Compound Interest
? The amount in T year = P (1 + R%)T = P 1+ R T
100
R R T–
Also, the interest in T year = P 1+ 100 T–P =P 1 + 100 1
Thus,
Compound amount = P 1+ R T and Compound interest = P 1 + R T– 1
100 100
1. When the interest is compounded annually but the rate being different in different
years, for example, R1% in the first year, R2% in the second year, R3% in the third
years, then the amount in 3 years is calculated as,
Compound Amount =P 1 + R1 1 + R2 1 + R3
100 100 100
Also, Compound Interest = P 1 + R1 1 + R2 1 + R3 –1
100 100 100
2. When the interest is compounded annually but the time is given in 'T' years and 'M'
months, then the amount is calculated as
Compound Amount =P 1 + R T 1 + MR
100 1200
Also, Compound Interest = P 1 + R T 1 + MR –1
100 1200
3.3 Interest compounded half-yearly and quarter-yearly
R
If the compound interest is payable half yearly, then rate = 2 % per half yearly and
time = 2T half years. R 2T
× 100
Now, the compound amount half yearly= P 1 + 2
Also, the compound interest half yearly = P 1 + 2 R 2T – 1
× 100
Similarly, if the compound interest is payable quarter-yearly (every 3 months),
then rate = R % per quarter-yearly and time = 4T quarter years.
4
So, the compound amount quarter-yearly =P 1 + R 4T
× 100
4
And the compound interest quarter-yearly = P 1 + 4 R 4T – 1
× 100
Worked-out examples
Example 1: Calculate the compound interest of Rs 7,000 at 10% per annum for 3 years
without using the formula.
Solution:
Here, principal (P) = Rs 7,000
Rate of interest (R) = 10% p.a.
Now, Time (T) = 3 years = PTR = 7,000 × 1 × 10
the interest in the first year (I1) 100 100
= Rs 700
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Compound Interest
? The principal for the second year = Rs 7,000 + Rs 700
Now, the interest in the second year (I2)
= Rs 7,700
= 7,700 × 1 × 10
100
= Rs 770
? The principal for the third year = Rs 7,700 + Rs 770
Again, the interest in the third year (I3)
= Rs 8,470
= 8,470 × 1 × 10
100
= Rs 847
? The compound interest (C.I.) = I1 + I2 + I3
= Rs 700 + Rs 770 + Rs 847 = Rs 2,317
Example 2: A sum of Rs 24,000 is deposited in a bank X at the rate of 5% p.a. simple
interest. If the sum would have been deposited in bank Y at the same rate of
compound interest, which bank would give more interest after 3 years and
by how much?
Solution:
Here, principal (P) = Rs 24,000, time (T) = 3 years and rate (R) = 5 % p.a.
Now, simple interest = PTR = Rs 24,000 × 3 × 5 = Rs 3,600
100 100
R
Also, compound interest = P 1 + 100 T– 1
= Rs 24,000 1 + 5 3 –1
100
= Rs 24,000 1 + 1 3– 1
20
= Rs 24,000 21 3 – 1
20
= Rs 24,000 21 × 21 × 21 – 8,000 = Rs 3 × 1261 = Rs 3,783
8,000
? Difference of compound and simple interest = Rs 3,783 – Rs 3,600 = Rs 183
Hence, bank Y would give more interest by Rs 183.
Example 3: Laxmi deposited Rs 16,000 at 10% p.a. interest compounded semi-annually
cfoorm1p12ouyenadresd.
Narayan deposited the equal sum at the same rate of interest
annually for the same duration of time. Who gets more interest
and by how much?
Solution:
Here, principal (P) = Rs 16,000, time (T) = 112 years and rate (R) = 10 % p.a.
R
Now, interest compounded semi annually (I1) = P 1 + 2 ×100 2T– 1
= Rs 16,000 1 + 10 3– 1
200
21 × 21 × 21
= Rs 16,000 20 × 20 × 20 –1
= Rs 2,522
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Compound Interest
Also, interest compounded annually (I2) Alternative process
1 +1210R0
=P 1+ R T –1 I2 = P 1 + R T 1 + MR –1
100 100 1200
= Rs 16,000 1+ 10 1 1 + 10 –1 = 16,000 1+ 10 1 1+ 6 × 10 –1
100 200 100 1,200
11 2210 –
= Rs 16,000 11 × 21 – 1 = 16,000 10 × 1
= Rs 16,000 10 20
231 – 200
200 = Rs 2,480
= Rs 2,480
? Difference between I1 and I2 = I1 – I2
= Rs 2,522 – Rs 2,480
= Rs 42
Hence, Laxmi will get more interest by Rs 42.
Example 4: The compound interest on the sum of money in 2 years at the rate of 5 % per
annum is Rs 364 more than simple interest. Find the sum.
Solution:
Here, time (T) = 2 years, rate (R) = 5 % and let the Principal be Rs P.
Now, C.I. = P 1+ R T–1 Answer checking:
100
=P 1+ 5 2 –1 = 41P P = Rs 1,45,600 R = 5% T = 2 years
100 400
C.I. = 1,45,600 1+ 5 2 –1 = Rs 14,924
100
Also, S.I = PTR = P×2×5 = P
100 100 10 S.I. = 1,45,600 × 2 × 5 = Rs 14,560
100
From the question, C.I. – S.I. = Rs 364
C.I. – S.I. = Rs 14,924 – Rs 14,560
41P P
or, 400 – 10 = Rs 364 = Rs 364 which is given in the
question.
or, P = Rs 364
400
or, P = Rs 1,45,600
Hence, the required sum is Rs 1,45,600.
Example 5: Simple interest on a sum of money for 2 years at 4% p.a. is Rs 800. Find the
compound interest on the same sum and at the same rate for 1 year if the
interest is reckoned half-yearly?
Solution:
Here, time (T) = 2 years, rate (R) = 4% p.a. and simple interest (S. I.) = Rs 800
Now, S. I. = PTR = P ×2× 4 = 2P
or, 100 100 25
800 = 2P
25
or, P = Rs 10,000
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Compound Interest
Again, interest compounded half-yearly:
C. I.= P 1+ R 2T– 1
2 ×100
= 10,000 1+ 4 2×1 – 1 = 10,000 51 × 51 –1 = 10,000 × 101 = Rs 404
200 50 × 50 2,500
Hence, the required compound interest is Rs 404.
Example 6: At what rate percent per annum compound interest will Rs 2,500 amounts to
Rs 2,704 in 2 years?
Solution:
Here, principal (P) = Rs 2,500, time (T) = 2 years and C.A. = Rs 2,704
Now, C.A. =P 1 + R T
100
or, 2704 Answer checking:
or, 2704 = 2500 1 + R 2
or, 2500 100
or, 52 2 P = Rs 2,500 T = 2 years R= 4% p.a.
50 100 + R 2
100 + R = 100 C.A. = Rs 2,500 1+ 4 2
100 100
= 100 + R 2 = Rs 2,704 which is given in the question.
R 100
= 52
50
= 5200 – 100 = 200 =4%
50 50
Hence, the required rate of interest is 4 % p.a.
Example 7: Mrs. Rai invests Rs 25,000 for 3 years at a certain rate of interest compounded
annually. At the end of one year the sum amounts to Rs 27,500. Calculate:
(i) the rate of interest (ii) the amount at the end of second year
(iii) the amount at the end of third year.
Solution:
Here, principal (P) = Rs 25,000, time (T) = 1 year and amount (A) = Rs 27,500
(i) Now, C.A. =P 1 + R T
100
27,500 = 25,000 1 + R 1
100
or, 1+ R = 27,500 = 11
100 25,000 10
or, R = 10%
?The rate of interest is 10% p.a.
(ii) Amount at the end of second year:
C.A.= P 1 + R T = 25,000 1 + 10 2 = 25,000 × 11 × 11 = Rs 30,250
100 100 10 10
(iii) Amount at the end of third year:
C.A.= P 1 + R T = 25,000 1 + 10 3 = 25,000 × 11 × 11 × 11 = Rs 33,275
100 100 10 10 10
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Compound Interest
Example 8: A floriculturist takes a loan of Rs 40,000 from an Agricultural bank. If the
rate of compound interest is 5 paisa per rupee per year, in how many years
does he pay the compound interest of Rs 6,305?
Solution:
Here, the rate of interest (R) = 5 paise per Re 1 per year
= 500 paise, i.e. Rs 5 per Rs 100 per year = 5 % per annum
Also, principal (P) = Rs 40,000 and C.I. = Rs 6,305
Now, C.I. = P 1 + R T– 1 Alternative process
100
(C.A.) = P + C. I
5 T– 1
or, 6,305 = 40,000 1+ 100 = Rs 40,000 + Rs, 6,305
or, 6,305 = 21 T– 1 = Rs 46,305 R
40,000 20 100
Now, C.A. =P 1 + T
or, 1261 = 21 T –1
8000 20
or, Rs 46,305 = 40,000 1 + 5 T
100
or, 21 T = 1261 + 1 21 T
20 8000 or, 46,305 = 20
40,000
or, 21 T = 9261 = 21 3 21 T
20 8000 20 or, 20 = 9,261
8000
or, T = 3 years 21 T 21 3
or, 20 = 20
So, the required time is 3 years.
or, T = 3 years
Example 9: Mr. Gupta deposited a sum of Rs 50,000 in a commercial bank for 3 years. If
the bank provided 4% interest compounded annually for the first year and
the rate of interest was gradually increased by 1% every year, how much
interest did he get in 3 years?
Solution:
Here, principal (P) = Rs 50,000
Rate of interest in the first year, R1 = 4 % p.a.
Rate of interest in the second year R2 = 5 %
Rate of interest in the third year R3 = 6 % p.a.
R1 R2 R3
Now, compound amount (C.A.) = P 1 + 100 1 + 100 1 + 100
= 50,000 1 + 4 1 + 5 1 + 6
100 100 100
= 50,000 × 104 × 105 × 106 = Rs 57,876
100 100 100
? Compound interest (C.I.) = C.A. – P
= Rs 57,876 – Rs 50,000
= Rs 7,876
Hence, he got the interest of Rs 7,876 in 3 years.
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Compound Interest
Example 10: The compound amount of a sum of money in 3 years is Rs 26,620 and in
4 years is Rs 29,282. Find the compound rate of interest per annum and the
sum.
Solution: Alternative process
Here, the C.A. in 3 years = Rs 26,620
Here, (C.A.)1 = Rs 26,620
or, P 1 + R 3 = Rs 26,620 (C.A.)2 = Rs 29,282
100
Now, (C.A.)2 – (C.A.)1 =Rs 29,282 – Rs 26,620
or, P 100 + R 3 = Rs 26,620 .......(i) = Rs 2,662
100
Here, Rs 2,662 is the simple interest (S.I.)
or, Also, the C.A. in 4 years = Rs 29,282 of Rs 26,620 for 1 year.
or, P 1 + R 4 = Rs 29,282 ? Rate (R) = I × 100
100 P×T
or, P 100 + R 4 = Rs 29,282 .......(ii) = 2,662 × 100 = 10 %
100 26,620 × 1
Again, (C.A.)1 =P 1 + R 3
100
Now, dividing equation (ii) by (i), we get,
10 3
Rs 26,620 = P 1 + 100
P 100 + R 4
P 100 or, P = Rs 20,000
100 + R = 29,282
3 26,620
100
or, 100 + R = 1.1
100
or, R = 10 % Answer checking:
Again, putting the value of R in equation (i), P = Rs 20,000 T = 3 years R= 10% p.a.
C.A. = Rs 20,000 1 + 10 3 = Rs 26,620
100
P 100 + 10 3 = Rs 26,620 10 4
100 C.A. = 20,000 1 + 100 = Rs 29,282
or, P 11 3 = Rs 26,620 which are given in the question.
10
or, P = Rs 20,000
Hence, the required sum is Rs 20,000 and the rate of interest is 10 % p.a.
Example 11: The yearly compound interest on a sum of money for two successive years
are Rs 320 and Rs 336 respectively. Calculate the rate of interest and the
original sum.
Solution:
Here, difference between the C. I. of two successive years= Rs 336 – Rs 320 = Rs 16
? Rs 16 is the simple interest of one year on Rs 320.
