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Class 8 R S Aggarwal Maths solutions

Contents

20. Volume and Surface Area of Solids Ex20A,B

22. Bar Graphs Ex22

23. Pie ChartEx 23A

24. Probability Ex 24A

 

 

VOLUME AND SURFACE AREA OF SOLIDS

EXERCISE 20A

Q1. [i] l = 22cm, b = 12cm, h = 7.5 cm volume of cuboid = ?, lateral surface = ?, total surface=?

volume of cuboid = (l x b x h )cm3 = 22 x 12 x 7.5 = 1980 cm3.

lateral surface = [2 (l +b )x h] = [2 x (22 + 12) x7.5] = 510 cm3.

total surface = 2(lb + bh +lh) = 2 x (264 + 90 + 165) = 1038 cm3.

Q2. Vol. of the tank = (2.75 x 1.80 x 1.40) m3 = 6.93 m3 = 6.93 x 1000ltr = 6930 ltr.

(since 1m3 = 1000 ltr)

Q3. Vol. of iron = 105 x 70 x 1.5 = 11025 cm3 x 8 = 88200gm = 88.200kg.

Q4. Area of the courtyard = 3750 m2. www.rsmaths99.com

Then, vol. of the courtyard = 3750 x 0.01 = 37.50 cm3.

If 1cm3 cost is Rs. 6.40 then 37.50 cm3 cost = 37.50 cm3 x 6.40 = Rs. 240.

Q5. l = 16m, b= 12.5m, h = 4.5m. vol. of the hall = 16 x 12.5 x 4.5 = 900 m3.

If 3.6 m3 required for 1 person

1 m3

900 m3 = 250 person.

Q6. L= 1.2m = 120cm, b = 72cm, h = 54cm.

vol. of the box = 120 x 72 x 54 = 466560cm3

vol. of the soap = 6 x 4.5 x 4 = 108 cm3.

Num. of bars we put into box = = 4320.

Q7. Size of the match box (vol.) = 4 x 2.5 x 1.5 = 15 cm3.

Vol. of the packet for 144 match box = 15 x 144 = 2160 cm3.

Num. of packet = = = 350.

Q8. Plank size(vol.) = 200cm x 25cm x 8cm = 40000 cm3.

Wooden block size(vol.) = 500 x 70 x 32 = 1120000 cm3.

No. of planks = = 28. www.rsmaths99.com

Q9. Vol. of bricks = 25cm x 13.5cm x 6cm = 2025 cm3.

Vol. of wall = 800cm x 540cm x 33cm = 14256000 cm3.

Required bricks = = 7040.

Q10. Vol. of wall = 1500 x 30 x 400 = 18000000 cm3.

Vol. of bricks = 22 x 12.5 x 7.5 = 2062.5 cm3.

Required bricks = = 8727.27 .

Now, part consists of mortat = x 18000000 = 15000000cm3.

Required bricks = = 727.27. www.rsmaths99.com

No. of bricks required in the wall = 8727.27 - 727.27 = 8000.

Q11. Vol. of the cistern = 11.2 x 6 x 5.8 = 389.76 m3.

Now, 1m3 = 1000 ltr. Then . 389.76 m3 = 389.76 x 1000 = 389760 ltr.

Area of iron steel required = Surface area of the cistern

= 2(11.2 x 6 + 6x5.8 + 11.2x 5.8) m2 = 2 x 166.96 = 333.92 m2 .

Q12. Thickness of the sheet = = m = mm = 0.05mm.

Q13. 2-hectare = 10000m2. and rainfall = 5cm = 0.05m.

Vol. of water fell on 2-hectare field = 0.05m x 20000 m2 = 1000m3.

Q14. Here, h = 2m, b = 45m, rate of flowing =L= 3km/h. = m/min = 50m/min

Vol. of the river = 2 x 45 x 50 = 4500 m3.

Q15. Here, l = 5m, b = 3.5m, vol.= 14 m3.

Now, vol. of the pit = l x b x h. www.rsmaths99.com

  14 = 5 x 3.5 x h => 14 = 17.5 x h => h = = 0.8m = 80cm.

