## Online Education for RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D

These Solutions are part of Online Education RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D.

Other Exercises

Question 1.
Solution:
(i) Radius of sphere = 3.5cm
(a) Volume = $$\frac { 4 }{ 3 }$$ πr3

Question 2.
Solution:
Let r be the radius of the sphere and volume = 38808 cm3
∴$$\frac { 4 }{ 3 }$$ πr3 = 38803
=> $$\frac { 4 }{ 3 }$$ x $$\frac { 22 }{ 7 }$$ r3 = 38803

Question 3.
Solution:
Let r be the radius of the sphere
∴ Volume = $$\frac { 4 }{ 3 }$$ πr3

Question 4.
Solution:
Surface area of a sphere = 394.24 m2
Let r be the radius, then 4πr2 = 394.24

Question 5.
Solution:
Surface area of sphere = 576π cm2
Let r be the radius, then 4r2 = 576π

Question 6.
Solution:
Outer diameter of shell = 12cm,
Outer radius (R) = $$\frac { 12 }{ 2 }$$ = 6cm
and inner diameter = 8cm

Question 7.
Solution:
Length of cuboid of (l) = 12cm
and height (h) = 9cm

Question 8.
Solution:
Radius of sphere (r) = 8cm
Volume = $$\frac { 4 }{ 3 }$$πr3

Question 9.
Solution:
Radius of solid sphere (R) = 3cm.
Volume = $$\frac { 4 }{ 3 }$$π(R)3 = $$\frac { 4 }{ 3 }$$π(3)3 cm3

Question 10.
Solution:
Radius of metallic sphere (R) = 10.5cm

Question 11.
Solution:
Diameter of a cylinder = 8cm
Radius (r) = $$\frac { 8 }{ 2 }$$ = 4cm

Question 12.
Solution:
Diameter of sphere = 6cm
Radius (R) = $$\frac { 6 }{ 2 }$$ = 3cm

Question 13.
Solution:
Diameter of sphere = 18cm
Radius (R) = $$\frac { 18 }{ 2 }$$ = 9cm.

Question 14.
Solution:
Diameter of the sphere = 15.6 cm
Radius (R) = $$\frac { 15.6 }{ 2 }$$ = 7.8 cm

Question 15.
Solution:
Diameter of the canonball = 28cm
Radius (R) = $$\frac { 28 }{ 2 }$$ = 14 cm

Question 16.
Solution:
Given,
Radius of spherical big ball (R) = 3cm

Question 17.
Solution:
Ratio in the radii of two spheres = 1:2
Let radius of smaller sphere = r then,
radius of bigger sphere = 2r

Question 18.
Solution:
Let r1 and r2 be the radii of two spheres

Question 19.
Solution:
Radius of the cylindrical tub = 12cm.
First level of water = 20cm
Raised water level = 6.75cm.

Question 20.
Solution:
Radius of the ball (r) = 9cm.
Volume of ball = $$\frac { 4 }{ 3 }$$πr³

Question 21.
Solution:
Given,

Question 22.
Solution:
Given,
Radius of hemispherical bowl (r) = 9cm

Question 23.
Solution:
External radius of spherical shell (R) = 9cm

Question 24.
Solution:
Inner radius (r) = 4 cm
Thickness of steel used = 0.5

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D are helpful to complete your math homework.

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## Online Education for RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G

These Solutions are part of Online Education RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2G.

Other Exercises

Factorize :

Question 1.
Solution:
x2 + 11x + 30

Question 2.
Solution:
x2 + 18x + 32

Question 3.
Solution:
x2 + 7x – 18

Question 4.
Solution:
x2 + 5x – 6

Question 5.
Solution:
y2 – 4y + 3

Question 6.
Solution:
x2 – 21x + 108

Question 7.
Solution:
x2 – 11x – 80

Question 8.
Solution:
x2 – x – 156

Question 9.
Solution:
z2 – 32z – 105

Question 10.
Solution:
40 + 3x – x2

Question 11.
Solution:
6 – x – x2

Question 12.
Solution:
7x2 + 49x + 84

Question 13.
Solution:
m2 + 17mn – 84n2

Question 14.
Solution:
5x2 + 16x + 3

Question 15.
Solution:
6x2 + 17x + 12

Question 16.
Solution:
9x2 + 18x + 8

Question 17.
Solution:
14x2 + 9x + 1

Question 18.
Solution:
2x2 + 3x – 90

Question 19.
Solution:
2x2 + 11x – 21

Question 20.
Solution:
3x2 – 14x + 8

Question 21.
Solution:
18x2 + 3x- 10

Question 22.
Solution:
15x2 + 2x – 8

Question 23.
Solution:
6x2 + 11x – 10

Question 24.
Solution:

