JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Jharkhand Board JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 12 Areas Related to Circles Exercise 12.3

Unless stated otherwise, use π = \(\frac{22}{7}\)

Question 1.
Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 1
Solution:
Given, PQ = 24 cm, PR = 7 cm.
We know any triangle drawn from diameter RQ to the circle is 90°.
Here, ∠RPQ = 90°
In right ΔRPQ, RQ2 = PR2 + PQ2 (By Pythagoras theorem)
RQ2 = 72 + 242
RQ2 = 49 + 576
RQ2 = 625
RQ = 25 cm
∴ Area of ΔRPQ = \(\frac{1}{2}\) × RP × PQ
= \(\frac{1}{2}\) × 7 × 24 = 84 cm2
∴ Area of semi-circle = \(\frac{\pi r^2}{2}=\frac{22}{7 \times 2}\left(\frac{25}{2}\right)^2\) (∵ r = \(\frac{\mathrm{PQ}}{2}=\frac{25}{2}\) cm)
= \(\frac{11 \times 625}{28}=\frac{6875}{28} \mathrm{~cm}^2\)
∴ Area of the shaded region = Area of the semi-circle – Area of right ΔRPQ
= \(\frac{6875}{28}-84\)
= \(\frac{6875-2352}{28}=\frac{4523}{28}\) cm2.

JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 2.
Find the area of the shaded region in the figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 2
Solution:
R – radius of the bigger circle, r – radius of the smaller circle.
Area of the shaded portion = Area of sector OAC – Area of sector OBD
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 3

Question 3.
Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 4
Solution:
Area of the shaded portion = Area of the square – 2 × Area of one semicircle
= (14 × 14) – 2 × \(\frac{\pi r^2}{2}\) [∵ r = 7]
= 14 × 14 – 2 × \(\frac{22}{7} \times \frac{7 \times 7}{2}\)
= 196 – 154 = 42 cm2.

Question 4.
Find the area of the shaded region in the figure, where a circular are of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 5
Solution:
Area of the shaded portion = Area of the circle of radius 6 cm + Area of equilateral ΔABC – Area of the sector
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 6

JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 7
Solution:
Area of the shaded portion = Area of the square – Area of the 4 quadrants – Area of the circle
= (4 × 4) – 4 × area of one quadrant – area of the circle
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 8

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 9
Solution:
Area of the design (shaded region) = Area of the circle – Area of ΔABC
Area of equilateral triangle ABC = \(\frac{\sqrt{3} a^2}{4}\)
In ΔABC, AL ⊥ BC.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 10
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 11

Alternative Method:
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 12
Area of shaded region
= 3[Area of segments]
= 3[Area of sector – Area of ΔOBC]
= 3[\(\pi r^2 \frac{\theta}{360^{\circ}}-\frac{1}{2}\) × BC × OL]

Question 7.
In the figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 13
Solution:
Area of the shaded region = Area of the square – Area of the 4 quadrants
= Area of the square – 4 × area of one quadrant
= (14)2 – 4 × \(\frac{1}{4}\)πr2
= (14)2 – 4 × \(\frac{1}{4} \times \frac{22}{7}\) × 7 × 7
= 196 – 154
= 42 cm2.

JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 8.
The figure depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m. long. If the track is 10 m wide, find:
(i) the distance around the track along its inner edge.
(ii) the area of the track.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 14
Solution:
i) Distance around the inner track
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 15
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 16

ii) Area of the track:
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 17
Area of the track = l × b + l × b + 2\(\left[\frac{\pi \mathrm{R}^2}{2}-\frac{\pi \mathrm{r}^2}{2}\right]\) r = 30 mts, R = (30 + 10) = 40 mts.
= 106 × 10 + 106 × 10 + 2 × \(\frac{\pi}{2}\)(R2 – r2)
= 1060 + 1060 + \(\frac{22}{7}\)[402 – 302]
= 2120 + \(\frac{22}{7}\)[1600 – 900]
= 2120 + \(\frac{22}{7}\)[700]
= 2120 + 2200
= 4320 m2.

Question 9.
In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 18
Solution:
Given,
OA = 7 cm
∴ OD = 7 cm
Now, area of smaller circle whose diameter (OD) = 7 cm is
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 19
Now,
Area of ΔABC = \(\frac{1}{2}\) × AB × OC
= \(\frac{1}{2}\) × 2 × OA × OC
= \(\frac{1}{2}\) × 14 × 7 (∵ OA = OC)
= 49 cm2.
Area of semi-circle ABCA = \(\frac{\pi r^2}{2}=\frac{22}{7 \times 2}(7)^2\)
= 77 cm2
∴ Area of segments BC and AC = Area of semi-circle – Area of AABC
= 77 – 49 = 28 cm2
∴ Area of total shaded region = Area of small circle + Area of segments BC and AC
= \(\frac{77}{2}\) + 28
= 38.5 + 28
= 66.5 cm2.

JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.73205).
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 20
Solution:
Area of the shaded portion = Area of equilateral ΔABC – Area of sector Axy – Area of sector Bxz – Area of sector Cyz.
π = 3.14, θ = 60°, r = ?, r = \(\frac{a}{2}\)
Area of the shaded portion = Area of the equilateral Δ – 3 × Area of one sector
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 21
Area of the shaded region = Area of ΔABC – \(\frac{3 \times \pi r^2 \theta}{360}\)
= 17320.5 – \(\frac{3 \times 3.14 \times 100 \times 100 \times 60}{360}\)
= 17320.5 – 1.57 × 10000
= 17320.5 – 15700.0
= 1620.5 sq.cm.

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 22
Solution:
Number of circular designs = 9
Radius of the circular design = 7 cm
There are three circles in one side of the square handkerchief.
∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm
Area of the square = 42 × 42 cm2 = 1764 cm2
Area of the circle = πr2 = \(\frac{22}{7}\) × 7 × 7 = 154 cm2
Total area of the design = 9 × 154 = 1386 cm2
Area of the remaining portion of the handkerchief = Area of the square – Total area of the design
= 1764 – 1386
= 378 cm2.

Question 12.
In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm find the area of the (i) quadrant OACB, (ii) shaded region.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 23
Solution:
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 24

Question 13.
In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 25
Solution:
Radius of the quadrant = Diagonal of the square (OB)
OB2 = OA2 + AB2
= 202 + 202
= 400 + 400
= 800
OB2 = 400 × 2
OB = \(\sqrt{400 \times 2}\) = 20\(\sqrt{2}\)
Area of the shaded region = Area of the quadrant OPBQ – Area of the square OABC
= \(\frac{1}{4}\) πr2 – (OA)2
= \(\frac{1}{4}\) × 3.14 × 20\(\sqrt{2}\) × 20\(\sqrt{2}\) – (20)2
= \(\frac{1}{4}\) × 3.14 × 400 × \(\sqrt{4}\) – 400
= \(\frac{1}{4}\) × 3.14 × 400 × 2 – 400
= 100 × 2 × 3.14 – 400
= 100 × 2(3.14 – 2)
= 200 × 1.14
= 228 sq.cm.

JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ∠AOB = 30°, find the area of the shaded region.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 26
Solution:
R = 21, r = 7, θ = 30°, π = \(\frac{22}{7}\)
Area of the shaded region = Area of sector OAB – Area of sector OCD
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 27

Question 15.
In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 28
Solution:
BC2 = 142 + 142
BC2 = 196 + 196 = 392
BC = \(\sqrt{392}\) = \(\sqrt{196 \times 2}\) = 14\(\sqrt{2}\) = d
r = \(\frac{14 \sqrt{2}}{2}=7 \sqrt{2}\)
Radius of the sector = 14.
Area of the shaded region Area of the semicircle BEC – Area of the segment BDCB
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 29

JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 16.
Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 30
Solution:
Area of the design = Area of sector DXB + Area of ΔDCB
Area of the segment DXB = Area of the sector DXBC – Area of ΔDCB
JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 31

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Jharkhand Board JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.4

Use π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 1

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
C.S.A. of the frustum = πl(r1 + r2)
2πr1 = 18
r1 = \(\frac{18}{2 \pi}=\frac{9}{\pi}\)
2πr2 = 6
r2 = \(\frac{6}{2 \pi}=\frac{3}{\pi}\)
C.S.A. of the frustum = xl(r1 + r2)
= π × \(4\left[\frac{9}{\pi}+\frac{3}{\pi}\right]\)
= π × 4 × \(\frac{12}{\pi}\)
= 48 cm2.3

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 2
Solution:
r1 = 4, r2 = 10, l = 15
Area of the material used = C.S.A. of the frustum + Area of the circular top
= πl(r1 + r2) + πr12
= \(\frac{22}{7}\) × 15 (4 + 10) + π × 42
= \(\frac{22}{7}\) × 15 × 14 + \(\frac{22}{7}\) × 16
= \(\frac{22}{7}\)(210 + 16)
= \(\frac{22}{7}\) × 226
= \(\frac{4972}{7}\)
= 710\(\frac{2}{7}\) cm2.

