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

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
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q1.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q1.2

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
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q2.1

Question 3.
Solution:
Let r be the radius of the sphere
∴ Volume = \(\frac { 4 }{ 3 } \) πr3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q3.1

Question 4.
Solution:
Surface area of a sphere = 394.24 m2
Let r be the radius, then 4πr2 = 394.24
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q4.1

Question 5.
Solution:
Surface area of sphere = 576π cm2
Let r be the radius, then 4r2 = 576π
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q5.1

Question 6.
Solution:
Outer diameter of shell = 12cm,
Outer radius (R) = \(\frac { 12 }{ 2 } \) = 6cm
and inner diameter = 8cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q6.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q6.2

Question 7.
Solution:
Length of cuboid of (l) = 12cm
Breadth (b) = 11cm
and height (h) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q7.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q7.2

Question 8.
Solution:
Radius of sphere (r) = 8cm
Volume = \(\frac { 4 }{ 3 } \)πr3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q8.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q8.2

Question 9.
Solution:
Radius of solid sphere (R) = 3cm.
Volume = \(\frac { 4 }{ 3 } \)π(R)3 = \(\frac { 4 }{ 3 } \)π(3)3 cm3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q9.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q9.2

Question 10.
Solution:
Radius of metallic sphere (R) = 10.5cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q10.1

Question 11.
Solution:
Diameter of a cylinder = 8cm
Radius (r) = \(\frac { 8 }{ 2 } \) = 4cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q11.1

Question 12.
Solution:
Diameter of sphere = 6cm
Radius (R) = \(\frac { 6 }{ 2 } \) = 3cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q12.1

Question 13.
Solution:
Diameter of sphere = 18cm
Radius (R) = \(\frac { 18 }{ 2 } \) = 9cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q13.1

Question 14.
Solution:
Diameter of the sphere = 15.6 cm
Radius (R) = \(\frac { 15.6 }{ 2 } \) = 7.8 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q14.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q14.2

Question 15.
Solution:
Diameter of the canonball = 28cm
Radius (R) = \(\frac { 28 }{ 2 } \) = 14 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q15.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q15.2

Question 16.
Solution:
Given,
Radius of spherical big ball (R) = 3cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q16.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q16.2

Question 17.
Solution:
Ratio in the radii of two spheres = 1:2
Let radius of smaller sphere = r then,
radius of bigger sphere = 2r
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q17.1

Question 18.
Solution:
Let r1 and r2 be the radii of two spheres
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q18.1

Question 19.
Solution:
Radius of the cylindrical tub = 12cm.
First level of water = 20cm
Raised water level = 6.75cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q19.1

Question 20.
Solution:
Radius of the ball (r) = 9cm.
Volume of ball = \(\frac { 4 }{ 3 } \)πr³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q20.1

Question 21.
Solution:
Given,
Radius of hemisphere of lead (r) = 9cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q21.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q21.2

Question 22.
Solution:
Given,
Radius of hemispherical bowl (r) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q22.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q22.2

Question 23.
Solution:
External radius of spherical shell (R) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q23.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q23.2

Question 24.
Solution:
Inner radius (r) = 4 cm
Thickness of steel used = 0.5
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q24.1

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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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G

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
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 1

Question 2.
Solution:
x2 + 18x + 32
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 2.1

Question 3.
Solution:
x2 + 7x – 18
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 3.1

Question 4.
Solution:
x2 + 5x – 6
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 4.1

Question 5.
Solution:
y2 – 4y + 3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 5.1

Question 6.
Solution:
x2 – 21x + 108
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 6.1

Question 7.
Solution:
x2 – 11x – 80
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 7.1

Question 8.
Solution:
x2 – x – 156
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 8.1

Question 9.
Solution:
z2 – 32z – 105
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 9.1

Question 10.
Solution:
40 + 3x – x2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 10.1

Question 11.
Solution:
6 – x – x2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 11.1

Question 12.
Solution:
7x2 + 49x + 84
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 12.1

Question 13.
Solution:
m2 + 17mn – 84n2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 13.1

Question 14.
Solution:
5x2 + 16x + 3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 14.1

Question 15.
Solution:
6x2 + 17x + 12
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 15.1

Question 16.
Solution:
9x2 + 18x + 8
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 16.1

Question 17.
Solution:
14x2 + 9x + 1
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 17.1

Question 18.
Solution:
2x2 + 3x – 90
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 18.1

Question 19.
Solution:
2x2 + 11x – 21
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 19.1

Question 20.
Solution:
3x2 – 14x + 8
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 20.1

Question 21.
Solution:
18x2 + 3x- 10
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 21.1

Question 22.
Solution:
15x2 + 2x – 8
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 22.1

Question 23.
Solution:
6x2 + 11x – 10
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 23.1

Question 24.
Solution:
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 24.1

Question 25.
Solution:
24x2 – 41x + 12
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 25.1

Question 26.
Solution:
2x2 – 7x – 15
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 26.1