Q16. Here, b = 90cm =0.90m, h = 40 cm = 0.40, water contain= 576 ltrs. L = ?

Since, 1000ltrs = 1m3

1ltr = m3

576 ltrs = = 0.567 m3.

Vol. of the tank = 0.90 x 0.40 x L => 0.576 = 0.36 x L => L = = 1.6m.

Q17. Here, l = 5m, h = 36cm = 0.36m, vol. = 1,35 m3, b=?

Vol. of wood =1.35 = 5 x 0.36 x b => b = o.75m. hence, width of beam = 75cm.

Q18. Here, vol. of the room = 378 m3 area of floor = l x b = 84m2, h = ?

Vol. of room = lx bx h => 378 = 84 x h www.rsmaths99.com

h = = 4.5 m. Hence height of the room = 4.5 m.

Q19. Here, l= 260m, b =140m, h = ?, vol. of swimming pool = 54600 m3.

Vol. of the swimming pool = l x b x h

54600 = 260 x 140 x h => h = 1.5m.

Q20. External length = 60cm, external breadth = 45cm and external height = 32cm.

External vol. = (60 x 45 x 32)cm3 = 86400cm3

Internal length = {60 (2.5 x 2)}cm = (60 - 5)cm = 55 cm.

Internal breadth = {45 (2.5 x 2)}cm = (45 - 5)cm = 40 cm.

Internal height = {32 -(2.5 x 2)}cm = (32 - 5)cm = 27 cm.

Internal vol. = {55 x 40 x 27}cm3 = 59400cm3 .

vol. of wood used to make the box = (external vol.)-(internal vol.)

(86400 59400)cm3 = 27000 cm3 .

Q21. Here, external L= 36cm, b = 25cm, h = 16.5cm. internal L = 36 3 = 33cm,

b = 25 3 = 22 , h = 16.5 1.5 = 15cm.

Vol. of iron = Ext. vol. int. vol. = 36 x 25 x 16.5 33 x 22 x 15

= 14850 10890 = 3960cm3 .

If 1 cm3 of iron weights = 8.5 gm = 0.0085kg.

weight of the empty box = 3960 cm3 x 0.0085kg = 33.66 kg.

Q22. Here, External L = 56cm, b = 39cm, h= 30 cm.

Internal L = 56 6 = 50cm, b= 39 6 = 33cm, h = 30 6 = 24cm.

Capacity = internal vol. = 50 x 30 x 24 cm3 = 39600cm3 .

Vol. of wood = Ent. Vol. Int. vol.

= 50 x 39 x 30 50 x 30 x 24 = 65520 39600 = 25920cm3 .

Q23. External L = 62cm, b= 30cm, h= 18cm. Internal L = 62 4 58cm. b = 30 4 = 26cm,

H = 18 4 = 14cm.

Capacity of iron = internal vol. = 58 x 26 x 14 = 2112 cm.

Q24. External L = 80cm, b= 65 cm, h = 45cm. www.rsmaths99.com

Internal L = 80 5 = 75cm, b = 65 5 = 60 cm, h = 45 5 = 40 cm.

Capacity = internal vol. = 75 x 60 x 40 = 180000cm3.

Now, vol. of box = ext. vol inter. Vol. = 234000- 180000cm2 = 54000cm3 .

If 100cm3 of wood weights = 8g = 0.008kg. then,

1 cm3 =

54000cm3 = = 4.32kg.

Q25. I] a = 7m. vol. = (edge)3 = (7)3 = 7 x 7 x 7 = 343 cm3.

LSA =4(side)2 = 4 x 72 = 4 x 49 = 196 cm2 .

TSA = 6(side)2 = 6 x 72 = 6 x 49 = 294 cm2

Q26. Surface area of the cube = (6a2 )cm2 = 1176 cm2

=> 6a2 = 1176 => a2 = 196 => a = 14 (14 x 14 = 196)

Vol. of the cube = a3 = 143 = 14 x 14 x 14 = 2744 cm3 .

Q27. Vol. of the cube = a3 = 729 cm3. a = 9 (9 x 9 x9 = 729)

Surface area = 6a2 = 6 x 9 x9 = 486cm2 .