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2. (Take π = 3.14).
Solution:
Volume of the milk = Volume of the container = Volume of the frustum
= \(\frac{1}{3}\)πh(r12 + r22 + r1r2)   [r1 = 20, r2 = 8, h = 16]
= \(\frac{1}{3}\) × 3.14 × 16[(20)2 + (8)2 + (20 × 8)]
= \(\frac{1}{3}\) × 3.14 × 16[400 + 64 + 160]
= \(\frac{1}{3}\) × 3.14 × 16 × 624 c.c. [1 litre 1000 cc]
= \(\frac{1}{3}\) × \(\frac{3.14 \times 16 \times 624}{1000}\) litres.
Cost of the milk = \(\frac{3.14 \times 16 \times 624}{3 \times 1000}\)
= \(\frac{3.14 \times 1664}{25}\)
= 3.14 × 66.56
= 208.99 = Rs. 209.
Area of the metal sheet used = πl(r1 + r2) + πr22 [r1 = 20, r2 = 8, h = 16]
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 3
Area of the metal sheet used = πl(r1 + r2) + πr22
= 3.14 × 20(20 + 8) + 3.14 × 82
= 3.14 × 20 × 28 + 3.14 × 64
= 3.14[(20 × 28) + 64]
= 3.14(560 + 64)
= 3.14 × 624 sq. cms.
Cost of the metal sheet = Area × Rate
= 3.14 × 624 × rate
= \(\frac{3.14 \times 624 \times 8}{100}\)
= 3.14 × 624 × 0.08
= 3.14 × 49.92
= 156.7488
= Rs. 156.75.

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{1}{6}\) cm, find the length of the wire.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 4

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Jharkhand Board JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.3

Take π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
R = 4.2, r = 6
Volume of the cylinder = Volume of the melted sphere
πr2h = \(\frac{4}{3}\)πR3
3πr2h = 4πR3
h = \(\frac{4 \pi \mathrm{R}^3}{3 \pi \mathrm{r}^2}\)
= \(\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6}\)
= 4 × 1.4 × 0.7 × 0.7
= 5.6 × 0.49
= 2.744 cm.
Height of the cylinder = 2.74 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
r1 = 6 cm, r2 = 8 cm, r3 = 10 cm.
Let the radius of the resulting sphere be r cm.
Its volume = \(\frac{4}{3}\)πR3
Volume of the resulting sphere = Volume of the sphere of radius r1 + Volume of the sphere of radius r2 + Volume of the sphere of radius r3
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 1
Radius of the resulting sphere = 12 cm.

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 2
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 3

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.
Solution:
Volume of the earth got by digging the well = πr2h
= \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14\)
= 99 cm.
R = 1.5 + 4
r = 1.5
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 4
Area of the circular embankment around the well = πR2 – πr2
= π[(5.5)2 – (1.5)2]
= π(5.5 + 1.5) (5.5 – 1.5)
= π(7) (4).
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 5
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 6

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
Volume of ice cream in the cylindrical container =
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 7
Alternative Method:
Number of cones will be = \(\frac{\text { Volume of cylinder }}{\text { Volume of cone }}\)
For the cylinder part,
Radius = \(\frac{12}{2}\) = 6 cm, Height = 15 cm
∴ Volume of cylinder = π × r2 × h = 540π
For the cone part,
Radius of conical part = \(\frac{6}{2}\) = 3 cm, Height = 12 cm
Radius of hemispherical part = \(\frac{6}{2}\) = 3 cm
Now,
Volume of cone = Volume of conical part + Volume of hemispherical part
\(\left(\frac{1}{3}\right)\) × π × r2 × h + \(\left(\frac{2}{3}\right)\) × π × r3
= 36π + 18π
= 54π.
∴ Number of cones = \(\frac{540 \pi}{54 \pi}\) = 10

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 8

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 9
Volume of sand in the bucket = πr2h
= π × 18 × 18 × 32 cc.
Volume of the conical heap of sand = \(\frac{1}{3}\)πr2h
\(\frac{1}{3}\)πr2h = π × 18 × 18 × 32
πr2h = 3 × π × 18 × 18 × 32
πr2 × 24 = 3 × π × 18 × 18 × 32
r2 = \(\frac{3 \times \pi \times 18 \times 18 \times 32}{\pi \times 24}\)
r2 = 182 × 22
r = 18 × 2 = \(\frac{72}{2}\) = 36 cm.
l2 = h2 + r2
= (24)2 + (36)2
= (24 × 24) + (36 × 36)
= (2 × 12 × 12 × 2) + (3 × 12 × 12 × 3)
= 122 (4 + 9)
= 122 × 13
l = 12\(\sqrt{13}\) cm.

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Quantity of water flowing in the canal in 1 hour = lbh
l = 10 km, b = 6 m, h = 1.5 m
10 × 1000 × 6 × 1.5 cubic metres.
The area this water can irrigate in 1 hour if 8 cm of standing water is needed
= \(\frac{10 \times 1000 \times 6 \times 1.5}{0.08}\)
8 cm = 0.08 mm
Area needed to irrigate in 30 mins. or \(\frac{1}{2}\) hr.
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 10

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
d = 20 cm, r = 10 cm = \(\frac{10}{100}\) = 0.1 mt, h = 3 km = 3 × 1000 m.
Volume of water that comes out of the pipe of diameter 20 cm
= πr2h
= π × 0.1 × 0.1 × 3000
= π × 30 cm.
In 1 hr. (60 mins.), volume of water that flows into the tank = π × 30
In 1 minute = \(\frac{\pi \times 30}{60}=\frac{\pi}{2}\) cm.
Volume of the cylindrical tank = πr2h = π × 5 × 5 × 2. d = 10, r = 5
\(\frac{\pi}{2}\) cubic metres of water will be filled in 1 minute
π × 5 × 5 × 2……. \(\frac{\pi \times 5 \times 5 \times 2 \times 2}{\pi}\) = 100 mins.

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Jharkhand Board JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.2

Unless stated otherwise, take π = \(\frac{22}{7}\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone equal to its radius. Find the volume of the solid in terms οf π.
Solution:
The given solid is a combination of a cone and a hemisphere.
We have radius of the cone r = Radius of the hemisphere = 1 cm and height of the cone h = 1 cm.
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 1
∴ Volume of the solid = Volume of the cone + Volume of the hemisphere
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 2

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
d = 3 cm, AF = 8 cm = H, Ax = Fy = 2 cm = h, r = \(\frac{3}{2}\)
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 3
Volume of the model = Volume of the cone ABC + Volume of the cylinder BCED + Volume of the cone DEF
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 4

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Question 3.
A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 5
Volume of 1 jamun = Volume of the cylinder + 2 × Volume of the hemisphere
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 6

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 7
Volume of wood in the stand = Volume of the cuboid – Volume of wood lost in making four conical depressions
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 8

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Volume of water in the cone = \(\frac{\pi r^2 h}{3}\)
= \(\frac{1}{3} \times \frac{22}{7}\) × 5 × 5 × 8 cc.
= \(\frac{22 \times 25 \times 8}{21}\) c.c.
When some lead shots are dropped into the vessel, volume of the water that flows out
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 9
= \(\frac{1}{4} \times \frac{22 \times 25 \times 8}{21}\)
= \(\frac{22 \times 25 \times 2}{21}\) cc
This is equal to the volume of the lead shots dropped.
Volume of a lead shot = \(\frac{4}{3}\)πr3
Let the no. of lead shots dropped be x.
The volume of x lead shots = Volume of water overflown
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 10
Number of lead shots dropped = 100.

