RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

Other Exercises

Question 1.
Write each of the following as percent: Solution—
(i) \(\frac { 7 }{ 25 }\)
(ii) \(\frac { 16 }{ 625 }\)
(iii) \(\frac { 5 }{ 8 }\)
(iv) 0.8
(v) 0.005
(vi) 3 : 25
(vii) 11 : 80
(viii) 111 : 125
(ix) 13 : 75
(x) 15 : 16
(xi) 0.18
(xii) \(\frac { 7 }{ 125 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 1
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 2

Question 2.
Convert the following percentages to fractions and ratios :
(i) 25%
(ii) 2.5%
(iii) 0.25%
(iv) 0.3%
(v) 125%
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 3
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 4

Question 3.
Express the following as decimal fractions :
(i) 27%
(ii) 6.3%
(iii) 32%
(iv) 0.25%
(v) 7.5%
(vi) \(\frac { 1 }{ 8 }\) %
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 5

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RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4

Other Exercises

Question 1.
The present population of a town is 28,000. If it increases at the rate of 5% per annum, what will be its population after 2 years ?
Solution:
Present population = 28000
Rate of increase (R) = 5% p.a.
Period (n) = 2 years
Population after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 1

Question 2.
The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.
Solution:
Present population = 125000
Rate of birth = 5.5%
and rate of death = 3.5%
Increase = 5.5 – 3.5 = 2% p.a.
Period = 3 years.
Population after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 2
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 3

Question 3.
The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
Solution:
Present population = 25000
Increase in first year = 4%
in second year= 5% and
in third year = 8%
Population after 3 years =
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 4

Question 4.
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.
Solution:
Three years ago,
Population of a town = 50000
Annual increase in population in first year = 4%
in second year = 5%
and in third year = 3%
Present population
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 5

Question 5.
There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago ?
Solution:
Let 3 years ago, population = P
Present population = 9261
Rate of increase (R) = 5% p.a.
Period (n) = 3 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 6

Question 6.
In a factory, the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.
Solution:
Production of scooters 3 years ago (P) = 40000
Present production (A) = 46305
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 7
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 8

Question 7.
The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago ?
Solution:
Let 3 years ago, the population of a city = P
Rate of growth (R) = 8% p.a.
Present population = 196830
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 9

Question 8.
The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.
Solution:
Population after 2 years = 22050
Rate of increase = 50 per thousand
Period (n) = 2 years
Let present population = P, then
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 10

Question 9.
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours ?
Solution:
Present count of bacteria = 13125000
In first hour increase = 10%
decrease in second hour = 8%
increase in third hour = 12%
Count of bacteria after 3 hours
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 11

Question 10.
The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000 ?
Solution:
On the last day of 1998,
Population of a town = 72000
In the first year, increase = 7%
In the second year, decrease = 10%
Population in the last day of 2000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 12

Question 11.
6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year ?
Solution:
Number of workers at the beginning = 6400
Period = 4 years.
At the end of 1st year, workers retrenched = 25%
At the end of second year, workers retrenched = 25%
At the end of third year, workers increased = 25%
Total number of workers during the 4 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 13

Question 12.
Aman started a factory with an initial investment of Rs 1,00,000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.
Solution:
Initial investment = Rs 100000
Loss in the first year = 5%
Profit in the second year = 10%
Profit in the third year = 12%
Investment at the end of 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 14

Question 13.
The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago ?
Solution:
Present population (A) = 175760
Increase rate = 40 per 1000
Period = 3 years
Let 3 years ago,
population was = P
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 15

Question 14.
The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years ?
Solution:
Production of Mixi in 1996 = 8000
Increase in next 2 years = 15%
Decrease in the third year = 5%
Production after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 16

Question 15.
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.
Solution:
Population of a city in 1999 = 6760000
Increase = 4%
(i) Population in 2001 is after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 17
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 18

Question 16.
Jitendra set up a factory by investing Rs 25,00,000. During the first two successive years his profits were 5% and f 10% respectively. If each year the profit was on previous year’s capital, compute his total profit.
Solution:
Investment in the beginning = Rs 25,00,000
Profit during the first 2 years = 5% and 10% respectively
Investment after 2 years will be
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 19
= Rs 28,87,500
Amount of profit = Rs 28,87,500 – Rs 25,00,000 = Rs 3,87,500

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3

Other Exercises

Find each of the following products (1-8)
Question 1.
5x2 x 4x3
Solution:
5x2 x 4x3 = 5 x 4 x x2 x x3
= 20x2 + 3 = 20xs

Question 2.
3a2 x 4b4
Solution:
-3a2 x 4b4 = -3 x 4 x a2b4
= -12a2b4

Question 3.
(-5xy) x (-3x2yz)
Solution:
(-5xy) x (-3x2yz)
= (-5) x (-3)xy x x2yz
= 15x1 + 2xy1+ 1z= 15x3y2z

Question 4.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 1
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 3

Question 5.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 4
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 5

Question 6.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 6
Solution:

Question 7.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 8
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 9

Question 8.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 10
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 11

Find each of the following products : (9-17)

Question 9.
(7ab) x (-5ab2c) x (6abc2)
Solution:
(7ab) x (-5ab2c) x (6abc2)
= 7 x (-5) x 6 x a x a x a x b x b2 x b x c x c2
=-210 x a1+1+1 x b1+2+1x c1+2
=-210 x a3b4c3

Question 10.
(-5a) x (-10a2) x (-2a3)
Solution:
(-5a) x (-10a2) x (-2a3)
= (-5) (-10) (-2) x a x a2 x a3
= -100a1 + 2 + 3 = -100a6

Question 11.
(-4x2) x (-6xy2) x (-3yz2)
Solution:
(-4x2) x (-6xy2) x (-3yz2)
= (-4) x (-6) x (-3) x2 x x x y2 x y xz2
= -72x2+1 x y2+1 x z2
= 72x3y3z3

Question 12.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 12
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 13

Question 13.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 14
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 15

Question 14.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 16
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 17

Question 15.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 18
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 19

Question 16.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 20
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 21
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 22

Question 17.
(2.3xy) x (0.1x) x (0.16)
Solution:
(2.3xy) x (0.1x) x (0.16)
= 2.3 x 0.1 x 0.16 x x x x x y
= 0.0368x1 +1 x y = 0.0368x2y

Express each of the following products as a monomials and verify the result in each case for x = 1 : (18 -26)

Question 18.
(3x) x (4x) x (-5x)
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 23

Question 19.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 24
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 25
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 26

Question 20.
(5x4) x (x2)3 x (2x)2
Solution:
(5x4) x (x2)3 x (2x)2
= 5x4 x x2 x 3 x 2x x 2x
= 5x4 * x6 x 4x2 = 5 x 4 x x4 + 6 + 2
= 20x12
Verification:
L.H.S. = (5x4) x (x2)3 x (2x)2
= 5 x (1)4 x [(1)2]3 x (2 x 1)2
= 5 x 1 x (1)2 x 3x (2)2
= 5 x 16 x 22 = 5 x 1 x 4 = 20
R.H.S. = 20x12 = 20 (1)12 = 20 x 1 = 20
∴ L.H.S. = R.H.S.

Question 21.
(x2)3 x (2x) x (-4x) x 5
Solution:
(x2)3 x (2x) x (-4x) x (5)
= x2 x 3 X 2x X (-4x) X 5
= x6 x 2x x (-4x) x 5 = 2 x (-4) x 5x6+1 +1
= -40x8
Verification
L.H.S. = (x2)3 x (2x) x (-4x) x (5)
= (12)3 x (2 x 1) x (-4 x 1) x 5
= 1x 2 x (- 4) x 5 = 16 x 2 x (-4) x 5
= 1 x 2 x (-4) x 5 = -40
R.H.S. = -40x8 = -40 x (1)8
= -40 x 1 = -40
∴ L.H.S. = R.H.S.

Question 22.
Write down the product of -8x2y6 and – 20xy Verify the product for x = 2.5, y = 1.
Solution:
Product of -8x2y6 and -20xy
= (-8x2y6) x (-20xy)
= -8 x (-20) x2 x x x y6 x y = 160x2 + 1 x y6 + 1
= 160x3y3
Verification.
L.H.S. = (-8x2y6) x (-20xy)
= -8 x (2.5)2 x (1) x (-20 x 2.5 x 1)
= -8 x 6.25 x 1 x -20 x 2.5
= (-50) x (-50) = 2500
R.H.S. = 160 x = 160 (2.5)3 x (1)7
= 160 x 15.625 x 1 =2500
∴ L.H.S. = R.H.S.