? Rate of interest= I × 100 = 16 × 100 = 5% p.a.
P×T 320 × 1
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Compound Interest
Again, the interest of the original sum (P) for the first year = Rs 320
P 1+ R T– 1 = Rs 320
100
5
or, 1 + 100 1– 1 = Rs 320
or, P = Rs 320
20
or, P = Rs 6,400
Hence, the rate of interest is 5% p.a. and the original sum is Rs 6,400.
Example 12: The compound interest calculated yearly on a certain sum of money for the
second year is Rs 1,650 and for the third year is Rs 1,815. Calculate the rate
of interest and the original sum of money.
Solution:
Here, the difference between the C. I. of two successive years = Rs 1,815 – Rs 1,650 = Rs 165
? Rs 165 is the interest on Rs 1,650.
Now, rate of interest (R) = I × 100 = 165 × 100 = 10% p.a.
P×T 1650 × 1
R T=P 10 1 11P
C.A. for the first year = P 1 + 100 1+ 100 = 10
So, 11P is the principal for the second year.
10
11P R T– 1 11P 10 1– 1 11P
? C. I. in the second year = 10 1 + 100 = 10 1 + 100 = 100
From the question,
11P = Rs 1,650
100
or, P = Rs 15,000
Hence, the rate of interest is 10% p.a. and the original sum is Rs 15,000.
Example 13: The simple interest of a certain sum of money for 2 years is Rs 1,600 and
the compound interest of the same sum at the same rate of interest for the
same duration of times is Rs 1,680. Find the rate of interest and the sum.
Solution:
Here, S. I. = Rs 1,600 and T = 2 years
Now, principal (P) = I × 100 = 1600 × 100 = 80000 ........ (i)
T×R 2×R R
Also, C. I. = P 1 + R T– 1
or, 1,680 = P 100
1 + R 2– 1 = (100 +R)2 – 10000
100 10000
= 10P000[(100 + R)2 – 10000]... (ii)
Putting the value of P from equation (i) in equation (ii), we get,
1,680 = 80000 [(100 + R)2 – 10000]
10000R
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Compound Interest
or, 210R = 10000 + 200R + R2 – 10000
or, R2 – 10R = 0
or, R = 10% p.a.
Again, from equation (i), P = 80000 = 80000 = Rs 8,000
R 10
Hence, the required rate of interest is 10% p.a. and principal is Rs 8,000.
Example 14: Suresh lent altogether 6,600 to Manoj and Pradeep for 2 years. Manoj
agreed to pay simple interest at 15 % p.a. and Pradeep agreed to pay
compound interest at the same rate. If Manoj paid Rs 112.50 more than
Pradeep as the interest, find how much did Suresh lend to each of them?
Solution:
Suppose the money lent to Pradeep = P1 = Rs x.
? the money lent to Manoj = P2 = Rs (6,600 – x)
Here, time (T) = 2 years and rate (R) = 15 % p.a.
Now, the simple interest to Manoj = P2TR = Rs (6,600 – x) × 2 × 15 = Rs 19,800 – 3x
100 100 10
Also, the compound interest to Pradeep
= P1 1 + R T– 1
100
Answer checking:
15
= Rs x 1 + 100 2– 1 S.I. paid by Manoj = 3600 × 2 × 15
100
3
= Rs x 1 + 20 2– 1 = Rs 1,080
= Rs x 23 2– 1 C.I. paid by Pradeep = 3000 1+ 15 2– 1
20 100
529 = Rs 967.50
400
= Rs x – 1 Now, Rs 1,080 – Rs 967.50 = Rs 112.50 which
= Rs x 529 – 400 = Rs 129x is given in the question.
400 400
According to the question,
or, 19,800 – 3x – 129x = Rs 112.50
10 400
? x = Rs 3,000
So, the money lent to Pradeep = x = Rs 3,000
The money lent to Manoj = 6,600 – x = 6,600 – 3,000 = Rs 3,600
Example 15: Mrs. Nepali borrowed Rs 75,000 from a commercial bank at the rate of
10% p.a. compounded annually for 2 years. After one year, the bank
changed it's policy to pay the interest compounded semi-annually at the
same rate. What is the percentage difference between the interest of the
first year and second year? Write reason with calculation.
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Compound Interest
Solution: Alternative process
Here, principal (P) = Rs 75,000 and rate (R) = 10% p.a.
Now, C.I. in the first year =P 1 + R T– 1 S.I. in the first year
100
= PTR = 75,000 × 1 × 10
100 100
= 75,000 1 + 10 1– 1 = Rs 7,500
100
= 75,000 11 – 10
10
= Rs 7,500
Then, the principal for the second year = Rs 75,000 + Rs 7,500
P' = Rs 82,500
Again. C. I. in the second year = P' 1 + 2 R 2T – 1
× 100
= 82,500 1+ 10 2– 1
200
= 82,500 21 × 21 – 400
400
= 206.25 × 41
= Rs 8,456.25
? The difference between the interest of second and first years = Rs 8,456.25 – Rs 7,500
= Rs 956.25
And, the difference of interest in percent = 956.25 × 100% = 12.75%
7,500
Since, the interest compounded semi-annually is paid at the end of every six months, it is
greater than the interest compounded annually by 12.75%.
EXERCISE 3.1
General section
1. a) If C.A. and C.I. are the compound amount and compound interest of a sum P
in T years at R% p.a. respectively, write the relationships among the following
variables.
(i) P, T, R, and CA (Compounded annually)
(ii) P, T, R, and CI (Compounded annually)
(iii) P, T, R, and CA (Compounded semi-annually)
(iv) P, T, R, and CI (Compounded semi-annually)
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Compound Interest
b) The compound amount on a sum P in T years M months at R % p.a. is CA. Write
the relation among P, T, M, R, and CA compounded annually.
c) If CI is the compound interest of a sum P in T years M months at R % p.a., write
the relation among P, T, M, R and CI compounded annually.
d) If CA is the compound amount of a sum P at the different rates R1 %, R2 %, and
R3% in the first, the second and the third years respectively. Write the relation
among CA, P, R1, R2, and R3 (compounded annually).
2. Calculate the compound interest without using the formula.
a) P = Rs 5,000, T = 2 years, R = 4% b) P = Rs 15,000, T = 3 years, R = 10%
3. a) Find the interest compounded annually and the simple interest of the sum of
Rs 2,000 in 1 year at 5 % p.a. Are both types of interest same? Write your conclusion.
b) Find the amount compounded annually.
(i) Principal (P) = Rs 2,400, time (T) = 2 years, rate (R) = 5 % p.a.
(ii) Principal (P) = Rs 3,125, time (T) = 112 years, rate (R) = 8 % p.a.
c) Find the interest compounded annually.
(i) Principal (P) = Rs 8,000, time (T) = 3 years, rate (R) = 5 % p.a.
(ii) Principal (P) = Rs 6,000, time (T) = 221 years, rate (R) = 10 % p.a.
d) Find the amount compounded half-yearly (semi-annually).
(i) Principal (P) = Rs 5,000, time (T) = 1 year, rate (R) =4 % p.a.
(ii) Principal (P) = Rs 16,000, time (T) = 121 year, rate (R) = 10 % p.a.
e) Find the interest compounded semi-annually.
(i) Principal (P) = Rs 2,500, time (T) = 1 year, rate (R) = 8 % p.a.
(ii) Principal (P) = Rs 16,000 time (T) 1 1 years, rate (R) = 10 % p.a.
2
Creative section - A
4. a) Find the difference between simple interest and the annual compound interest on
Rs 9,600 for 2 years at the rate of 5 % per annum.
b) Mrs. Pariyar borrowed Rs 50,000 from Mrs. Limbu for 3 years at 10% p.a. simple
interest. She immediately deposited this sum of money in a commercial bank
at the same rate of interest compounded annually for the same period of time.
Calculate her profit at the end of 3 years.
c) Find the difference between compound interest compounded semi-annually and
1
simple interest on Rs 8,000 at 10 % per annum in 1 2 years.
d) Sunayana borrowed a sum of Rs 25,000 at 12 % p.a. simple interest for 1 year
6 months and lent to Bishwant at the same rate of compound interest compounded
half-yearly for the same interval of time. How much profit did she make?
e) Devasis deposited Rs 5,000 at 8 % p.a. compound interest in a bank. Find the
difference between the amounts compounded yearly and half-yearly in two years.
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Compound Interest
f) Find the difference between compound interest compounded semi-annually and
1
the interest compounded annually on Rs 16,000 at 10 % p.a. in 1 2 years.
g) Pratik borrowed Rs 1,50,000 from Bishu at the rate of 21 % per year. At the end
of nine months how much compound interest compounded half-yearly should he
pay? [Hints: 9 months = 6 months + 3 months = 1 year + 3 months]
2
5. a) Find the amount compounded annually on Rs 9,000 for 2 years if the rates of
interest for two years are 8 % and 10 % respectively.
b) Find the compound interest compounded annually on Rs 1,00,000 for 3 years if
the rates of interest in the first, second and the third years are 4 %, 5 % and 7 %
respectively.
c) Mr. Tharu borrowed a sum of Rs 1,00,000 from Agricultural Development Bank for
3 years to upgrade his poultry farming. The bank charged 5% interest compounded
annually for the first year and the rate of interest was gradually increased by 1%
every year. How much interest did he pay at the end of third year?
6. a) Simple interest on a sum of money for 2 years at 5% p.a. is Rs 960. Find the
compound interest on the same sum and at the same rate for 1 year, if the interest
is reckoned half-yearly.
b) A person paid Rs 6,200 compound interest on a certain sum of money that he
1
borrowed from a bank at 10% p.a. compounded annually for 1 2 years. Find the
compound interest on the same sum and at the same rate for the same period of
time, if the interest had to be paid semi-annually.
7. a) The compound interest on the sum of money at 8 % p.a. for 2 years is more
than the simple interest on the same sum at the same rate for the same time by
Rs 76.80. Find the sum.
b) Harka Tamang borrowed a certain sum of money from Roshan Shrestha at the rate
of 9 % p.a. simple interest for 2 years. He immediately lent this money to Raju Jha
at the same rate of compound interest for the same period of time. If the interest
paid by Harka is Rs 243 less than the interest paid by Raju, calculate the sum.
c) If the compound interest on a sum of money compounded semi-annually in one
year at 5 % per annum is Rs 20 more than the compound interest on the same sum
compounded annually in the same time and at the same rate, find the sum.
d) The difference between the annual and semi-annual compound interest on a sum
1
of money is Rs 63 at the rate of 10 % per annum for 1 2 years. Find the sum.
e) If the sum of simple interest and compound interest at the rate of 8% p.a. for 2 years
is Rs 816, find the Principal.
f) The compound interest of a certain sum for 2 years is Rs 738 and the simple interest
of the same sum for the same interval of time at the same rate of interest is Rs 720.
Find the sum and the rate of interest.
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Compound Interest
8. a) In how many years does Rs 5,000 amount to Rs 6,272 at 12% per annum interest
compounded annually?
b) Mrs. Dhital deposited Rs 10,000 in her fixed deposit account of a bank at 10% p.a.
compound interest. In how many years would she withdraw a compound amount
of Rs 13,310?
c) Rajesh took a loan of Rs 16,000 from his friend Gopal at the rate of 5 paisa per
rupee per year interest compounded annually. In how many years would he pay a
compound interest of Rs 2,522 to Gopal?
9. a) At what rate percent per annum compound interest does Rs 3,125 amount to
Rs 3,380 in 2 years?
b) A farmer borrowed Rs 1,00,000 from Agricultural Development Bank to promote
his fish farming. If he paid a compound interest of Rs 33,100 at the end of 3 years,
find the rate of interest compounded annually charged by the bank.
10. a) If the compound amounts of a sum of money in 2 years and 3 years are Rs 17,640
and Rs 18,522 respectively, find the rate of interest and the sum.
b) If a sum becomes Rs 6,655 in 3 years and Rs 7,320.50 in 4 years interest being
compounded annually, find the rate of interest and the sum.
c) If a sum of money becomes Rs 1,40,450 in 1 year and Rs 1,48,877 in 112 years,
interest being compounded semi-annually, calculate the rate of interest and the
sum.