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surrounded by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14).
Solution:
R = 12, r = 4.
Common factor of 144 and 64 is 16 (HCF)
HCF of 220 and 60 is 20.
1 kg = 1000 gms.
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 11
Volume of the given solid = Volume of the bigger cylinder + Volume of the surmounted cylinder
= πR2H + πr2h
= 3.14 × 122 × 220 + 3.14 × 82 × 60
= 3.14 × 144 × 220 + 3.14 × 64 × 60
= 3.14 × 16 × 20(9 × 11 + 4 × 3)
= 3.14 × 320(99 + 12)
= 3.14 × 320 × 111 c.c.
Mass of the solid = Volume × density
= 3.14 × 320 × 111 × 8 gm
= \(\frac{3.14 \times 320 \times 111 \times 8}{1000}\)kg.
= \(\frac{3.14 \times 32 \times 888}{100}\)
= 3.14 × 32 × 8.88
= 892.26 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
h = 180 (cylinder), H = 120 (Cone).
Volume of the solid given = Volume of the cone + Volume of the hemisphere
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 12
Volume of water left in the cylinder = Volume of water in the cylinder – Volume of the solid immersed
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 13
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 14

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
r = \(\frac{8.5}{2}\), R = \(\frac{2}{2}\) = 1
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 15
Volume of water in the vessel = Volume of water in the spherical part + Volume of water in the cylindrical neck.
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 16
= 3.14 × 110.354 = 346.51 cm3.
The child’s answer is wrong.

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Jharkhand Board JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.1

Unless stated otherwise, take π = \(\frac{22}{7}\)

Question 1.
Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of the cube = a3 = 64 cm3
Side of the cube \(\sqrt[3]{64}\) = a = 4 cm.
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 1
Surface area of the cuboid = 2(lb + bh + lh)
= 2[(8 × 4) + (4 × 4) + (8 × 4)]
= 2[32 + 16 + 32]
= 2 × 80
= 160 cm2.

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 2
π = \(\frac{22}{7}\), radius of the hemisphere = 7 cm,
height of the hemisphere = 7 cm,
height of the cylinder, h = 13 – 7 = 6 cm.
Inner area of the vessel = Inner area of the hemisphere vessel + Inner area of the cylinder
= 2πr2 + 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 7 + 2 × \(\frac{22}{7}\) × 7 × 6
= 2 × 22 × 7+2 × 22 × 6
= 2 × 22(7 + 6)
= 44 × 13 = 572 cm2.

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 3
π = \(\frac{22}{7}\), r = 3.5, l = ?
Total surface area of the toy = C.S.A. of the cone + C.S.A. of the hemisphere
= πrl + 2πr2
lv = r2 + h2
= (3.5)2 + (12)2
= 12.25 + 144
= 156.25
l = \(\sqrt{156.25}\)
= 12.5 cm.

h = Ax – Ox
= 15.5 – 3.5
= 12 cm.

Surface area of the toy = πrl + 2πr2
= \(\frac{22}{7}\) × 3.5 × 12.5 + 2 × \(\frac{22}{7}\) × 3.5 × 3.5
= \(\frac{22}{7}\) × 3.5[12.5 + 2 × 3.5]
= 22 × 0.5[12.5 + 7]
= 11[12.5 + 7]
= 11 × 19.5 = 214.5 cm2.

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 4
Greatest diameter = 7 cm.
Surface area of the block = T.S.A. of the cube – Base area of the hemisphere + C.S.A. of the hemisphere
= 6 × 72 – πr2 + 2πr2
= 6 × 49 + πr2
= 6 × 49 + \(\frac{22}{7}\) × 3.5 × 3.5
= 294 + 11 × 3.5
= 294 + 38.5
= 332.5 cm2.

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 5
Surface area of the remaining solid = Surface area of the cube + Surface area of the hemisphere
= 6l2 + 2πr2
= 6l2 + 2π\(\left(\frac{l}{2}\right)^2\)
= 6l2 + 2π\(\frac{l^2}{4}=\frac{l^2}{4}\)(24 + 2π) sq. units.

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 6
Height of the cylindrical portion = 14 – 2.5 – 2.5 = 9m = h
r = 2.5 m.
Surface area of the capsule = Surface area of the cylindrical portion + Areas of the hemispherical regions
= 2πrh + 2πr2 + 2πr2
= 2πr(h + r + r)
= 2 × \(\frac{22}{7}\) × 2.5 (9 + 2.5 + 2.5)
= 2 × \(\frac{22}{7}\) × 2.5 (9 + 5)
=2× \(\frac{22}{7}\) × 2.5 × 14
= 44 × 5
= 220 mm2.

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per m2. (Note that the base of the tent will not be covered with canvas.)
Solution:
r = \(\frac{4}{2}\) = 2m, h = 2.1, l = 2.8
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 7
Area of the canvas used = C.S.A. of the cylindrical portion + C.S.A. of the conical region
= 2πrh + πrl
= πr(2h + l)
= \(\frac{22}{7}\) × 2(2 × 2.1 + 2.8)
= \(\frac{22}{7}\) × 2(4.2 + 2.8)
= \(\frac{22}{7}\) × 2 × 7
= 44 m2.
Cost of the canvas = Area × Rate
= 44 × 500
= Rs. 22000.

JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Height of conical part = Height of cylindrical part h = 2.4 cm.
Diameter of cylindrical part = 1.4 cm, so, the radius of cylindrical part r = 0.7 cm
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 8
Slant height of cylindrical part l = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(0.7)^2+(2.4)^2}\)
= \(\sqrt{0.49+5.76}\)
= \(\sqrt{6.25}\)
= 2.5
The total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of base of cylinder
= 2πrh + πrl + πr2
= 2 × \(\frac{22}{7}\) × 0.7 × 2.4 + \(\frac{22}{7}\) × 0.7 × 2.5 + \(\frac{22}{7}\) × 0.7 × 0.7
= 4.4 × 2.4 + 2.2 × 2.5 + 2.2 × 0.7
= 10.56 + 5.50 + 1.56
= 17.60 cm2.

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:
JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 9
Total surface area of the article = C.S.A. of the cylinder + Surface area of the hemisphere at the top + Surface area of the hemisphere at the bottom
= 2πrh + 2πr2 + 2πr2
= 2πr(h + r + r)
= 2 × \(\frac{22}{7}\) × 3.5 (10 + 3.5 + 3.5)
= 2 × 22 × 0.5 × 17
= 22 × l × 17
= 374 cm2.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 1
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 2
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 3

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg) No. of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:

Weight (in kg) Frequency Cumulative frequency
36 – 38 0 0
38 – 40 3 3
40 – 42 2 5
42 – 44 4 9
44 – 46 5 14 = cf
46 – 48 14 = f 28
48 – 50 4 32
50 – 52 3 35
n = 35

\(\frac{\mathrm{n}}{2}=\frac{35}{2}=17.5\)
Plot the points (38, 0) (40, 3) (42, 5) (44, 9) (46, 14) (48, 28) (50, 32) (52, 35)
Median = l + \(\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]\) × h
= 46 + \(\left[\frac{17.5-14}{14}\right]\) × 2
= 46 + \(\frac{3.5 \times 2}{14}\)
= 46 + 0.5= 46.5.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 4

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 5
Change the distribution to a more than type distribution, and draw its ogive.
Solution:

Production yield (in kg/hec) Number of farms c.f.
More than 50 2 100
More than 55 8 98
More than 60 12 90
More than 65 24 78
More than 70 38 54
More than 75 16 16

∴ Co-ordinate points are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 6

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the media, mean and mode of the data and compare them.

Monthly consumption (in units) No. of consumers
65 – 85 4
85 – 105 5
105 – 125 13
125 – 145 20
145 – 165 14
165 – 185 8
185 – 205 4

Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 1
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 2
Mean is 137 units.
Median is 137 units.
Mode is 135.76 units.
The three measures are approximately the same.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

Class interval Frequency
0 – 10 5
10 – 20 x
20 – 30 20
30 – 40 15
40 – 50 y
50 – 60 5
Total 60

Solution:

Class interval Frequency Cumulative frequency
0 – 10 5 5
10 – 20 x 5 + x
20 – 30 20 25 + x
30 – 40 15 40 + x
40 – 50 y 40 + x + y
50 – 60 5 45 + x + y
60

n = 60, 45 + x + y = 60
x + y = 60 – 45
x + y = 15
The median is 28.5. It lies in the class interval 20 – 30.
∴ l = 20, f = 20, cf = 5 + x, h = 10
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 3

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years) No. of policyholders
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100

Solution:

Class interval No. of policyholders c.f.
Below 20 2 2
20 – 25 4 6
25 – 30 18 24
30 – 35 21 45
35 – 40 33 78
40 – 45 11 89
45 – 50 3 92
50 – 55 6 98
55 – 60 2 100
n = 100 \(\frac{n}{2}\) = 50

l = 35, \(\frac{n}{2}\) = 50, cf = 45, f = 33, h = 5
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 4

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length (in mm) Number of leaves
118 – 126 3
127 – 135 5
136 – 144 9
145 – 153 12
154 – 162 5
163 – 171 4
172 – 180 2

Find the median length of the leaves.
(Hint: The data need to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, ….., 171.5 – 180.5).
Solution:
The data have to be converted to continuous classes for finding the median since the formula. assumes continuous classes.