Question 25.
Solution:
24x2 – 41x + 12

Question 26.
Solution:
2x2 – 7x – 15

Question 27.
Solution:
6x2 – 5x – 21

Question 28.
Solution:
10x2 – 9x – 7

Question 29.
Solution:
5x2 – 16x – 21

Question 30.
Solution:
2x2 – x – 21

Question 31.
Solution:
15x2 – x – 28

Question 32.
Solution:
8a2 – 27ab + 9b2

Question 33.
Solution:
5x2 + 33xy -14y2

Question 34.
Solution:
3x3 – x2 – 10x

Question 35.
Solution:

Question 36.
Solution:

Question 37.
Solution:
√2x2 + 3x + √2

Question 38.
Solution:
√5x2 + 2x – 3√5

Question 39.
Solution:
2a2 + 3√3x + 3

Question 40.
Solution:
2√3x² + x – 5√3

Question 41.
Solution:
5√5x2 + 20x + 3√5

Question 42.
Solution:
7√x² – 10x – 4√2

Question 43.
Solution:
6√3 x2 – 47x + 5√3

Question 44.
Solution:
7x2 + 2√14x + 2

Question 45.
Solution:
2(x + y)2 – 9(x + y) – 5

Question 46.
Solution:
9(2a-b)2-4(2a-b)-13

Question 47.
Solution:
7(x – 2y)2 – 25 (x – 2y) + 12

Question 48.
Solution:
4x4 + 7x2-2

Hope given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2G are helpful to complete your math homework.

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## Online Education for RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11B.

Other Exercises

Question 1.
Solution:
(i) O is the centre of the circle
∠OAB = 40°, ∠OCB = 30°
Join OB.

Question 2.
Solution:
O is the centre of the cirlce and ∠AOB = 70°
∵ Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.
∵ ∠ACB = $$\frac { 1 }{ 2 }$$ ∠AOB = $$\frac { 1 }{ 2 }$$ x 70°
=> ∠ACB = 35°
or ∠OCA = 35°
In ∆OAC,
OA = OC (radii of the same circle)
∴ ∠OAC = ∠OCA = 35° Ans.

Question 3.
Solution:
In the figure, O is the centre of the circle. ∠PBC = 25°, ∠APB =110°
∠ APB + ∠ BPC = 180° (Linear pair)
=> 110° + ∠ BPC = 180°

Question 4.
Solution:
O is the centre of the circle
∠ABD = 35° and ∠B AC = 70°
BOD is the diameter of the circle
∠BAD = 90° (Angle in a semi circle)
But ∠ADB + ∠ABD + ∠BAD = 180° (Angles of a triangle)
=> ∠ADB + 35° + 90° = 180°
=> ∠ADB + 125° = 180°
=> ∠ADB = 180° – 125° = 55°
But ∠ACB = ∠ADB (Angles in the same segment of the circle)
∠ACB = 55° Ans.

Question 5.
Solution:
O is the centre of a circle and ∠ACB = 50°
∴ arc AB subtends ∠ AOB at the centre and ∠ ACB at the remaining part of the circle.
∴ ∠ AOB = 2 ∠ ACB
= 2 x 50° = 100
∴ OA = OB (radii of the same circle)
∴ ∠ OAB = ∠ OBA (Angles opposite to equal sides)
Now in ∆ OAB,
∠ OAB + ∠ OBA + ∠ AOB = 180°
=> ∠ OAB + ∠ OAB + ∠ AOB = 180° (∠OAB = ∠OBA)
=> 2 ∠ OAB + 100°= 180°
=> 2 ∠ OAB = 180° – 100° = 80°
=> ∠OAB = $$\frac { { 80 }^{ o } }{ 2 }$$ = 40°
Hence, OAB = 40° Ans.

Question 6.
Solution:
(i) In the figure,
∠ABD = 54° and ∠BCD = 43°
∠BAD = ∠BCD (Angles in the same segment of a circle)

Question 7.
Solution:
Chord DE || diameter AC of the circle with centre O.
∠CBD = 60°
(Angles in the same segment of a circle)

Question 8.
Solution:
In the figure,
chord CD || diameter AB of the circle with centre O.
∠ ABC = 25°
Join CD and DO.
AB || CD
∠ ABC = ∠ BCD (alternate angles)

Question 9.
Solution:
AB and CD are two straight lines passing through O, the centre of the circle and ∠AOC = 80°, ∠CDE = 40°
∠ CED = 90° (Angle in a semi circle)

Question 10.
Solution:
O is the centre of the circle and ∠AOB = 40°, ∠BDC = 100°
Arc AB subtends ∠AOB at the centre and ∠ ACB at the remaining part of the circle
∠ AOB = 2 ∠ ACB

Question 11.
Solution:
Chords AC and BD of a circle with centre O, intersect each other at E at right angles.
∠ OAB = 25°. Join OB.
In ∆ OAB,
OA = OB (radii of the same circle)

Question 12.
Solution:
In the figure, O is the centre of a circle ∠ OAB = 20° and ∠ OCB = 55° .
In ∆ OAB,

Question 13.
Solution:
Given : A ∆ ABC is inscribed in a circle with centre O and ∠ BAC = 30°
To Prove : BC = radius of the circle
Const. Join OB and OC

Question 14.
Solution:
In a circle with centre O and PQ is its diameter. ∠PQR = 65°, ∠SPR = 40° and ∠PQM = 50°
(i) ∠PRQ = 90° (Angle in a semicircle) and ∠PQR + ∠RPQ + ∠PQR = 180° (Angles of a triangle)

Hope given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11B are helpful to complete your math homework.