Question 23.
Evaluate : (3.2x6y3) x (2.1x2y2) when x = 1 and y = 0.5.
Solution:
3.2x6y3 x 2.1x2y2
= 3.2 x 2.1 x x6+2 x y3+2
= 6.72x8y5 = 6.72 x (1)8 x (0.5)5
= 6.72 x 1 x 0.03125
= 0.21

Question 24.
Find the value of (5x6) x (-1.5x2y3) x (-12xy2) when x = 1, y = 0.5.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 27

Question 25.
Evaluate : (2.3a5b2) x (1.2a2b2) when a = 1, b = 0.5.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 28

Question 26.
Evaluate : (-8x2y6) x (-20xy) for x = 2.5 and y = 1.
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 29

Express each of the following products as a monomials and verify the result for x = 1,y = 2: (27-31)

Question 27.
(-xy3) x (yx3) x (xy)
Solution:
(-xy3) x (yx3) x (xy)
= -x x xx x x yx y x y = -x1 + 3 + 1 x y3 + 1 + = -x5y5
Verification:
L.H.S. = (-xy3) x (yx3) x (xy)
= (-1 x 23) x [2 x (1)3] x (1 X 2)
= (-1 x 8) x (2 x 1) x (1 x 2)
= -8 x 2 x 2 = -32
R.H.S. =-x5y5  = -(1)5 (2)5
= -1 x 32 =-32
∴ L.H.S. = R.H.S.

Question 28.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 30
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 31
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 32

Question 29.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 33
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 34

Question 30.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 35
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 36

Question 31.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 37
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 38
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 39

Question 32.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 40
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 41
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 42

Question 33.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 43
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 44

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

Other Exercises

Making use of the cube root table, find the cubes root of the following (correct to three decimal places) 
Question 1.
7
Solution:
\(\sqrt [ 3 ]{ 7 }\) =1.913 (From the table)

Question 2.
70
Solution:
\(\sqrt [ 3 ]{ 70 }\) =4.121 (From the table)

Question 3.
700
Solution:
\(\sqrt [ 3 ]{ 700 } =\sqrt [ 3 ]{ 7\times 100 } \)= 8.879 (from \(\sqrt [ 3 ]{ 10x }\))

Question 4.
7000
Solution:
\(\sqrt [ 3 ]{ 7000 } =\sqrt [ 3 ]{ 70\times 100 }\) = 19.13 (from \(\sqrt [ 3 ]{ 100x }\))

Question 5.
1100
Solution:
\(\sqrt [ 3 ]{ 1100 } =\sqrt [ 3 ]{ 11\times 100 }\) = 10.32 (from \(\sqrt [ 3 ]{ 100x }\))

Question 6.
780
Solution:
\(\sqrt [ 3 ]{ 780 } =\sqrt [ 3 ]{ 78\times 100 }\) = 9.205 (from \(\sqrt [ 3 ]{ 10x }\))

Question 7.
7800
Solution:
\(\sqrt [ 3 ]{ 7800 } =\sqrt [ 3 ]{ 78\times 100 }\) = 19.83 (from \(\sqrt [ 3 ]{ 100x }\))

Question 8.
1346
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 1

Question 9.
250
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 2
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 3

Question 10.
5112
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 4

Question 11.
9800
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 5

Question 12.
732
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 6

Question 13.
7342
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 7
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 8

Question 14.
133100
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 9

Question 15.
37800
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 10

Question 16.
0.27
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 11

Question 17.
8.6
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 12
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 13

Question 18.
0.86
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 14

Question 19.
8.65
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 15

Question 20.
7532
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 16

Question 21.
833
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 17
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 18

Question 22.
34.2
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 19

Question 23.
What is the length of the side of a cube whose volume is 275 cm3. Make use of the table for the cube root.
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 20

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

Question 1.
Rakesh can do a piece of work in 20 days. How much work can he do in 4 days ?
Solution:
Rakesh can do it in 20 days = 1
his 1 day’s work = \(\frac { 1 }{ 20 }\)
and his 4 days work = \(\frac { 1 }{ 20 }\) x 4 = \(\frac { 1 }{ 5 }\) th work

Question 2.
Rohan can paint \(\frac { 1 }{ 3 }\) of a painting in 6 days. How many days will he take to complete the painting ?
Solution:
Rohan can paint \(\frac { 1 }{ 3 }\) of painting in = 6 days
he will complete the painting in = \(\frac { 6 x 3 }{ 1 }\) = 18 days

Question 3.
Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together ?
Solution:
Anil’s 1 day’s work = \(\frac { 1 }{ 5 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 1

Question 4.
Mohan takes 9 hours to mow a large lawn. He and Sohan together can mow it in 4 hours. How long will Sohan take to mow the lawn if he works alone ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 2

Question 5.
Sita can finish typing a 100 page document in 9 hours, Mita in 6 hours and Rita in 12 hours. How long will they take to type a 100 page document if they work together?
Solution:
Sita can do a work in 1 hour = \(\frac { 1 }{ 9 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 3

Question 6.
A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 4

Question 7.
A and B can do a piece of work in 18 days; B and C in 24 days and A and C in 36 days. In what time can they do it, all working together ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 5

Question 8.
A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 6
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 7

Question 9.
A, B and C can reap a field in 15\(\frac { 3 }{ 4 }\) days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 8
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 9

Question 10.
A and B can polish the floors of a building in 10 days A alone can do \(\frac { 1 }{ 4 }\) th of it in 12 days. In how many days can B alone polish the floor ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 10

Question 11.
A and B can finish a work in 20 days. A alone can do \(\frac { 1 }{ 5 }\) th of the work in 12 days. In how many days can B alone do it ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 11
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 12

Question 12.
A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 13

Question 13.
A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 14
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 15

Question 14.
Aasheesh can paint his doll in 20 minutes and his sister Chinki can do so in 25 minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 16
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 17

Question 15.
A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 18
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 19

Question 16.
6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job ?
Solution:
6 men can complete the work in = 7 days
1 man will complete the same work in = 7 x 6 days (Less men, more days)
21 men will finish the work in = \(\frac { 7 x 6 }{ 21 }\) days (More men, less days) = 2 days

Question 17.
8 men can do a piece of work in 9 days. In how many days will 6 men do it ?
Solution:
8 men can do a work in = 9 days
1 men will do the work in = 9 x 8 days (Less men, more days)
6 men will do the work in = \(\frac { 9 x 8 }{ 6 }\) days (More men, less days)
= \(\frac { 72 }{ 6 }\) = 12 days

Question 18.
Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets?
Solution:
Reema can weave 35 baskets in = 25 days
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 20

Question 19.
Neha types 75 pages in 14 hours. How many pages will she type in 20 hours ?
Solution:
Neha types pages in 14 hours = 75 pages
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 21

Question 20.
If 12 boys earn Rs. 840 in 7 days, what will 15 boys earn in 6 days ?
Solution:
12 boys in 7 days earn an amount of = Rs. 840
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 22

Question 21.
If 25 men earn Rs. 1000 in 10 days, how much will 15 men earn in 15 days ?
Solution:
25 men can earn in 10 days = Rs. 1000
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 23

Question 22.
Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days ?
Solution:
Ashu can copy a book in 18 days working in a day = 8 hours
He will copy the book in 1 day working = 8 x 18 hours a day (Less days, more hours a day)
He will copy the book in 12 days working in a day = \(\frac { 8 x 18 }{ 12 }\) hours
(More days, less hours a day)
= \(\frac { 144 }{ 12 }\) = 12 hours a day

Question 23.
If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour.
Solution:
135 garlands in 3 hours are prepared by = 9 girls
1 garland in 3 hours will be prepared by
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 24

Question 24.
A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together ?
Solution:
First tap’s 1 hour work to fill the cistern = \(\frac { 1 }{ 8 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 25

Question 25.
Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 26

Question 26.
A pipe can fill a cistern in 10 hours. Due to a leak in the bottom, it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 27

Question 27.
A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 28
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 29

Question 28.
A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern of both the tap and the pipe are opened together ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 30

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RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3

Other Exercises

Question 1.
On what sum will the compound interest at 5% p.a. annum for 2 years compounded annually be Rs 164 ?
Solution:
Let Principal (P) = Rs 100
Rate (R) = 5% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 1
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 2

Question 2.
Find the principal of the interest compounded annually at the rate of 10% for two years is Rs 210.
Solution:
Let principal (P) = Rs 100
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 3
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 4

Question 3.
A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.
Solution:
Amount (A) = Rs 756.25
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 5

Question 4.
What sum will amount to Rs 4913 in 18 months, if the rate of interest is 12\(\frac { 1 }{ 2 }\) % per annum, compounded half-yearly.
Solution:
Amount (A) = Rs 4,913
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 6

Question 5.
The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.
Solution:
Let sum (P) = Rs 100
Rate (R) = 15% p.a.
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 7
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 8
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 9

Question 6.
Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1,290 as interest compounded annually, find the sum she borrowed.
Solution:
C.I. = Rs 1,290
Rate (R) = 15% p.a.
Period (n) = 2 years
Let sum (P) = Rs 100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 10