Class interval No. of leaves Cumulative frequency (cf)
117.5 – 126.5 3 3
126.5 – 135.5 5 8
135.5 – 144.5 9 17
144.5 – 153.5 12 29
153.5 – 162.5 5 34
162.5 – 171.5 4 38
171.5 – 180.5 2 40

n = 40, \(\frac{n}{2}\) = 20
The median lies in the class interval 144.5 – 153.5.
l = 144.5, \(\frac{\mathrm{n}}{2}=\frac{40}{2}\) = 20, cf = 17, f = 12, h = 9.
Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\) × h
= 144.5 + \(\left[\frac{20-17}{12} \times 9\right]\)
= 144.5 + \(\left[\frac{27}{12}\right]\)
= 144.5 + 2.25
= 146.75 mm.

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:

Lifetime (in hours) Number of lamps
1500 – 2000 14
2000 – 2500 56
2500 – 3000 60
3000 – 3500 86
3500 – 4000 74
4000 – 4500 62
4500 – 5000 48

Find the median lifetime of a lamp.
Solution:

Lifetime in hours (CI) No. of lamps (l) Cumulative frequency
1500 – 2000 14 14
2000 – 2500 56 70
2500 – 3000 60 130
3000 – 3500 86 216
3500 – 4000 74 290
4000 – 4500 62 352
4500 – 5000 48 400

The median lies in the class interval 3000 – 3500.
\(\frac{\mathrm{n}}{2}=\frac{400}{2}=200\)
l = 3000, \(\frac{n}{2}\) = 200, cf = 130, f = 86, h = 500.
Median = l + \(\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]\) × h
= 3000 + \(\left[\frac{200-130}{86}\right]\) × 500
= 3000 + \(\left[\frac{70}{86}\right]\) × 500
= 3000 + \(\frac{35000}{86}\)
= 3000 + 406.976
Median life of a lamp is 3406.98 hours.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:

No. of letters No. of surnames
1 – 4 6
4 – 7 30
7 – 10 40
10 – 13 16
13 – 16 4
16 – 19 4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 5
Hence the modal size of the surnames is 7.88.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 6
Solution:

Weight in kg. No. of students Cumulative frequency (c.f.)
40 – 45 2 2
45 – 50 3 5
50 – 55 8 13
55 – 60 6 19
60 – 65 6 25
65 – 70 3 28
70 – 75 2 30

\(\frac{n}{2}\) = 15
The median lies in the class 55 – 60.
l = 55, \(\frac{n}{2}\) = 15, c.f. = 13, f = 6, h = 5.
Median = l + \(\frac{1}{2}\) × h
= 55 + \(\left[\frac{15-13}{6}\right]\) × 5
= 55 + \(\frac{2}{6}\) × 5
= 55 + \(\frac{5}{3}\)
= 55 + 1.666 = 56.666
∴ Median = 56.67 kg.
Hence, the median weight of the students is 56.67 kg.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 1
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
The class interval having the maximum frequencies is 35 – 45.
f1 = 23, l = 35, h = 10, f0 = 21, f2 = 14
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 2
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 3
Maximum number of patients admitted in the hospital are of the age 36.8 years. The average age of the patient admitted to the hospital is 35.37 years.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 4
Determine the modal lifetimes of the components.
Class interval having the maximum frequency is 60 – 80.
f1 = 61, f0 = 52, f2 = 38, l = 60, h = 20.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 5

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in Rs.) No. of families
1000 – 1500 24
1500 – 2000 40
2000 – 2500 33
2500 – 3000 28
3000 – 3500 30
3500 – 4000 22
4000 – 4500 16
4500 – 5000 7

Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 6
Step deviation method: \(\bar{x}\) = a + \(\left[\frac{\sum f_i u_i}{\sum f_i}\right]\)h
= 3250 + \(\left[\frac{-235}{200}\right]\) × 500
= 3250 – \(\left[-\frac{1175}{2}\right]\)
= 3250 – 587.5
= 2662.5.
Mean expenditure Rs. 2662.50.
l = 1500, f1 = 40, f2 = 33, f0 = 24, h = 500
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 7
Modal monthly expenditure = 1847.83.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2

Question 4.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher No. of states/U.T.
15 – 20 3
20 – 25 8
25 – 30 9
30 – 35 10
35 – 40 3
40 – 45 0
45 – 50 0
50 – 55 2

Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 8
l = lower limit of the CI = 30, f1 = 10, f0 = 9, f2 = 3, h = 5
Mode = \(l+\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right] \times \mathrm{h}\)
= \(=30+\left[\frac{10-9}{20-9-3}\right] \times 5\)
= \(30+\left[\frac{1}{8} \times 5\right]=30+\frac{5}{8}\)
= 30 + 0.625 = 30.625.
Most states/UT’s have a student-teacher ratio of 30.6. On an average this ratio is 29.2.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored Number of batsmen
3000 – 4000 4
4000 – 5000 18
5000 – 6000 9
6000 – 7000 7
7000 – 8000 6
8000 – 9000 3
9000 – 10000 1
10000 – 11000 1

Find the mode of the data.
Solution:
l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 9

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 10
Solution:
Class interval having the maximum frequency is 40 – 50. f1 = 20, f0 = 12, f2 = 11, l = 40, h = 10.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 11

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 1
Which method did you use for finding the mean, and why?
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 2
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 3

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 4
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 5
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 6

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 7
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 8
752 + 20f = 792 + 18.1
20f – 18f = 792 – 752
2f = 40
f = \(\frac{40}{2}\) = 20.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 9
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 10
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 11

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 12
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 13

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 14
Find the mean daily expenditure on food by a suitable method.
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 15
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 16

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1

Question 7.
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO2 (in ppm) Frequency
0.00 – 0.04 4
0.04 – 0.08 9
0.08 – 0.12 9
0.12 – 0.16 2
0.16 – 0.20 4
0.20 – 0.24 2

Find the mean concentration of SO2 in the air.
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 17
Mean concentration of SO2 in air = 0.099 ppm

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 18
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 19
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 20
Mean number of days a student was absent is 12.48.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 21
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 22

JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1

Jharkhand Board JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 15 Probability Exercise 15.1

Question 1.
Complete the following statements:
i) Probability of an event E+ Probability of the event ‘not E’ =
ii) The probability of an event that cannot happen is …………….. Such an event is called ……………
iii) The probability of an event that is certain to happen is …………. Such an event is called …………….
iv) The sum of the probabilities of all the elementary events of an experiment is ……………
v) The probability of an event is greater than or equal to ………………. and less than or equal to ………………
Solution:
i) 1,
ii) 0, impossible event,
iii) 1, sure or certain,
iv) 1,
v) 0, 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
i) A driver attempts to start a car. The car starts or does not start.
ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
iii) A trial made to answer a true-false question. The answer is right or wrong.
iv) A baby is born. It is a boy or a girl.
Solution:
(iii) and (iv) have equally likely outcomes. Only two possibilities are there in each of these cases.

JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1=

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
When a coin is tossed, head or tail are equally likely possible events. So the result of an individual coin toss is unpredictable.

Question 4.
Which of the following cannot be the probability of an event:
A) \(\frac{2}{3}\)
B) -1.5
C) 15%
D) 0.7?
Solution:
(B) Because, probability of an event cannot be negative.

Question 5.
If P(E)= 0.05, what is the probability of ‘not E’?
Solution:
P(E) = 0.05.
[P(\(\overline{\mathrm{E}}\)) = Probability of not an event]
We know that P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(E) = 1 – 0.05 = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
i) an orange flavoured candy?
ii) a lemon flavoured candy?
Solution:
i) P(orange flavoured candy) = 0. Impossible event.
ii) P(Lemon flavoured candy) = 1. Sure event.

JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1=

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let E = Event of 2 students not having the same birthday
∴ P(E) = 0.992
∴ P(E) + P(\(\overline{\mathrm{E}}\)) = 1
∴ 0.992 + P(\(\overline{\mathrm{E}}\)) = 1
⇒ P(\(\overline{\mathrm{E}}\)) = 1 – 0.992
= 0.008
So, the probability of two students having the same birthday is 0.008.