Creative Section - B
11. a) The yearly compound interest on a sum of money for two successive years are
Rs 400 and 440 respectively. Calculate the rate of interest and the original sum.
b) The compound interest calculated yearly on a certain sum of money for the second
year is Rs 1,320 and for the third year is Rs 1,452. Calculate the rate of interest and
the original sum of money.
c) The compound interest of a sum of money in 1 year and 2 years are Rs 450 and
Rs 945 respectively. Find the rate of interest compounded yearly and the sum.
d) The compound interest of a sum of money in 1 year is Rs 350 and in 2 years is
Rs 724.50. Find the rate of interest compounded yearly and the sum.
12. a) Krishna lend altogether Rs 10,000 to Gopal and Radha for 2 years. Gopal agrees to
pay simple interest at 12 % p.a. and Radha agrees to pay compound interest at the
rate of 9 % p.a. If Radha paid Rs 596.70 more than Gopal as the interest, find how
much did Krishna lend to each.
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Compound Interest
b) Suntali deposited Rs 9,000 altogether in her saving account and fixed deposit
account in a bank. Saving account gives her 5% p.a. interest compounded annually
and fixed deposit account gives 10% p.a. interest compounded half-yearly. If she
got Rs 160 more interest from fixed deposit account at the end of one year, find how
much money did she deposit in her each account?
c) A bank has fixed the rate of interest 10% p.a. semi-annually compound interest in
account X and 12% p.a., annually compound interest in account Y. If you are going
to deposit Rs 50,000 for 2 years, in which account do you deposit and why? Give
your reason with calculation.
d) A business person borrowed Rs 1,20,000 from a commercial bank at the rate of
10% p.a. compounded annually for 2 years. After one year the bank changed it's
policy to pay the interest compounded semi-annually at the same rate. What is the
percentage difference between the interest of the first year and second year? Give
reason with calculation.
e) A person deposited Rs 80,000 in a bank at the rate of 12% p.a. interest compounded
semi-annually for 2 years. After one year the bank revised its policy to pay the
interest compounded annually at the same rate. What is the percentage difference
between the interest of the second year due to the revised policy? Give reason with
calculation.
f) A housewife deposited Rs 10,000 on saving account at 5% p.a. interest compounded
yearly and another sum on fixed deposit account at 8% p.a. interest compounded
half yearly. After one year the interest on fixed deposit account was Rs 152.80 more
than the interest on the saving account, find the total amount of money in her two
accounts at the end of the year.
Project Work
13. a) Make the groups of your friends and visit different Banks and Finance Companies
and collect their brochures. Study about the bank deals, bonuses, promotions offers
mentioned in their brochures. Which plans and offers do you think the best and
exciting to deposit your money?
b) Make the groups of your friends and visit nearby banks or finance companies.
Collect the various information about the rate of interest in different types of
accounts such as current account, saving account and fixed deposit account. In
which account are they providing more interest?
c) Ask your parents or any other family members whether they have any bank
accounts. If so, what type of account do they have and how much rate of interest
are they getting?
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Unit Population Growth and Depreciation
4
4.1 Population growth - introduction
Population growth is the increase in the number of individuals in a population. Many
of the world's countries have seen a sharp rise in population. However, it is found to
remain nearly constant or slightly decreasing in some other countries.
The way of increasing population is exactly similar to the way of increasing compound
amount. Because, every year population increases from the previously increased
population.
Suppose the population of a country or a place at a certain time = P
the rate of growth of population = R % per annum,
the population of the country or place after T years = Pt,
Then, pt = P 1 + R T
100
The growth of population is also affected by the number of deaths, in-migrants and
out-migrants at the end of the given period of time.
In this case, T – no. of deaths + Min – Mout
The actual population in T years (Pt) = P 1 + R
100
Where Min and Mout are the number of in-migrants and out-migrants respectively at the
end of the given period of time.
Similarly, the increased population in T years = Pt – P
= P 1 + R T
100
–P
=P 1+ R T–1
100
If the rate of growth of population every year is different, then the population after T
years is calculated by using the following formula.
Pt = P 1 + R1 1 + R2 ... 1 + RT
100 100 100
Where R1, R2 ... RT are the rates of growth of population in the first year, second years,
and ... Tth years.
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Population Growth and Depreciation
Worked-out examples
Example 1: The population of a municipality before 2 years was 60,000 and the rate
of annual growth of population is 2%. If the numbers of in-migrants and
out-migrants at the end of 2 years were 750 and 410 respectively and 620
people died within this time interval, find the present population of the
municipality.
Solution:
Here, the population of the municipality before 2 years (P) = 60,000
The rate of growth of population (R) = 2%
The number of in-migrants (Min) = 750
The number of out-migrants (Mout) = 410
The number of deaths (D) = 620
Now, the present population of the municipality =P 1 + R T
100
– D + Min – Mout
= 60,000 1 + 2 2
100
– 620 + 750 – 410
= 60,000 u 51 u 51 – 280 = 62,424 – 280 = 62,144
50 u 50
Hence, the present population of the municipality is 62,144.
Example 2: The present population of a Metropolitan city is 16,87,500. If the rate
of growth of population is 4% per year, find the increased population in
2 years.
Solution:
Here, the present population of the city (P) = 16,87,500
the rate of growth of population (R) = 4% p.a.
Time (T) = 2 years
Now, the increased population = P 1 + R T– 1
100
= 16,87,500 1+ 4 2–1
100
26 2
= 16,87,500 25 – 1
= 16,87,500 × 51 = 1,37,700
625
Hence, the increased population of the Metropolitan city in 2 years is 1,37,700.
Example 3: Two years ago the population of a village was 40,000. If the birth rate is
Solution: 4.5% p.a. and death rate is 2% p.a., find the present population of the
village.
Here, the population of the village before 2 years (P) = 40,000
The annual growth rate of the population (R) = 4.5% – 2% = 2.5%
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Population Growth and Depreciation
Now, the present population (Pt) =P 1 + R T
100
= 40,000 1 + 2.5 2
100
= 40,000 u 41 u 41 = 42,025.
40 40
Hence, the present population of the village is 42,025.
Example 4: The population of a village increases every year by 5%. At the end of two years,
if 1,025 people migrated to other places and the population of the village
remained 10,000, what was the population of the village in the beginning?
Solution:
Here, the total population including the migrated number = 10,000 + 1,025 = 11,025
We know that Pt =P 1 + R T
or, 100
11,025 =P 1 + 5 2
100
or, 11,025 = P 21 × 21
20 × 20
or, P = 11025 × 20 × 20 = 10,000
21 × 21
Hence, the population of the village in the beginning was 10,000.
Example 5: The population of a town before 3 years was 1,50,000. If the annual growth
rates of the population in the last 3 years were 2%, 4% and 5% respectively
every year, find the population of the town at the end of 3 years.
Solution:
Here, the population of the town before 3 years (P) = 1,50,000
The annual growth rates in every year were R1 = 2%, R2 = 4% and R3 = 5%
Now, the population of the town at the end of 3 years is
Pt =P 1 + R1 1 + R2 1 + R3
100 100 100
= 1,50,000 1 + 2 1 + 4 1 + 5
100 100 100
= 1,50,000 u 51 u 26 u 21 = 1,67,076
50 25 20
Hence, the population of the town at the end of 3 years is 1,67,076
Example 6: In the beginning of 2074 B.S., the population of a village was 50,000. The
population growth rate of the village is 2% p.a. In the beginning of 2076 B.S.
420 people died due to earth quake. Find the population of the village in the
beginning of 2077 B.S.
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Population Growth and Depreciation
Solution:
Here, Population of the village (P) = 50,000
Time (T) = 2 years
Population of the village at the end of 2 years (Pt) = ?
We have Pt = P 1 + R T
100
= 50,000 1 + 2 2 102 × 102 = 52,020
100 100 × 100
= 50,000
After 2 years, 420 people died.
So, the population of the village in the beginning of 2076 B.S. = 52,020 – 420 = 51,600
Again, the population in the beginning of 2077 B.S. =P 1 + R T
100
= 51,600 1 + 2 1
100
= 51,600 102 = 52,632
100
Hence, the population of the village in the beginning of 2077 B.S. is 52,632.
Example 7: In the beginning of 2074 B.S. the population of a city was 5,00,000 and it was
6,65,500 at the end of 2076 B. S. What will be the population of the city at
the end of 2078 B.S. at the same rate of growth of the population?
Solution:
Here, the population of the city in the beginning of 2074 B.S. (P) = 5,00,000
Population of the city at the end of 2076 B.S. ((PTt)) = 6,65,500
Time = 3 years
The rate of growth of population (R) = ?
We have Pt =P 1 + R T 1 + R 3
100 100
= 5,00,000
or, 6,65,500 = 5,00,000 100 + R 3
100
or, 100 + R 3 = 665500 = 11 3
100 500000 10
or, 100 + R = 11
100 10
or, R = 10% = 6,65,500
Again, the population of the city in the beginning of 2077 B.S. (P)
Population of the city at the end of 2078 B.S. (Pt) = ?
Time (T) = 2 years and Rate (R) = 10% p.a.
Then, Pt = P 1 + R T 1 + 10 2 = 665500 × 11 × 11 = 8,05,255
100 100 10 × 10
= 665500
Hence, the population of the city at the end of 2078 will be 8,05,255.
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Population Growth and Depreciation
EXERCISE 4.1
General Section
1. a) If the initial population of a place is P and the rate of population growth is R% p.a.,
write the formula to find the population of the place in T years.
b) If the initial population of a region is P and the rate of growth of population is
R% p.a., write the formula to find the increased population after T years.
c) The rates of annual growth of population of a town in 3 years are R1%, R2%,
and R3%. If the present population of the town is P, write the formula to find the
population after 3 years.
2. a) The population of a town in the beginning of 2076 B.S. was 32,400. Within a year the
population increased 3% by birthrate and 2% by immigration. Find the population
of the village in the beginning of 2077 B.S.
b) The population of a village increased from 10,000 to 11,000 in one year. Find the
rate of growth of population.
c) One year ago, the population of a village was 10,000. If the present population of
the village is 10,210, find the population growth rate.
Creative Section
3. a) The present population of a village is 30,000. If it is increased at the rate of 10 %
per annum, what will be the population after 2 years?
b) According to the students enrollment statistics published in 2075 B.S., the number
of admitted students in 2074 B.S. in a region was 3,20,000. If the annual rate of
growth of the number of students is 5%, how many students were admitted in
2077 B.S.?
c) 3 years ago, the population of a village was 16,000. The rate of population growth
of that village is 5%. What is the population at present?
d) After two years the population of a town will be 33,620 at the population growth
rate of 2.5 % p.a. Find the present population of the town.
e) A village has population 1,75,760. If its population growth rate is 4 % p.a., find its
population before 3 years.
4. a) 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?
b) The population of a town is 96,250. In how many years would it be 1,04,104 if the
population increases at the rate of 4% every year?
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Population Growth and Depreciation
c) At present the population of a town is 80,000. At what growth rate of population
would it be 88,200 in two years?
d) The population of a municipality in the beginning of 2074 B.S. was 1,80,000 and at
the end of 2076 B.S. was 2,39,580. Find the rate of growth of population per year.
5. a) The population of a town was 3,75,000 three years ago and the annual growth
rate is 2%. if the number of in-migrants and out-migrants at the end of 3 years
were 1,480 and 875 respectively, and 2,750 people died within the times. Find the
present population of the town.
b) The population of a village increases every year by 5%. At the end of two years,
if 460 people were migrated to other village and the population of the village
remained 26,000, what was the population of the village in the beginning?
c) The population of a town increases every year by 10%. At the end of two years, if
5,800 people were added by migration and the total population of the town became
30,000, what was the population of the town in the beginning?
d) The population of a village increases every year by 2%. If 950 people migrated to
other places at the end of two years and the population of the village remained
9,454, what was the population of the village in the beginning?
e) In the beginning of 2074 B.S., the population of a town was 1,00,000 and the rate
of growth of population is 2% every year. If 8,000 people migrated there from
different places in the beginning of 2075 B.S., what is the population of the town
in the beginning of 2077 B.S.?
f) In the beginning of the year 2017, the population of a village was 25,000 and the rate
of growth of population is 10% every year. In the beginning of 2018, if 500 people
migrated to other places, what is the population of the town in the beginning of
2021?