Question 7.
The interest on a sum of Rs 2,000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.
Solution:
Sum (P) = Rs 2,000
C.I. = Rs 163.20
Amount (A) = P + C.I. = Rs 2000 + Rs 163.20 = Rs 2163.20
Rate (R) = 4%
Let period = n years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 11
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 12

Question 8.
In how much time would Rs 5,000 amount to Rs 6,655 at 10% per annum compound interest ?
Solution:
Principal (P) = Rs 5,000
Amount (A) = Rs 6,655
Rate (R) = 10%
Let period = n years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 13

Question 9.
In what time will Rs 4,400 becomes Rs 4,576 at 8% per annum interest compounded half-yearly ?
Solution:
Principal (P) = Rs 4,400
Amount (A) = Rs 4,576
Rate (R) = 8% or 4% half-yearly
Let period = n half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 14
n = 1
Period = 1 half year

Question 10.
The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.
Solution:
Difference between C.I. and S.I. = Rs 20
Rate (R) = 4% p.a.
Period (n) = 2 years
Let principal (P) = Rs 100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 15
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 16

Question 11.
In what time will Rs 1,000 amount to Rs 1,331 at 10% per annum compound interest.
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1,331
Rate (R) = 10% p.a.
Let period = n year
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 17

Question 12.
At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years ?
Solution:
Principal (P) = Rs 640
Amount (A) = Rs 774.40
Period (n) = 2 years.
Let R be the rate of interest p.a.
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 18
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 19

Question 13.
Find the rate percent per annum if Rs 2000 amount to Rs 2,662 in 1\(\frac { 1 }{ 2 }\) years, interest being compounded half-yearly ?
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2,662
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 20

Question 14.
Kamala borrowed from Ratan a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years, she received Rs 210 as compound interest, but paid Rs 200 only as simple interest. Find the sum and the rate of interest.
Solution:
Simple interest = Rs 200
and compound interest = Rs 210.
Period = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 21

Question 15.
Find the rate percent per annum, if Rs 2,000 amount to Rs 2,315.25, in an year and a half, interest being compounded six monthly.
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2315.25
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 22
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 23

Question 16.
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
Solution:
Let Principal (P) = Rs 100
then Amount (A) = Rs 200
Period (n) = 3 years
Let R be the rate % p.a.
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 24

Question 17.
Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half- yearly.
Solution:
Let Principal (P) = Rs 100
Then Amount (A) = Rs 400
Period (n) = 2 years or 4 half years
Let R be the rate % half-yearly, then
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 25
Rate % = 41.42% half yearly and 82.84% p.a.

Question 18.
A certain sum amounts to Rs 5,832 in 2 years at 8% compounded interest. Find the sum.
Solution:
Amount (A) = Rs 5,832
Let P be the sum
Rate (R) = 8% p.a.
Period (n) = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 26

Question 19.
The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.
Solution:
Let sum (P) = Rs 100
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 27
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 28

Question 20.
The difference in simple interest and compound interest on a certain sum of money at 6\(\frac { 2 }{ 3 }\) % per annum for 3 years is Rs 46. Determine the sum.
Solution:
Let sum (P) = Rs 100
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 29
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 30
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 31

Question 21.
Ishita invested a sum of Rs 12,000 at 5% per annum compound interest. She received an amount of Rs 13,230 after n years, Find the value of n.
Solution:
Principal (P) = Rs 12,000
Amount (A) = Rs 13,230
Rate (R) = 5% p.a.
Period = n years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 32

Question 22.
At what rate percent per annum will a sum of Rs 4,000 yield compound interest of Rs 410 in 2 years ?
Solution:
Principal (P) = Rs 4,060
C.I. = Rs 410
Amount (A) = Rs 4,000 + 410 = Rs 4,410
Let rate = R % p.a.
Period (n) = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 33

Question 23.
A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.
Solution:
Amount (A) = Rs 10,404
Rate (R) = 2% p.a.
Period (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 18
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 19

Question 24.
In how much time will a sum of Rs 1,600 amount to Rs 1852.20 at 5% per annum compound interest ?
Solution:
Principal (P) = Rs 1,600
Amount (A) = Rs 1852.20
Rate (R) = 5% p.a.
Let n be the time
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 20

Question 25.
At what rate percent will a sum of Rs 1,000 amount to Rs 1102.50 in 2 years at compound interest ?
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1102.50 .
Period (n) = 2 years
Let R be the rate of interest p.a.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 21

Question 26.
The compound interest on Rs 1,800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.
Solution:
Principal (P) = Rs 1,800
C.I. = Rs 378
Amount (A) = P + C.I. = Rs 1,800 + 378 = Rs 2,178
Rate (R) = 10% p.a.
Let n be the period in years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 22
Comparing, we get:
n = 2
Period = 2 years

Question 27.
What sum of money will amount to Rs 45582.25 at 6\(\frac { 3 }{ 4 }\) % per annum in two years, interest being compounded annually ?
Solution:
Amount (A) = Rs 45582.25
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 23

Question 28.
Sum of money amounts to Rs 4,53,690 in 2 years at 6.5% per annum compounded annually. Find the sum.
Solution:
Amount (A) = Rs 4,53,690
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 24
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 25
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 26

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2

Other Exercises

Question 1.
Compute the amount and the compound interest in each of the following by using the formula when :
(i) Principal = Rs 3,000, Rate = 5%, Time = 2 years
(ii) Principal = Rs 3,000 Rate = 18%, Time = 2 years
(iii) Principal = Rs 5,000 Rate = 10 paise per rupee per annum, Time = 2 years
(iv) Principal = Rs 2,000, Rate = 4 paise per rupee per annum, Time = 3 years.
(v) Principal = Rs 12,800, Rate = 7\(\frac { 1 }{ 2 }\) %, Time = 3 years
(vi) Principal = Rs 10,000, Rate = 20% per annum compounded half-yearly, time = 2 years
(vii) Principal = Rs 1,60,000, Rate = 10 paise per rupee per annum compounded half- yearly, Time = 2 years.
Solution:
(i) Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Time (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 1
and compound interest (C.I) = A – P = Rs 3307.50 – Rs 3,000 = Rs 307.50
(ii) Principal (P) = Rs 3,000
Rate (R) = 18% p.a.
Time (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 2
and compound interest (C.I.) = A – P = Rs 4177.20 – Rs 3,000 = Rs 1177.20
(iii) Principal (P) = Rs 5,000
Rate (R) =10 paise per rupee or 10% p.a.
Time (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 3
C.I. = A – P = Rs 6,050 – Rs 5,000 = Rs 1,050
(iv) Principal (P) = Rs 2,000
Rate (R) = 4 paise per rupee or 4% p.a.
Time (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 4
C.I. = A – P = Rs 2249.73 – Rs 2,000 = Rs 249.73
(v) Principal (P) = Rs 12,800
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 5
C.I. = A – P = Rs 15901.40 – Rs 12,800 = Rs 3101.40
(vi) Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Time = 2 years or 4 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 6
C.I. = A – P = Rs 14,641 – Rs 10,000 = Rs 4,641
(vii) Principal (P) = Rs 1,60,000
Rate (R) = 10 paise per rupee or 10% p.a. or 5% half-yearly
Time (n) = 2 years or 4 half-years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 7
C.I. = A – P = Rs 1,94,481 – Rs 1,60,000 = Rs 34,481

Question 2.
Find the amount of Rs 2,400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.
Solution:
Principal (P) = Rs 2,400
Rate (R) = 20%
Time (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 8

Question 3.
Rahman lent Rs 16,000 to Rasheed at the rate of 12\(\frac { 1 }{ 2 }\) % per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.
Solution:
Principal (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 9

Question 4.
Meera borrowed a sum of Rs 1,000 from Sita for two years. If the rate of interest is 10% compounded annually find the amount that Meera has to pay back.
Solution:
Amount of loan (P) = Rs 1,000
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 10

Question 5.
Find the difference between the compound interest and simple interest. On a sum of Rs 50,000 at 10% per annum for 2 years.
Solution:
Principal (P) = Rs 50,000
Rate (R) = 10% p.a.
Period (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 11
Difference between C.I. and S.I. = Rs 10,500 – Rs 10,000 = Rs 500

Question 6.
Amit borrowed Rs 16,000 at 17\(\frac { 1 }{ 2 }\) % per annum simple interest on the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years ?
Solution:
Amount of loan (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 12
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 13
C.I. = A – P = Rs 22,090 – Rs 16,000 = Rs 6,090
Now gain = C.I. – S.I = Rs 6,090 – 5,600 = Rs 490

Question 7.
Find the amount of Rs 4,096 for 18 months at 12\(\frac { 1 }{ 2 }\) % per annum, interest being compounded semi-annually ?
Solution:
Principal (P) = Rs 4,096
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 14

Question 8.
Find the amount and the compound interest on Rs 8,000 for 1\(\frac { 1 }{ 2 }\) years at 10% per annum, compounded half-yearly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 10% p.a. or 5% half yearly
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 15
and C.I. = A – P = Rs 9,261 – Rs 8,000 = Rs 1,261

