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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.

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:

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.

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:

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:

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.

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:

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:

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

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.

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

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:

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.

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:

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:

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.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village:

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).

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:


Mean is 137 units.
Median is 137 units.
Mode is 135.76 units.
The three measures are approximately the same.

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

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

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.

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:

Hence the modal size of the surnames is 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

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.

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:

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


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.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical

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.

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:

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

Modal monthly expenditure = 1847.83.

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:

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

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.

Solution:
Class interval having the maximum frequency is 40 – 50. f₁ = 20, f₀ = 12, f₂ = 11, l = 40, h = 10.

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.

Which method did you use for finding the mean, and why?
Solution:

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory:

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

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.

Solution:

752 + 20f = 792 + 18.1
20f – 18f = 792 – 752
2f = 40
f = \(\frac{40}{2}\) = 20.

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.

Solution:

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:

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:

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:

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.

Solution:


Mean number of days a student was absent is 12.48.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:

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.

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.

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) = \(\frac{100}{180}+\frac{50}{180}+\frac{20}{180}\)
= \(\frac{170}{180}=\frac{17}{18}\).

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?

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 = \(\frac{n\left(E_6\right)}{n(S)}=\frac{1}{52}\).

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) = \(\frac{18}{90}=\frac{1}{5}\).

Question 19.
A child has a die whose six faces show the letters as given below:

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:

Solution:


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.

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 = \(\frac{1}{2}\).

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., \(\frac{1}{2}\).

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 S_n is to be found.
S_n = \(\frac{n}{2}\)[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 S_n is to be found.
S_n = \(\frac{n}{2}\)[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 S_n is to be found.
S_n = \(\frac{n}{2}\)[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.

Thus, the sum of first 11 terms of the given AP is \(\frac{33}{20}\).

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.
a_n = a + (n – 1)d
∴ 84 = 7 + (n – 1) (3\(\frac{1}{2}\))
∴ 77 = \(\frac{7}{2}\)(n – 1)
∴ (n – 1) = 22
∴ n = 23
Now,
S_n = \(\frac{n}{2}\)(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.
a_n = a + (n – 1)d
∴ 10 = 34 + (n – 1)(-2)
∴ -24 = -2(n – 1)
∴ (n – 1) = 12
∴ n = 13
Now,
S_n = \(\frac{n}{2}\)(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.
a_n = a + (n – 1)d
∴ -230 = -5 + (n – 1)(-3)
∴ -225 = -3 (n – 1)
∴ n – 1 = 75
∴ n = 76
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{76}{2}\)[(-5) + (-230)]
= 38(-235) = -8930
Thus, the required sum is -8930.

Question 3.
In an AP:
1. Given a = 5, d = 3, a_n = 50, find n and S_n.
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, S_n = 90, find n and a_n.
7. Given a = 8, a_n = 62, S_n = 210, find n and d.
8. Given a_n = 4, d = 2, S_n = -14, find n and a.
9. Given a = 3, n = 8, S_n = 192, find d.
10. Given l = 28, S_n = 144, and there are total 9 terms. Find a.
Solution:
1. a = 5, d = 3, a_n = 50, n = ? S_n = ?
a_n = a + (n – 1)d
∴ 50 = 5 + (n – 1)3
∴ 45 = 3(n – 1)
∴ 15 = n – 1
∴ n = 16
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₆ = \(\frac{16}{2}\)(5 + 50)
∴ S₁₆ = 8 × 55
∴ S₁₆ = 440

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

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

4. a₃ = 15, S₁₀ = 125, d = ?, a₁₀ = ?
a_n = a + (n – 1)d
∴ a₃ = a + 2d
∴ a + 2d = 15 …….(1)
S_n = \(\frac{n}{2}\)[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.
a_n = a + (n – 1)d
∴ a₁₀ = a + 9d
∴ a₁₀ = 17 + 9(-1)
∴ a₁₀ = 8

5. d = 5, S₉ = 75, a = ?, a₉ = ?
S_n = \(\frac{n}{2}\)[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 = –\(\frac{35}{3}\)
a_n = 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, S_n = 90, n = ?, a_n = ?
S_n = \(\frac{n}{2}\)[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
a_n = a + (n – 1)d
∴ a₅ = a + 4d
∴ a₅ = 2 + 4(8)
∴ a₅ = 34

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

8. a_n = 4, d = 2, S_n = -14, n = ?, a = ?
a_n = a + (n – 1)d
∴ 4 = a + (n – 1) (2)
∴ 4 = a + 2n – 2
∴ a = 6 – 2n
S_n = \(\frac{n}{2}\)[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, S_n = 192, d = ?
S_n = \(\frac{n}{2}\)[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, S_n = 144, n = 9, a = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ 144 = \(\frac{9}{2}\)(a + 28)
∴ 32 = (a + 28)
∴ a = 4

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; S_n = 636, n = ?
S_n = \(\frac{n}{2}\)[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; S_n = 400; n = ?; d = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ 400 = \(\frac{n}{2}\)(5 + 45)
∴ 800 = n (50)
∴ n = 16
l = a_n = 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 = a_n = 350; d = 9; n = ?; S_n = ?
a_n = a + (n – 1)d
∴ 350 = 17 + (n – 1)9
∴ 333 = 9 (n – 1)
∴ n – 1 = 37 ∴ n = 38
Again, S_n = \(\frac{n}{2}\)(a + l)
∴ S₃₈ = \(\frac{38}{2}\)(17 + 350)
∴ S₃₈ = 19 × 367
∴ S₃₈ = 6973
Thus, there are 38 terms and their sum is 6973.