6. a) The population of a town before 3 years was 1,75,000. If the annual growth rates
of the population in the last 3 years were 2%, 4%, and 5% respectively every year,
find the population of the town at the end of 3 years.
b) The population of a village decreased by 5% in 2075 B.S. and by 10% in
2076 B.S. What would be the population of the town in the beginning of 2077 B.S.
if its population in the beginning of 2075 B.S. was 32,000?
7. a) The population of a town in the beginning of 2019 was 1,68,000. If the annual rate
of growth of population is 2.5%, find the increased population at the end of 2020.
b) The present population of a state is 97,65,625. If the rate of growth of population is
4% p.a., find the increased population after 2 years.
8. a) The rate of growth of a plant is 5% every month. If the height of the plant in the
beginning of Baisakh 2077 is 20 cm, find its height at the end of Asar 2077.
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Population Growth and Depreciation
b) The growth rate of a certain type of useful bacteria is 10% per day. During a research
work in a laboratory, if the bacteria are grown upto the number of 2.662 × 1012 in
3 days, how many bacteria were there 3 days ago?
c) A house owner made an agreement to increase the house rent by 10% every year. If
the rent of the house this year is Rs 16,000, find the house rent after 3 years.
d) Due to the annual increment of price of land in a rapidly growing area by 25% the
present value of a piece of land is Rs 12,50,000 per ropani. How much was the
value of the land before 4 years?
Project work
9. a) Make the different groups of your friends. Visit different parts of your Ward and
collect the statistics of the present population of your Ward. Visit to the concerned
Ward Administration Office and get the statistics of the population of the Ward in
the last census report. Calculate the rate of growth of population of your Ward.
b) How many students were there in your school before 2 years and how many
students are there in this year? Collect these data from your school administration
office and calculate the growth rate of students in your school.
c) Visit the available website and search the rate of growth of population of various
countries in the world. Compare these rates with the rate of growth of population
of our country.
4.2 Depreciation
When any asset such as furniture, vehicles, machinery items, etc. are being used for
some time, their values are decreased. The reduction of value of an item due to its
constant use is known as depreciation.
Depreciation may be simple or compound. In simple depreciation, the amount of
reduction is constant every year from every depreciated value.
For example, suppose the original cost of a machine is Rs 36,000 and every year it
is depreciated by Rs 3,600.
Then, after 1 year its price = Rs 36,000 – Rs 3,600 = Rs 32,400
After, 2 years its price = Rs 32,400 – Rs 3,600 = Rs 28,800
After, 3 years its price = Rs 28,800 – Rs 3,600 = Rs 25,200 and so on.
On the other hand, in case of compound depreciation, the amount of reduction is
calculated every year from every depreciated value. The compound depreciation is
calculated in the similar way of calculation of compound interest, but here the value
is gradually decreasing.
? Pt = P 1 – R T
100
Where, P is the original price, R is the rate of depreciation, T is the time period for
depreciation and Pt is the depreciated value after T years.
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Population Growth and Depreciation
Again, the depreciated amount = P – Pt
=P–P 1 – R T
100
=P 1– 1 – R T
100
Similarly, if R1, R2, R3, ... be the rates of depreciation in the first, second, third, ... years
respectively, then
the depreciated value (Pt) = P 1 – R1 1 – R2 1 – R3 ...
100 100 100
Worked-out examples
Example 1: Sangita bought a new scooter for Rs 2,10,000. After 2 years, she sold it to
her friend Anjali at 10% per year compound depreciation. How much did
Anjali pay for the scooter?
Solution:
Here, the initial cost of the scooter (P) = Rs 2,10,000
The time period for depreciation (T) = 2 years
The rate of compound depreciation (R) = 10% per year
Now, the depreciation cost (Pt) = P 1 – R T 1 – 10 2
100 100
= Rs 2,10,000
= Rs 2,10,000 9 2
10
= Rs 1,70,100
Hence, Anajali paid Rs 1,70,100 for the scooter.
Example 2: Mr. Dhurmus purchased a taxi for Rs 18,00,000. He earned a profit of
Rs 5,40,000 in 3 years. If he sold it after 3 years at 5% p.a. compound
depreciation, find his profit or loss.
Solution:
Here, the original cost of the taxi (P) = Rs 18,00,000.
The time period for depreciation (T) = 3 years
The rate of compound depreciation (R) = 5% p.a.
Now, the depreciated cost (Pt) = P 1 – R T
100
= Rs 18,00,000 1 – 5 3
100
= Rs 18,00,000 u 19 u 19 u 19 = Rs 15,43,275
20 20 20
Again, the profit in 3 years = Rs 5,40,000.
? The value of the taxi with profit = Rs 15,43,275 + Rs 5,40,000 = Rs 20,83,275
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Population Growth and Depreciation
But the original cost of the taxi = Rs 18,00,000
? Profit = Rs 20,83,275 – Rs 18,00,000 = Rs 2,83,275
Hence, his profit is Rs 2,83,275.
Example 3: The compound depreciation of shares of a business company for 2 years
Solution: is at the rate of 2% p.a. If the present value of certain number of shares is
Rs 24,010, how many shares at Rs 100 per share were sold before 2 years?
Here the value of shares 2 years before = Rs P.
The rate of depreciation for 2 years = 2% p.a.
Now, the present value of the shares (Pt) = Rs 24,010
or, P 1 – R T = Rs 24,010
100
or, P 1 – 2 2 = Rs 24,010
100
or, Pu 49 u 49 = Rs 24,010
50 50
or, P = Rs 24,010 × 50 × 50 = Rs 25,000
49 × 49
Again, the number of shares = Total value = 25,000 = 250
Rate of cost 100
So, the required number of shares sold before 2 years were 250.
Example 4: Bishwant purchased a mountain bike for Rs 32,000. If he used it for 3 years
and sold to Sunayana for Rs 19,652, find the rate of compound depreciation.
Solution:
Here, the initial cost of the bike (P) = Rs 32,000
The cost of bike after 3 years (Pt) = Rs 19,652
Time (T) = 3 years
We have, Pt = P 1 – R T 100 – R 3
100 100
= 32,000
or, 19652 = 100 – R 3
32000 100
or, 17 3 = 100 – R 3
20 100
or, 17 = 100 – R
20 100
or, 1700 = 2,000 – 20R
or, R = 15%
Hence, the rate of compound depreciation is 15% p.a.
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Population Growth and Depreciation
EXERCISE 4.2
General Section
1. a) If the original value of a printing machine is P and the rate of compound depreciation
is R% p.a., write the formula to calculate the depreciated value of the machine after
T years.
b) Write the formula to calculate the amount of deprecation of a motorbike in T year,
if the initial price of the bike is P and the rate of compound depreciation is R% p.a.
c) The rates of compound depreciation of a printing machine in 3 years are R1%, R2%,
and R3%. If the initial value of the machine is P, write the formula to calculate its
value after 3 years.
2. a) The value of a machine is Rs 4,50,000. If its value is depreciated by 7% every year,
find its value after 1 year.
b) After the depreciation at the rate of 15% p.a., the value of a vehicle became
Rs 13,26,000 after one year. Find the original value of the vehicle.
c) At what rate of depreciation is the price of a building reduced to Rs 22,84,200 from
Rs 24,30,000 in 1 year?
Creative Section
3. a) If the value of a laptop which was bought for Rs 61,200 depreciates at 10% annually,
find its value after 2 years.
b) A set of office furniture costing Rs 96,000 is depreciated at the rate of 5% per year.
What will be its cost after 3 years?
c) The value of a machine is depreciated every year by 10%. What will be the value of
the machine worth Rs 1,80,000 after 3 years?
d) Mr. Ghising bought a tripper for Rs 28,08,000. He used it for transporting
construction materials and earned Rs 7,80,000 in 2 years. If he sold it at 15% p.a.
compound depreciation, find his profit or loss.
4. a) If the cost is depreciated at the rate of 12% per annum, the cost of a photocopy
machine becomes Rs 69,696 after 2 years. Find the original price of the machine.
b) If the cost is depreciated at the rate of 20% per annum, the cost of a tractor after
3 years becomes Rs 9,21,600. Calculate the original price of the tractor.
c) Mrs. Kandel sold a scooter for Rs 1,12,999 in the system of compound depreciation
at the rate of 15% p.a. If she had purchased it 3 years before, how much did she pay
for it by that time?
d) The compound depreciation of shares of business company for 2 years is at the
rate of 2.5% p.a. If the present value of certain number of shares is Rs 45,630, how
many shares at Rs 200 per share were sold before 2 years?
5. a) The original value of a computer is Rs 96,000. If it is depreciated by 15% every year
with the compound depreciation, by how much is its value depreciated in 3 years?
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Population Growth and Depreciation
b) Mr. Mahato bought a building for Rs 25,20,000. He sold it after 3 years at the rate
of compound depreciation of 10% p.a. By how much was the value of the building
depreciated?
6. a) The original price of a furniture is Rs 40,000. If the rate of depreciation is 5% per
annum, after how many years will its price become Rs 36,100?
b) A machine was bought for Rs 4,00,000 some years ago and now its value is
Rs 1,96,000. If the value of the machine is depreciated at 30% p.a. compound
depreciation when was the machine bought?
c) A few years ago a man bought a computer for Rs 1,25,000. In the recent year, he
sold it for Rs 64,000 at 20% annual rate of depreciation. How long did he use the
computer?
d) Maya Limbu bought a television for Rs 64,000. In the recent year, she sold it for
Rs 46,240 at 15% annual rate of compound depreciation. How long did she use the
television?
7. a) A mobile costing Rs 6,000 is depreciated per year and after 2 years its price becomes
Rs 5,415. Find the rate of depreciation.
b) If the value of an electric item depreciated from Rs 18,000 to Rs 14,580 in two
years, find the rate of depreciation.
c) Binju bought an iPad for Rs 1,20,000 and after using it for 3 years, she sold it for
Rs 61,440. Find the rate of compound depreciation of the iPad.
d) The value of a motorbike is depreciated from Rs 2,50,000 to Rs 1,70,368 in 3 years,
find the rate of depreciation.
8. a) Mr. Chhetri bought a scanner machine 3 years before for Rs 40,000. If the value of
the machine is depreciated by 5 %, 8 % and 10 % in the first, the second and the
third year respectively. At what price did he sell the machine at the end of 3 years?
b) The present value of an asset is Rs 7,20,000. If its value is deprecated by 7% in the
first year, 10% in the second year and 15% in the third year, find its value after
3 years.
Project work
9. a) Make the different groups of your friends. Visit different showrooms or shops where
the second-hand vehicles are sold. Also visit different companies and organsations
where the second-hand electronic items, electrical machines, vehicles, etc. are in
sale. Ask the original price and the second-hand price of those items and calculate
the rate of depreciation. Also calculate the profit or loss percent by selling those
items.
b) Visit the available website and search the rates of depreciation of different types
of materials such as machinery items, vehicles, furniture, buildings, etc. Do these
materials have the same rate of deprecation?
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Objective Questions
Tick the correct alternatives.
1. If the discount amount given in an item is 25% of its selling price, then the actual rate of
discount is
(A) 1632% (B) 20% (C) 25% (D) 30%
2. A stationer allows equal rate of discount in each item. If a customer pays Rs. 405 for a
book printing Rs 450 on it, at what price can a student buy the pen costing Rs 50 in the
same shop?
(A) Rs 45 (B) Rs 50 (C) Rs 55 (D) Rs 60
3. The advertised price of a television is x, what will be its selling price after deduction of
two successive discounts at the rates of y% and z%?
(A) x 1 – y+z (B) x 1 – y+z (C) x 1 – y 1 – z (D) x 1 + y–z
100 200 100 100 100
4. What will be the cost of a bag costing Rs 1800 after levying 13% VAT?
(A) Rs 2043 (B) Rs 2034 (C) Rs 2340 (D) Rs 3420
5. If the sale tax is 10% of marked price, what is the actual sale tax rate?
(A) 1191 % (B) 9111% (C) 1131 (D) 91110
6. In a shopping complex 5% discount is given in every item. On the occasion of New Year,
if it announces next 5% cash back as discount, how much should a customer have to pay
for a refrigerator with its catalog price Rs 20000?