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Solution:
Total number of balls, n(S) = 3 + 5 = 8.
Let E = Event of drawing 1 red ball
∴ n(E) = 3
(i) Probability of drawing a red ball = \(\frac{n(E)}{n(S)}=\frac{3}{8}\)
(ii) Probability of not drawing a red ball = 1 – P(Drawing a red ball)
= \(1-\frac{3}{8}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total possible outcomes = 5 + 8 + 4 = 17.
P(R) = \(\frac{5}{17}\), P(W) = \(\frac{8}{17}\)
P(Not green) = P(R + W) = \(\frac{5}{17}+\frac{8}{17}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a Rs. 5 coin?
Solution:
Total possible outcomes: 100 + 50 + 20 + 10 = 180.
P(50 paise coin) = \(\frac{100}{180}=\frac{5}{9}\)
P(Not Rs. 5 coin) = \(\frac{100}{180}+\frac{50}{180}+\frac{20}{180}\)
= \(\frac{170}{180}=\frac{17}{18}\).

JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1=

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Total number of fish in an aquarium = 5 male fish + 8 female fish = 13 fish
∴ Probability of taking out a male fish = \(\frac{\text { Number of male fish }}{\text { Total number of fish }}=\frac{5}{13}\)

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at
i) 8?
ii) an odd number?
iii) a number greater than 2?
iv) a number less than 9?
JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 1
Solution:
Total possible outcomes = 8.
i) P(8) = \(\frac{1}{8}\)
ii) P(odd number) = \(\frac{4}{8}=\frac{1}{2}\)
iii) P(no. > 2) = \(\frac{6}{8}=\frac{3}{4}\)
iv) P(no. < 9) = \(\frac{8}{8}\) = 1

Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number, (ii) a number lying between 2 and 6, (iii) an odd number.
Solution:
Total possible outcomes 1, 2, 3, 4, 5, 6 = 6
P(Prime number) (2, 3, 5) = \(\frac{3}{6}=\frac{1}{2}\)
P(Number between 2 and 6) = \(\frac{3}{6}=\frac{1}{2}\)
P(Odd number) = \(\frac{3}{6}=\frac{1}{2}\)

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
i) a king of red colour
ii) a face card
iii) a red face card
iv) the jack of hearts
v) a spade
vi) the queen of diamonds.
Solution:
Total number of cards in one deck, n(S) = 52.
i) Let E1 = Event of getting a king of red colour
∴ n(E1) = 2
(∵ In a deck of cards, 26 cards are red and 26 cards are black. There are four kings in a deck in which two are red and two are black)
Probability of getting a king of red colour = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{52}=\frac{1}{26}\)

ii) Let E2 = Event of getting a face card
∴ n(E2) = 12
(∵ In a deck of cards, there are 12 face cards – 4 king, 4 jack, 4 queen)
Probability of getting a face card = \(\frac{n\left(E_2\right)}{n(S)}=\frac{12}{52}=\frac{3}{13}\)

iii) Let E3 = Event of getting a red face card
∴ n(E3) = 6
(∵ In a deck of cards, there are 12 face cards – 6 red, 6 black)
Probability of getting a red face card = \(\frac{n\left(E_3\right)}{n(S)}=\frac{6}{52}=\frac{3}{26}\)

iv) Let E4 = Event of getting a jack of hearts
∴ n(E4) = 1
(∵ There are four jacks in a deck- 1 heart, 1 club, 1 spade, 1 diamond)
Probability of getting a jack of hearts = \(\frac{n\left(E_4\right)}{n(S)}=\frac{1}{52}\)

v) Let E5 = Event of getting a spade
∴ n(E5) = 13
(∵ In a deck of cards, there are 13 spades, 13 clubs, 13 hearts, 13 diamonds)
Probability of getting a spade = \(\frac{n\left(E_5\right)}{n(S)}=\frac{13}{52}=\frac{1}{4}\)

vi) Let E6 = Event of getting a queen of diamond
∴ n(E6) = 1
(∵ In 13 diamond cards, there is only one queen)
Probability of getting a queen of diamond = \(\frac{n\left(E_6\right)}{n(S)}=\frac{1}{52}\).

JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1=

Question 15.
Five cards – the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.
i) What is the probability that the card is the queen?
ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
i) Total possible outcomes = 5
P(Queen card) = \(\frac{1}{5}\)
ii) If the queen card is put aside, total possible outcomes = 4.
iii) P(ace) = \(\frac{1}{4}\)
iv) P(queen) = \(\frac{0}{2}\) = 0.

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Total possible outcomes = 132 + 12 = 144
No. of good pens = 132.
P(good pen) = \(\frac{132}{144}=\frac{11}{12}\)

Question 17.
i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
i) Total possible outcomes = 20
P(Defective bulbs) = \(\frac{4}{20}=\frac{1}{5}\)

ii) Total possible outcomes = 20 – 1 = 19
No. of defective bulbs = 4
No. of good bulbs = 15
P(Not defective bulb) = \(\frac{15}{19}\)

Question 18.
A box contains 90 dises which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.
Solution:
S = {1, 2, 3, 4, 5, … 90}
∴ Total possible outcomes n(S) = 90
i) Number of 2-digit numbers = 90 – 9 = 81
P(a 2-digit number) = \(\frac{81}{90}=\frac{9}{10}\)

ii) Event A = {A perfect square number}
A = (1, 4, 9, 16, 25, 36, 49, 64, 81) = n(A) = 9
Probability of the event P(A) = \(\frac{n(A)}{n(S)}=\frac{9}{90}\)

iii) A number divisible by 5, i.e., multiples of 5.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90 = 18.
P(a no. divisible by 5) = \(\frac{18}{90}=\frac{1}{5}\).

JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1=

Question 19.
A child has a die whose six faces show the letters as given below:
JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 2
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Total possible outcomes = 6
No. of A’s = 2
No. of D’s = 1
P(A) = \(\frac{2}{6}=\frac{1}{3}\)
P(D) = \(\frac{1}{6}\)

Question 20.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
i) she will buy it?
ii) she will not buy it?
Solution:
Total number of possible outcomes = 144
No. of good pens = 144 – 20 = 124.
P(of buying) = \(\frac{124}{144}=\frac{31}{36}\)
P(of not buying) = \(\frac{20}{144}=\frac{5}{36}\)

Question 21.
(i) Two dice, one blue and one grey, are thrown at the same time. Write down all possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8, (ii) 13, (iii) less than or equal to 12? Complete the following table:
JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 3
Solution:
JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 4
JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 5
1) Sum of 2 dice = 2 (1 + 1)
P(Sum 2) = \(\frac{1}{36}\)

2) Sum of 2 dice = 3 (1 + 2) (2 + 1)
P(Sum 3) = \(\frac{2}{36}\)

3) Sum 4 (1, 3) (2, 2) (3, 1)
P(Sum 4) = \(\frac{3}{36}\)

4) Sum 5 (1, 4) (2, 3) (3, 2) (4, 1)
P(Sum 5) = \(\frac{4}{36}\)

5) Sum 6 (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)
P(Sum 6) = \(\frac{5}{36}\)

6) Sum 7 (1,6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)
P(Sum 7) = \(\frac{6}{36}\)

7) Sum 8 (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)
P(Sum 8) = \(\frac{5}{36}\)

8) Sum 9 (3, 6) (4, 5) (5, 4) (6, 3)
P(Sum 9) = \(\frac{4}{36}\)

9) Sum 10 (4, 6) (5, 5) (6, 4)
P(Sum 10) = \(\frac{3}{36}\)

10) Sum 11 (5, 6) (6,5)
P(Sum 11) = \(\frac{2}{36}\)

11) Sum 12 (6,6)
P(Sum 12) = \(\frac{1}{36}\)

ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.
Solution:
Total possible outcomes of throwing the two dice, S =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6,3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36

a) Let E1 = Sum of two dice is 3 = {(1, 2), (2, 1)}
n(E1) = 2
∴ P(E1) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{36}\)

b) Let E2 = Sum of two dice is 4 = {(1, 3), (2, 2), (3, 1)}
n(E2) = 3
∴ P(E2) = \(\frac{n\left(E_2\right)}{n(S)}=\frac{3}{36}\)

c) Let E3 = Sum of two dice is 5 = {(1, 4), (2, 3), (3,2), (4, 1)}
n(E3) = 4
∴ P(E3) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{4}{36}\)

d) Let E4 = Sum of two dice is 6 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
n(E4) = 5
∴ P(E4) = \(\frac{5}{36}\)

e) Let E5 = Sum of two dice is 7 = {(1, 6), (2, 5), (3, 4), (4,3), (5, 2), (6, 1)}
n(E5) = 6
∴ P(E5) = \(\frac{n\left(E_5\right)}{n(S)}=\frac{6}{36}\)

f) Let E6 = Sum of two dice is 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6,2)}
n(E6) = 5
∴ P(E6) = \(\frac{n\left(E_6\right)}{n(S)}=\frac{5}{36}\)

g) Let E7 = Sum of two dice is 9 = {(3, 6), (4, 5), (5, 4), (6,3)}
n(E7) = 4
∴ P(E7) = \(\frac{n\left(E_7\right)}{n(S)}=\frac{4}{36}\)

h) Let E8 = Sum of two dice is 10 = {(4, 6), (5, 5), (6, 4)}
n(E8) = 3
∴ P(E8) = \(\frac{n\left(E_8\right)}{n(S)}=\frac{3}{36}\)

i) Let E9 = Sum of two dice is 11 = {(6,5), (5, 6)}
n(E9) = 2
∴ P(E9) = \(\frac{n\left(E_9\right)}{n(S)}=\frac{2}{36}\)

j) Let E10 = Sum of two dice is 12 = {(6, 6)}
n(E10) = 1
∴ P(E10) = \(\frac{n\left(E_{10}\right)}{n(S)}=\frac{1}{36}\)
No. The eleven events are not equally likely.

JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1=

Question 22.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Total possible outcomes (H + T)3
= H3 + 3H2T + 3HT2 + T3
HHH HHT HTH THH HTT THT TTH TTT = 8.
Possible losses HHT HTH THH HTT THT TTH = 6
P(of losses) = \(\frac{6}{8}=\frac{3}{4}\)

Question 23.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
Solution:
i) Total number of cases, n(S) = 62 = 36
Let \(\overline{\mathrm{E}}\) = Event that 5 will come up either time
= {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
⇒ n(\(\overline{\mathrm{E}}\)) = 11
and E = Event that 5 will not come up either time
n(E) = 36 – 11 = 25
∴ Probability that 5 will not come up either time = \(1-\frac{11}{36}=\frac{36-11}{36}\)
= \(\frac{25}{36}\)
ii) Probability that 5 will come up at least once = 12 – 1 = \(\frac{1}{2}\).

JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1=

Question 24.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Solution:
i) Incorrect: We can classify the outcomes like this but they are not then ‘equally likely’. The reason is that ‘one of each’ can result in two ways – from head on first coin and tail on the second coin or from tail on the first coin and head on the second coin. This makes it twice as likely as 2 heads or 2 tails.

ii) Correct. The two outcomes considered in the question are equally likely. Both have the same probability. i.e., \(\frac{1}{2}\).

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.3

Question 1.
Find the sum of following APs:
1. 2, 7, 12, ……, to 10 terms.
2. -37, -33, -29, ……, to 12 terms.
3. 0.6, 1.7, 2.8, ……, to 100 terms.
4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots\) to 11 terms.
Solution:
1. For the given AP 2, 7, 12, ……., a = 2.
d = 7 – 2 = 5, n = 10 and Sn is to be found.
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
S10 = \(\frac{10}{2}\)[4 + (10 – 1) 5]
= 5(49) = 245
Thus, the sum of first 10 terms of the given AP is 245.

2. For the given AP -37, -33, -29, a = -37, d = (-33) – (-37) = 4, n = 12 and Sn is to be found.
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
S12 = \(\frac{12}{2}\)[-74 + (12 – 1)4]
= 6(-30) = – 180
Thus, the sum of first 12 terms of the given AP is -180.

3. For the given AP 0.6, 1.7, 2.8,…… a = 0.6, d = 1.7 – 0.6 = 1.1, n = 100 and Sn is to be found.
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S100 = \(\frac{100}{2}\)[1.2 + (100 – 1) (1.1)]
= 50[1.2 + 108.9]
= 50 × 110.1 = 5505
Thus, the sum of first 100 terms of the given AP is 5505.

4.
JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 1
Thus, the sum of first 11 terms of the given AP is \(\frac{33}{20}\).

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 2.
Find the sums given below:
1. 7 + 10\(\frac{1}{2}\) + 14 + … + 84
2. 34 + 32 + 30 + … + 10
3. (-5) + (-8) + (-11) + … + (-230)
Solution:
1. 7 + 10\(\frac{1}{2}\) + 14 + …… + 84
Here, a = 7; d = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\); last term l = 84.
Let the last term be nth term.
an = a + (n – 1)d
∴ 84 = 7 + (n – 1) (3\(\frac{1}{2}\))
∴ 77 = \(\frac{7}{2}\)(n – 1)
∴ (n – 1) = 22
∴ n = 23
Now,
Sn = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91 = 1046\(\frac{1}{2}\)
Thus, the required sum is 1046\(\frac{1}{2}\).

2. 34 + 32 + 30 + … + 10
Here, a = 34; d = 32 – 34 = (-2); last term l = 10.
Let the last term be nth term.
an = a + (n – 1)d
∴ 10 = 34 + (n – 1)(-2)
∴ -24 = -2(n – 1)
∴ (n – 1) = 12
∴ n = 13
Now,
Sn = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{13}{2}\)(34 + 10)
= 13 × 22 = 286
Thus, the required sum is 286.

3. (-5) + (-8) + (-11) + … + (-230)
Here, a = (-5); d = (-8) – (-5) = (-3):
last term l = (-230).
Let the last term be nth term.
an = a + (n – 1)d
∴ -230 = -5 + (n – 1)(-3)
∴ -225 = -3 (n – 1)
∴ n – 1 = 75
∴ n = 76
Now,
Sn = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{76}{2}\)[(-5) + (-230)]
= 38(-235) = -8930
Thus, the required sum is -8930.

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 3.
In an AP:
1. Given a = 5, d = 3, an = 50, find n and Sn.
2. Given a = 7, a13 = 35, find d and S13.
3. Given a12 = 37, d = 3, find a and S12.
4. Given a3 = 15, S10 = 125, find d and a10.
5. Given d = 5, S9 = 75, find a and a9.
6. Given a = 2, d = 8, Sn = 90, find n and an.
7. Given a = 8, an = 62, Sn = 210, find n and d.
8. Given an = 4, d = 2, Sn = -14, find n and a.
9. Given a = 3, n = 8, Sn = 192, find d.
10. Given l = 28, Sn = 144, and there are total 9 terms. Find a.
Solution:
1. a = 5, d = 3, an = 50, n = ? Sn = ?
an = a + (n – 1)d
∴ 50 = 5 + (n – 1)3
∴ 45 = 3(n – 1)
∴ 15 = n – 1
∴ n = 16
Sn = \(\frac{n}{2}\)(a + l)
∴ S16 = \(\frac{16}{2}\)(5 + 50)
∴ S16 = 8 × 55
∴ S16 = 440

2. a = 7, a13 = 35, d = ?, S13 = ?
an = a + (n – 1)d
a13 = a + (13 – 1) d
∴ 35 = 7 + 12d
∴ 28 = 12d
∴ d = \(\frac{28}{12}\)
∴ d = \(\frac{7}{3}\)
Sn = \(\frac{n}{2}\)(a + l)
∴ S13 = \(\frac{13}{2}\)(17 + 35)
∴ S13 = 13 × 21
∴ S13 = 273

3. a12 = 37, d = 3, a = ?, S12 = ?
an = a + (n – 1)d
∴ a12 = a + 11d
∴ 37 = a + 11 (3)
∴ a = 4
Sn = \(\frac{n}{2}\)(a + l)
∴ S12 = \(\frac{12}{2}\)(4 + 37)
∴ S12 = 246

4. a3 = 15, S10 = 125, d = ?, a10 = ?
an = a + (n – 1)d
∴ a3 = a + 2d
∴ a + 2d = 15 …….(1)
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S10 = \(\frac{10}{2}\)[2a + 9d]
∴ S125 = 5(2a + 9d)
∴ 2a + 9d = 25 …….(2)
Solving equations (1) and (2), we get
d = -1 and a = 17.
an = a + (n – 1)d
∴ a10 = a + 9d
∴ a10 = 17 + 9(-1)
∴ a10 = 8

5. d = 5, S9 = 75, a = ?, a9 = ?
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S9 = \(\frac{9}{2}\)[2a + (9 – 1) d]
∴ 75 = \(\frac{9}{2}\)[2a + 8(5)]
∴ 75 = 9(a + 20)
∴ \(\frac{25}{3}\) = a + 20
∴ a = \(\frac{25}{3}\) – 20
∴ a = –\(\frac{35}{3}\)
an = a + (n – 1) d
∴ a9 = a + 8d
∴ a9 = (-\(\frac{35}{3}\)) + 8(5)
∴ a9 = –\(\frac{35}{3}\) + 40
∴ a9 = \(\frac{85}{3}\)