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## Online Education for RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B

These Solutions are part of Online Education RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B.

Other Exercises

Question 1.
Solution:
Radius of the base of a cylinder (r) = 5cm.
and height (h) = 21cm

Question 2.
Solution:
Diameter of the base of the cylinder = 28cm
Radius = $$\frac { 1 }{ 2 }$$ x 28 = 14 cm
Height (h) = 40cm.

Question 3.
Solution:
Radius of cylinder (r) = 10.5cm
Height (h) = 60cm.

Question 4.
Solution:
Diameter of cylinder = 20cm
Radius (r) = $$\frac { 20 }{ 2 }$$ = 10cm

Question 5.
Solution:
Curved surface area of cylinder = 4400 cm²
Circumference of its base = 110 cm

Question 6.
Solution:
The ratio of the radius and height of a cylinder = 2:3
Volume =1617 cm³
and height = 3x.

Question 7.
Solution:
Total surface area of the cylinder = 462 cm²
Curved surface area = $$\frac { 1 }{ 3 }$$ x 462 = 154

Question 8.
Solution:
Total surface area of solid
cylinder = 231 cm²

Question 9.
Solution:
Sum of radius and height = 37m.
and total surface area = 1628 m²

Question 10.
Solution:
Total surface area = 616 cm²
Curved surface area = $$\frac { 616X1 }{ 2 }$$ = 308

Question 11.
Solution:
Volume of gold = 1 cm³
diameter of wire = 0.1 mn

Question 12.
Solution:
Ratio in the radii of two cylinders = 2:3
and ratio in the heights = 5:3
If r1 and r2 and the radii and h1 and h2 are the heights, then

Question 13.
Solution:
Side of square = 12cm
and height = 17.5cm

Question 14.
Solution:
Diameter of cylindrical bucket = 28cm
Radius (r) = $$\frac { 28 }{ 8 }$$ = 14cm
Height (h) = 72cm.

Question 15.
Solution:
Length of pipe (l) = 1m = 100cm
diameter of pipe = 3cm.
Inner radius = $$\frac { 3 }{ 2 }$$ cm

Question 16.
Solution:
Internal diameter of cylindrical tube = 10.4 cm
Radius (r) = $$\frac { 10.4 }{ 2 }$$ = 5.2cm.

Question 17.
Solution:
Length of barrel (h) = 7cm
Diameter = 5mm.

Question 18.
Solution:
Diameter of pencil = 7mm
.’. Radius (R) = $$\frac { 7 }{ 2 }$$ mm = $$\frac { 7 }{ 20 }$$ cm.
and diameter of graphite in it = 1mm

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B are helpful to complete your math homework.

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## Online Education for RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A.

Other Exercises

Question 1.
Solution:
A number in the form of $$\frac { p }{ q }$$ where p and q are integers and q ≠ 0, is called a rational number

are all rational numbers.

Question 2.
Solution:
The given rational number are represented on a number line on given below :

Question 3.
Solution:
We know that, if a and b are two rational numbers, then a rational number between a and b will be

Question 4.
Solution:
Here, n = 3, x = $$\frac { 1 }{ 5 }$$, y = $$\frac { 1 }{ 4 }$$

Question 5.
Solution:
Here, n=5, x = $$\frac { 2 }{ 5 }$$, y = $$\frac { 3 }{ 4 }$$

Question 6.
Solution:
Here, n = 6, x = 3, y = 4

Question 7.
Solution:
Here, n = 16, x = 2.1, y = 2.2

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A are helpful to complete your math homework.

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## Online Education for RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A.

Other Exercises

Question 1.
Solution:
(i) Length of cuboid (l) = 12cm
and height (h) = 4.5cm

Question 2.
Solution:
Length of closed rectangular cistern (l) = 8m
and depth (b) = 2.5m.
(i) .’. Volume of cistern = l.b.h.
= 8 x 6 x 2.5 m³ = 120m³
(ii) Total surface area = 2(lb + bh + hl)
= 2(8 x 6 + 6 x 2.5 + 2.5 x 8) cm²
= 2(48 + 15 + 20)
= 2 x 83 m²
= 166 m² Ans.