(A) Rs 18000 (B) Rs 22000 (C) Rs 18005 (D) Rs 18050
7. The marked price of a jacket is Rs 2500. If the shopkeeper allows certain discount and
sells it for Rs 2260 with 13% value added tax, what is the rate of discount?
(A) 20% (B) 15% (C) 13% (D) 10%
8. The price of a mobile excluding and including value added tax are Rs 4950 and Rs 5445
respectively. If the rate of discount is same as that of value added tax then its marked
price is
(A) Rs 4000 (B) Rs 4400 (C) Rs 5500 (D) Rs 5445
9. A tourist bought a Nepali art with 13% VAT. If he got back Rs 520 at airport while leaving
Nepal, what was the actual price of the art?
(A) Rs 3000 (B) Rs 4000 (C) Rs 5000 (D) Rs 6000
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10. If 1 £ = NRs 119 and 176$ = 100£ then how much Nepali rupees can be exchanged with
132$?
(A) NRs 9825 (B) NRs 8952 (C) NRs 5298 (D) NRs 8925
11. If P be the sum, R be the rate of interest, and T be the time duration in years, the
compounded amount compounded half yearly is
(A) P 1 + R 2T (B) P 1 + R 2T P 1 + R T P 1 – R 2T
200 100 200 200
(C) (D)
12. Sita invested Rs. 66500 in a bank. In two years how much compound interest will she
get, if the first year rate of interest was 10% and second year was 2% more than first year?
(A) Rs. 18245 (B) Rs. 15428 (C) Rs. 18524 (D) Rs. 14825
13. If the difference between compound and simple interest for two years at 5% rate per
annum is Rs. 400 then the principal is …
(A) Rs. 1000 (B) Rs. 10000 (C) Rs. 2000 (D) Rs. 20000
14. The present population of a rural municipality is 5440. If the population of the place
increases 3% by birth and 2% by immigration every year, then the population of the
place after 1 year will be:
(A) 5712 (B) 5217 (C) 7125 (D) 7215
15. The initial value of a machine is V. What will be its value after 3 years if the rates of
depreciation in three successive are r1%, r% and r3 % respectively?
r1 r1 r1 r1 + r2 + r3 3
100 100 100 100
(A) V 1 – V 1 – V 1 – (B) V 1 –
r1 3 r1 3 r1 3 r1 r1 r1
100 100 100 100 100 100
(C) V 1 – 1 – 1 – (D)V 1– 1 – 1 – 1 –
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Unit Mensuration (I): Area of Plane Surface
5
5.1 Area of triangles and quadrilaterals - review
The table given below helps to recall the area of triangles and quadrilaterals.
Name of figure Diagram Formula to find area
Triangle A = 1 × b × h or
2
A = s (s – a) (s – b) (s – c)
Rectangle A=l×b
Square
dl A = l2 or
Parallelogram l
A = 1 × d2
2
A=b×h
A = b × h or
Rhombus d1 d A = 1 × d1 × d2
Trapezium 2 2
a
A = 1 h (a + b)
2
Kite d1 A = 1 d1 × d2
Quadrilateral d2 2
p1 A = 1 d (p1 + p2)
d p2 2
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Mensuration (I): Area of Plane Surface
5.2 Perimeter and area of plane figures - review
Perimeter of a plane figure is the total of the length of its boundary. The unit of
perimeter is the same as the unit of length, for example centimeter (cm), metre (m),
etc. are the units of perimeter.
Perimeter of a triangle = a + b + c, where, a, b and c are the length of sides of a triangle.
Perimeter of a rectangle = 2(l + b), where, l is the length and b is the breadth of a rectangle.
Perimeter of a square = 4l, where, l is the length of each side of a square.
Perimeter of a circle = 2Sr, where, r is the radius of a circle.
Area of a plane figure is the measure of the plane surface enclosed by its boundary
line. The unit of area is square centimetre (cm2) square metre (m2), etc.
It is noted that 'square metre' and 'metre square' have different meanings. 7 square
metre means the area is 7 m2. However, 7 metre square means a square each of whose
side is 7 m long.
5.3 Area of triangle
We can find the area of different types of triangles by using special formulae which are
generalised on the basis of the given measurements of various parts of the triangles.
(i) Area of a triangle in terms of its base and height (altitude)
In the adjoining ' ABC, AD A BC. So, AD is the height and BC is the base of the
triangle ABC. AA
Here, area of ' ABC = 1 base × height
2
1
= 2 BC × AD h
= 1 b × h B D CD Bb C
2
(ii) Area of a right-angled triangle
In the given right-angled ' ABC right angled at B, the perpendicular AB is the
height of the triangle and BC is its base.
Area of right-angled ' ABC = 1 base × height A
2 p
1 B
= 2 base × perpendicular
= 1 BC × AB
2
Cb
(iii) Area of an equilateral triangle
In the given equilateral triangle ABC, AB = BC = CA = a (suppose). AD
perpendicular to BC is drawn. A
In an equilateral triangle, perpendicular drawn from a
vertex to its opposite side bisects the side.
? BD = DC = a aa
2 h
By using Pythagoras Theorem in right-angled ' ABD,
AD = AB2 – BD2 = a2 – a2 = 32 a B a D a C
4 2 a 2
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Mensuration (I): Area of Plane Surface
Now, area of ' ABC = 1 base × height
2
= 1 BC × AD = 1 a × 23 a = 43 a2
2 2
Thus, area of equilateral triangle = 43 (side)2
(iv) Area of an isosceles triangle
In the given isosceles triangle ABC, AB = AC = a and BC = b (suppose).
AD perpendicular to BC is drawn.
A
We know that, in an isosceles triangle, perpendicular
drawn from the vertical angle to the base bisects the base.
Therefore, BD = DC = b a a
2
By using Pythagoras Theorem in right-angled ' ABD,
AD = AB2 – BD2 = a2 – b2 = 4a22– b2 Bb D bC
4 22
Now, area of ' ABC = 1 base × height
2
1
= 2 BC × AD
= 1 b × 4a22– b2 = 1 b 4a2 – b2
2 4
Thus, area of an isosceles triangle = 1 b 4a2 – b2
4
(v) Area of scalene triangle in terms of its three sides (Heron's formula)
In the adjoining scalene triangle ABC, suppose the A
sides BC = a, CA = b and AB = c. Let's draw ADA%C.
Therefore, AD = h is the height of 'ABC. ch b C
Let the length of DC be x cm. B a–x D x
? The length of BD = (a – x) cm
Here, the perimeter of ' ABC = 2s = BC + CA + AB a
=a+b+c
? 2s = a + b + c
or, sri=ghta-a+ngb2le+d c which is called the semi-perimeter of the triangle.
Now, in ' ACD, by using Pythagoras Theorem,
AD2 = AC2 – DC2
or, h2 = b2 – x2 .................(i)
Also, in right-angled ' ABD,
AD2 = AB2 – BD2
or, h2 = c2 – (a – x)2 .................(ii)
From equations (i) and (ii), we get,
c2 – (a – x)2 = b2 – x2
or, c2 – a2 + 2ax – x2 = b2 – x2
or, x = a2 + b2 – c2 ................. (iii)
2a
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Mensuration (I): Area of Plane Surface
Again, from the equations (i) and (iii), we get,
h2 = b2 – a2 + b2 – c2 2
2a
= b + a2 + b2 – c2 b – a2 + b2 – c2
2a 2a
= 2ab + a2 + b2 – c2 2ab – a2 – b2 + c2
2a 2a
= (a + b)2 – c2 c2 – (a – b)2
2a 2a
(a + b + c) (a + b – c) (c + a – b) (c – a + b)
= 2a 2a
= (a + b + c) (a + b + c – 2c) (a + b + c – 2b) (a + b + c – 2a)
4a2
Now, substituting, a + b + c = 2s, we get,
h2 = 2s (2s – 2c) (2s – 2b) (2s – 2a)
4a2
16s (s – a) (s – b) (s – c)
= 4a2
h = 2 s(s – a) (s – b) (s – c)
a ' ABC =
Then, area of 1 base × height
2
1
= 2 BC × h
= 1 a × 2 s(s – a) (s – b) (s – c)
2 a
= s(s – a) (s – b) (s – c)
Thus, when three sides of a triangle are given, its area A = s(s – a) (s – b) (s – c)
Worked-out examples
Example 1: Find the area of the triangle and the quadrilateral given below.
a) A b) 13 cm C
B
14 cm 13 cm 12 cm
4 cm
B 15 cm C
Solution: A 3 cm D
a) In 'ABC, s = 14 + 15 + 13 cm = 21 cm
2
? Area of 'ABC = s (s – a) (s – b) (s – c)
= 21 (21 – 15) (21 – 13) (21 – 14)
= 21 × 6 × 8 × 7 = 7 × 3 × 3 × 2 × 2 × 4 × 7
= 7 × 3 × 2 × 2 = 84 cm2.
b) In 'ABD, by Pythagoras theorem, we have,
BD = AB2 + AD2 = 42 + 32 = 5 cm
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Mensuration (I): Area of Plane Surface
Now, area of 'ABD = 1 × AB × AD = 1 × 4 × 3 = 6 cm2
2 2
In 'BDC, s = 13 + 12 + 5 cm = 15 cm
2
? Area of 'BDC = s (s – a) (s – b) (s – c)
= 15 (15 – 13) (15 – 12) (15 – 5) = 15 × 2 × 3 × 10 = 30 sq. cm.
Hence, area of quad. ABCD = Area of ('ABD + 'BDC) = 6 sq. cm. + 30 sq. cm = 36 sq. cm.
Example 2: Find the area of the given parallelogram, in which AB = DC = 41 cm and
AD = BC = 14 cm and diagonal BD = 28 cm. D C
Solution:
The semi-perimeter of 'ABD is 15 cm 28 cm
s = 41 cm + 28 cm + 15 cm = 42 cm A 41 cm B
2
Now, area of 'ABD = s (s – a) (s – b) (s – c)
= 42 (42 – 41) (42 – 28) (42 – 15) = 42 × 1 × 14 × 27 = 126 cm2
? Area of parm. ABCD = 2 × area of ' ABD = 2 × 126 cm2 = 252 cm2
Example 3: The adjoining field is in the shape of a quadrilateral D
with the longer diagonal 136 m. The lengths of the C
perpendiculars from the opposite vertices on this
diagonal are 15.4 m and 24.6 m. Find the area of the field. Q
P
Solution:
B
Let ABCD be the field and AC is it's longer diagonal. A
Here, BP A AC, DQ A AC, AC = 136 m, BP = 24.6 m and DQ = 15.4 m
Now, area of the field ABCD = Area of ( 'ABC + 'ACD)
= 1 AC × BP + 1 AC × DQ
2 2
1
= 2 AC(BP + DQ)
= 1 × 136 (24.6 + 15.4) = 2,720 m2
2
Hence, the required area of the field is 2,720 m2.
Example 4: The perimeter of a triangle is 48 cm. If the ratios of its sides are 4 : 5 : 3, find
its area.
Solution:
Here, the perimeter of the triangle = 48 cm
? The semi-perimeter of the triangle = 48 cm = 24 cm
2
Again, let the sides of the triangle are 4x cm, 5x cm and 3x cm.
? 4x + 5x + 3x = 48 cm
or, 12x = 48 cm
or, x = 48 cm = 4 cm.
12
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Mensuration (I): Area of Plane Surface
Now, 4x = 4 u 4 cm = 16 cm, 5x = 5 u 4 cm = 20 cm and 3x = 3 u 4 cm = 12 cm.
? a = 16 cm, b = 20 cm and c = 12 cm.
? Area of the triangle = s (s – a) (s – b) (s – c)
= 24 (24 – 16) (24 – 20) (24 – 12) = 24 × 8 × 4 × 12= 96 cm2.
Hence, the required area of the triangle is 96 cm2.
Example 5: Find the equal sides of an isosceles triangle whose area is 672 cm2 and the
base is 28 cm.
Solution:
Here, the area of the isosceles triangle is 672 cm2.
Let, two equal sides of the triangle are a and b and the base is c.
? The semi-perimeter of the triangle = a+b+c = a + a+ 28 = 2a + 28 = (a + 14) cm.