6. a = 2, d = 8, Sn = 90, n = ?, an = ?
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 90 = \(\frac{n}{2}\)[4 + (n – 1)8]
∴ 90 = \(\frac{n}{2}\)[8n – 4]
∴ 90 = n (4n – 2)
∴ 4n2 – 2n – 90 = 0
∴ 2n2 – n – 45 = 0
∴ 2n2 – 10n + 9n – 45 = 0
∴ 2n (n – 5) + 9 (n – 5) = 0
∴ (n – 5) (2n + 9) = 0
∴ n – 5 = 0 or 2n + 9 = 0
∴ n = 5 or n = –\(\frac{9}{2}\)
Since n is a positive integer, n ≠ –\(\frac{9}{2}\)
∴ n = 5
an = a + (n – 1)d
∴ a5 = a + 4d
∴ a5 = 2 + 4(8)
∴ a5 = 34

7. a = 8, an = 62, Sn = 210, n = ? d = ?
Sn = \(\frac{n}{2}\)(a + l)
∴ Sn = \(\frac{n}{2}\)(a + an)
∴ 210 = \(\frac{n}{2}\)(8 + 62)
∴ 420 = n (70)
∴ n = 6
an = a + (n – 1)d
∴ a6 = a + 5d
∴ 62 = 8 + 5d
∴ 54 = 5d
∴ d = \(\frac{54}{5}\)

8. an = 4, d = 2, Sn = -14, n = ?, a = ?
an = a + (n – 1)d
∴ 4 = a + (n – 1) (2)
∴ 4 = a + 2n – 2
∴ a = 6 – 2n
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ -14 = \(\frac{n}{2}\)[2 (6 – 2n) + (n – 1) (2)] (by (1))
∴ -14 = \(\frac{n}{2}\)[12 – 4n + 2n – 2]
∴ -14 = \(\frac{n}{2}\)[-2n + 10]
∴ -14 = n(-n + 5)
∴ -14 = -n2 + 5n
∴ n2 – 5n – 14 = 0
∴ (n – 7)(n + 2) = 0
∴ n – 7 = 0 or n + 2 = 0
∴ n = 7 or n = -2
Since n is a positive integer, n ≠ -2.
∴ n = 7
By (1), a = 6 – 2n
∴ a = 6 – 2(7)
∴ a = -8

9. a = 3, n = 8, Sn = 192, d = ?
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 192 = \(\frac{8}{2}\)[6 + (8 – 1) d]
∴ 192 = 4[6 + 7d]
∴ 48 = 6 + 7d
∴ 42 = 7d
∴ d = 6

10. l = 28, Sn = 144, n = 9, a = ?
Sn = \(\frac{n}{2}\)(a + l)
∴ 144 = \(\frac{9}{2}\)(a + 28)
∴ 32 = (a + 28)
∴ a = 4

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 4.
How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?
Solution:
Here, a = 9; d = 17 – 9 = 8; Sn = 636, n = ?
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 636 = \(\frac{n}{2}\)[18 + (n – 1)8]
∴ 636 = \(\frac{n}{2}\)[10 + 8n]
∴ 636 = n[4n + 5]
∴ 4n2 + 5n – 636 = 0
Here, a = 4; b = 5; c = -636
b2 – 4ac = (5)2 – 4(4)(-636)
= 25 + 10176
= 10201
∴ \(\sqrt{b^2-4 a c}=\sqrt{10201}=101\)
Then, n = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ n = \(\frac{-5 \pm 101}{8}\)
∴ n = \(\frac{96}{2}\) or n = \(\frac{-106}{8}\)
∴ n = 12 or n = \(-\frac{53}{4}\)
As n denotes the numbers of terms, it is a positive integer.
∴ n = \(-\frac{53}{4}\) is not possible.
∴ n = 12
Thus, 12 terms of the AP 9, 17, 25,… must be taken to give a sum of 636.

Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Here, a = 5; l = 45; Sn = 400; n = ?; d = ?
Sn = \(\frac{n}{2}\)(a + l)
∴ 400 = \(\frac{n}{2}\)(5 + 45)
∴ 800 = n (50)
∴ n = 16
l = an = a + (n – 1)d
∴ a16 = a + 15d
∴ 45 = 5 + 15d
∴ 40 = 15d
∴ d = \(\frac{40}{15}\)
∴ d = \(\frac{8}{3}\)
Thus, the number of terms is 16 and the common difference is \(\frac{8}{3}\).

Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
Here, a = 17; l = an = 350; d = 9; n = ?; Sn = ?
an = a + (n – 1)d
∴ 350 = 17 + (n – 1)9
∴ 333 = 9 (n – 1)
∴ n – 1 = 37 ∴ n = 38
Again, Sn = \(\frac{n}{2}\)(a + l)
∴ S38 = \(\frac{38}{2}\)(17 + 350)
∴ S38 = 19 × 367
∴ S38 = 6973
Thus, there are 38 terms and their sum is 6973.

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Here, a22 = 149; d = 7; S22 = ?
an = a + (n – 1) d
∴ a22 = a + 21d
∴ 149 = a + 21 × 7
∴ a = 149 – 147
∴ a = 2
Now, Sn = \(\frac{n}{2}\)(a + l)
∴ S22 = \(\frac{22}{2}\)(2 + 149)
∴ S22 = 11 × 151
∴ S22 = 1661
Thus, the sum of first 22 terms of the given AP is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Here, a2 = 14; a3 = 18; S51 = ?
an = a + (n – 1)d
∴ a2 = a + d = 14 ……..(1)
∴ a3 = a + 2d = 18 ……..(2)
Solving equations (1) and (2), we get
d = 4 and a = 10.
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S51 = \(\frac{51}{2}\)[20 + 50 × 4]
∴ S51 = 51 × 110
∴ S51 = 5610
Thus, the sum of first 51 terms of the given AP is 5610.

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Here, S7 = 49; S17 = 289: Sn = ?
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S7 = \(\frac{7}{2}\)[2a + 6d]
∴ 49 = 7(a + 3d)
∴ a + 3d = 7 ……..(1)
Again S17 = \(\frac{17}{2}\)[2a + 16d]
∴ 289 = 17(a + 8d)
∴ a + 8d = 17 ……..(2)
Solving equations (1) and (2), we get d = 2 and a = 1.
∴ Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ Sn = \(\frac{n}{2}\)[2 + (n – 1)2]
∴ Sn = \(\frac{n}{2}\)[2 + 2n – 2]
∴ Sn = \(\frac{n}{2}\)(2n)
∴ Sn = n2
Thus, the sum of first n terms of the given AP is n2.

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 10.
Show that a1, a2, ……., an, ……. form an AP where a is defined as below:
1. an = 3 + 4n
2. an = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
1. an = 3 + 4n
a1 = 3 + 4(1) = 7,
a2 = 3 + 4(2) = 11,
a3 = 3 + 4(3) = 15,
a4 = 3 + 4(4) = 19 and so on.
Here, a4 – a3 = a3 – a2 = a2 – a1 = 4.
Hence, ak+1 – ak remains the everywhere.
Hence, a1, a2, a3, ….. defined as an = 3 + 4n form an AP in which a = 7 and d = 4.
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S15 = \(\frac{15}{2}\)[14 + 14 × 4]
∴ S15 = 15 × 35
∴ S15 = 525
The sum of first 15 terms of the given AP is 525.

2. an = 9 – 5n
a1 = 9 – 5(1) = 4,
a2 = 9 – 5(2) = -1,
a3 = 9 – 5(3) = -6,
a4 = 9 – 5(4) = -11 and so on.
Here, a4 – a3 = a3 – a2 = a2 – a1 = -5.
Hence, ak+1 – ak remains the same everywhere.
Hence, a1, a2, a3, ….. defined as an = 9 – 5n form an AP in which a = 4 and d = -5.
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S15 = \(\frac{15}{2}\)[8 + 14 (-5)]
∴ S15 = \(\frac{15}{2}\)(-62)
∴ S15 = 15 × (-31)
∴ S15 = -465
The sum of first 15 terms of the given AP is -465.