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 = ?
a_n = a + (n – 1) d
∴ a₂₂ = a + 21d
∴ 149 = a + 21 × 7
∴ a = 149 – 147
∴ a = 2
Now, S_n = \(\frac{n}{2}\)(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 = ?
a_n = 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.
S_n = \(\frac{n}{2}\)[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: S_n = ?
S_n = \(\frac{n}{2}\)[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.
∴ S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S_n = \(\frac{n}{2}\)[2 + (n – 1)2]
∴ S_n = \(\frac{n}{2}\)[2 + 2n – 2]
∴ S_n = \(\frac{n}{2}\)(2n)
∴ S_n = n²
Thus, the sum of first n terms of the given AP is n².

Question 10.
Show that a₁, a₂, ……., a_n, ……. form an AP where a is defined as below:
1. a_n = 3 + 4n
2. a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
1. a_n = 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, a_k+1 – a_k remains the everywhere.
Hence, a₁, a₂, a₃, ….. defined as a_n = 3 + 4n form an AP in which a = 7 and d = 4.
S_n = \(\frac{n}{2}\)[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. a_n = 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, a_k+1 – a_k remains the same everywhere.
Hence, a₁, a₂, a₃, ….. defined as a_n = 9 – 5n form an AP in which a = 4 and d = -5.
S_n = \(\frac{n}{2}\)[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
S_n = 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, S_n = 4n – n²
∴ S_n-1 = 4(n – 1) – (n – 1)²
= 4n – 4 – n² + 2n – 1
= -n² + 6n – 5
Now nth term a_n = S_n – S_n-1
∴ a_n = (4n – n²) – (-n² + 6n – 5)
∴ a_n = 4n – n² + n² – 6n + 5
∴ a_n = -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.
S_n = \(\frac{n}{2}\)(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.
S_n = \(\frac{n}{2}\)(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.

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.
a_n = a + (n – 1) d
49 = 1 + (n – 1)²
2(n – 1) = 48
n – 1 = 24
n = 25
Now, S_n = \(\frac{n}{2}\)(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.
S_n = \(\frac{n}{2}\)[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.
S_n = \(\frac{n}{2}\)[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.

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.
S_n = \(\frac{n}{2}\)[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 = \(\frac{22}{7}\))

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.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S_n = [1 + (13 – 1) (0.5)]
∴ S_n = \(\frac{13}{2}\) × 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?

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.
S_n = \(\frac{n}{2}\)[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.
a_n = 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.

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).

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.
S_n = \(\frac{n}{2}\)[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.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a, the nth term of the AP:

Solution:
1. Here, a = 7, d = 3, n = 8 and a_n is to be found.
We have a_n = a + (n – 1)d
a₈ = 7 + (8 – 1) 3 = 7 + 21 = 28

2. Here, a = -18, n = 10, a_n = a₁₀ = 0 and d is to be found.
a_n = a + (n – 1)d
∴ 0 = – 18 + (10 – 1)d
∴ 18 = 9d ∴ d = 2

3. Here, d = -3, n = 18, a_n = a₁₈ = -5 and a is to be found.
an = a + (n – 1)d
∴ -5 = a + (18 – 1)(-3)
∴ -5 = a – 51
∴ a = 51 – 5 ∴ a = 46

4. Here, a = -18.9, d = 2.5, a_n = 3.6 and n is to be found.
a_n = a + (n – 1)d
∴ 3.6 = – 18.9 + (n – 1)(2.5)
∴ 22.5 = 2.5 (n – 1)
∴ (n – 1) = \(\frac{22.5}{2.5}\)
∴ n – 1 = 9 ∴ n = 10

5. Here, a = 3.5, d = 0, n = 105 and a_n is to be found.
a_n = a + (n – 1) d
∴ a₁₀₅ = 3.5 + (105 – 1) (0)
∴ a₁₀₅ = 3.5

Question 2.
Choose the correct choice in the following and justify:
1. 30th term of the AP: 10, 7, 4, ……, is
(A) 97
(B) 77
(C) -77
(D) -87
2. 11th term of the AP: -3, –\(\frac{1}{2}\), 2, ….. is
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
Solution:
1. For the given AP 10, 7, 4,… a = 10,
d = 7 – 10 = -3 and n = 30
a_n = a + (n – 1)d
∴ a₃₀ = 10 + (30 – 1) (-3)
∴ a₃₀ = 10 – 87
∴ a₃₀ = -77
Thus, the correct choice is (C) -77.

2. For the given AP -3, –\(\frac{1}{2}\), 2, ….., a = -3.
d = –\(\frac{1}{2}\) – (-3) = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) and n = 11.
a_n = a + (n – 1)d
∴ a₁₁ = -3 + (11 – 1)(\(\frac{5}{2}\))
∴ a₁₁ = – 3 + 25
∴ a₁₁ = 22
Thus, the correct choice is (B) 22.