Question 9.
Kamal borrowed Rs 57,600 from LIC against her policy at 12\(\frac { 1 }{ 2 }\) % per annum to build a house. Find the amount that she pays LIC after 1\(\frac { 1 }{ 2 }\) years if the interest is calculated half-yearly.
Solution:
Amount of loan (P) = Rs 57,600
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 16

Question 10.
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64,000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
Solution:
Price of house (P) = Rs 64,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 17
Compound interest (C.I.) = A – P = Rs 68,921 – Rs 64,000 = Rs 4,921

Question 11.
Rakesh lent out Rs 10,000 for 2 years at 20% per annum compounded annually. How much more he could earn if the interest be compounded half-yearly ?
Solution:
Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 18
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 19
C.I. = Rs 14,400 – Rs 10,000 = Rs 4,400
Now difference in C.I. = Rs 4,641 – Rs 4,400 = Rs 241

Question 12.
Romesh borrowed a sum of Rs 2,45,760 at 12.5% per annum compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest but compounded semi-annually. Find his gain after 2 years.
Solution:
In first case,
Principal (P) = Rs 2,45,760
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 20
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 21
C.I. = A – P = Rs 313203.75 – Rs 2,45,760 = Rs 67443.75
Gain = 67443.75 – Rs 65,280 = Rs2163.75

Question 13.
Find the amount that David would receive if he invests Rs 8,192 for 18 months at 12\(\frac { 1 }{ 2 }\) % per annum, the interest being compounded half-yearly.
Solution:
Principal (P) = Rs 8,192
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 22
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 23

Question 14.
Find the compound interest on Rs 15,625 for 9 months at 16% per annum, compounded quarterly.
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16% p.a. or 4% quarterly
Period (n) = 9 months or 3 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 24
Compound interest = A – P = Rs 17,576 – Rs 15,625 = Rs 1,951

Question 15.
Rekha deposited Rs 16,000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.
Solution:
Principal (P) = Rs 16,000
Rate (R) = 20% p.a. or 5% quarterly
Period (n) = one year or 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 25
C.I. = A – P = Rs 19448.10 – Rs 16,000 = Rs 3448.10

Question 16.
Find the amount of Rs 12,500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.
Solution:
Principal (P) = Rs 12,500
Rate (R1) = 15% p.a. for first year
R2 = 16% p.a. for second year
Period = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 26

Question 17.
Ramu borrowed Rs 15,625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment he will have to make after 2\(\frac { 1 }{ 4 }\) years ?
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16%
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 27
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 28

Question 18.
What will Rs 1,25,000 amount to at the rate of 6% if interest is calculated after every 4 months for one year ?
Solution:
Principal (P) = Rs 1,25,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 29

Question 19.
Find the compound interest at the rate of 5% for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 12,000 as simple interest.
Solution:
In first case,
S.I. = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 30
= Rs 80,000
In second case,
Principal (P) = Rs 80,000
Rate (R) = 5% p.a.
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 31
C.I. = A – P = Rs 92,610 – 80,000 = Rs 12,610

Question 20.
A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.
Solution:
Let Sum (P) = Rs x
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half years
In first case,
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 32
Interests = A – P = Rs 146.41 – Rs 100 = Rs 46.41
Now difference in interests = Rs 46.41 – Rs 44.00 = Rs 2.41
If difference is 2.41 then sum is 100 If difference is Rs 482, then sum
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 33

Question 21.
Simple interest on a sum of money for 2 years at 6\(\frac { 1 }{ 2 }\) % per annum is Rs 5,200. What will be the compound interest on the sum at the same rate for the same period ?
Solution:
In first case,
S.I. = Rs 5,200
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 34
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 35
Compound interest = A – P = Rs 45,369 – Rs 40,000 = Rs 5,369

Question 22.
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1,200 as simple interest.
Solution:
In first case,
S.I. = Rs 1,200
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 36
In second case,
Principal (P) = Rs 8,000
Rate (R) = 5% p.a.
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 37
= Rs 9,261
C.I. = A – P = Rs 9,261 – Rs 8000 = Rs 1,261

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2

Other Exercises

Question 1.
Add the following algebraic expressions
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 1
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 2
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 3
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 4
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 5
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 6

Question 2.
Subtract:
(i) -5xy from 12xy
(ii) 2a2 from -7a2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 8
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 9
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 10
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 11

Question 3.
Take away :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 12
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 13
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 14
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 15
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 16
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 17
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 18
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 19

Question 4.
Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and – 4X + 9y- 11z.
Solution:
Sum of x – 3y + 2z and – 4x + 9y – 11z
= x – 3y + 2z + (- 4x + 9y – 11z)
= x – 3y + 2z – 4x + 9y – 11z
= x – 4x – 3y + 9y + 2z – 11z
= – 3x + 6y – 9z
Now (-3x + 6y – 9z) – (3x – 4y – 7z)
= -3x + 6y – 9z – 3x + 4y + 7z
= -3x – 3x + 6y + 4y -9z +7z
= -6x + 10y – 2z

Question 5.
Subtract the sum of 3l- 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3nand -3l + m + 4n2.
Solution:
Sum of 9l + 2m – 3n2 and -3l + m + 4n2
= 9l + 2m – 3 n2 + (-3l) + m + 4n2
= 9l + 2m – 3n2 – 3l + m + 4n2
= 9l- 3l+ 2m + m – 3 n2 + 4n2
= 6l + 3m + n2
and sum of 3l – 4m – 7n2 and 2l +3m- 4n2
= 3l- 4m – 7n2 + 2l+ 3m- 4n2
= 3l + 2l – 4m + 3m- 7n2 – 4n2
= 5l -m- 11n2
Now (6l + 3m + n2) – (5l – m – 11n2)
= 6l + 3m + n2 – 5l + m + 11n2
= 6l – 5l + 3m + m + n2 + 11n2
= l + 4m+ 12n2

Question 6.
Subtract the sum of 2x – x2 + 5 and -4x – 3 + 7x2 from 5.
Solution:
5 – (2x-x2 + 5-4x-3 + 7x2)
= 5 – (2x – 4x- x2 + 7x2 + 5-3)
= 5 – (-2x + 6x2 + 2)
= 5 + 2x – 6x2 – 2
= – 6x2+2x+3
= 3 + 2x – 6x2

Question 7.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 20
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 21
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 22
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 23

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RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

Other Exercises

Question 1.
Find the compound interest when principal = Rs 3,000, rate = 5% per annum and time = 2 years.
Solution:
Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 1
Amount after one year = Rs 3,000 + Rs 150 = 3,150
and principal for the second year = Rs 3,150
and interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 2
Compound interest for two years = Rs 150 + Rs 157.50 = Rs 307.50

Question 2.
What will be the compound interest on Rs 4,000 in two years when rate of interest is 5% per annum ?
Solution:
Principal (P) = Rs 4,000
Rate (R) = 5% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 3
Amount after one year = Rs 4,000 + Rs 200 = Rs 4,200
Principal for the second year = Rs 4,200
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 4
Compound interest for 2 years = Rs 200 + Rs 210 = Rs 410

Question 3.
Rohit deposited Rs 8,000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years ?
Solution:
Principal (P) = Rs 8,000
Rate (R) = 15% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 5
Amount after first year = Rs 8,000 + RS 1,200 = Rs 9,200
or Principal for the second year = Rs 9,200
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 6
Amount after 2 years = Rs 9,200 + Rs 1,380 = Rs 10,580
or Principal for the third year = Rs 10,580
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 7
Compound for the 3 years = Rs 1,200 + Rs 1,380 + Rs 1,587 = Rs 4,167

Question 4.
Find the compound interest on Rs 1,000 at the rate of 8% per annum for 1\(\frac { 1 }{ 2 }\) years when interest is compounded half-yearly ?
Solution:
Principal (P) = Rs 1,000
Rate (R) = 8% p.a.
Period (T) = 1\(\frac { 1 }{ 2 }\) years = 3 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 8
Amount after one half-year = Rs 1,000 + Rs 40 = 1,040
Or principal for the second half-year = Rs 1,040
Interest for the second half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 9
Amount after second half-year = Rs 1,040 + 41.60 = Rs 1,081.60
Or principal for the third half-year = Rs 1081.60
Interest for the third half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 10
Compound interest for the third half-years or 1\(\frac { 1 }{ 2 }\) years
= Rs 40 + Rs 41.60 + Rs 43.264 = Rs 124.864

Question 5.
Find the compound interest on Rs 1,60,000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.
Solution:
Principal (P) = Rs 1,60,000
Rate (R) = 20% p.a. or 5% quarterly
Period (T) = 1 year or 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 11
Amount after first quarter = Rs 1,60,000 + 8,000 = 1,68,000
Or principal for the second quarter = Rs 1,68,000
Interest for the second quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 12
Amount after the second quarter = Rs 1,68,000 + Rs 8,400 = 1,76,400
Or principal for the third quarter = Rs 1,76,400
Interest for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 13
Amount after third quarter = Rs 1,76,400 + 8,820 = Rs 1,85,220
or Principal for the fourth quarter = Rs. 1,85,220
Interest for the fourth quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 14
Total compound interest for the 4 quarters = Rs 8,000 + Rs 8,400 + Rs 8,820 + 9,261 = Rs 34,481