Q28. Vol. of the block = 2.25 x 1.5 x 0.27 = 911250 cm3 .

Vol. of the cube = (edge)3 = 45 x 45 x 45 = 91125 cm3 .

No. of cubes formed = = 10. www.rsmaths99.com

Q29. Let the side of the cube = 10cm then vol. = (edge)3 = 10 x 10 x 10 = 1000 cm3 .

If the double the side then, new vol. = 2(edge)3 = 20 x 20 x 20 = 8000 cm3 .

Now change in vol. = = = 8 times.

Surface area =(6 a2) = (6 x 102) = 6 x 10 x 10 = 600.

New surface area = 6 ( 2 x 10)2 = 6 x 20 x 20 = 2400.

Now change in vol. = = 4 times.

Q30. If the cost is Rs. 500 then the vol. = 1m3

Rs.1 www.rsmaths99.com

Rs.256 x 256 = 0.512m3 = 512000cm3 .

Vol. of a cube = a3 => 512000 = a3 => a = = 80cm.

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EXERCISE 20B

volume of the cylinder = r2 h cubic unit

Curved (lateral ) surface area of the cylinder(CSA) = 2rh sq unit.

Total surface area of cylinder (TSA)= 2r(h + r)

Q1. [i] vol. of cylinder = r2 h = x 7 x 7 x 50 = 7700cm3.

[ii] CSA = 2rh = 2 x x 7 x 50 = 2200 cm3.

[iIi] TSA =2r(h + r) = 2 x x 7 x (50 + 7) = 2508cm2 .

Q2. Here, r = 1.5m, h = 10.5m,

Vol. of cylinder = r2 h = x 1.5 x 1.5 x 10.5 = 74.25m3.

Since, 1 m3 = 1000L www.rsmaths99.com

74.25m3 = 74.25 x 1000 = 74250 liters.

Q3. Vol. of cylinder = r2 h = x 10 x 10 x 7 = 0.22m3.

Now, vol. of 1m3 wood weight = 225kg.

0.22m3 = 225 x 0.22 = 49.5kg.

Q4. Here, h = ?, vol. of cyl. = 1.54m3 , d = 140cm, then r = 70cm = 0.70m.

Vol. of cylinder = r2 h

1.54 = x 0.70 x 0.70 x h h = = 1m.

Q5. Here, h = 1m, vol. = 3850cm3 = 38.50m3 , d = 2r =?

Vol. of cylinder = r2 h => 3850 = x r2 x 100

=> r2 = = = = 3.5. d =2r = 2 x 3.5 = 7 cm.

Q6. Here, d = 14m, then r = 7m, h = 5m.

TSA = 2r(h + r) = 2 x x 7 (5 + 7) = 44 x 12 = 528 m2.

Q7. Circumference = 88cm, h = 60cm, vol. =?, CSA = ?

Let the radius of the cyl = r cm.

Then circumference of the box = (2r) cm = 88 cm

=> 2 x x r = 88 => r = = 14cm. www.rsmaths99.com

Vol. of the cyl. = r2 h = x 14 x 14 x 60 = 36960 cm3 .

CSA =2rh = 2 x x 14 x 60 = 5280cm3 .

Q8. Here, CSA = 2rh = 220m2 , h = 14m, vol. of the cyl. = ?.

  220 = 2 x x r x 14 => r = = = 2.5 m.

  Vol. of the cyl. = r2 h = x 2.5 x 2.5 x 14 = 275 m3.

Q9. Here, h = 8cm, vol. = 1232cm2 , CSA = ?, TSA =?

Vol. of the cyl. = r2 h. www.rsmaths99.com

  1232 = x r2 x 8 => r2 = = 49 r = 7cm.

  CSA = 2rh = 2 x x 7 x 8 = 352 cm2.

  TSA = 2r(h + r) 2 x x 7 (15) = 660 cm2 .

Q10. Let ratio of common multiple be x. then radius = 7xcm, h = 2xcm.

Given, vol. of cyl. = 8316 cm3

  r2 h = 8316 => x 7x2 x 2x = 8316. => => x 7 x 7 x 2x3 = 8316.

  x3 = = 27 = 33 => x = 3.

radius = 7x = 7 x 3 = 21cm, h = 2x = 2 x 3 = 6 cm.