Question 3.
In the following APs, find the missing terms in the boxes:

Solution:
For the given AP, first term = a = 2 and third term = a + 2d = 26.
a = 2 and a + 2d = 26 gives d = 12.
Then, second term = a + d = 2 + 12 = 14
Thus, the missing term in the box is

2. For the given AP,
second term = a + d = 13 …….(1)
fourth term = a + 3d = 3 …….(2)
Solving equations (1) and (2) we get d = -5 and a = 18.
Now, first term = a = 18 and third term = a + 2d
= 18 + 2(-5) = 8.
Thus, the missing terms in the boxes are

Alternative Method:
Let the terms of the given AP be a₁, a₂, a₃, a₄.
Here, a₂ = 13 and a₄ = 3.
Now, a₄ – a₃ = a₃ – a₂ = d
∴ 3 – a₃ = a₃ – 13
∴ 2a₃ = 16
∴ a₃ = 8
Again, a₂ – a₁ = a₃ – a₂
∴ 13 – a₁ = 8 – 13
∴ 13 – a₁ = -5
∴ a₁ = 18
Thus, the missing terms in the boxes are

3. For the given AP
first term a = 5 ……….(1)
fourth term = a + 3d = 9\(\frac{1}{2}\) ……….(2)
From equations (1) and (2), we get a = 5 and d = 1\(\frac{1}{2}\).
Now, second term = a + d = 5 + 1\(\frac{1}{2}\) = 6\(\frac{1}{2}\)
and third term = a + 2d = 5 + 2 (1\(\frac{1}{2}\)) = 8
Thus, the missing terms in the boxes are

4. For the given AP
first term a = -4 ……….(1)
sixth term = a + 5d = 6 ……….(2)
From equations (1) and (2), we get
a = -4 and d = 2
Now, second term = a + d(-4) + 2 = -2,
third term = a + 2d = (-4) + 2 (2) = 0.
fourth term = a + 3d = (-4) + 3(2) = 2 and
fifth term = a + 4d = (-4) + 4(2) = 4.
Thus, the missing terms in the boxes are

5. For the given AP,
second term = a + d = 38 ……….(1)
sixth term = a + 5d = -22 ……….(2)
Solving equations (1) and (2), we get
d = -15 and a = 53.
Now, first term = a = 53,
third term = a + 2d = 53 + 2(-15) = 23,
fourth term = a + 3d = 53 + 3(-15) = 8 and
fifth term = a + 4d = 53 + 4(-15) = -7.
Thus, the missing terms in the boxes are

Question 4.
Which term of the AP: 3, 8, 13, 18, … is 78 ?
Solution:
Suppose nth term of the AP 3, 8, 13, 18, … is 78.
Here, a = 3, d = 8 – 3 = 5, a_n = 78 and n is to be found.
a_n = a + (n – 1)d
∴ 78 = 3 + (n – 1)5
∴ 75 = 5 (n-1)
∴ 15 = n – 1 ∴ n = 16
Thus, the 16th term of the AP 3, 8, 13, 18, …….., is 78.

Question 5.
Find the number of terms in each of the following APs:
1. 7, 13, 19, …….., 205
2. 18, 15\(\frac{1}{2}\), 13, ……., -47
Solution:
1. For the given finite AP 7, 13, 19, ….. 205, a = 7, d = 13 – 76 and last term l = 205.
Let us consider that the last term is the nth term.
a_n = a + (n – 1)d
∴ 205 = 7 + (n – 1)6
∴ 198 = 6 (n – 1)
∴ n – 1 = 33
∴ n = 34
Thus, there are 34 terms in the given finite AP.

2. For the given finite AP 18, 15\(\frac{1}{2}\), 13….. -47, a = 18, d = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\) and last term l = -47.
Let us consider that the last term is the nth term.
a_n = a + (n – 1) d
∴ -47 = 18 + (n-1) (-\(\frac{5}{2}\))
∴ -65 = –\(\frac{5}{2}\)(n – 1)
∴ n – 1 = 26
∴ n = 27
Thus, there are 27 terms in the given finite AP.

Question 6.
Check whether -150 is a term of the AP: 11, 8, 5, 2…
Solution:
If possible, let -150 be the nth term of the AP 11, 8, 5, 2,…
Here, a = 11; d = 8 – 11 = -3 and a_n = -150.
a_n = a + (n – 1)d
∴ -150 = 11 + (n – 1)(-3)
∴ -161 = -3 (n – 1)
∴ n – 1 = \(\frac{161}{3}\)
∴ n = \(\frac{164}{3}\)
∴ But n must be positive integer for an AP.
Hence, -150 cannot be a term of the AP 11, 8, 5, 2…..

Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
For any AP, a_n = a + (n – 1)d.
a₁₁ = a + 10 d
∴ a + 10 d = 38 ………(1)
∴ a₁₆ = a + 15d
∴ a + 15d = 73 ………(2)
Solving equations (1) and (2), we get
d = 7 and a = -32.
Now, 31st term = a₃₁ = a + 30d
= -32 + 30(7)
= -32 + 210
= 178
Thus, the 31st term of the given AP is 178.
Note: d = \(\frac{a_{16}-a_{11}}{16-11}=\frac{73-38}{5}=\frac{35}{5}=7\) can also be used.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
The given finite AP has 50 terms, hence its last term is a₅₀.
So, a₃ = 12 and a₅₀ = 106.
Now, a_n = a + (n – 1)d.
That gives, a₃ = a + 2d = 12 ………(1)
and a₅₀ = a + 49d = 106 ………(2)
Solving equations (1) and (2), we get
d = 2 and a = 8.
Now, 29th term = a₂₉ = a + 28d
∴ a₂₉ = 8 + 28 (2)
∴ a₂₉ = 64
Thus, the 29th term of the given AP is 64.

Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
For the given AP, a₃ = 4 and a9 = -8
We know, a_n = a + (n – 1)d.
∴ a₃ = a + 2d = 4 ………(1)
and a₉ = a + 8d = -8 ………(2)
Solving equations (1) and (2), we get
d = -2 and a = 8.
Now, let nth term of the AP be 0.
a_n = a + (n – 1)d
∴ 0 = 8 + (n – 1) (-2)
∴ 2(n – 1) = 8
∴ n – 1 = 4
∴ n = 5
Thus, the 5th term of the given AP is zero.

Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
For the given AP,
a₁₇ + 16d = a + 9d + 7a [∵ a_n = a + (n – 1) d]
∴ 7d = 7
∴ d = 1.
Thus, the common difference of the given AP is 1.

Question 11.
Which term of the AP:3, 15, 27, 39, …….. will be 132 more than its 54th term?
Solution:
For the given AP 3, 15, 27, 39, …….., a = 3
and d = 15 – 3 = 12.
Suppose nth term of the AP is 132 more than its 54th term.
∴ a_n = a₅₄ + 132
∴ a + (n – 1)d = a + 53d + 132
∴ 3 + (n – 1) (12) = 3 + 53(12) + 132
∴ 12(n – 1) = 12 (53 + 11)
∴ 12(n – 1) = 12 × 64
∴ n – 1 = 64
∴ n = 65
Thus, the 65th term of the given AP is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let the first term of two given APs be a₁ and a₂ respectively and let the same common difference be d.
Also, let a₁ > a₂
Then, 100th term of the first AP = a₁ + 99d
(a_n = a + (n – 1)d)
100th term of the second AP = a₂ + 99d.
The difference between their 100th terms is 100.
∴ (a₁ + 99d) – (a₂ + 99d) = 100 (a₁ > a₂)
∴ a₁ – a₂ = 100 ……(1)
Now, 1000th term of the first AP = a₁ + 999d
1000th term of the second AP = a₂ + 999d.
Then, the difference between their 1000th terms
= (a₁ + 999d) – (a₂ + 999d)
= a₁ – a₂
= 100 (by (1))
Thus, the difference between the 1000th terms of the two APs is 100.

Question 13.
How many three digit numbers are divisible by 7?
Solution:
The list of three digit numbers divisible by 7 is as below:
105, 112, 119, …….., 987, 994.
These numbers form a finite AP with a = 105, d = 112 – 105 = 7 and last term l = 994.
Suppose the last term of the AP is its nth term.
∴ l = a_n
∴ 994 = a + (n – 1)d
∴ 994 = 105 + (n – 1)7
∴ 7(n – 1) = 889
∴ n – 1 = 127
∴ n = 128
Hence, there are 128 terms in the AP.
Hence, 128 three digit numbers are divisible by 7.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The multiples of 4 lying between 10 and 250 give rise to following finite AP:
12, 16, 20, ….., 244, 248.
Here, a = 12, d = 16 – 12 = 4 and last term l = 248.
If the last term is the nth term of the AP then l = a_n.
∴ l = a + (n – 1) d
∴ 248 = 12 + (n – 1)4
∴ 236 = 4 (n – 1)
∴ n – 1 = 59
∴ n = 60
Thus, there are 60 terms in the AP.
Thus, 60 multiples of 4 lie between 10 and 250.

Question 15.
For what value of n, are the nth terms of two APs 63, 65, 67, …… and 3, 10, 17,… equal?
Solution:
For the first AP 63, 65, 67, ….., a = 63, d = 65 – 63 = 2.
Then, nth term of the first AP a is given by a_n = a + (n – 1)d = 63 + (n – 1)(2).
For the second AP 3, 10, 17, ……, A = 3, D = 10 – 3 = 7.
Then, nth term of the second AP An is given by
A_n = A + (n – 1) D = 3 + (n – 1) (7).
Now, a_n = A_n
∴ 63 + (n – 1) (2) = 3 + (n – 1)(7)
∴ 63 – 3 = (n – 1)(7 – 2)
∴ 60 = 5(n – 1)
∴ n – 1 = 12
∴ n = 13
Thus, for n = 13, the nth term of two given APs are equal.

Question 16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
For the given AP a₃ = 16 and a₇ = a₅ + 12.
For any AP, a_n = a + (n – 1) d.
∴ a + 2d = 16 and a + 6d = a + 4d + 12
a + 6d = a + 4d + 12 gives 2d = 12,
i.e., d = 6.
Substituting d = 6 in a + 2d = 16, we get a = 4.
Then, the required AP is 4, 4 + 6, 4 + 2 (6), 4 + 3(6), …..
Hence, the required AP is 4, 10, 16, 22, ……

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ….., 253.
Solution:
For the given finite AP 3, 8, 13, …., 253,
a = 3, d = 8 – 3 = 5 and last term l = 253.
Let the last term be its nth term.
∴ l = a_n
∴ l = a + (n – 1)d
∴ 253 = 3 + (n – 1)(5)
∴ 250 = 5 (n – 1)
∴ n – 1 = 50
∴ n = 51
Thus, there are in all 51 terms in the AP.
Now, the 20th term from the last term is (51 – 20 + 1)th term = 32nd term from the beginning.
a₃₂ = a + 31d
∴ a₃₂ = 3 + 31(5)
∴ a₃₂ = 158
Thus, the 20th term from the last term is 158.

Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
For any AP, a_n = a + (n – 1) d.
a₄ = a + 3d, a₈ = a + 7d, a₆ = a + 5d and a₁₀ = a + 9d.
Now, a₄ + a₈ = 24 (Given)
∴ (a + 3d) + (a + 7d) = 24
∴ 2a + 10d = 24
∴ a + 5d = 12 ……..(1)
Again, a₆ + a₁₀ = 44 (Given)
∴ (a + 5d) + (a + 9d) = 44
∴ 2a + 14d = 44
∴ a + 7d = 22 …….(2)
Solving equations (1) and (2), we get d = 5 and a = -13.
Then, a₂ = a + d = 13 + (5) = -8 and
a₃ = a + 2d = -13 + 2(5) = -3.
Thus, the first three terms of the AP are -13, 8, 3.

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7,000?
Solution:
Subba Rao’s income in first year = ₹ 5,000
His income in second year = ₹ 5,000 + ₹ 200
= ₹ 5200
His income in third year = ₹ 5200 + ₹ 200
= ₹ 5400
and so on.
These numbers of his income (in rupees) form the AP 5000, 5200, 5400, …….
Here, a = 5000; d = 5200 – 5000 = 200;
a_n = 7000 and n is to be found.
a_n = a + (n – 1)d
∴ 7000 = 5000 + (n – 1)(200)
∴ 2000 = 200(n – 1)
∴ n – 1 = 10
∴ n = 11
Thus, Subba Rao’s income will reach ₹ 7000 in 11th year, i.e., in the year 2005.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution:
Ramkali’s savings in first week = ₹ 5
her savings in second week = ₹ 5 + ₹ 1.75
= ₹ 6.75,
her savings in third week = ₹ 6.75 + ₹ 1.75
= ₹ 8.50.
and so on.
Thus, the weekly savings (in rupees) of Ramkalt form the AP 5, 6.75, 8.50, …..
Here, a = 5; d = 6.75 – 5 = 1.75; a_n = 20.75 and n is to be found.
a_n = a + (n – 1)d
∴ 20.75 = 5 + (n – 1)(1.75)
∴ 1.75(n – 1) = 15.75
∴ n – 1 = 9
∴ n = 10
Thus, if Ramkali’s weekly savings is ₹ 20.75 in nth week, then n = 10.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
1. The taxi fare after each km, when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
2. The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
3. The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
4. The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.
Solution:
1. Here, the fare for 1 km = ₹ 15
the fare for 2 km = ₹ 15 + ₹ 8 = ₹ 23,
the fare for 3 km = ₹ 15 + 2 (₹8) = ₹ 31,
the fare for 4 km = ₹ 15 + 3 (₹8) = ₹ 39, and so on.
The list of numbers formed is 15, 23, 31, 39, …………..
Here, a₂ – a₁ = 23 – 15 = 8.
a₃ – a₂ = 31 – 23 = 8.
a₄ – a₃ = 39 – 31 = 8, and so on.
Thus, a_k+1 – a_k is the same every time.
Hence, the list of numbers forms an AP with a = 15 and d = 8.

2. Let the volume of air present in the cylinder at the beginning be V units. Then, volume of air remaining in the cylinder after first attempt = \(\frac{3}{4}\)V units. Also, volume of air remaining in the cylinder after second attempt = \(\left(\frac{3}{4}\right)^2\)V units. Here, the list of numbers formed is V, \(\frac{3}{4}\)V, \(\left(\frac{3}{4}\right)^2\)V, ………
Now, a₂ – a₁ = \(\frac{3}{4}\)V – V = –\(\frac{1}{4}\)V
a₃ – a₂ = \(\left(\frac{3}{4}\right)^2\)V – \(\frac{3}{4}\)V
= \(V\left(\frac{9}{16}-\frac{3}{4}\right)\)
= \(– \frac{3}{16}\)V
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the list of numbers does not form an AP.

3. Cost of digging first metre = ₹ 150
Cost of digging the second metre = ₹ 150 + ₹ 50
= ₹ 200
Cost of digging the third metre = ₹ 200 + ₹ 50
= ₹ 250
Cost of digging the fourth metre = ₹ 250 + ₹ 50
= ₹ 300
The list of numbers formed is 150, 200, 250, 300,…
Here, a₂ – a₁ = 200 – 150 = 50,
a₃ – a₂ = 250 – 200 = 50,
a₄ – a₃ = 300 – 250 = 50, and so on.
Thus, a_k+1 – a_k is the same every time. Hence, the list of numbers forms an AP with a = 150 and d = 50.

4. The formula of compound interest is known to us.
A = \(P\left(1+\frac{R}{100}\right)^T\)
Here, P = ₹ 10,000; R = 8% and T = 1, 2, 3, 4, ……….
Amount at the end of 1st year = ₹ 10000 (1.08).
Amount at the end of 2nd year = ₹ 10000 (1.08)².
Amount at the end of 3rd year = ₹ 10000 (1.08)³.
The list of numbers formed is 10000 (1.08), 10000 (1.08)2, 10000 (1.08), ………..
a₂ – a₁ = 10000 (1.08)² – 10000 (1.08)³
= 10000 (1.08) (1.08 – 1)
= 10000 (1.08) (0.08)
a₃ – a₂ = 10000 (1.08)³ – 10000 (1.08)²
= 10000 (1.08)² (1.08 – 1)
= 10000 (1.08)² (0.08)
Thus, a₂ – a₁ ≠ a₃ – a₂
Hence, the list of numbers does not form an AP.