Question 6.
Swati took a loan of Rs 16,000 against her insurance policy at the rate of 12\(\frac { 1 }{ 2 }\) % per annum. Calculate the total compound interest payable by Swati after 3 years.
Solution:
Amount of loan or principal (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 15
Amount after first year = Rs 16,000 + Rs 2,000 = Rs 18,000
Principal for the second year = Rs 18,000
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 16
Amount after second year = Rs 18,000 + 2,250 = Rs 20,250
Principal for the third year = Rs 20,250
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 17
Compound for 3 years = Rs 2,000 + Rs 2,250 + 2531.25 = Rs 6,781.25

Question 7.
Roma borrowed Rs 64,000 from a bank for 1\(\frac { 1 }{ 2 }\) years at the rate of 10% per annum. Compute the total compound interest payable by Roma after 1\(\frac { 1 }{ 2 }\) years, if the interest is compounded half-yearly.
Solution:
Principal (sum borrowed) (P) = Rs 64,000
Rate (R) = 10% p.a. or 5% half-yearly
Period (T) = 1\(\frac { 1 }{ 2 }\) years or 3 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 18
Amount after first half-year = Rs 64,000 + Rs 3,200 = Rs 67,200
Or principal for the second half-year = Rs 67,200
Interest for the second half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 19
Amount after second half-year = Rs 6,7200 + 3,360 = Rs 70,560
Or principal for the third half-year = Rs 70,560
Interest for the third half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 20
Total compound interest for 3 half-years
or 1\(\frac { 1 }{ 2 }\) years = Rs 3,200 + Rs 3,360 + Rs 3,528 = Rs 10,088

Question 8.
Mewa Lai borrowed Rs 20,000 from his friend RoopLal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
Solution:
Principal (P) = Rs 20,000
Rate (R) = 18% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 21
In second case
Interest for the first year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 22
Amount after one year = Rs 20,000 + Rs 3,600 = Rs 23,600
Or principal for the second year = Rs 23,600
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 23
Interest for two years = Rs 3,600 + 4,248 = Rs 7,848
Gain = Rs 7,848 – Rs 7,200 = Rs 648

Question 9.
Find the compound interest on Rs 8,000 for 9 months at 20% per annum compounded quarterly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 20% p.a. or 5% p.a. quarterly
Period (T) = 9 months or 3 quarters
Interest for the first quarterly
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 24
Amount after first quarter = Rs 8,000 + Rs 400 = Rs 8,400
Or principal for second quarter = Rs 8,400
Interest for the second quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 25
Amount after second quarter = Rs 8,400 + Rs 420 = Rs 8,820
Or principal for the third quarter = Rs 8,820
Interest for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 26
Compound interest for 9 months or 3 quarters = Rs 400 + Rs 420 + Rs 441 = Rs 1,261

Question 10.
Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs. 200 as simple interest.
Solution:
Simple interest = Rs 200
Rate (R) = 10% p.a.
Period (T) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 27
Now in second case,
Principal CP) = Rs 1,000
Rate (R) = 10% p.a.
Period (T) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 28
Amount after one year = Rs 1,000+ Rs 100 = Rs 1,100
Or principal for the second year = Rs 1,100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 29
Total interest for two years = Rs 100 + Rs 110 = Rs 210

Question 11.
Find the compound interest on Rs 64,000 for 1 year at the rate of 10% per annum compounded quarterly.
Solution:
Principal (P) = Rs 64,000
Rate (R) = 10% p.a. or \(\frac { 5 }{ 2 }\) % quarterly
Period (T) = 1 year = 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 30
Amount after first quarter = Rs 64,000 + Rs 1,600 = Rs 65,600
Or principal for the second quarter = Rs 65,600
Interest for the second quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 31
Amount after second quarter = Rs 65,600 + Rs 1,640 = Rs 67,240
Or principal for the third year = Rs 67,240
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 32
= Rs 1,681
Amount after third quarter = Rs 67,240 + Rs 1,681 = Rs 68,921
Or principal for the fourth quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 33
Total compound interest for 4 quarters or one year
= Rs 1,600 + Rs 1,640 + Rs 1,681 + Rs 1723.025 = Rs 6644.025

Question 12.
Ramesh deposited Rs 7,500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months ?
Solution:
Principal (P) = Rs 7,500
Rate (R) = 12% p.a. or 3% quarterly
Time (T) = 9 months or 3 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 34
Amount after one quarter = Rs 7,500 + Rs 225 = Rs 7,725
Or Principal for second quarter = Rs 7,725
Interest for the second quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 35
Amount after second quarter = Rs 7,725 + Rs 231.75 = Rs 7956.75
Or principal for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 36
Total amount he received after 9 months = Rs 7956.75 + Rs 238.70 = Rs 8195.45

Question 13.
Anil borrowed a sum of Rs 9,600 to install a hand pump in his dairy. If the rate of interest is 5\(\frac { 1 }{ 2 }\) % .per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.
Solution:
Principal (P) = Rs 9,600
Rate of interest (R) = 5\(\frac { 1 }{ 2 }\) % = \(\frac { 11 }{ 2 }\) % p.a.
Period (T) = 3 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 37
Amount after one year = Rs 9,600 + Rs 528 = Rs 10,128
Or principal for second year = Rs 10,128
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 38
Amount after second year = Rs 10,128 + Rs 557.04 = Rs 10685.04
or Principal for the third year = Rs 10685.04
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 39
Total compound interest = Rs 528 + Rs 557.04 + Rs 587.68 = Rs 1672.72

Question 14.
Surabhi borrowed a sum of Rs 12,000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years.
Solution:
Sum of money borrowed (P) = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 40
Amount after one year = Rs 12,000 + Rs 600 = Rs 12,600
Or principal for the second year = Rs 12,600
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 41
Amount after second year = Rs 12,600 + Rs 630 = Rs 13,230
Or Principal for the third year = Rs 13,230
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 42
Total compound interest for 3 years = Rs 600 + Rs 630 + Rs 661.50 = Rs 1891.50

Question 15.
Daljit received a sum of Rs 40,000 as a loan from a finance company. If the rate of interest is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years.
Solution:
Amount of loan (P) = Rs 40,000
Rate (R) = 7% p.a.
Period = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 43
Amount after one year = Rs 40,000 + Rs 2,800 = Rs 42,800
Or principal for the second year = Rs 42,800
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 44
Total interest paid after two years = Rs 2,800 + 2,996 = Rs 5,796

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1

RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III (Special Types of Quadrilaterals) Ex 17.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1

Other Exercises

Question 1.
Given below is a parallelogram ABCD. Complete each statement along with the definition or property used:
(i) AD = ………
(ii) ∠DCB = ……….
(iii) OC’ = …….
(iv) ∠DAB + ∠CDA = …….
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 1
Solution:
(i) AD = BC
(ii) ∠DCB = ∠ADC
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 2
(iii) OC = OA
(iv) ∠DAB + ∠CDA = 180°

