. (a) 194.86 cm² (b) 50 cm² (c) 48.8 cm² (d) 168 cm² (e) 120 cm²
(f) 93.53 cm² 7. (a) 12 cm (b) 24 cm (c) 43.3 cm² (d) 60 cm² (e) 29.76 cm² (f) 12cm (g) 10cm

8. (a) 84 cm² (b) 30 cm, 30 cm (c) 5 cm, 12cm, 13 cm 9. (a) Rs. 889 (b) Rs. 434
10. (a) 6600 sq. ft., Rs. 33,000 (b) 1260 cm2, Rs. 50,400, 11. (a) Rs. 4 (b) Rs. 80

Oasis School Mathematics-10 93

Unit Cylinder, Sphere,
Hemisphere and Cone
6

6.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.
• Take a rectangular sheet of paper and roll it. What type of structure is formed?
• Take a wooden log and identify whether its surface is flat or curved?
• Take a ball and identify whether its surface is curved or flat.
• What is the shape of a birthday cup? Identify its curved surface and flat surface.

6.2 Cylinder

Let's collect objects like a glue stick, circular pipe, wooden log, etc.

Glue stick Circular pipe Wooden log

Discuss their shape and similarities. All these objects are cylindrical in shape. Hence,
they are cylindrical objects.

Look at this figure. B' O' A'

This is a figure of a cylinder. h
OA or O'A' represents its radius.
AA' or BB' represents its height. BO A

OO' represents its axis. Radius

Activity: Take a rectangular sheet of paper, fix it on a stick and rotate it about the stick.
What structure is formed?

Surface area of cylinder

Take a rectangular sheet of paper ABCD. Roll it along AD such that A coincides with B

and D coincides with C. B
A



hh

D C r
Fig (ii)
Fig (i)

94 Oasis School Mathematics-10