Question 2.
Write first four terms of the AP when the first term a and the common difference d are given as follows:
1. a = 10, d = 10
2. a = -2, d = 0
3. a = 4, d = -3
4. a = -1, d = \(\frac{1}{2}\)
5. a = -1.25, d = -0.25
Solution:
1. a = 10, d = 10
First term a = 10
Second term = First term + d
= 10 + 10 = 20
Third term = Second term + d
= 20 + 10 = 30
Fourth term = Third term + d
= 30 + 10 = 40
Thus, the required first four terms of the AP are 10, 20, 30, 40.

2. a = -2, d = 0
First term = a = -2
Second term = First term + d
= -2 + 0 = -2
Third term = Second term + d
= -2 + 0 = -2
Fourth term = Third term + d
= -2 + 0 = -2
Thus, the required first four terms of the AP are -2, -2, -2, -2.

3. a = 4, d = -3
First term = a = 4
Second term = First term + d
= 4 + (-3) = 1
Third term = Second term + d
= 1 + (-3) = -2
Fourth term = Third term + d
= (-2) + (-3) = -5
Thus, the required first four terms of the AP are 4, 1, -2, -5.

4. a = -1, d = \(\frac{1}{2}\)
First term = a = -1
Second term = First term + d
= -1 + \(\frac{1}{2}\) = –\(\frac{1}{2}\)
Third term = Second term + d
= \(-\frac{1}{2}+\frac{1}{2}\) = 0
Fourth term = Third term + d
= 0 + \(\frac{1}{2}\) = \(\frac{1}{2}\)
Thus, the required first four terms of the AP are -1, – \(\frac{1}{2}\), 0, \(\frac{1}{2}\).

5. a = -1.25, d = -0.25
The general form of an AP is a, a + d, a + 2d, a + 3d, …….. Then,
First term = a = -1.25
Second term = a + d
= -1.25 + (-0.25) = -1.50
Third term = a + 2d
= -1.25 + 2(-0.25) = -1.75
Fourth term = a + 3d
= -1.25 + 3(-0.25) = -2.00
Thus, the required first four terms of the AP are -1.25, -1.50, -1.75, -2.00.

Question 3.
For the following APs, write the first term and the common difference:
1. 3, 1, 1, -3, ……..
2. -5, -1, 3, 7, …….
3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
4. 0.6, 1.7, 2.8, 3.9, …
Solution:
1. 3, 1, -1, -3,….
First term a = 3
Common difference d = a₂ – a₁ = 1 – 3 = -2

2. -5, 1, 3, 7,…..
First term a = -5
Common difference d = (-1) – (-5) = 4

3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
First term a = \(\frac{1}{3}\)
Common difference d = \(\frac{5}{3}-\frac{1}{3}=\frac{4}{3}\)

4. 0.6, 1.7, 2.8, 3.9, ……
First term a = 0.6
Common difference d = 1.7 – 0.6 = 1.1

Question 4.
Which of the following are APs? If they form an AP find the common difference d and write three more terms:
1. 2, 4, 8, 16, …….
2. 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …….
3. -1.2, -3.2, -5.2, -7.2, …….
4. -10, -6, -2, 2, …….
5. 3, 3 + 2\(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\), ….
6. 0.2, 0.22, 0.222, 0.2222, ……..
7. 0, -4, -8, -12, ……..
8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \cdots\)
9. 1, 3, 9, 27, ……
10. a, 2a, 3a, 4a, …….
11. a, a², a³, a⁴, …….
12. \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\), ……
13. \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\), …..
14. 1², 3², 5², 7², …….
15. 1², 5², 7², 7², …….
Solution:
2, 4, 8, 16, ….
a₂ – a₁ = 4 – 2 = 2,
a₃ – a₂ = 8 – 4 = 4,
a₄ – a₃ = 16 – 8 = 8
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

2. 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …
a₂ – a₁ = \(\frac{5}{2}-2=\frac{1}{2}\)
a₃ – a₂ = \(3-\frac{5}{2}=\frac{1}{2}\)
a₄ – a₃ = \(\frac{7}{2}-3=\frac{1}{2}\)
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = \(\frac{1}{2}\)
Next three terms are given by-
a₅ = \(\frac{7}{2}+\frac{1}{2}=4\),
a₆ = \(4+\frac{1}{2}=\frac{9}{2}\) and
a₇ = \(\frac{9}{2}+\frac{1}{2}=5\)

3. -1.2, -3.2, -5.2, -7.2, …..
a₂ – a₁ = -3.2 – (-1.2) = -2
a₃ – a₂ = -5.2 – (-3.2) = -2
a₄ – a₃ = -7.2 – (-5.2) = -2
Here, a_k+1 – a_k is the same everywhere. So, the given list of numbers forms an AP with d = -2.
Next three terms are given by-
a₅ = -7.2 + (-2) = -9.2.
a₆ = -9.2 + (-2) = -11.2 and
a₇ = -11.2 + (-2) = -13.2

4. -10, -6, -2, 2, ……..
a₂ – a₁ = (-6) – (-10) = 4.
a₃ – a₂ = (-2) – (-6) = 4.
a₄ – a₃ = 2 – (-2) = 4
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = 4.
Next three terms are given by-
a₅ = 2 + 4 = 6,
a₆ = 6 + 4 = 10 and
a₇ = 10 + 4 = 14

5. 3, 3+\(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\), …
a₂ – a₁ = 3 + \(\sqrt{2}\) – 3 = \(\sqrt{2}\),
a₃ – a₂ = (3 + 2\(\sqrt{2}\)) – (3 + \(\sqrt{2}\)) = \(\sqrt{2}\)
a₄ – a₃ = (3 + 3\(\sqrt{2}\)) – (3 + 2\(\sqrt{2}\)) = \(\sqrt{2}\)
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = \(\sqrt{2}\).
Next three terms are given by-
a₅ = (3 + 3\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 4\(\sqrt{2}\),
a₆ = (3 + 4\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 5\(\sqrt{2}\) and
a₇ = (3 + 5\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 6\(\sqrt{2}\)

6. 0.2, 0.22, 0.222. 0.2222,…
a₂ – a₁ = 0.22 – 0.2 = 0.02,
a₃ – a₂ = 0.222 – 0.22 = 0.002
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the given list of numbers does not form an AP.