TSA = 2r(h + r) = 2 x x 21(21 + 6) = 66 x 2 x 27 = 3564 cm2 .

Q11. Given, CSA of cyl. = 4400cm2 , circumference of base = 2r = 110.

  r = x 7 = cm. www.rsmaths99.com

CSA = 4400 =>2rh = 4400 => 110 x h = 4400.

h = = 40 cm.

vol. = r2 h = x x x 40 = 22 x 35 x 50 = 38500 cm3 .

Q12. Vol. of cuboid = (l x b x h ) = 5 x 5 x 14 = 350cm2 .

Vol. of cyl. = r2 h = x 3.5 x 3.5 x 12 = 462cm2 .

Difference = 462cm2 - 350cm2 = 112 cm2. Hence, the capacity of cyl is greater, and

It is 112 cm2 greater than the square can.

Q13. Here, d= 48cm then, r = 0.24m, h = 7m. www.rsmaths99.com

CSA = 2rh = 2 x x 0.24 x 7 = 10.56m2.

cost of painting for 1m2 = Rs. 250.

10.56m2 = Rs. 10.56m2 x 250 = Rs. 26.40 x 15 = Rs. 396.

Q14. Vol. of rectangular vessel = 22 x 16 x 14 = 4928 cm3.

Vol. of cyl. = r2 h = x 8 x 8 x h = = = 24.5cm

Q15. Let length of wire = h cm. then vol. of wire = vol. of cylinder

  r2 h = R2 H => ( )2 x h = x x 11

  h = x 20 x 20 = 1100cm = 11m.

Q16. Vol. of wire = vol. of cube www.rsmaths99.com

  r2 h = a3

  x x x h = x x

  h = = = 338.8 cm.

Q17. Let the height of platform be h1m. Then,

Vol. of plot = vol. of cyl.

28 x 11 x h1 = x x x 20. => h1 = = = 2.5m.

Q18. Let the height of embankment be h1 m . then,

Vol. of embankment = vol. of cyl. Well

  (R2 r2)h1 = r2 h => (142 - 72) x h1 = 72 x 12.

=> (14 7)(14 + 7) x h1 = 7 x 7 x 12 => h1 = = 4m.

height of embankment = 4m. www.rsmaths99.com

Q19. Here, d = 84cm, then r = 42cm = 0.42m, h = 1m. Then,

CSA = 2rh = 2 x x 0.42 x 1 = 2.64m2 .

Since, it covers 2.64m2 in one revolution, then in 750 revolution = 2.64m2 x 750

= 1980m2 . hence, the area of the road = 1980m2 .

Q20. Thickness of the cylinder = 1.5 cm

External Diameter = 12 cm , then radius = r = 6 cm

And internal radius = 4.5 cm, Height = 84 cm

Now, Total Vol. = r2 h = x 6 x 6 x 84 = 9505 cm3

Inner Vol. = r2 h = x 4.5 x 4.5 x 84 = 5346 cm3

Now, Vol. of the metal = total vol. inner vol. = 9504 5346 = 4158 cm3

Weight of iron = 4158 x 7.5 = 31185 g = 31.185 kg [given 1 cm3 = 7.5 g]

Q21. Length = 1m = 100 cm www.rsmaths99.com

Inner diameter = 12 cm, Radius = 6 cm

Now, Inner vol. = r2 h = x 7 x 7 x 100 = 15400 cm2

Vol. of the tube = total vol. Inner vol. = 15400 11314.286 = 4085.714 cm3

Density of the tube = 7.7g/cm3

Weight of the tube = vol. x density = 4085.714 x 7.7 = 31459.9978g = 31.459 kg

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Exercise 22 www.rsmaths99.com

Q1.

Subject

Hindi

English

Maths

Science

Social Science

Marks

43

56

80

65

50

Q2. www.rsmaths99.com

Sports

Cricket

Football

Tennis

Badminton

Swimming

No. of students

75

35

50

25

65

 

Q3. www.rsmaths99.com

Years

2005-06

2006-07

2007-08

2008-09

2009-10

No. of students

800

975

1100

1400

1625

 

The height of the bar are:

No. of students in 2005-06 = =32 small divisions.