Question 3.
Solution:
Length of room (l) = 9m
and height (h) = 6.5m

Question 4.
Solution:
Length of pit (l) = 20m

Question 5.
Solution:
Length of wall (l) = 8m.
Width (b) = 22.5 cm = $$\frac { 225 }{ 10X100 } =\frac { 9 }{ 40 } m$$
and height (h) = 6m.
Volume of wall = l.b.h.

Question 6.
Solution:
Length of wall (l) = 15m.
Width (b) = 30cm = $$\frac { 30 }{ 100 } =\frac { 3 }{ 10 } m$$
Height (h) = 4m

Question 7.
Solution:
Outer length of opened cistern = 1.35m = 135 cm
Breadth = 1.08 m = 108 cm
Depth = 90cm
Thickness of iron = 2.5cm.

Question 8.
Solution:
Depth of river = 2m
width = 45m.
Length of current in 60 minutes = 3km

Question 9.
Solution:
Total cost of box = Rs. 1620
Rate per sq. m = Rs. 30

Question 10.
Solution:
Length of room (l) = 10m
Height (h) = 5m

Question 11.
Solution:
Length of hall (l) = 20m
and height (h) = 4.5m.
Volume of the air inside the hall

Question 12.
Solution:
Length of class room (l) = 10m
Width (b) = 6.4 m
Height (h) = 5m.

Question 13.
Solution:
Volume of cuboid = 1536 m³
Length (l) = 16m
Ratio in breadth and height = 3:2
their height (h) = 2x

Question 14.
Solution:
Length of cuboid (l) = 14 cm
Breadth (b) = 11 cm .
Let height (h) =x cm
Surface area = 2(lb + bh + hl)

Question 15.
Solution:
(a) Edge of cube (a) = 9m .
(i) volume = a³ = (9)³ m³ = 729 m³
(ii) Lateral surface area = 4a²

Question 16.
Solution:
Total surface area of a cube = 1176 cm²
Let each edge he ‘a’
then 6a² =1176

Question 17.
Solution:
Lateral surface area of a cube = 900 cm²
Let ‘a’ be the edge of the cube

Question 18.
Solution:
Volume of a cube = 512 cm³
Let ‘a’ be its edge, then

Question 19.
Solution:
Edge of first-cube = 3 cm.
Volume = (3)³ = 27 cm³

Question 20.
Solution:
Area of ground = 2 hectares
= 2 x 10000 = 20000 m²
Height of rain falls 5cm = $$\frac { 5 }{ 100 }$$m
∴ Volume of rain water = 20000 x $$\frac { 5 }{ 100 }$$ m³
= 1000 m³ Ans.

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A.

Question 1.
Solution:
Number of trials = 500 times
Let E be the no. of events in each case, then
∴No. of heads (E1) = 285 times
and no. of tails (E2) = 215 times
∴ Probability in each case will be
∴(i)P(E1) = $$\frac { 285 }{ 500 }$$ = $$\frac { 57 }{ 100 }$$ = 0.57
(ii) P(E2) = $$\frac { 215 }{ 500 }$$ = $$\frac { 43 }{ 100 }$$ = 0.43

Question 2.
Solution:
No. of trials = 400
Let E be the no. of events in each case, then
No. of 2 heads (E1) = 112
No. of one head (E2) = 160 times
and no. of O. head (E3) = 128 times
∴ Probability in each case will be:
∴ (i)P(E1) = $$\frac { 112 }{ 400 }$$ = $$\frac { 28 }{ 100 }$$ = 0.28
(ii)P(E2) = $$\frac { 160 }{ 400 }$$ = $$\frac { 40 }{ 100 }$$= 0.40
(iii) P(E3) = $$\frac { 128 }{ 400 }$$ = $$\frac { 32 }{ 100 }$$ = 0.32 Ans.

Question 3.
Solution:
Number of total trials = 200
Let E be the no. of events in each case, then
No. of three heads (E1) = 39 times
No. of two heads (E2) = 58 times
No. of one head (E3) = 67 times
and no. of no head (E4) = 36 times
∴ Probability in each case will be .
(i) P(E1) = $$\frac { 39 }{ 200 }$$ = 0.195
(ii) P(E3) = $$\frac { 67 }{ 200 }$$ = 0.335
(iii) P(E4) = $$\frac { 36 }{ 200 }$$ = $$\frac { 18 }{ 100 }$$ = 0.18
(iv) P(E2) = $$\frac { 58 }{ 200 }$$ = $$\frac { 29 }{ 100 }$$ = 0.29

Question 4.
Solution:
Solution No. of trials = 300 times
Let E be the no. of events in each case, then
No. of outcome of 1(E1) = 60
No. of outcome of 2(E2) = 72
No. of outcome of 3(E3) = 54
No. of outcome of 4(E4) 42
No. of outcome of 5(E5) = 39
No. of outcome of 6(E6) = 33
The probability of
(i) P(E3) = $$\frac { 54 }{ 300 }$$ = $$\frac { 18 }{ 100 }$$ = 0.18
(ii) P(E6) = $$\frac { 33 }{ 100 }$$ = $$\frac { 11 }{ 100 }$$= 0.11
(iii) P(E5) = $$\frac { 39 }{ 300 }$$ = $$\frac { 13 }{ 100 }$$ = 0.13
(iv) P(E1) = $$\frac { 60 }{ 300 }$$ = $$\frac { 20 }{ 100 }$$= 0.20 Ans.