7. 0, -4, -8, -12, ….
a₂ – a₁ = (-4) – 0 = -4,
a₃ – a₂ = (-8) – (-4) = -4,
a₄ – a₃ = (-12) – (-8) = -4
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = -4.
Next three terms are given by-
a₅ = (-12) + (-4) = -16,
a₆ = (-16) + (-4) = -20 and
a₇ = (-20) + (-4) = -24.

8.

Here, a_k+1 – a_k is the same everywhere. Hence, the given list of numbers forms an AP with d = 0.
Next three terms are given by-

9. 1, 3, 9, 27, ….
a₂ – a₁ = 3 – 1 = 2,
a₃ – a₂ = 9 – 3 = 6
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

10. a, 2a, 3a, 4a…
a₂ – a₁ = 2a – a = a,
a₃ – a₂ = 3a – 2a = a.
a₄ – a₃ = 4a – 3a = a
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of unknown numbers forms an AP with d = a.
Next three terms are given by-
a₅ = 4a + a = 5a,
a₆ = 5a + a = 6a and
a₇ = 6a + a = 7a

11. a, a², a³, a⁴,….
a₂ – a₁ = a² – a = a (a – 1),
a₃ – a₂ = a³ – a² = a² (a – 1)
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the given list of unknown numbers does not form an AP.

12. \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\), ….
We know, \(\sqrt{8}\) = \(\sqrt{4 \times 2}\) = 2\(\sqrt{2}\).
\(\sqrt{18}\) = \(\sqrt{9 \times 2}\) = 3\(\sqrt{2}\) and
\(\sqrt{32}\) = \(\sqrt{16 \times 2}\) = 4\(\sqrt{2}\).
Hence, the given list of numbers is
\(\sqrt{2}\), 2\(\sqrt{2}\), 3\(\sqrt{2}\), 4\(\sqrt{2}\), ….
a₂ – a₁ = 2\(\sqrt{2}\) – \(\sqrt{2}\) = \(\sqrt{2}\),
a₃ – a₂= 3\(\sqrt{2}\) – 2\(\sqrt{2}\) = \(\sqrt{2}\)
a₄ – a₃ = 4\(\sqrt{2}\) – 3\(\sqrt{2}\) = \(\sqrt{2}\)
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = \(\sqrt{2}\).
Next three terms are given by-
a₅ = 4\(\sqrt{2}\) + \(\sqrt{2}\) = 5\(\sqrt{2}\) = \(\sqrt{50}\).
a₆ = 5\(\sqrt{2}\) + \(\sqrt{2}\) = 6\(\sqrt{2}\) = \(\sqrt{72}\) and
a₇ = 6\(\sqrt{2}\) + \(\sqrt{2}\) = 7\(\sqrt{2}\) = \(\sqrt{98}\).

13. \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\), ……
a₂ – a₁ = \(\sqrt{6}\) – \(\sqrt{3}\) = \(\sqrt{3}\)(\(\sqrt{2}\) – 1)
a₃ – a₂ = \(\sqrt{9}\) – \(\sqrt{6}\) = \(\sqrt{3}\)(\(\sqrt{3}-\sqrt{2}\))
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

14. 1², 3², 5², 7², ….
a₂ – a₁ = 3² – 1² = 9 – 1 = 8,
a₃ – a₂ = 5² – 3² = 25 – 9 = 16
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

15. 1², 5², 7², 7²,…
a₂ – a₁ = 5² – 1² = 25 – 1 = 24,
a₃ – a₂ = 7² – 5² = 49 – 25 = 24,
a₄ – a₃ = 73 – 7² = 73 – 49 = 24.
Here. a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = 24.
Next three terms are given by-
a₅ = 73 + 24 = 97,
a₆ = 97 + 24 = 121 and
a₇ = 121 + 24 = 145.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
1. 2x² – 3x + 5 = 0
2. 3x² – 4\(\sqrt{3}\)x + 4 = 0
3. 2x² – 6x + 3 = 0
Solution:
The given equation is of the form
ax² + bx + c = 0; where a = 2, b = -3 and c = 5.
Then, the discriminant
b² – 4ac = (-3)² – 4(2)(5)
= 9 – 40
= -31 <0
So, the given equation has no real roots.