Question 11.
If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
For the given AP
Sn = 4n – n2
∴ S1 = 4(1) – (1)2 = 4 – 1 = 3,
S2 = 4 (2) – (2)2 = 8 – 4 = 4,
S3 = 4(3) – (3)2 = 12 – 9 = 3,
S9 = 4 (9) – (9)2 = 36 – 81 = -45,
S10 = 4 (10) – (10)2 = 40 – 100 = -60
Now, the first term = a – a1 = S1 = 3
The sum of first two terms S2 = 4
The second term a2 = S2 – S1 = 4 – 3 = 1
The third term a3 = S3 – S2 = 3 – 4 = -1
The tenth term a10 = S10 – S9
= -60 – (-45) = -15
Now, Sn = 4n – n2
∴ Sn-1 = 4(n – 1) – (n – 1)2
= 4n – 4 – n2 + 2n – 1
= -n2 + 6n – 5
Now nth term an = Sn – Sn-1
∴ an = (4n – n2) – (-n2 + 6n – 5)
∴ an = 4n – n2 + n2 – 6n + 5
∴ an = -2n + 5

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The first 40 positive integers divisible by 6 form the AP 6, 12, 18, ……, 240.
Here, a = 6; d = 12 – 6 = 6; n = 40 and l = 240.
Sn = \(\frac{n}{2}\)(a + l)
∴ S40 = \(\frac{40}{2}\)(6 + 240)
∴ S40 = 20 × 246
∴ S40 = 4920
Thus, the required sum is 4920.

Alternate method:
Required sum
= 6 + 12 + 18 + … + 240
= 6(1 + 2 + 3 + … + 40)
= 6 × \(\frac{40 \times 41}{2}\) (1 + 2 + 3 + …. + n = \(\frac{n(n+1)}{2}\))
= 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 form the AP 8, 16, 24, ….., 120.
Here, a = 8, d = 16 – 8 = 8, n = 15 and l = 120.
Sn = \(\frac{n}{2}\)(a + l)
∴ S15 = \(\frac{15}{2}\)(8 + 120)
∴ S15 = 15 × 64
∴ S15 = 960
Thus, the required sum is 960.

Alternate method:
Required sum
= 8 + 16 + 24 + … + 120
= 8(1 + 2 + 3 + … + 15)
= 8 × \(\frac{15 \times 16}{2}\) (1 + 2 + 3 + … + n = \(\frac{n(n+1)}{2}\))
= 960.

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 form the AP 1, 3, 5, ….., 49.
Here, a = 1, d = 3 – 1 = -2, l = 49.
Let the last term be the nth term.
an = a + (n – 1) d
49 = 1 + (n – 1)2
2(n – 1) = 48
n – 1 = 24
n = 25
Now, Sn = \(\frac{n}{2}\)(a + l)
∴ S25 = \(\frac{25}{2}\)(1 + 49)
∴ S25 = 25 × 25
∴ S25 = 625
Thus, the required sum is 625.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc.. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
The sums (in rupees) of penalty for delay of completion form the AP 200, 250, 300, …..
Here, a = 200; d = 250 – 200 = 50; n = 30 as the contractor has delayed the work by 30 days. The total penalty amount (in rupees) to be paid by the contractor will be given by S30.
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S30 = \(\frac{30}{2}\)[400 + (30 – 1)50]
∴ S30 = 15 × 1850
∴ S30 = 27750
Thus, the contractor has to pay a penalty of ₹ 27,750.

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If cach prize is 20 less than its preceding prize, find the value of each of the prizes.
Solution:
₹ 700 is to be distributed as seven prizes such that each prize is ₹ 20 less than its preceding prize. Let the highest prize, i.e., the first prize be ₹ a. Then, the second prize will be of ₹ a – 20, the third prize will be of ₹ a – 40 and so on up to seven prizes. Hence, the amount (in rupees) of these prizes form a finite AP with seven terms as a, a – 20, a – 40, a – 60, a – 80, a – 100 and a – 120.
Here, the first term = a; d = (a – 20) – a = -20;
n = 7 and the sum of all the terms = S7 = 700.
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 700 = \(\frac{7}{2}\)[2a + (7 – 1) (-20)]
∴ 200 = 2a + 6(-20)
∴ 200 = 2a – 120
∴ 2a = 320
∴ a = 160
Then, a – 20 = 140; a – 40 = 120; a – 60 = 100; a – 80 = 80; a – 100 = 60 and a – 120 = 40.
Thus, the values (in rupees) of those seven prizes are 160, 140, 120, 100, 80, 60 and 40.

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class. will plant, will be the same as the class, in which they are studying, e.g.. a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
The number of trees that the three sections of Class I will plant = 1 + 1 + 1 = 3.
The number of trees that the three sections of Class II will plant = 2 + 2 + 2 = 6.
This system will continue till Class XII.
The number of trees that the three sections of Class XII will plant = 12 + 12 + 12 = 36.
Thus, the number of trees that will be planted will form a finite AP with 12 terms as 3, 6, 9, ….., 36.
Here, a = 3, d = 6 – 3 = 3, n = 12 and S12 will give the total number of trees that will be planted.
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S12 = \(\frac{12}{2}\)[6 + (12 – 1)3]
∴ S12 = 6 × 39
∴ S12 = 234
Thus, 234 trees will be planted by the students.

Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, as shown in the given figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = \(\frac{22}{7}\))
JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 2
Hint: Length of successive semicircles is l1, l2, l3, l4, …with centres at A, B, A, B, ….., respectively.]
Solution:
We know that the length of a semicircle = πr, where r is the radius.
Length of 1st semicircle with centre A and radius 0.5 cm = l1 = π × 0.5 cm.
Length of 2nd semicircle with centre B and radius 1 cm = l2 = π × 1 cm.
Length of 3rd semicircle with centre A and radius 1.5 cm = l3 = π × 1.5 cm.
This system continues till 13 semicircles are drawn.
Then, the 13th semicircle will be drawn with centre A and radius 6.5 cm. Length of 13th semicircle with centre A and radius 6.5 cm = l13 = π × 6.5 cm.
Now, the total length of the spiral
= l1 + l2 + l3 + … + l13
= (π × 0.5) + (π × 1) + (π × 1.5) + … + (π × 6.5)
= π(0.5 + 1 + 1.5 + … + 6.5)
The sum inside the brackets is the sum of all the 13 terms of the finite AP 0.5, 1, 1.5, ….., 6.5.
For this AP a = 0.5; d = 1 – 0.5 = 0.5 and n = 13.
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ Sn = [1 + (13 – 1) (0.5)]
∴ Sn = \(\frac{13}{2}\) × 7
Hence, the total length of the spiral
= π\(\left(\frac{13}{2} \times 7\right)\)
= \(\frac{22}{7} \times \frac{13}{2} \times 7\)
= 143 cm
Thus, the total length of the spiral made up of thirteen consecutive semicircles is 143 cm.

Question 19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the given figure). In how many rows are the 200 logs placed and how many logs are in the top row?
JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 3
Solution:
The number of logs stacked in the first row from the bottom = 20.
The number of logs stacked in the second row from the bottom = 19.
The number of logs stacked in the third row from the bottom = 18.
This system continues till all the 200 logs are stacked.
Thus, the number of logs stacked in the rows form the finite AP 20, 19, 18,….. up ton-terms and the sum of those n terms is 200. Here, a = 20 and d = 19 – 20 = -1.
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 200 = \(\frac{n}{2}\)[40 + (n – 1) (-1)]
∴ 400 = n(40 – n + 1)
∴ 400 = n(41 – n)
∴ 400 = 41n – n2
∴ n2 – 41n + 400 = 0
∴ n2 – 16n – 25n + 400 = 0
∴ n (n – 16) – 25 (n – 16) = 0
∴ (n – 16) (n – 25) = 0
n – 16 = 0 or n – 25 = 0
n = 16 or n = 25
Here, both the answers are admissible. Hence, we verify by the value of 16th term and 25th term.
an = a + (n – 1) d
∴ a16 = 20 + 15(-1) = 5
∴ a25 = 20 + 24(-1) = -4
Thus, for the 25th row, the number of logs in the row becomes negative. This is inadmissible.
Hence, n ≠ 25.
∴ n = 16 and a16 = 5.
Thus, the 200 logs are placed in 16 rows and in the top row there are 5 logs.

JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see the given figure).
JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 4
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)].
Solution:
The distance (in metres) to be covered to pick up first potato = 2 × 5 = 10.
The distance (in metres) to be covered to pick up second potato = 2 × (5 + 3) = 16.
The distance (in metres) to be covered to pick up third potato = 2 × (5 + 3 + 3) = 22.
Thus, the distances to be covered to pick up 10 potatoes form the finite AP 10, 16, 22, …. up to 10 terms.
Here, a = 10, d = 16 – 10 = 6, n = 10 and Sn will give the total distance (in metres) that the competitor has to run.
Sn = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S10 = \(\frac{10}{2}\)[20 + (10 – 1)6]
∴ S10 = 5 × 74
∴ S10 = 370
Thus, the total distance that the competitor has to run is 370 metres.