Question 5.
Solution:
Let E be the no. of events in each case.
No. of ladies who like coffee (E1) = 142
No. of ladies who like coffee (E2) = 58
Probability of
(1) P(E1) = $$\frac { 142 }{ 200 }$$ = $$\frac { 71 }{ 100 }$$ = 0.71
(ii) P(E2) = $$\frac { 58 }{ 200 }$$ = $$\frac { 29 }{ 100 }$$ = 0.29 Ans.

Question 6.
Solution:
Total number of tests = 6
No. of test in which the students get more than 60% mark = 2
Probability will he
P(E) = $$\frac { 2 }{ 6 }$$ = $$\frac { 1 }{ 3 }$$Ans.

Question 7.
Solution:
No. of vehicles of various types = 240
No. of vehicles of two wheelers = 64.
Probability will be P(E) = $$\frac { 84 }{ 240 }$$ = $$\frac { 7 }{ 20 }$$ = 0.35 Ans.

Question 8.
Solution:
No. of phone numbers are one page = 200
Let E be the number of events in each case,
Then (i) P(E5) = $$\frac { 24 }{ 200 }$$ = $$\frac { 12 }{ 100 }$$ = 0.12
(ii) P(E8) = $$\frac { 16 }{ 200 }$$ = $$\frac { 8 }{ 100 }$$ = 0.08 Ans.

Question 9.
Solution:
No. of students whose blood group is checked = 40
Let E be the no. of events in each case,
Then (i) P(E0) = $$\frac { 14 }{ 40 }$$ = $$\frac { 7 }{ 20 }$$ = 0.35
(ii) P(EAB) = $$\frac { 6 }{ 40 }$$ = $$\frac { 3 }{ 20 }$$ = 0.15 Ans.

Question 10.
Solution:
No. of total students = 30.
Let E be the number of elements, this probability will be of interval 21 – 30
P(E) = $$\frac { 6 }{ 30 }$$ = $$\frac { 1 }{ 5 }$$ = 0.2 Ans.

Question 11.
Solution:
Total number of patients of various age group getting medical treatment = 360
Let E be the number of events, then
(i) No. of patient which are 30 years or more but less than 40 years = 60.
P(E) = $$\frac { 60 }{ 360 }$$ = $$\frac { 1 }{ 6 }$$
(ii) 50 years or more but less than 70 years = 50 + 30 = 80
P(E) = $$\frac { 80 }{ 360 }$$ = $$\frac { 2 }{ 9 }$$
(iii) Less than 10 years = zero
P(E) = $$\frac { 0 }{ 360 }$$ = 0
(iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360

Hope given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F.

Other Exercises

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Mean = 8

Question 6.
Solution:
Mean = 28.25

Question 7.
Solution:
Mean = 16.6

Question 8.
Solution:
Mean = 50

Question 9.
Solution:

Question 10.
Solution:
Let assumed mean = 67

Question 11.
Solution:
Here h = 1, Let assumed mean (A) = 21

Question 12.
Solution:
Here h = 400 and let assumed mean (A) = 1000

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H.

Other Exercises

Question 1.
Solution:
Arranging the given data in ascending order :
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
We see that 6 occurs in maximum times.
Mode = 6 Ans.

Question 2.
Solution:
Arranging in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
We see that 25 occurs in maximum times.
Mode = 25 Ans.

Question 3.
Solution:
Arranging in ascending order we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Here, we see that 9 occurs in maximum times.
Mode = 9 Ans.

Question 4.
Solution:
Arranging in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Here, we see that 50 occurs in maximum times.
Modal score = 50 scores Ans.