Question 2.
The following figures are parallelograms. Find the degree values of the unknowns x,y, z.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 3
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 4
Solution:
In a parallelogram, opposite angles are equal and sum of adjacent angle is 180°.
(i) In parallelogram ABCD,
∠B = 100°
∠A = ∠C = 180° (Co-interior angles)
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 5
⇒ x + 100° = 180°
⇒ x = 180° – 100°
⇒ x = 80°
But ∠A = ∠C and ∠B = ∠D (Opposite angles)
A = x
⇒ z = x
⇒ z = 80°
and ∠D = ∠B
⇒ y = 100°
x = 80°, y = 100° and z = 80°
(ii) In parallelogram PQRS, side PQ is produced to T.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 6
∠S = 50°
∠PQR = ∠S (Opposite angles)
w = 50°
But ∠P + ∠PQR = 180° (Sum of adjacent angles)
⇒ x + 50° = 180°
⇒ x = 180° – 50° = 130°
⇒ But ∠P = ∠R (Opposite angles)
x = y
⇒ y = 130°
But w + z = 180° (A linear pair)
⇒ 50° + z = 180°
⇒ z = 180° – 50° = 130°
⇒ x = 130°, y = 130°, ∠ = 130
(iii) In parallelogram LMNP, PM is its diagonal ∠NPM = 30°, ∠PMN = 90°
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 7
PN || LM (Opposite sides of a parallelogram) and PM is its transversal
∠NPM = ∠PML
⇒ 30° = x
x = 30°
In ∆PMN,
∠P = 30°, ∠M = 90°
But ∠P + ∠M + ∠N = 180° (Sum of angles of a triangle)
⇒ 30° + 90° + z = 180°
⇒ 120° + z = 180°
⇒ z = 180° – 120° = 60°
But ∠L = ∠N (Opposite angles of a parallelogram)
y = z
⇒ y = 60°
Hence x = 30°, y = 60° and ∠ = 60°
(iv) In rhombus ABCD, diagonals AC and BD bisect each other at right angles.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 8
x = 90°
In ∆OCD,
∠O + ∠C + ∠D = 180°
⇒ 90° + 30° + y = 180°
⇒ 120° + y = 180°
⇒ y = 180° – 120° = 60°
y = 60°
CD || AB, BD is the transversal
y = z (Alternate angles)
z = 60°
Hence x = 90°, y = 60° and z = 60°
(v) In parallelogram PQRS, side QR is produced to T
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 9
∠Q = 80°
∠P + ∠Q = 180° (Sum of adjacent angles)
⇒ x + 80° = 180°
⇒ x = 180° – 80° = 100°
∠Q = ∠S (Opposite angles of a parallelogram)
⇒ 80° = y
⇒ y = 80°
PQ || SR and QRT is transversal
∠TRS = ∠RQP (Corresponding angles)
⇒ ∠ = 80°
Hence x = 100°, y = 80° and z = 80°
(vi) In parallelogram TUVW, UW is its diagonal
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 10
∠TUW = 40° and ∠V = 112°
∠T = ∠V (Opposite angles)
y = 112°
In ∆TUW,
∠T + ∠V + ∠W = 180° (Sum of angles of a parallelogram)
⇒ y + 40° + x = 180°
⇒ 112° + 40° + x = 180°
⇒ 152° + x = 180°
⇒ x = 180° – 152° = 28°
UV || TW and UW is its transversal
∠WUV = ∠TWU (Alternate angles)
⇒ z = x
⇒ z = 28°
Hence x = 28°, y = 112°, z = 28°

Question 3.
Can the following figures be parallelograms. Justify your answer.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 11
Solution:
(i) In quadrilateral PLEH
∠H = 100°, ∠L = 80°
But there are opposite angles
∠H ≠ ∠L
PLEH is not a parallelogram.
(ii) In quadrilateral GNIR,
RI = 8 cm, GN = 8 cm, RG = 5 cm and IN = 5 cm
PI = GH and RG = IN
But there are opposite sides of the quadrilateral.
GNIR is a parallelogram.
(iii) In quadrilateral BEST,
BS and ET are its diagonals
But these diagonal do not bisect each other.
BEST is not a parallelogram

Question 4.
In the adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the geometrical truths you use to find them.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 12
Solution:
In the figure, HOPE is a parallelogram in which HG is produced and HP is the diagonal
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 13
∠EHP = 40° and ∠POQ = 70°
But ∠POQ + ∠POH = 180° (Linear pair)
⇒ 70° + w = 180°
⇒ w = 180° – 70° = 110°
But ∠E = ∠POE (Opposite angles of a parallelogram)
x = 110°
HE || OP and HP is its transversal.
∠EHP = ∠HPO (Alternate angles)
⇒40° = y
⇒ y = 40°
In ∆PHO,
Ext. ∠POQ = ∠PHO + ∠HPO
⇒ 70° = z + y
⇒ 70° = z + 40°
⇒ z = 70° – 40° = 30°
Hence x = 110°, y = 40°, z = 30°

Question 5.
In the following figures GUNS and RUNS are parallelograms. Find x and y.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 14
Solution:
(i) In parallelogram GUNS,
Opposite sides are parallel and equal
3x = 18
⇒ x = 6
and 3y – 1 = 26
⇒ 3y = 26 + 1 = 27
⇒ y = 9
x = 6, y = 9
(ii) Diagonals of a parallelogram bisect each other.
y – 7 = 20
⇒ y = 20 + 7 = 27
and x – 27 = 16
⇒ x – 27 = 16
⇒ x = 16 + 27 = 43
x = 43, y = 27

Question 6.
In the following figure RISK and CLUE are parallelograms. Find the measure of x.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 15
Solution:
In the figure, RISK and CLUE are parallelograms
∠K = 120° and ∠L = 70°
In parallelogram RISK
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 16
RK || IS
∠RKS = ∠ISU (Corresponding angles)
⇒ ∠ISU = 120°
In parallelogram CLUE,
∠E = ∠L (Opposite angles of a parallelogram)
∠E = 70° (∠L = 70°)
Now in ∆EOS,
Ext. ∠ISU = x + ∠E
⇒ 120° = x + 70°
⇒ x = 120° – 70° = 50°
x = 50°

Question 7.
Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 17
∠A = (3x – 2)° and ∠C = (50 – x)°
∠A = ∠C (Opposite angles of a parallelogram)
⇒ 3x – 2° = 50° – x
⇒ 3x + x = 50° + 2°
⇒ 4x = 52°
x = 13°

Question 8.
If an angle of a parallelogram is two- third of its adjacent angle, find the angles of the parallelogram.
Solution:
Let the parallelogram is ABCD and ∠A of a parallelogram ABCD be x
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 18
Hence other angles will be
∠C = ∠A = 108° (Opposite angles)
and ∠D = ∠B = 72° (Opposite angles)
Hence ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°

Question 9.
The measure of one angle of a parallelogram is 70°. What are the measures of the remaining angles ?
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 19
∠A = 70°
But ∠A + ∠B= 180° (Sum of adjacent angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70°
⇒ ∠B = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 10.
Two adjacent angles of a parallelogram are as 1 : 2. Find the measures of all the angles of the parallelogram.
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 20
∠A : ∠B = 1 : 2
Let ∠A = x, then ∠B = 2x
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ x + 2x = 180°
⇒ 3x = 180°
∠A = x = 60°
and ∠B = 2x = 2 x 60°= 120°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 60° and ∠D = 120°
Hence ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°

Question 11.
In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 21
∠D = 135°
But ∠A + ∠D= 180° (Sum of adjacent angles)
∠A + 135° = 180°
∠A = 180° – 135° = 45°
But ∠B = ∠D (Opposite angles)
∠B = 135°
Hence ∠A = 45° and ∠B = 135°

Question 12.
ABCD is a parallelogram in which ∠A = 70°, compute ∠B, ∠C and ∠D.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 22
∠A = 70°
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70° = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 13.
The sum of two opposite angles of a parallelogram is 130°. Find all the angles of the parallelogram.
Solution:
Let in parallelogram
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 23
∠A + ∠C = 130°
But ∠A = ∠C (Opposite angles)
⇒ ∠C = \(\frac { 130 }{ 2 }\) = 65°
∠B + ∠D = 180° (Sum of adjacent angles)
⇒ 65° + ∠B = 180°
⇒ ∠B = 180° – 65° = 115°
⇒ ∠B = 115°
But ∠D = ∠B (Opposite angles)
∠D = 115°
Hence ∠A = 65°, ∠B = 115°, ∠C = 65° and ∠D = 115°

Question 14.
All the angles of a quadrilateral are equal to each other. Find the measure of each. Is the quadrilateral a parallelogram ? What special type of parallelogram is it ?
Solution:
All the angles of a quadrilateral are equal and sum of the four angles = 360°
Each angle will be = \(\frac { 360 }{ 4 }\) = 90°
Let in quadrilateral ABCD,
∠A = ∠B = ∠C = ∠D = 90°
It is a parallelogram as opposite angles are equal
i.e., ∠A = ∠C and ∠B = ∠D.
Each angle is of 90°
This parallelogram is a rectangle.

Question 15.
Two adjacent sides of a parallelogram are 4 cm and 3 cm respectively. Find its perimeter.
Solution:
Length of two adjacent sides = 4 cm and 3 cm
i.e., l = 4 cm and b = 3 cm
Perimeter = 2 (l + b) = 2 (4 + 3) cm = 2 x 7 = 14 cm

Question 16.
The perimeter of a parallelogram is 150 cm. One of its sides is greater ii.au the other by 25 cm. Find the length of the sides of the parallelogram.
Solution:
Perimeter of a parallelogram = 150 cm
Let l he the longer side and b be the shorter side
l = b + 25 cm.
⇒ 2 (l + b) = 150
⇒ l + b = 75
⇒ b + 25 + b = 75
⇒ 2b = 75 – 25 = 50
⇒ b = 25
l = b + 25 = 25 + 25 = 50
Sides are 50 cm, 25 cm

Question 17.
The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.
Solution:
In a parallelogram shorter side (b) = 4.8 cm.
longer side (l) = 4.8 + \(\frac { 1 }{ 2 }\) x 4.8
= 4.8 + 2.4 = 7.2 cm
Perimeter = 2 (l + b) = 2 (7.2 + 4.8) cm = 2 x 12.0 = 24 cm