2 2 2
? s – a = a + 14 – a = 14 cm
s – b = a + 14 – a = 14 cm Alternative process
s – c = a + 14 – 28 = a – 14 cm 1
Area of an isosceles triangle = 4 b (4a2 – b2)
Now, the area of triangle = 672
or, 1 × 28 4a2 – 282 = 672
or, s (s – a) (s – b) (s – c) = 672 4
or, (a + 14) × 14 × 14 (a – 14) = 672 or, 4a2 – 784 = 96
or, 14 a2 – 196 = 672 or, 4a2 – 784 = 9216
or, a2 – 196 = 48 or, 4a2 = 10000
or, a2 = 2500
Squaring both sides, we get, or, a = 50 cm
( a2 – 196 )2 = (48)2
or, a2 – 196 = 2304
or, a2 = 2304 + 196
or, a2 = 2500
? a = 50 cm
Also, b = a= 50 cm.
So, two equal sides of the isosceles triangle are 50 cm each.
Example 6: A field is in the shape of a rhombus. The diagonals of the field are 350 m and
240 m respectively. Find the area of the field in hectare. (1 hectare = 10,000 m2)
Solution: DC
In a rhombus, its diagonals bisect each other perpendicularly. O
AB
? In ' ABC, BO A AC and in ' ADC, DO A AC
Now, the area of the rhombus ABCD = Area of ('ABC + 'ADC)
= 1 AC u BO + 1 AC u DO
2 2
= 1 AC (BO + DO)
2
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Mensuration (I): Area of Plane Surface
= 1 AC u BD
2
= 1 u 350 m u 240m = 42,000 m2
2
10,000 m2 = 1 hectare
? 42,000 m2 = 42,000 hectare = 4.2 hectares.
10,000
Hence, the required area of the field is 4.2 hectares.
Example 7: The perimeter of a right-angled triangle is 12 cm and its area is 6 cm2. Find
Solution: the sides of the triangle.
Let, p, b and h are the perpendicular, base and hypotenuse of the right-angled triangle.
Here, p + b + h = 12
or, p + b + p2 + b2 = 12
or, p2 + b2 = 12 – (p + b) hp
Squaring both sides, we get,
p2 + b2 = [12 – (p + b)]2
b
or, p2 + b2 = 122 – 2.12.(p + b) + (p + b)2
or, p2 + b2 = 144 – 24p – 24b + p2 + 2pb + b2
or, 24p + 24b – 2pb = 144 ............................ (i)
1
Again, the area of the right angled triangle = 2 base u perpendicular
or, 6 = 1 bup
2
or, pb = 12
Putting the value of pb in equation (i), we get,
24p + 24b – 2 u 12 = 144
or, 24p + 24b = 144 + 24
or, 24 (p + b) = 168
or, p + b = 7 ............................ (ii)
Again, we have, (p – b)2 = (p + b)2 – 4pb
= 72 – 4 u 12
= 49 – 48
=1
? p–b =1 ............................ (iii)
Now, adding equations (ii) and (iii), we get,
p+b+p–b =7+1
or, 2p = 8
or, p = 4
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Mensuration (I): Area of Plane Surface
Putting the value of p in equation (ii), we get,
4+b =7
or, b = 7 – 4 = 3
Also, h = p2 + b2 = 42 + 32 = 16 + 9 = 25 = 5
So, the required sides of the right angled triangle are 4 cm, 3 cm and 5 cm.
Example 8: The perimeter of a triangle is 24 cm. If its area is 24 cm2 and its one of the
three sides is 10 cm, find the sides of the triangle.
Solution:
Let the unknown sides of the triangle are a and b.
Here, the perimeter of the triangle = 24 cm
? The semi-perimeter of the triangle = 24 cm = 12 cm
2
Also, a + b + c = 24, then a + b = 24 – c = 24 – 10 = 14,
? b = 14 - a
Now, the area of the triangle = s (s – a) (s – b) (s – c)
or, 24 = 12 (12 – a) (12 – 14 + a) (12 – 10)
or, 24 = 12 (12 – a) (a – 2) 2
or, 576 = 24(12a – 24 – a2 + 2a)
or, a2 – 14a + 48 = 0
or, a2 – 8a – 6a + 48 = 0
or, (a – 8) (a – 6) = 0
? a = 8 or 4
When a = 8cm, b = 6 cm and when a = 6 cm, b = 8 cm.
Hence, the required sides of the triangle are 6 cm and 8 cm.
Example 9: An umbrella is made by stitching 8 triangular pieces 50 cm 50 cm 50 cm 50 cm
of cloth, each measuring 50 cm × 20 cm × 50 cm. If cm
the rate of the cloth is Rs 0.75 per sq. cm, find the
cost of the cloth required to make the umbrella. 50
Solution: 20 cm 20 cm 20 cm 20 cm
Here, two equal sides of each triangular piece of cloth (a) = 50 cm
Base of each triangular piece of cloth (b) = 20 cm
Now, the area of each isosceles triangular piece of cloth = 1 × b 4a2 – b2
4
1
= 4 × 20 4 × (50)2 – (20)2
= 5 10,000 – 400
= 5 × 40 6 cm2 = 200 6 cm2
? The area of 8 isosceles triangular pieces = 8 × 200 6 cm2
= 1600 × 2.45 cm2 = 3,920 cm2
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Mensuration (I): Area of Plane Surface
Now, the rate of cost of cloth = Rs 0.75 per sq. cm
? The total cost of the cloth = Area of cloth × rate
= 3,920 × Rs 0.75 = Rs 2,940
Hence, the cost of the cloth required to make the umbrella is Rs 2,940.
EXERCISE 5.1
General section
1. a) If the base and height of a triangle are p cm and q cm respectively, find its area.
b) If the base and perpendicular of a right-angled triangle are x cm and y cm
respectively, find its area.
c) The length of each of three equal sides of an equilateral triangle is k cm, find its
area.
d) The length of each of two equal sides of an isosceles triangle is m cm and the base
is n cm, find its area.
e) If the lengths of three sides of a triangle are x cm, y cm and z cm respectively, find
its area.
2. a) Find the area of the following triangles:
(i) A X
(ii) P (iii)
10 cm 15 cm
8 cm
14 cm
B 10Dcm C R 8 cm Q Y 13 cm Z
b) Find the perimeter and area of a triangle whose sides are of lengths 12 cm, 16 cm
and 20 cm.
c) Find the area of an equilateral triangle of each side 12 cm.
d) Each of two equal sides of an isosceles triangle is 6 cm and its base is 4 cm. Find
its area.
e) If the area of an equilateral triangle is 25 3 cm2, find the length of its side.
f) If the perimeter of an equilateral triangle is 12 cm, find its area.
g) The area of an isosceles triangle is 240 cm2. If the base of the triangle is 20 cm, find
the length of its equal sides.
Creative section - A
3. a) Derive that the area of an equilateral triangle is 34 (side)2.
b) Derive that the area of an isosceles triangle is 14 q 4p2 – q2 , where p is the length
of each of two equal sides and q is the length of the base.
c) If x, y, z are the three sides and s is the semi-perimeter of a triangle, derive that its
area is s(s – x) (s – y) (s – z).
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Mensuration (I): Area of Plane Surface
4. Find the area of the following figures.
a) D b) D 14 cm C c) D 21 cm C
15 cm C 20 cm
13 cm
15 cm
16 cm
12 cm
13 cm
A 14 cm B
A 20 cm B A 16 cm B
d) E 1 cm
e) F 6 6 cm E2 14cm f)
D
1 cm D 25 cm
C 25 cm
13 cm
1 cm 25 cm
A 1 cm B
25 cm
25 cm
25 cm
25 cm
C 14 cm14 cm 14 cm 14 cm 14 cm 14 cm
3 cm
A 4 cm B 21 cm
5. a) The adjoining figure is a trapezium ABCD. Find D C
(i) its area (ii) the length of sides BC and AD. 20 cm
b) In the given figure, ABCD is a piece of land. Find A 16 cm B
D 14 m C
(i) The lengths of sides BD and BC.
(ii) the area of the land. 12 m
A 9m B
D C
c) A piece of land given alongside is in the shape of a 39.8 m
quadrilateral with the longer diagonal 340 m. The 340 m
length of the perpendiculars from the opposite vertices
on this diagonal are 39.8 m and 40.2 m. Find the area of 40.2 m B
the land in hectares. (1 hectare = 10,000 m2) A
6. a) The shape of a piece of land is a parallelogram whose adjacent sides are 12 m and
9 m and the corresponding diagonal is 15 m. Find the area of the land.
b) Two adjacent sides of a parallelogram are 8 cm and 6 cm and its area is 48 cm2.
Find the length of its diagonal.
c) A rhombus-shaped piece of land has diagonals 20 m and 12 m. Find the area of the
land.
d) A garden is in the shape of a rhombus whose each side is 15 m and its one of two
diagonals is 24 m. Find the area of the garden.
7. a) If the ratio of the sides of a triangle are 13 : 14 : 15 and its perimeter is 84 cm, find
the area of the triangle.
b) The perimeter of a triangle is 84 cm and its area is 336 cm2. If one of its sides is
30 cm, find the length of other two sides.
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Mensuration (I): Area of Plane Surface
c) The perimeter of a triangular garden is 18 m. If its area is 135 m2 and one of the
three sides is 8 m, find the remaining two sides.
d) The perimeter of a right-angled triangle is 24 cm and its area is 24 cm2. Find the
sides of the triangle.
Creative section - B
8. a) An umbrella is made up of 6 isosceles triangular pieces of cloths. The measurement
of the base of each triangular piece is 28 cm and two equal sides are 50 cm each.
If the rate of cost of the cloth is Rs 0.50 per sq. cm., find the cost of the cloths
required to make the umbrella.
b) A floral design on a floor of a room is made up of 16 triangular
pieces of tiles. The sides of each piece of tile are 28 cm, 15 cm
and 41 cm. Find the cost of polishing the tiles at Rs 1.50 per
sq. cm.
c) A real estate agent divided the adjoining triangular 50 m 20 m 105 m
land into two 'Kitta' in which one is in the shape
of a trapezium. If he sold the shaded 'Kitta' at 32 m
Rs 2,10,000 per 'aana', how much did the buyer pay
for it? (1 aana = 31.79 sq. m)
85 m
d) ABCD is a field in the shape of a trapezium D 100 m C
whose parallel sides are 100 m and 40 m and the
nonparallel sides are 56 m and 52 m. Find the cost of 56 m
ploughing the field at Rs 7.50 per sq, m. 52 m
(Hint: Draw BE // AD to meet CD at E and BF A CD at F)
A 40 m B
Project work D
9. a) The given quadrilateral ABCD
is a kitchen garden. Measure
the length of it's each side and
the diagonal in centimeters by
using a ruler. If the scale of the A
drawing is 1:500, find the area C
of the garden.
B
b) Measure the dimensions of your school compound or playground and find its
perimeter and area.
c) Cut chart papers into a few number of triangles and quadrilateral shapes. Measure
the dimensions of each triangular and quadrilateral piece. Then, find the area of
each shape.
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Unit Mensuration (II): Cylinder and Sphere
6
6.1 Surface area and volume of solid objects
Solid objects are three dimensional (3-D) objects. The surface area of a solid object is a
measure of the total area that the surface of the object occupies.
The volume of a solid object is a measure of the total space that the whole body of the
object occupies. In the case of a regular solid object, its volume is calculated as the
product of the area of base and the height of the object. Volume is measured in cubic
units.
The table given below shows the area and volume of cylinder and sphere at a glance.
Here, C.S.A. and T.S.A. represents the curved surface (or lateral surface) area and total
surface area respectively.
Name of Diagram C. S. A. or T. S. A. Volume
solids L. S. A.
r
Cylinder h 2Srh 2Sr(r+ h) Sr2h
Hollow Rr 2Sh(R + r) 2S(R+r)(h+R–r) Sh(R + r) (R – r)
cylinder h
Half cylinder d=2r 1 Sr2h
2
(A horizontal h h rh(S + 2) rh(S + 2) + Sr2
cylindrical
segment)
Sphere Surface area = 4Sr2 4 Sr3 or
Hemisphere 3
1 Sd3
6
r 2
3
2Sr2 3Sr2 Sr3 or
1 Sd3
12
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Mensuration (II): Cylinder and Sphere
6.2 Area and volume of a cylinder
Let's take a rectangular sheet of paper and bend it as shown in the figure. In this way,
we will get a curved surface figure with two circular faces. It is called a cylinder.