Question 5.
Solution:
Arranging in ascending order, we get:
10, 10, 11, 11, 12, 12, 13, 14,15, 17
Here, number of terms is 10, which is even
∴ Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (5th term + 6th term)

Question 6.
Solution:

Question 7.
Solution:
Writing its cumulative frequency table

Question 8.
Solution:
Writing its cumulative frequency table

Here, number of items is 40 which is even.
∴ Median = $$\frac { 1 }{ 2 } \left[ \frac { 40 }{ 2 } th\quad term+\left( \frac { 40 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (20th term + 21th term)
= $$\frac { 1 }{ 2 }$$ (30 + 30) = $$\frac { 1 }{ 2 }$$ x 60 = 30
Mean= $$\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$$ = $$\frac { 1161 }{ 40 }$$ = 29.025
∴Mode = 3 median – 2 mean = 3 x 30 – 2 x 29.025 = 90 – 58.05 = 31.95

Question 9.
Solution:
Preparing its cumulative frequency table we get:

Here number of terms is 50, which is even

Question 10.
Solution:
Preparing its cumulative frequency table :

Question 11.
Solution:
Preparing its cumulative frequency table we have,

Question 12.
Solution:
Preparing its cumulative frequency table we have,

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H 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.

## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G.

Other Exercises

Question 1.
Solution:
Arranging in ascending order, we get:
2,2,3,5,7,9,9,10,11
Here, number of terms is 9 which is odd.
∴ Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 9+1 }{ 2 }$$ th term = 5th term = 7 Ans.
(ii) Arranging in ascending order, we get: 6, 8, 9, 15, 16, 18, 21, 22, 25
Here, number of terms is 9 which is odd.
∴ Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 9+1 }{ 2 }$$ th term = 5th term = 16 Ans.
(iii) Arranging in ascending order, we get: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, number of terms is 11 which is odd.
∴ Median = $$\frac { 11+1 }{ 2 }$$ th term = $$\frac { 12 }{ 2 }$$ th term = 6th term = 16 Ans.
(iv) Arranging in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, number of terms is 13, which is odd.
Median = $$\frac { 13+1 }{ 2 }$$ th term = $$\frac { 14 }{ 2 }$$ th term = 7th term = 4 Ans.

Question 2.
Solution:
Arranging in ascending order, we get 9, 10, 17, 19, 21, 22, 32, 35
Here, number of terms is 8 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [4th term + 5th term] = $$\frac { 1 }{ 2 }$$ (19 + 21) = $$\frac { 1 }{ 2 }$$ x 40 = 20
(ii) Arranging in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here number of terms is 10 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (60 + 63) = $$\frac { 1 }{ 2 }$$ x 123 = 61.5 Ans.
(iii) Arranging in ascending order we get
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here number of terms is 12 which is even.
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 12 }{ 2 } th\quad term+\left( \frac { 12 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (6th term + 7th term) = $$\frac { 1 }{ 2 }$$ (15 + 17)= $$\frac { 1 }{ 2 }$$ x 32
= 16 Ans.

Question 3.
Solution:
Arranging the given data in ascending order, we get :
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
∴ Median = $$\frac { 15+1 }{ 2 }$$ th term = $$\frac { 16 }{ 2 }$$ th term = 8th term = 23
∴ Median score = 23 Ans.

Question 4.
Solution:
Arranging in ascending order, we get:
143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1
Here, number of terms is 9 which is odd.
Median = $$\frac { 9+1 }{ 2 }$$ th term = $$\frac { 10 }{ 2 }$$ th term = 5th term = 147.3 cm
Hence, median height = 147.3 cm Ans.

Question 5.
Solution:
Arranging in ascending order, we get:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here number of terms is 8 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$[4th term + 5th term]
= $$\frac { 1 }{ 2 }$$ (13.4 + 14.3) = $$\frac { 1 }{ 2 }$$ (27.7) = 13.85
∴ Median weight = 13.85 kg. Ans.

Question 6.
Solution:
Arranging in ascending order, we get:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, number of terms is 10 which is even.
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [5th term + 6th term ] = $$\frac { 1 }{ 2 }$$ (40 + 44) = $$\frac { 1 }{ 2 }$$ x 84 = 42 .
∴ Median age = 42 years.

Question 7.
Solution:
The given ten observations are 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41
These are even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [5th term + 6th term ] = $$\frac { 1 }{ 2 }$$(x + 1 + x + 3) = $$\frac { 1 }{ 2 }$$(2x + 4)
= x + 2
But median is given = 24
∴ x + 2 = 24 => x = 24 – 2 = 22
Hence x = 22.

Question 8.
Solution:
Preparing the cumulative frequency table, we have:

Here, number of terms (n) = 41, which is odd,
Median = $$\frac { 41+1 }{ 2 }$$ th term = $$\frac { 42 }{ 2 }$$ th term = 21st term = 50 (∵ 20th to 28th term = 50)
Hence median weight = 50 kg Ans.

Question 9.
Solution:
Arranging first in ascending order, we get:

Now preparing its cumulative frequency table

Here, number of terms is 37 which is odd.
Median = $$\frac { 37+1 }{ 2 }$$ th term = $$\frac { 38 }{ 2 }$$ th term = 19 th term = 22 (∵18th to 21st = 22)
Hence median – 22 Ans.