Question 18.
Two adjacent angles of a parallelogram are (3x – 4)° and (3x + 10)°. Find the angles of the parallelogram.
Solution:
Let in parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 24
∠A = (3x + 4)° and ∠B = (3x + 10)°
But ∠A + ∠B = 180° (Sum of adj adjacent angles)
⇒ 3x – 4 + 3x + 10 = 180°
⇒ 6x + 6° = 180°
⇒ 6x = 180° – 6° = 174°
⇒ x = 29°
∠A = 3x – 4 = 3 x 29 – 4 = 87° – 4° = 83°
∠B = 3x + 10 = 3 x 29° + 10° = 87° + 10° = 97°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 83° and ∠D = 97°
Hence ∠A = 83°, ∠B = 97°, ∠C = 83° and ∠D = 97°

Question 19.
In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC = 30°, ∠BDC = 10° and ∠CAB = 70°.
Find : ∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC and ∠DBA.
Solution:
In parallelogram ABCD, diagonal AC and ED bisect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 25
∠ABC = 30°, ∠CAB = 70° and ∠BDC = 10°
∠ADC = ∠ABC = 30° (Opposite angles)
and ∠ADB = ∠ADC – ∠BDC = 30° – 10° = 20°
AB || DC and AC is the transversal
∠ACD = ∠CAB = 70°(Altemate angles)
AB || DC and BD is transversal
∠CDB = ∠ABD = 10°(Altemate angles)
In ∆ABC
∠CAB + ∠ABC + ∠BCA = 180° (Sum of anlges of a triangle)
⇒ 70° + 30° + ∠BCA = 180°
⇒ 100° + ∠BCA = 180°
∠BCA = 180° – 100° = 80°
∠BCD = ∠BCA + ∠ACD = 80° + 70° = 150°
∠BCD = ∠DAB (Opposite angles)
⇒ ∠DAB = 150° and ∠CAD = 150° – 70° = 80°
In ∆OCD,
∠ODC + ∠OCD + ∠COD = 180° (Angles of a triangle)
⇒ ∠ACD + ∠ACD + ∠COD = 180°
⇒ 70° + 10° + ∠COD = 180°
⇒ 80° + ∠COD = 180°
⇒ ∠COD = 180° – 80° = 100°
∠COD = 100°
But ∠AOD + ∠COD = 180° (Linear pair)
⇒ ∠AOD + 100° = 180°
⇒ ∠AOD = 180° – 100° = 80°
⇒ ∠AOD = 80°
But ∠AOB = ∠COD
and ∠BOC = ∠AOD (Vertically opposite angles)
∠AOB = 100° and ∠BOC = 80°
Hence ∠DAB = 150°, ∠ADC = 30°, ∠BCD = 150°
∠AOD = 80° ∠DOC = 100°
∠BOC = 80° ∠AOB = 100°
∠ACD = 70°, ∠CAB = 70°, ∠AD = 20°, ∠ACB = 80°, ∠DBC = 20°, ∠DBA = 10

Question 20.
Find the angles marked with a question mark shown in the figure.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 26
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 27
CE ⊥ AB and CF ⊥ AD
∠BCE = 40°
In ∆BCE,
∠BCE + ∠CEB + ∠EBC = 180° (Sum of angles of a triangle)
⇒ 40° + 90° + ∠EBC = 180°
⇒ 130° + ∠EBC = 180°
⇒ ∠EBC = 180° – 130° = 50°
or ∠B = 50°
But ∠D = ∠B (Opposite angles)
∠D = 50° or ∠ADC = 50°
Similarly in ∆DCF,
∠DCF + ∠CFD + ∠FDC = 180°
⇒ ∠DCF + 90° + 50° = 180°
⇒ ∠DCF + 140° = 180°
⇒ ∠DCF = 180° – 140° = 40°
But ∠C + ∠B= 180° (Sum of adjacent angles)
⇒ ∠BCE + ∠ECF + ∠DCF + ∠B = 180°
⇒ 40° + ∠ECF + 40° + 50° = 180°
⇒ ∠ECF + 130° = 180°
⇒ 40° + ∠ECF + 40° + 50° = 180°
⇒ ∠ECF + 130° = 180°
⇒ ∠ECF = 180° – 130° = 50°
∠ECF = 50°

Question 21.
The angle between the altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 28
∠A is an obtused angle
AE ⊥ BC and AF ⊥ DC
∠EAF = 60°
In quadrilateral AECF,
∠EAF + ∠F + ∠C + ∠E = 360° (Sum of angles of a quadrilateral)
⇒ 60° + 90° + ∠C + 90° = 360°
⇒ 240° + ∠C = 360°
⇒ ∠C = 360° – 240° = 120°
∠C = 120°
But ∠A = ∠C (Opposite angles)
∠A = 120°
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ 120° + ∠B = 180°
⇒ ∠B = 180° – 120° = 60°
∠B = 60°
But ∠D = ∠B (Opposite angles)
∠D = 60°
Hence ∠A = 120°, ∠B = 60°, ∠C = 120° and ∠D = 60°

Question 22.
In the figure, ABCD and AEFG are parallelograms. If ∠C = 55°, what is the measure of ∠F ?
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 29
Solution:
In the parallelogram ABCD,
∠A = ∠C …..(i) (Opposite angles of a parallelogram)
Similarly in parallelogram AEFG,
∠A = ∠F …(ii)
From (i) and (ii),
∠C = ∠F = 55° (∠C = 55°)
Hence ∠F = 55°

Question 23.
In the figure, BDEF and DCEF are each a parallelogram. It is true that BD = DC ? Why or why not ?
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 30
Solution:
In parallelogram BDEF,
BD = EF ……(i) (Opposite sides of a parallelogram)
Similarly, in parallelogram DCEF
DC = EF
From (i) and (ii),
BD = DC
Hence it is true that BD = DC.

Question 24.
In the figure, suppose it is known that DE = DF. Then is ∆ABC isosceles ? Why or why not ?
Solution:
In parallelogram BDEF,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 31
∠B = ∠E ……(i) (Opposite angles of a parallelogram)
Similarly, in parallelogram DCEF,
∠C = ∠F ……(ii)
But DE = DF (Given)
In ∆DEF
∠E = ∠F
From (i) and (ii),
∠B = ∠C
AC = AB (Sides opposite to equal angles)
∆ABC is an isosceles triangle.

Question 25.
Diagonals of parallelogram ABCD intersect at O as shown in the figure. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:
(i) OB = OD
(ii) ∠OBY = ∠ODX
(iii) ∠BOY = ∠DOX
(iv) ∆BOY = ∆DOX
Now, state if XY is bisected at O.
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 32
diagonals AC and BP intersect each other at O
O is the mid-point of AC and BD.
Through O, XY is draw such that X lies on AD and Y, on BC.
(i) OB = OD (O is mid-point of BD)
(ii) AD || BC and BD is transversal
∠OBY = ∠ODX (Alternate angles)
(iii) ∠BOY = ∠DOX (Vertically opposite angles)
(iv) Now in ∆BOY and ∆DOX,
OB = OD
∠OBY = ∠ODX
∠BOY = ∠DOX
∆BOY = ∆DOX (ASA axiom)
OY = OX (c.p.c.t.)
Hence XY is bisected at O.

Question 26.
In the figure ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same.
(i) ∠A = ∠C
(ii) ∠FAB = \(\frac { 1 }{ 2 }\) ∠A
(iii) ∠DCE = \(\frac { 1 }{ 2 }\) ∠C
(iv) ∠CEB = ∠FAB
(v) CE || AF.
Solution:
In parallelogram ABCD,
CE is the bisector of ∠C and and AF is the bisector of ∠A.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 33
(i) ∠A = ∠C (Opposite angles of a parallelogram)
(ii) AF is the bisector of ∠A
∠FAB = \(\frac { 1 }{ 2 }\) ∠A
(iii) CE is the bisector of ∠C
∠DCE = \(\frac { 1 }{ 2 }\) ∠C
(iv) From (i), (ii) and (iii)
∠FAB = ∠DCE
(v) ∠FAB = ∠DCE
But these are opposite angles of quadrilateral AECF
AB or AE || DC or FC
AECF is a parallelogram
CE || AF
Hence proved

Question 27.
Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM ? Why or why not ?
Solution:
In parallelogram ABCD, diagonals AC and BD intersect each other at O.
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 34
O is the mid-point of AC and BD.
AL ⊥ BD and CM ⊥ BD.
In ∆ALO and ∆CMO
∠L = ∠M (Each 90°)
∠AOL = ∠COM (Vertically opposite angles)
AO = CO (O is mid-point of AC)
∆ALO = ∆CMO (AAS axiom)
AL = CM (c.p.c.t.)
Hence proved

Question 28.
Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF. What type of quadrilateral is BFDE ?
Solution:
In parallelogram ABCD,
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 35
AC is its diagonal. E and F are points on AC such that AE = CF
Join EB, BF, FD and DE .
Join also diagonal BD which intersects AC at O
O is the mid-point of AC and BD
AO = OC
But AE = CF
⇒ AO – AE = CO – CF
⇒ EO = OF
But BO = OD (O is mid-point of BD)
Diagonals EF and BD of quadrilateral bisect each other at O.
BFDE is a parallelogram.