Question 27.
Solution:
6x2 – 5x – 21
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 27.1

Question 28.
Solution:
10x2 – 9x – 7
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 28.1

Question 29.
Solution:
5x2 – 16x – 21
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 29.1

Question 30.
Solution:
2x2 – x – 21
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 30.1

Question 31.
Solution:
15x2 – x – 28
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 31.1

Question 32.
Solution:
8a2 – 27ab + 9b2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 32.1

Question 33.
Solution:
5x2 + 33xy -14y2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 33.1

Question 34.
Solution:
3x3 – x2 – 10x
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 34.1

Question 35.
Solution:
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 35.1

Question 36.
Solution:
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 36.1

Question 37.
Solution:
√2x2 + 3x + √2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 37.1

Question 38.
Solution:
√5x2 + 2x – 3√5
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 38.1

Question 39.
Solution:
2a2 + 3√3x + 3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 39.1

Question 40.
Solution:
2√3x² + x – 5√3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 40.1

Question 41.
Solution:
5√5x2 + 20x + 3√5
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 41.1

Question 42.
Solution:
7√x² – 10x – 4√2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 42.1

Question 43.
Solution:
6√3 x2 – 47x + 5√3
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 43.1

Question 44.
Solution:
7x2 + 2√14x + 2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 44.1

Question 45.
Solution:
2(x + y)2 – 9(x + y) – 5
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 45.1

Question 46.
Solution:
9(2a-b)2-4(2a-b)-13
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 46.1

Question 47.
Solution:
7(x – 2y)2 – 25 (x – 2y) + 12
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 47.1

Question 48.
Solution:
4x4 + 7x2-2
RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2G 48.1

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

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.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q1.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q1.2
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q1.3

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°
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q3.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q3.2

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)
∠BAD = 43°
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q6.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q6.2

Question 7.
Solution:
Chord DE || diameter AC of the circle with centre O.
∠CBD = 60°
∠CBD = ∠ CAD
(Angles in the same segment of a circle)
∠CAD = 60°
Now in ∆ ADC,
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q7.1

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)
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q8.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q8.2

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)
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q9.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q9.2

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
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q10.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q10.2

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)
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q11.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q11.2

Question 12.
Solution:
In the figure, O is the centre of a circle ∠ OAB = 20° and ∠ OCB = 55° .
In ∆ OAB,
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q12.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q12.2

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
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q13.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q13.2

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)
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q14.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q14.2
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B Q14.3

Hope given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11B 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.

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

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
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B Q1.1

Question 2.
Solution:
Diameter of the base of the cylinder = 28cm
Radius = \(\frac { 1 }{ 2 } \) x 28 = 14 cm
Height (h) = 40cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B Q2.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B Q2.2

Question 3.
Solution:
Radius of cylinder (r) = 10.5cm
Height (h) = 60cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B Q3.1

Question 4.
Solution:
Diameter of cylinder = 20cm
Radius (r) = \(\frac { 20 }{ 2 } \) = 10cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 04.1

Question 5.
Solution:
Curved surface area of cylinder = 4400 cm²
Circumference of its base = 110 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 05.1

Question 6.
Solution:
The ratio of the radius and height of a cylinder = 2:3
Volume =1617 cm³
Let radius = 2x
and height = 3x.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 06.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 06.2

Question 7.
Solution:
Total surface area of the cylinder = 462 cm²
Curved surface area = \(\frac { 1 }{ 3 } \) x 462 = 154
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 07.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 07.2

Question 8.
Solution:
Total surface area of solid
cylinder = 231 cm²
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 08.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 08.2

Question 9.
Solution:
Sum of radius and height = 37m.
and total surface area = 1628 m²
Let r be the radius
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 09.1

Question 10.
Solution:
Total surface area = 616 cm²
Curved surface area = \(\frac { 616X1 }{ 2 } \) = 308
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 010.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 010.2

Question 11.
Solution:
Volume of gold = 1 cm³
diameter of wire = 0.1 mn
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 011.1

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
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 012.1

Question 13.
Solution:
Side of square = 12cm
and height = 17.5cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 013.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 013.2

Question 14.
Solution:
Diameter of cylindrical bucket = 28cm
Radius (r) = \(\frac { 28 }{ 8 } \) = 14cm
Height (h) = 72cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 014.1

Question 15.
Solution:
Length of pipe (l) = 1m = 100cm
diameter of pipe = 3cm.
Inner radius = \(\frac { 3 }{ 2 } \) cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 015.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 015.2

Question 16.
Solution:
Internal diameter of cylindrical tube = 10.4 cm
Radius (r) = \(\frac { 10.4 }{ 2 } \) = 5.2cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 016.1

Question 17.
Solution:
Length of barrel (h) = 7cm
Diameter = 5mm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 017.1

Question 18.
Solution:
Diameter of pencil = 7mm
.’. Radius (R) = \(\frac { 7 }{ 2 } \) mm = \(\frac { 7 }{ 20 } \) cm.
and diameter of graphite in it = 1mm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 018.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 018.2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13B 018.3

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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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A

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
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 1
are all rational numbers.