Question 10.
Solution:
first arranging in ascending order we get

Now preparing its cumulative frequency table,we find:

Here, number of terms is 43, which if odd.
Median = $$\frac { 43+1 }{ 2 }$$ th term = $$\frac { 44 }{ 2 }$$ th term = 22nd term = 25 25 (∵ 11th to 26th = 25)

Question 11.
Solution:
Arranging in ascending order,we get

Now preparing its cumulative frequency table, we find :

Here, number of terms = 50 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 50 }{ 2 } th\quad term+\left( \frac { 50 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (154 + 155) = $$\frac { 1 }{ 2 }$$ (309) = 154.5 (∵ 22nd to 25th = 154, 26th to 34th= 155)

Question 12.
Solution:
Arranging in ascending order, we get:

Now, preparing its cumulative frequency table.

Here, number of terms is 60 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (30th term + 31st term)
= $$\frac { 1 }{ 2 }$$ (20 + 23) = $$\frac { 1 }{ 2 }$$ x 43 = 21.5 (∵ 18th to 30th term = 20, 31st term to 34th = 23)
Hence median = 21.5 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G 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.

## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14E

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E.

Other Exercises

Question 1.
Solution:
Mean marks of 7 students = 226
∴Total marks of 7 students = 226 x 1 = 1582
Marks obtained by 6 of them are 340, 180, 260, 56, 275 and 307
∴ Sum of marks of these 6 students = 340 + 180 + 260 + 56 + 275 + 307 = 1418
∴ Marks obtained by seventh student = 1582 – 1418 = 164 Ans.

Question 2.
Solution:
Mean weight of 34 students = 46.5 kg.
∴ Total weight of 34 students = 46.5 x 34 kg = 1581 kg
By including the weight of teacher, the mean weight of 35 persons = 500 g + 46.5 kg
= 46.5 + 0.5 = 47.0 kg
∴ Total weight = 47.0 x 35 = 1645 kg
∴ Weight of the teacher = (1645 – 1581) kg = 64 kg Ans.

Question 3.
Solution:
Mean weight of 36 students = 41 kg
∴ Total weight of 36 students = 41 x 36 kg = 1476 kg
After leaving one student, number of students = 36 – 1 =35
Their new mean = 41.0 – 200g = (41.0 – 0.2) kg = 40.8 kg.
∴ Total weight of 35 students = 40.8 x 35 = 1428 kg
∴ Weight of leaving student = (1476 – 1428) kg = 48 kg Ans.

Question 4.
Solution:
Average weight of 39 students = 40 kg
∴ Total weight of 39 students = 40 x 39 = 1560 kg
By admitting of a new student, no. of students = 39 + 1 =40
and new mean = 40 kg – 200 g = 40 kg – 0.2 kg = 39.8 kg
∴Weight of 40 students = 39.8 x 40 kg = 1592 kg
∴Weight of new student = 1592 – 1560 kg = 32 kg Ans.

Question 5.
Solution:
Average salary of 20 workers = Rs. 7650
∴Their total salary = Rs. 7650 x 20 = Rs. 153000
By adding the salary of the manager, their mean salary = Rs. 8200
∴Their total salary = Rs. 8200 x 21 = Rs. 172200
∴Salary of the manager = Rs. 172200 – Rs. 153000 = Rs. 19200 Ans.

Question 6.
Solution:
Average wage of 10 persons = Rs. 9000
Their total wage = Rs. 9000 x 10 = Rs. 90000
Wage of one person among them = Rs. 8100
and wage of new member = Rs. 7200
∴ New total wage = Rs. (90000 – 8100 + 7200) = Rs. 89100
Their new mean wage = Rs.$$\frac { 89100 }{ 10 }$$ = Rs. 8910

Question 7.
Solution:
Mean consumption of petrol for 7 months of a year = 330 litres
Mean consumption of petrol for next 5 months = 270 litres
∴Total consumption for first 7 months = 7 x 330 = 2320 l
and total consumption for next 5 months = 5 x 270 = 1350 l
Total consumption for 7 + 5 = 12 months = 2310 + 1350 = 3660 l
Average consumption = $$\frac { 3660 }{ 12 }$$ = 305 liters per month.

Question 8.
Solution:
Total numbers of numbers = 25
Mean of 15 numbers = 18
∴Total of 15 numbers = 18 x 15 = 270
Mean of remaining 10 numbers = 13
∴Total = 13 x 10 = 130
and total of 25 numbers = 270 + 130 = 400
∴Mean = $$\frac { 400 }{ 25 }$$ = 16 Ans.

Question 9.
Solution:
Mean weight of 60 students = 52.75 kg.
Total weight = 52.75 x 60 = 3165 kg
Mean of 25 out of them = 51 kg.
∴Their total weight = 51 x 25 = 1275 kg
∴ Total weight of remaining 60 – 25 = 35 students = 3165 – 1275 = 1890 kg 1890
Mean weight = $$\frac { 1890 }{ 35 }$$ = 54 kg Ans.