Question 29.
In a parallelogram ABCD, AB = 10 cm, AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find the length of CF.
Solution:
AB || DC
RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 36
∠DEA = ∠EAB (Alternate angles)
= ∠DAE (EA is bisector of ∠A)
In ∆DAE,
∠DEA = ∠DAE
AD = DE = 6 cm
But DE = AB = 10 cm.
EC = DC – DE = 10 – 6 = 4 cm
AD || BC or BF and AF is transversal
∠DAE = ∠EFC (Alternate angle)
But ∠DAE = ∠DEA Prove
= ∠FEC (DEA = FEC vertically opposite angles)
In ∆ECF,
CE = CF = 4 cm (CE = 4 cm)

 

Hope given RD Sharma Class 8 Solutions Chapter 17 Understanding Shapes III Ex 17.1 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

Other Exercises

Question 1.
Find :
(i) 22% of 120
(ii) 25% of Rs. 1000
(iii) 25% of 10 kg
(iv) 16.5% of 5000 metres
(v) 135% of 80 cm
(vi) 2.5% of 10000 ml
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 1

Question 2.
Find the number ‘a’, if
(i) 8.4% of a is 42
(ii) 0.5% of a is 3
(iii) \(\frac { 1 }{ 2 }\) % of a is 50
(iv) 100% of a is 100
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 2
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 3

Question 3.
x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.
Solution:
x = 5% of y, y = 24% of z.
x = 480
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 4

Question 4.
A coolie deposits Rs. 150 per month in his post office Saving Bank account. If this is 15% of his monthly income, find his monthly income.
Solution:
Let his monthly income = Rs. x
15% of x = Rs. 150
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 5

Question 5.
Asha got 86.875% marks in the annual examination. If she got 695 marks, find the number of marks of the Examination.
Solution:
Let total marks of the examination = x
86.875% of x = 695
=> 86.875 x \(\frac { 1 }{ 100 }\) x x = 695
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 6

Question 6.
Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened ?
Solution:
Let the school opened for = x days = 90% of x = 216
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 7

Question 7.
A garden has 2000 trees. 12% of these are mango trees, 18% lemon and the rest are orange trees. Find the number of orange trees.
Solution:
Number of total trees = 2000
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 8
Rest trees = 2000 – (240 + 360) = 2000 – 600 = 1400
Number of orange trees = 1400

Question 8.
Balanced diet should contain 12% of protein, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food in take.
Solution:
Balance diet contains
Protein = 12%
Fats = 25%
Carbohydrates = 63%
Number of total calories = 2600
Number of calories of proteins = 12% of 2600 = \(\frac { 12 }{ 100 }\) x 2600 = 312
Number of calories of fats = 25% of 2600 = \(\frac { 25 }{ 100 }\) x 2600 = 650
Number of calories of carbohydrates = 63% of 2600 = \(\frac { 63 }{ 100 }\) x 2600 = 1638

Question 9.
A cricketer scored a total of 62 runs in 96 balls. He hits 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in :
(i) Sixes
(ii) Fours
(iii) Twos
(iv) Singles
Solution:
Total score of a cricketer = 62 runs
(z) Number of sixes = 3
Run from 3 sixes = 3 x 6 = 18
Percentage = \(\frac { 18 }{ 62 }\) x 100 = 29.03%
(ii) Number of fours = 8
Total run from 8 fours = 4 x 8 = 32
Percentage = \(\frac { 32 }{ 62 }\) x 100 = 51.61%
(iii) Number of twos = 2
Total score from 2 twos = 2 x 2 = 4
Percentage = \(\frac { 4 }{ 62 }\) x 100 = \(\frac { 400 }{ 62 }\) = 6.45%
(iv) Number of single run = 8
Percentage = \(\frac { 8 }{ 62 }\) x 100 = \(\frac { 800 }{ 62 }\) = 12.9%

Question 10.
A cricketer hits 120 runs in 150 balls during a test match. 20% of the runs came in 6’s, 30% in 4’s, 25% in 2’s and the rest in 1’s. How many runs did he score in :
(i) 6’s
(ii) 4’s
(iii) 2’s
(iv) singles
What % of his shots were scoring ones ?
Solution:
Total runs scored by a cricketer =120
(i) Number of runs from sixes (6’s) = 20% of 120
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 9

Question 11.
Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest ?
Solution:
Total earning from investment = Rs. 187
Percent earning = 22%
Let his investment = x
Then 22% of x = Rs. 187
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 10

Question 12.
Rohit deposits 12% his income in a bank. He deposited Rs. 1440 in the bank during 1997. What was his total income for the year 1997 ?
Solution:
Deposit in the bank = Rs. 1440
Percentage = 12% of his total income
Let his total income = Rs. x
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 11

Question 13.
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur ?
Solution:
(i) In gunpowder,
Nitre = 75%
Sulphur = 10%
Let total amount of gunpowder = x kg
Nitre = 9 kg
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 12

Question 14.
An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy ?
Solution:
In an alloy,
Number of parts of tin = 15
and number of parts of copper = 105
Total parts = 15 + 105 = 120
Percentage of copper in the alloy = \(\frac { 105 }{ 120 }\) x 100 = 87.5%

Question 15.
An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.
Solution:
In an alloy,
Copper = 32%
Nickel = 40%
Rest is zinc = 100 – (32 + 40) = 100 – 72 = 28%
Mass of zinc in 1 kg = 28% of 1 kg = \(\frac { 28 }{ 100 }\) x 100 gm = 280 gm.

Question 16.
A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride ?
Solution:
Distance travelled before first stop = 122 km
Let total journey = x km
10% of x = 122
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 13

Question 17.
A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female ?
Solution:
Total numbers of teachers = 30
Number of male teachers = 12
Number of female teacher = 30 – 12 = 18
Percentage of female teachers = \(\frac { 18 x 100 }{ 30 }\) = 60%

Question 18.
Aman’s income is 20% less than that of Anil. How much percent is Anil’s income more than Aman’s income ?
Solution:
Let Anil’s income = Rs. 100
Then Aman’s income = Rs, 100 – 20 = Rs. 80
Now, difference of both’s incomes = 100 – 80 = Rs. 20
Anil income is Rs. 20 more than that of Aman’s
Percentage = \(\frac { 20 x 100 }{ 80 }\) = 25%

Question 19.
The value of a machine depreciates every year by 5%. If the present value of the machine be Rs. 100000, what will be its value after 2 years ?
Solution:
Present value of machine = Rs. 100000
Rate of depreciation per year = 5%
Period = 2 years
Value of machine after 2 years
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 14

Question 20.
The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years ?
Solution:
Present population of the town = 60000
Increase annually = 10%
Period = 2 years
Population after 2 years will be
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 15

Question 21.
The population of a town increases 10% annually. If the present population is 22000, find its population a year ago.
Solution:
Let the population of the town a year ago was = x
Increase in population = 10%
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 16

Question 22.
Ankit was given an increment of 10% on his salary. His new salary is Rs. 3575. What was his salary before increment ?
Solution:
Let the salary of Ankit before increment = x
Increment given = 10% of the salary
Salary after increment will be
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 17

Question 23.
In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase ?
Solution:
Let price of petrol before budged = Rs. 100
Increase = 10%
Price after budget = Rs. 100 + 10 = Rs. 110
Let the consumption of petrol before budget = 100 l
Price pf 100 l = Rs. 110
Now of new price is Rs. 110, consumption = 100 l
are of new price will be 100, then
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 18

Question 24.
Mohan’s income is Rs. 15500 per month. He saves 11% of his income. If his income increases by 10% then he reduces his saving by 1%, how much does he save now ?
Solution:
Mohan’s income = Rs. 15500
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 19
We see that the savings is same
There is no change in savings.

Question 25.
Shikha’s income is 60% more than that of Shalu. What percent is Shalu’s income less than Shikha’s ?
Solution:
Let Shalu’s income = Rs. 100
Then Shikha’s income will be = Rs. 100 + 60 = Rs. 160
Now difference in their incomes = Rs. 160 – 100 = Rs. 60
Shalu’s income is less than Shikha’s income by Rs. 60
Percentage less = \(\frac { 60 x 100 }{ 160 }\) = \(\frac { 75 }{ 2 }\) % = 37.5%

Question 26.
Rs. 3500 is to be shared among three people so that the first person gets 50% of the second who in turn gets 50% of the third. How much will each of them get ?
Solution:
Let the third person gets = Rs. x
Then second person will get
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 20
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 21

Question 27.
After a 20% hike, the cost of Chinese Vase is Rs. 2000. What was the original price of the object ?
Solution:
Let the original price of the vase = Rs. x
Hike in price = 20%
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 22
Original price of the vase = Rs. 1666.66

Hope given RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.