From the above figures, it is clear that,
the curved surface area of a cylinder = 2Srh
the total surface area of a cylinder = 2Srh + Sr2 + Sr2 = 2Srh + 2Sr2 = 2Sr (r + h)
Where, r is the radius of a circular face and h is the height of the cylinder.
Similarly, the volume of a cylinder = Area of circular base × height
= Sr2 × h = Sr2h.
6.3 Curved surface area, total surface area, and volume of a hollow
cylinder
Rr
The figure alongside is a hollow cylinder (or a pipe). Let R be the
external radius, r the internal radius and h be the height of the hollow
cylinder.
Here, the external curved surface area = 2SRh h
The internal curved surface area = 2Srh
? The total curved surface area = 2SRh + 2Srh = 2Sh(R + r)
Again, the area of a ring-shaped circular base = SR2 – Sr2 = S(R + r) (R – r)
? The area of two ring-shaped circular bases = 2S(R + r) (R – r)
Now, the total surface area of a hollow cylinder = 2Sh(R + r) + 2S(R + r) (R – r)
= 2S(R + r) (h + R – r)
Volume of material contained by a hollow cylinder:
From the above diagram,
The external volume of the cylinder = SR2h
The internal volume of the cylinder = Sr2h
? Volume of the material contained by the cylinder = SR2h – Sr2h
= Sh(R2 – r2)
= Sh (R + r) (R – r)
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Mensuration (II): Cylinder and Sphere
6.4 Half cylinder or semicylinder d=2r
hh
When a cylinder is cut into half along it's axis it is called a
half cylinder or a semicylinder. It is a horizontal cylindrical
segment.
A half cylinder has a curved surface, a rectangular flat
surface and two semi-circular bases.
Length of the rectangular flat surface = height = h
Breadth of the rectangular flat surface = diameter of circular base = 2r
? Area of the rectangular flat surface = length × breadth = h × 2r = 2rh
Also, the area of curved surface of a half cylinder = 1 × 2Srh = Srh
2
1
And, the area of two semi-circles = 2 × 2 × Sr2 = Sr2
Now, lateral surface area of a half cylinder = Area of (curved surface + flat surface)
= Srh + 2rh
= rh(S + 2)
Total surface area of a half cylinder = lateral surface area + area of 2 semi-circles
= rh(S + 2) + Sr2
Also, volume of a half cylinder = 1 Sr2h
2
6.5 Area and volume of a sphere
A sphere is a perfectly round geometrical object in r
three-dimensional space. It is generated by revolving a
circular lamina about any of it's diameters. A football, a
cricket ball, etc. are said to have the shape of a sphere.
Total surface area of a sphere
Let's take a hollow tennis ball and cut off it into two equal halves. Again, cut off each
half in such a way that they can be flattened and spread on the squared graph paper and
find the approximate area of each half. Now, the sum of the areas of two halves is the
surface area of the sphere.
Again, divide this area by the square of its diameter. It will be found that the value so
obtained is approximately equal to 22 or S.
7
Thus, surface area of a sphere = S.
(diameter)2
or, A = S.
d2
or, A = Sd2 = S × (2r)2 = = 4Sr2
? The surface area of a sphere = Sd2 or 4Sr2.
Where, d is the diameter and r is the radius of the curved surface of the sphere.
Vedanta Excel in Mathematics - Book 10 88 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Mensuration (II): Cylinder and Sphere
Alternative process to verify surface area of a sphere = 4Sr2 r h=2r
Let the radius of the given sphere is equal to the radius of the cylinder
and the diameter of the sphere is equal to the height of the cylinder.
Then, the sphere could easily be fitted in the cylinder.
Here, surface area of the sphere = curved surface area of the cylinder
= 2Srh
Now, let's replace h by 2r, we get surface area of the sphere = 2Sr × 2r = 4Sr2.
Volume of a sphere
Let's place a spherical object on a meter scale and press it from two
sides by two wooden cubes as shown in the figure. Now, find the
diameter of the sphere in the ruler.
Suppose, its diameter is 7 cm.
Take a measuring cylinder filled with water and mark the water
level. Tie the sphere with a thread and immerse completely into
water and find the volume of the sphere.
Volume of sphere = final water level – initial water level 200cc
100cc
Here, diameter (d) = 7 cm (say)
Volume (V) = 179.5
V = 179.5
d 7
6V = 179.5 × 6 = 3.14 (Approx.) = 22 (Approx.) = S
d3 73 7
Thus, 6V =S
or, d3
6V = Sd3
or, V = Sd3
6
Also, d = 2r, so V = S(2r)3 = 4 Sr3
6 3
Thus, the volume of a sphere = Sd3 or 4 Sr3
6 3
6.6 Hemisphere and great circle r
When a sphere is cut into two halves, one of the half
portions is called hemisphere. A hemisphere has a
plane circular face and a curved surface. The radius
of the circular face is equal to the radius of the sphere.
The circular face of a hemisphere is called a great circle.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 89 Vedanta Excel in Mathematics - Book 10
Mensuration (II): Cylinder and Sphere
The area of a great circle of a sphere = Sr2 or Sd2
4
The circumference of the great circle = 2Sr or Sd
Also, the surface area of a sphere = 4Sr2 1 × 4Sr2
? The curved surface area of a hemisphere = 2
= 2Sr2 or S d2
2
And, the total surface area of a hemisphere = Area of curved surface + Area of great circle
= 2Sr2 + Sr2
= 3Sr2 or 3Sd2
4
1
Similarly, the volume of a hemisphere = 2 of the volume of sphere
= 1 × 4 Sr3
2 3
2 Sd3
= 3 Sr3 or 12
6.7 Surface area and volume of a hollow hemispherical object
The object alongside is a hollow hemispherical vessel. R r
Let the outer radius of the vessel = R
The inner radius of the vessel = r
Now, area of the outer surface of the vessel = 2SR2
And, area of the inner surface of the vessel = 2Sr2
Also, the area of the ring at the top = SR2 – Sr2 R
Then, the surface area of the vessel = 2SR2 + 2Sr2 + SR2 – Sr2
= 3SR2 + Sr2
= S(3R2 + r2)
Again, the external volume of a hollow hemisphere = 2 SR3 r
3
2
The internal volume of the hollow hemisphere = 3 Sr3
Then, volume of material contained by the hollow hemisphere = 2 SR3 – 2 Sr3
3 3
2
= 3 S(R3 – r3)
6.8 Volume of material contained by a hollow sphere
The hollow spherical shell is like a football, basketball, etc. Let R be the external
radius and r the internal radius of a hollow spherical shell.
Then, the external volume of the shell = 4 SR3
3
The internal volume of the shell = 4 Sr3
3
? The volume of the material contained by the shell = 4 SR3 - 4 Sr3
3 3
= 4 S (R3 – r3)
3
Vedanta Excel in Mathematics - Book 10 90 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Mensuration (II): Cylinder and Sphere
6.9 Area and Volume of cylinder having hemispherical ends 2Srh
The figure given alongside is a cylinder with two hemispherical
ends.
Let r be the radius of the cylinder as well as of the hemisphere
and h be the height of the cylinder.
Here, the curved surface area of the cylinder = 2Srh
The curved surface area of two hemispheres = 2 × 2Sr2 = 4Sr2
? The total surface area of the solid = 2Srh + 4Sr2
= 2Sr (h + 2r)
Similarly, volume of the solid = Volume of cylinder + Volume of 2 hemispheres
= Sr2h + 4 Sr3
3
Worked-out examples
Example 1: The diameter of the circular base of a solid cylindrical rod is 2.8 cm and it
is 50 cm long. Find
(i) its curved surface area (ii) total surface area (iii) volume
Solution:
Here, diameter of the cylindrical rod (r) = 2.8 cm
So, the radius of the rod (cry)l=ind22e.8r = 1.4 cm
Length (or height) of the (h) = 50 cm
(i) Now, the curved surface area = 2Srh = 2 × 22 × 1.4 × 50 cm2
7
= 440 cm2
(ii) Also, the total surface area = 2Sr (r + h)
= 2 × 22 × 1.4 (1.4 + 50) cm2
7
= 452.32 cm2
(iii) And, the volume of the cylinder = Sr2h
= 22 × 1.4 × 1.4 × 50 cm3 = 308 cm3
7
Example 2: Find the (i) area of curved surface (ii) total surface area and (iii) volume of a
cylinder having the perimeter of the circular base 44 cm and height 40 cm.
Solution:
Here, perimeter of circular base = 44 cm
or, 2Sr = 44 cm
or, r = 7 cm
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 91 Vedanta Excel in Mathematics - Book 10
Mensuration (II): Cylinder and Sphere
(i) Now, the curved surface area = 2Srh
= 2 × 22 × 7 × 40 cm2 = 1,760 cm2
7
(ii) Also, total surface area = 2Sr (r + h)
= 2 × 22 × 7 (7 + 40) cm2 = 2,068 cm2
7
(iii) And, the volume of the cylinder = Sr2h
= 22 × 7 × 7 × 40 cm3 = 6,160 cm3
7
Example 3: The total surface area of a cylinder of radius 7 cm is 1,100 cm2. Find the
volume of the cylinder.
Solution:
Here, radius of the cylinder (r) = 7 cm and the total surface area (T. S. A.) = 1,100 cm2
Now, the total surface area of cylinder = 1,100 cm2
or, 2Sr(r + h) = 1,100 cm2
or, 2× 22 × 7(7 + h) = 1100 cm2
7
or, 7 + h = 25 cm
or, h = 25 – 7 = 18 cm
Again, volume of the cylinder = Sr2h
= 22 × 7 × 7 × 18 = 2,772 cm3
7
Hence, the required volume of the cylinder is 2,772 cm3.
Example 4: A roller of diameter 100 cm and length 140 cm takes 450 complete
Solution: revolutions to level a playground. Calculate the cost of levelling the ground
at Rs 7.50 per square metre.
Here, diameter of the roller = 100 cm ? Radius of the roller (r) = 100 = 50 cm
2
Length of the roller (h) = 140 cm
Now, area covered by the roller in 1 revolution = C. S. A. of the roller
= 2Srh
= 2 ×272 × 50 × 140 = 44,000 cm2
? Area covered by the roller in 450 revolutions = 450 × 44,000 cm2
= 450 × 44,000 m2 = 1,980 m2
100 × 100
? Area of the playground = 1,980 m2
Then, cost of levelling the playground = 1,980 × Rs 7.50 = Rs 14,850
Hence, the required cost of levelling the playground is Rs 14,850.
Vedanta Excel in Mathematics - Book 10 92 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Mensuration (II): Cylinder and Sphere
Example 5: The external and the internal radii of a hollow cylindrical pipe 70 cm long
are 5.4 cm and 5.1 cm respectively. Find (i) the volume of the material
contained by the vessel and (ii) the cost of polishing the whole pipe at
10 paisa per square cm.
Solution:
Here, the external radius of the pipe (R) = 5.4 cm
The internal radius of the pipe (r) = 5.1 cm
The height of the pipe (h) = 70 cm
Now, the volume of the material = Sh(R + r) (R – r)
= 22 × 70 (5.4 + 5.1) (5.4 – 5.1) cm3
7
= 693 cm3
Hence, the required volume of the material contained by the vessel is 693 cm3.
Again, the T.S.A. of the hollow pipe = 2S(R + r) (h + R – r)
= 2 × 22 (5.4 + 5.1) (70 + 5.4 – 5.1)
7
= 4,639.8 cm2
Now, the cost of polishing the pipe = 4,639.8 × 10 paisa = 46,398 paisa = Rs 463.98
Hence, the required cost of polishing the pipe is Rs 463.98.
Example 6: The internal diameter and height of a cylindrical bucket are 14 cm and 35 cm
respectively and it is filled with water completely. If the water is poured into
a rectangular glass tray with internal length 28 cm and breadth 17.5 cm and
it is completely filled with water, find the height of the tray.