Question 10.
Solution:
Average increase of 10 oarsman = 1.5 kg.
∴ Total increased weight =1.5 x 10 = 15kg
Weight of out going oarsman = 58 kg .
∴ Weight of new oarsman = 58 + 15 = 73 kg Ans.

Question 11.
Solution:
Mean of 8 numbers = 35
∴Total of 8 numbers = 35 x 8 = 280
After excluding one number, mean of 7 numbers = 35 – 3 = 32
Total of 7 numbers = 32 x 7 = 224
Hence excluded number = 280 – 224 = 56 Ans.

Question 12.
Solution:
Mean of 150 items = 60
Total of 150 items = 60 x 150 = 9000
New total = 9000 + 152 + 88 – 52 – 8 = 9000 + 240 – 60 = 9180
∴New mean = $$\frac { 9180 }{ 150 }$$ = 61.2 Ans.

Question 13.
Solution:
Mean of 31 results = 60
∴Total of 31 results = 60 x 31 = 1860
Mean of first 16 results = 58
∴Total of first 16 results = 58 x 16 = 928
and mean of last 16 results = 62
∴Total of last 16 results = 62 x 16 = 992
∴16th result = (928 + 992) – 1860 = 1920 – 1860 = 60 Ans.

Question 14.
Solution:
Mean of 11 numbers = 42
∴Total of 11 numbers = 42 x 11 = 462
Mean of first 6 numbers = 37
∴Total of first 6 numbers = 37 x 6 = 222
Mean of last 6 numbers = 46
∴Total of last 6 numbers = 46 x 6 = 276
∴6th number = (222 + 276) – 462 = 498 – 462 = 36 Ans.

Question 15.
Solution:
Mean weight of 25 students = 52 kg
∴ Total weight of 25 students = 52 x 25 = 1300 kg
Mean weight of first 13 students = 48 kg
∴ Total weight of first 13 students = 48 x 13 kg = 624 kg
Mean of last 13 students = 55 kg
∴Total of last 13 students = 55 x 13 kg = 715 kg
∴Weight of 13th student = (624 + 715) – 1300 = 1339 – 1300 = 39 kg Ans.

Question 16.
Solution:
Mean of 25 observations = 80
Total of 25 observations = 80 x 25 = 2000
Mean of another 55 observations = 60
∴ Total of these 55 observations = 60 x 55 = 3300
Total number of observations = 25 + 55 = 80
and total of 80 numbers = 2000 + 3300 = 5300
Mean of 80 observations = $$\frac { 5300 }{ 80 }$$ = 66.25 Ans.

Question 17.
Solution:
Marks in English = 36
Marks in Hindi = 44
Marks in Mathematics = 75
Marks in Science = x
∴Total number of marks in 4 subjects = 36 + 44 + 75 + x = 155 + x
Average marks in 4 subjects = 50
∴Total marks = 50 x 4 = 200
∴155 + x = 200
=> x = 200 – 155
=> x = 45
Hence, marks in Science = 45 Ans.

Question 18.
Solution:
Mean of monthly salary of 75 workers = Rs. 5680
Total salary of 75 workers = Rs. 5680 x 75 = Rs. 426000
Mean salary of 25 among them = Rs. 5400
Total of 25 workers = Rs. 5400 x 25 = Rs. 135000
Mean salary of 30 among them = Rs. 5700
∴Total of 30 among them = Rs. 5700 x 30 = Rs. 171000
∴ Total salary of 25 + 30 = 55 workers = Rs. 135000 + 171000 = Rs. 306000
∴Total salary of remaining 75 – 55 = 20 workers = Rs. 426000 – 306000 = Rs. 120000
∴ Mean of remaining 20 workers = Rs. $$\frac { 120000 }{ 20 }$$ = Rs. 6000 Ans.

Question 19.
Solution:
Let distance between two places = 60 km
∴Time taken at the speed of 15 km/h = $$\frac { 60 }{ 15 }$$ = 4 hours
and time taken at speed of 10 km/h for coming back = $$\frac { 60 }{ 10 }$$ = 6 hours
Total the taken = 4 + 6 = 10
hours and distance covered = 60 + 60 = 120 km
∴Average speed = $$\frac { 120 }{ 10 }$$ =12 km/hr.

Question 20.
Solution:
No. of total students = 50
No. of boys = 40
∴ No. of girls = 50 – 40 = 10
Average weight of class = 44kg
∴Total weight of 50 students = 44 x 50 = 2200 kg
Average weight of 10 girls = 40 kg
∴Total weight = 40 x 10 = 400 kg
Total weight of 40 boys = 2200 – 400 = 1800 kg
∴Average weight of 40 boys = $$\frac { 1800 }{ 40 }$$kg = 45 kg Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E 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.