Question 2.
Solution:
The given rational number are represented on a number line on given below :
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 2
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 3
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 4

Question 3.
Solution:
We know that, if a and b are two rational numbers, then a rational number between a and b will be
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 5
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 6
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 7

Question 4.
Solution:
Here, n = 3, x = \(\frac { 1 }{ 5 } \), y = \(\frac { 1 }{ 4 } \)
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 8
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 9

Question 5.
Solution:
Here, n=5, x = \(\frac { 2 }{ 5 } \), y = \(\frac { 3 }{ 4 } \)
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 10
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 11
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 12

Question 6.
Solution:
Here, n = 6, x = 3, y = 4
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 13
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 14

Question 7.
Solution:
Here, n = 16, x = 2.1, y = 2.2
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 15
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 16
RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1A 17

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A 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.

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

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
Breadth (b) = 8cm
and height (h) = 4.5cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.3
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.4

Question 2.
Solution:
Length of closed rectangular cistern (l) = 8m
breadth (b) = 6m
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
Breadth (b) = 8m
and height (h) = 6.5m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q3.1

Question 4.
Solution:
Length of pit (l) = 20m
Breadth (b) = 6m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q4.1

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.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q5.1

Question 6.
Solution:
Length of wall (l) = 15m.
Width (b) = 30cm = \(\frac { 30 }{ 100 } =\frac { 3 }{ 10 } m\)
Height (h) = 4m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q6.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q6.2

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.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q7.1

Question 8.
Solution:
Depth of river = 2m
width = 45m.
Length of current in 60 minutes = 3km
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q8.1

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

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q8.2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q8.3

Question 10.
Solution:
Length of room (l) = 10m
Breadth (b) = 10m
Height (h) = 5m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q10.1

Question 11.
Solution:
Length of hall (l) = 20m
Breadth (b) = 16m
and height (h) = 4.5m.
Volume of the air inside the hall
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q11.1

Question 12.
Solution:
Length of class room (l) = 10m
Width (b) = 6.4 m
Height (h) = 5m.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q12.1

Question 13.
Solution:
Volume of cuboid = 1536 m³
Length (l) = 16m
Ratio in breadth and height = 3:2
Let breadth (b) = 3x
their height (h) = 2x
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q13.1

Question 14.
Solution:
Length of cuboid (l) = 14 cm
Breadth (b) = 11 cm .
Let height (h) =x cm
Surface area = 2(lb + bh + hl)
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q14.1

Question 15.
Solution:
(a) Edge of cube (a) = 9m .
(i) volume = a³ = (9)³ m³ = 729 m³
(ii) Lateral surface area = 4a²
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q15.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q15.2

Question 16.
Solution:
Total surface area of a cube = 1176 cm²
Let each edge he ‘a’
then 6a² =1176
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q1.16.1

Question 17.
Solution:
Lateral surface area of a cube = 900 cm²
Let ‘a’ be the edge of the cube
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q17.1

Question 18.
Solution:
Volume of a cube = 512 cm³
Let ‘a’ be its edge, then
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q18.1

Question 19.
Solution:
Edge of first-cube = 3 cm.
Volume = (3)³ = 27 cm³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13A Q19.1

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

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:
No. of ladies on whom survey was made = 200.
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

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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q1.1

Question 2.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q2.1

Question 3.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q3.1

Question 4.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q4.1

Question 5.
Solution:
Mean = 8
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q5.1

Question 6.
Solution:
Mean = 28.25
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q6.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q6.2

Question 7.
Solution:
Mean = 16.6
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q7.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q7.2

Question 8.
Solution:
Mean = 50
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q8.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q8.2

Question 9.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q9.1

Question 10.
Solution:
Let assumed mean = 67
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q10.1

Question 11.
Solution:
Here h = 1, Let assumed mean (A) = 21
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q11.1

Question 12.
Solution:
Here h = 400 and let assumed mean (A) = 1000
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q12.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F Q12.2

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

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)
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q5.1

Question 6.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q6.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q6.2

Question 7.
Solution:
Writing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q7.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q7.2

Question 8.
Solution:
Writing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q8.1
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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q9.1
Here number of terms is 50, which is even
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q9.2

Question 10.
Solution:
Preparing its cumulative frequency table :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q10.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q10.2

Question 11.
Solution:
Preparing its cumulative frequency table we have,
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q11.1
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q11.2

Question 12.
Solution:
Preparing its cumulative frequency table we have,
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H Q12.1
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G

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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q8.1
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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q9.1
Now preparing its cumulative frequency table
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q9.2
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
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q10.1
Now preparing its cumulative frequency table,we find:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q10.2
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
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q11.1
Now preparing its cumulative frequency table, we find :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q11.2
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:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q12.1
Now, preparing its cumulative frequency table.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G Q12.2
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.

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RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14E

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.

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