No. of students in 2005-06 = =39 small divisions.

No. of students in 2005-06 = =44 small divisions.

No. of students in 2005-06 = =56 small divisions.

No. of students in 2005-06 = 5=65 small divisions.

OR www.rsmaths99.com

Q13. I] It shows the marks obtained by a students in five subjects.

Ii] mathematics

Iii] Hindi

Iv] 56

Q14. I] It gives the no. of families containing 2,3,4,5 members each

Ii] 40

iii] None www.rsmaths99.com

iv] Family having 3 members, 3 members

Q15. [i] Mount Everest, 8800m [ii] 44:41 [iii] 8800m, 8200m, 8000m, 7500m, 6000m.

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23. PIE CHART EXERCISE 23A

Q1. The monthly income of a family is Rs. 14400. The monthly expenditure of the family

on various items is given below. www.rsmaths99.com

Items

Rent

Food

Clothing

Education

Savings

Expenditure

(in Rs.)

4000

5400

2800

1800

400

Solution: Total money = Rs. 14400.

Central angle for a component =

Calculation of central angles

Item

Amount(in Rs.)

Central angle

Rent

4000

Food

5400

Clothing

2800

Education

1800

Savings

400

 

 

PROBABILITY EXERCISE -24A.

Probability of occurrence of an event P(E) =

www.rsmaths99.com

Q1. I] H, T II] HH, HT, TH, TT III] 1, 2, 3, 4, 5, 6 IV] 52.

Q2. Total no. of all possible outcome = 2. No. of tail = 1.

P(E) = = .

Q3. Total no. of all possible outcomes = HH, HT, TH, TT = 4

I] getting both tails means = 1

P(getting both heads) = .

Ii] getting at least one tail means HT, TH, TT. No of such outcomes= 3

P(getting at least 1 heads) = .

Iii] getting at the most 1 tail means HT, TH, TT. No of such outcomes= 3

P(getting at the most 1 heads) = . www.rsmaths99.com

Q4. Total no. of balls = 4 + 5 = 9. Now, P(E) =

I] P(getting white ball) = . ii] P(getting white ball) = .

Q5. Total no. of balls = 5 + 6 + 4 = 15. Now, P(E) = . Then,

I] P(getting green ball) = . ii] P(getting white ball) = =

Iii] P(getting white ball) = = .

Q6. Here, total no. of outcome = 10 + 20 = 30. Now,

P(getting prize) = = .

Q7. Here, total no. of outcome = 100. No. of defective outcome = 8.

I] P(getting defective)= = = .

ii] No. of non-defective outcome = 100 8 = 92.

P(getting non-defective) = = .

Q8. Here, total no. of outcome= 6. (1,2,3,4,5,6).

[i] P(getting 2) = = . www.rsmaths99.com

[ii] P(getting no. less than 3) = = .

[iii] P(getting a composite no.) = = .

[iv] P(getting not less than 4) = = .

Q9. Here, total no. of all possible outcome= 200.

Here, no. of dislike coffee outcome = 118.

P(getting dislike coffee) = = .

Q10. Here, total no. of all possible outcome= 19.

Now, prime nos. = 2, 3, 5, 7, 11, 13, 17, 19. No. of prime nos. = 8.

P(getting a prime no.) = .

Now, even nos. = 2, 4, 6, 8, 10, 12, 14, 16, 18. No. of even nos.= 9.

P(getting a even no.) = .

Q11. Here, total no. of all possible outcome= 52. I] Now, no. of king = 4.

P(getting a king) = = . ii] Now, no. of spade = 13.

P(getting spade) = = . iii] Now, no. of red queen= 2.

P(getting red queen) = = . iv] Now, no of a black 8 = 2.

P(getting red queen) = = . www.rsmaths99.com

Q12. Here, total no. of all possible outcome= 52.

I] Now, no. of a 4 = 4. P(getting a king) = = .

Ii] Now, no. of a queen = 4. P(getting a queen) = = .

Iii] Now, no. of a black card= 26. P(getting a black card) = = .

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