Solution:
Here, the diameter of the cylindrical bucket = 14 cm. So, radius (r) = 7 cm
Height of the bucket (h) = 35 cm
Now, the internal volume of the bucket = Sr2h = 22 × 7 × 7 × 35 = 5,390 cm3
7
Thus, the volume of water = internal volume of bucket = 5,390 cm3
Again, volume of the rectangular glass tray = volume of water
l × b × h = 5,390
or, 28 × 17.5 × h = 5,390
or, h = 5,390 = 11 cm
28 × 17.5
Hence, the height of the rectangular glass tray is 11 cm.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 93 Vedanta Excel in Mathematics - Book 10
Mensuration (II): Cylinder and Sphere
Example 7: If the circumference of the great circle of a sphere is 66 cm, calculate,
(i) the surface area of the sphere (ii) the volume of the sphere
Solution:
Here, the circumference of the great circle = 66 cm
or, 2Sr = 66 cm
or, 2 × 22 × r = 66 cm
7
or, r = 10.5 cm
? The radius of the sphere = the radius of the great circle = 10.5 cm.
(i) Now, the surface area of the sphere = 4 Sr2
= 4 × 22 × 10.5 × 10.5 cm2 = 1,386 cm2
7
4 4 22
(ii) Again, the volume of the sphere = 3 Sr3 = 3 × 7 × 10.5 × 10.5 × 10.5 cm3
= 4,851 cm3
Hence, the required surface area of the sphere is 1,386 cm2 and its volume is 4,851 cm3.
Example 8: Find the curved surface area, total surface area and volume of a hemisphere
of diameter 14 cm.
Solution: 14
2
Here, the diameter of the hemisphere = 14 cm, so, its radius = = 7 cm
Now, the curved surface area of the hemisphere = 2Sr2
= 2 × 22 × 7 × 7
7
= 308 cm2
Also, the total surface area of the hemisphere = 3Sr2
= 3 × 22 × 7 × 7
7
= 462 cm2
And, the volume of the hemisphere = 2 Sr3
3
2 22
= 3 × 7 × 7 × 7 × 7 = 718.66 cm3
Hence, the curved surface area of the hemisphere is 308 cm2, the total surface area is
462 cm2 and the volume is 718.66 cm3.
Example 9: Three metallic spheres of radii 1 cm, 6 cm, and 8 cm are melted and
reformed into a single sphere. Find the radius of the new sphere.
Solution:
Here, the radius of the first sphere (r1) = 1 cm
= 6 cm
The radius of the second sphere (r2) = 8 cm
The radius of the third sphere (r3)
Now, the volume of the first sphere (V1) = 4 Sr13 = 4 S × 13
3 3
Vedanta Excel in Mathematics - Book 10 94 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Mensuration (II): Cylinder and Sphere
The volume of the second sphere (V2) = 4 Sr23 = 4 S × 63
The volume of the third sphere 3 3
4 4
(V3) = 3 Sr33 = 3 S × 83
Again, the volume of the new single sphere (V)= V1 + V2 + V3
or, 4 Sr3 = 4 S × 13 + 4 S × 63 + 4 S × 83
3 3 3 3
or, 4 Sr3 = 4 S (13 + 63 + 83)
3 3
or, r3 = 729
? r = 9 cm
Hence, the radius of the new sphere is 9 cm.
Example 10: A solid metallic sphere of radius 4.5 cm is melted and drawn into a
cylindrical wire of diameter 5 mm. Find the length of the wire.
Solution:
Here, the radius of the cylindrical wire (r) = 5 mm = 2.5 mm = 0.25 cm
2
Now, the volume of the sphere = 4 SR3 = 4 S × (4.5 cm)3 = 121.5 S cm3
3 3
Also, the volume of the cylindrical wire= Sr2h = S × (0.25 cm)2 h = 0.0625 Sh cm2
The volume of the cylindrical wire = the volume of the sphere
or, 0.0625 Sh cm2 = 121.5 S cm3
or, h = 121.5 S cm = 1944 cm = 19.44 m
0.0625 S
Hence, the required length of the wire is 19.44 m.
Example 11: The figure alongside is a cylinder with two r=3.5cm
hemispherical ends. Calculate:
40cm
(i) the surface area and (ii) the volume of the solid.
Solution:
Here, the area of the two hemispherical ends = 2 × 2Sr2
= 4 × 22 × 3.5 × 3.5 cm2 = 154 cm2
7
22
Again, the curved surface area of the cylinder = 2Srh = 2 × 7 × 3.5 × 40 cm2 = 880 cm2
? The total area of the solid = 154 cm2 + 880 cm2 = 1034 cm2
Now, the volume of the solid = volume of cylinder + volume of 2 hemisphere
= Sr2h + 4 Sr3
3
22 4 22
= 7 × 3.5 × 3.5 × 40 cm3 + 3 × 7 × 3.5 × 3.5 × 3.5 cm3
= 1,540 cm3 + 179.67 cm3 = 1,719.67 cm3.
Hence, the required area of the solid is 1,034 cm2, and its volume is 1,719.67 cm3.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 95 Vedanta Excel in Mathematics - Book 10
Mensuration (II): Cylinder and Sphere
Example 12: A water tank is made up of the combination of a cylinder and a hemisphere.
The height of the tank is 12.1 m and the area of the base is 13.86 m2. If the
tank is filled with water at the rate of 50 paisa per litre, find the cost of
filling the tank with water.
Solution: 2.1 m
Here, area of the circular base of the tank = 13.86 m2
or, Sr2 = 13.86 m2 12.1 m
10 m
or, 22 × r2 = 13.86 m2
7
or, r2 = 4.41 m2
or, r = 2.1 m
Now, volume of the tank
= Volume of the cylinder + volume of the hemisphere Direct process
= Sr2h + 2 Sr3 V = Sr2h + 32Sr3 = Sr2h + 23Sr2.r
3
2 22 2
= Sr2(h + 3 × r) = 7 × 2.1 × 2.1 × (10 + 3 × 2.1) = 13.86 × 10 + 2 . 13.86 × 2.1
= 158.004 m3 3
= 13.86 × 11.4 = 158.004 m3
We know that, 1m3 = 1000 litres
158.004m3 = 158.004 × 1000 litres = 1,58,004 litres
Now, the cost of filling water = 1,58,004 × 50 paisa
= 79,00,200 paisa = Rs 79,002
Hence, the cost of filling the tank with water is Rs 79,002.
EXERCISE 6.1
General section
1. a) If x be the radius and y be the height of a cylinder, write the formulae to find it's
(i) curved surface area (C.S.A.) (ii) Total surface area (T.S.A.) (iii) volume
b) If p be the radius of a sphere, write the formula to find its (i) surface area
(ii) volume
c) A and a are the external and internal radii and b be the height of a hollow cylinder
respectively. Write the formula to find it's (i) curved surface area (C.S.A.) (ii) total
surface area (T.S.A.) (iii) volume of material contained by the cylinder. d=2 m
d) Write the formulae to find (i) C.S.A. (ii) T.S.A. and
(iii) volume of the given half-cylinder.
lm
e) If the external and internal radii of a hollow spherical shell are x cm and y cm
respectively write the formula to find the volume of material contained by the
shell.
x cm
f) Write the formula to find (i) C.S.A. (ii) T.S.A. and
(iii) Volume of the given hemisphere.
Vedanta Excel in Mathematics - Book 10 96 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Mensuration (II): Cylinder and Sphere
x y
g) Write the formulae to find (i) the surface area and (ii) volume
of the given hollow hemispherical shell.
2. a) The radius of the circular base of a cylinder is 7 cm and its height is 15 cm. Find
its (i) curved surface area (C.S.A) (ii) total surface area (T.S.A) (iii) volume
b) If the diameter of a solid cylinder with length 70 cm is 14 cm, find its curved
surface area, total surface area and volume.
3. a) Find the (i) curved surface area (C.S.A) (ii) total surface area (T.S.A.) and
(iii) volume of a cylinder with height 30 cm and the perimeter of the circular base
is 88 cm.
b) If the circumference of the base of cylinder is 44 cm and the sum of its radius and
height is 27 cm, find its total surface area.
c) The sum of the height and the radius of the base of a cylinder is 34 cm. If the total
surface area of the cylinder is 2,992 cm2, find the radius.
d) Find the radius of a cylinder if its curved surface area and height are 616 cm2 and
14 cm respectively.
e) If the volume of a cylinder 30 cm high is 4,620 cm3, find its radius.
4. a) If the volume of a cylindrical solid is 1,320 m3 and base area is 264 m2, find its
height.
b) The volume of a cylindrical vessel is 1.54 litre and area of its base is 77 cm2, find
its height.
c) Find the height of a cylindrical water vessel of diameter 1.4 m if it holds 770 l of
water.
d) A cylindrical water tank contains 4,62,000 litres of water and its radius is 3.5 m,
find the height of the tank.
e) The volume of a cylinder is 539 cm3 and the circumference of the base is 22 cm,
find its height.
f) Find the height of a cylinder whose diameter is 42 cm and the area of its curved
surface is 11,880 sq. cm.
g) If the height and radius of a cylinder are equal and its curved surface area is
308 cm2, find its height.
5. a) The volume of a cylindrical tank is 936 cubic meter. If its height is 6 m, determine
the base area of the tank.
b) If the volume of a cylindrical can is 1.54 litre and the height is 20 cm, find the area
of its base.
6. a) If the radius of a sphere is 21 cm, find its surface area and volume.
b) The diameter of a sold spherical ball is 35 cm, find its surface area and the volume.
c) Find (i) curved surface area (ii) the total surface area and 21 cm
(iii) volume of the adjoining solid hemisphere.
d) If the perimeter of the great circle is S cm, find the volume of the hemisphere.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 97 Vedanta Excel in Mathematics - Book 10
Mensuration (II): Cylinder and Sphere
e) If the circumference of the great circle is 44 cm, find the total surface area of the
hemisphere.
f) The total surface area of a hemisphere is 243 S cm2, find its volume.
7. a) A solid metallic sphere of radius 7 cm is cut into two halves. Find the total surface
area of the two hemisphere so formed.
b) A solid metallic sphere of diameter 42 cm is cut into two halves. Find the total
surface area of the two hemispheres so formed.
8. a) If the volume of a spherical object is 9S cm3, find its diameter.
2
b) If the total surface area of a solid sphere is 616 cm2, what will be its radius?
c) If the volume of a hemisphere is 19,404 cm3, find its radius.
Creative section - A
9. a) The perimeter of the circular base of a cylinder is 132 cm and it is 40 cm high.
Find the following.
(i) its curved surface area, (ii) total surface area (iii) volume
b) If the curved surface area of a cylinder with height 15 cm is 1,320 cm2, find
(i) its total surface area (ii) volume
c) The curved surface area of a cylinder is 1,320 cm2 and the circumference of its
base is 66 cm. Find the total surface area and volume of the cylinder.
d) The volume of a cylindrical can is 3.08 litre. If the area of its base is 154 cm2, find
its curved surface area.
e) The total surface area of a cylindrical can, whose height is equal to the radius of
10. a) the base is 2,464 cm2, find its volume.
The area of curved surface of a solid cylinder is equal to 2 of the total surface area.
3
If the total surface area is 924 cm2, find the volume of the cylinder.
b) The number of volume of a cylinder is half of its number of the total surface area.
If the radius of its base is 7 cm, find the volume.
c) By how many times does the volume of a cylinder increase when its diameter is
doubled?
11. a) The sum of the radius of the base and the height of a solid cylinder is 37 cm. If the
total surface area of the cylinder is 1,628 cm2, calculate its curved surface area and
volume.
b) The circumference of the base of a cylindrical drum is 44 cm and the sum of its
radius and height is 27 cm, find the volume of the cylinder.
12. a) A hollow cylindrical metallic pipe is 21 cm long. If the external and internal
diameters of the pipe are 12 cm and 8 cm respectively, find the volume of metal
used in making the pipe.
b) The external and the internal radii of a cylindrical pipe of length 50 cm are 3 cm
and 2.6 cm respectively. Find the volume of material contained by the pipe.
13. a) The internal radius of a cylindrical bucket of height 50 cm is 21 cm. It is filled with
water completely. If the water is poured into a rectangular vessel with internal
length 63 cm and breadth 44 cm and it is completely filled with water, find the
height of the vessel.
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