2. Comparing the given equation with the standard quadratic equation
ax² + bx + c = 0, we get a = 3, b = -4\(\sqrt{3}\) and c = 4.
Then, the discriminant
b² – 4ac = (-4\(\sqrt{3}\))² – 4(3)(4)
= 48 – 48
= 0
So, the given equation has equal real roots.
The roots are \(-\frac{b}{2 a},-\frac{b}{2 a}\)
i.e., \(-\frac{-4 \sqrt{3}}{2(3)},-\frac{-4 \sqrt{3}}{2(3)}\), i.e., \(\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

3. The given equation is of the form
ax² + bx + c = 0, where a = 2, b = -6 and c = 3.
Then, the discriminant
b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24
= 12
So, the given equation has two distinct roots.
The roots are given by
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
\(=\frac{6 \pm \sqrt{12}}{2(2)}\)
\(=\frac{6 \pm 2 \sqrt{3}}{4}=\frac{3 \pm \sqrt{3}}{2}\)
Thus, the roots of the given equation are \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3-\sqrt{3}}{2}\)

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots:
1. 2x² + kx + 3 = 0
2. kx (x – 2) + 6 = 0
Solution:
1. Comparing the given equation with the standard quadratic equation, we have
a = 2, b = k and c = 3.
Then, the discriminant = b² – 4ac
= (k)² – 4 (2) (3)
= k² – 24
If the equation has two equal roots, then the discriminant = 0
∴ k² – 24 = 0
∴ k² = 24
∴ k = ±\(\sqrt{24}\)
∴ k ± 2\(\sqrt{6}\)

2. kx(x – 2) + 6 = 0
∴ kx² – 2kx + 6 = 0
Here, a = k, b = -2k and c = 6.
Then, the discriminant = b² – 4ac
= (-2k)² – 4(k)(6)
= 4k² – 24k
If the equation has two equal roots, then the discriminant = 0
∴ 4k² – 24k = 0
∴ 4k (k – 6) = 0
∴ k = 0 or k = 6
But k = 0 is not possible because if k = 0, the equation reduces to 6 = 0, not a quadratic equation.
∴ k = 6

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Solution:
Considering that the required mango grove can be designed, let the breadth of the mango grove be x m.
Then, the length of the mango grove is 2x m. Area of rectangular mango grove
= Length × Breadth
= 2x × x
= 2x² m²
The required area is 800 m².
∴ 2x² = 800
∴ 2x² – 800 = 0
∴ x² – 400 = 0
If the above equation has real roots, then it is possible to design the required mango grove.
Here, a = 1, b = 0 and c = -400.
Then, the discriminant = b² – 4ac
= (0)² – 4(1)(-400)
= 1600 > 0
Hence, the equation has real roots. So, it is possible to design the mango grove with required measures.
Now, x² – 400 = 0
∴ (x + 20) (x – 20) = 0
∴ x + 20 = 0 or x – 20 = 0
∴ x = -20 or x = 20
Since x is the breadth of the rectangular mango grove, x = -20 is not possible.
∴ x = 20 and 2x = 40
Thus, the length of the mango grove is 40 m and its breadth is 20 m.

Question 4.
Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the present ages of two friends be x years and (20 – x) years.
Four years ago, their respective ages were (x – 4) years and (20 – x – 4) years, i.e., (16 – x) years.
Then, according to given,
(x – 4) (16 – x) = 48
∴ 16x – x² – 64 + 4x = 48
∴ -x² + 20x – 64 – 48 = 0
∴ -x² + 20x – 112 = 0
∴ x² – 20x + 112 = 0
Here, a = 1, b = -20 and c = 112.
Then, the discriminant
b² – 4ac = (-20)² – 4(1)(112)
= 400 – 448
= -48 < 0
Hence, the equation has no real roots. So, the given situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth.
Solution:
Let the length of the rectangular park be x m.
Perimeter of rectangular park = 2 (Length + Breadth)
∴ 80 = 2(x + Breadth)
∴ 40 = x + Breadth
∴ Breadth = (40 – x) m
Now, area of rectangular park = Length × Breadth
∴ 400 = x (40 – x)
∴ 400 = 40x – x²
∴ x² – 40x + 400 = 0
Here, a = 1, b = -40 and c = 400.
Then, the discriminant = b² – 4ac
= (-40)² – 4(1)(400)
= 1600 – 1600
= 0
Hence, the equation has equal real roots. So, it is possible to design a rectangular park with given measures.
x² – 40x + 400 = 0
∴ x² – 20x – 20x + 400 = 0
∴ x(x – 20) – 20(x – 20) = 0
∴ (x – 20) (x – 20) = 0
∴ x – 20 = 0 or x – 20 = 0
∴ x = 20 or x = 20
Thus, the length of the rectangular park = x = 20 m and the breadth of the rectangular park = 40 – x = 40 – 20 = 20 m.
Note: Here the shape of the park turns out to be a square, but as we know, every square is a rectangle.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
1. 2x² – 7x + 3 = 0
2. 2x² + x – 4 = 0
3. 4x² + 4\(\sqrt{3}\)x + 3 = 0
4. 2x² + x + 4 = 0
Solution:
1. 2x² – 7x + 3 = 0
Dividing throughout by 2, we get

Thus, the roots of the given quadratic equation are 3 and \(\frac{1}{2}\).

2. 2x² + x – 4 = 0
Dividing throughout by 2, we get

Thus, the roots of the given quadratic equation are \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)

3. 4x² + 4\(\sqrt{2}\)x + 3 = 0
4x² + 4\(\sqrt{3}\)x + (\(\sqrt{3}\))² = 0
(2x + \(\sqrt{3}\))² = 0
(2x + \(\sqrt{3}\)) (2x + \(\sqrt{3}\)) = 0
2x + \(\sqrt{3}\) = 0 or 2x + \(\sqrt{3}\) =0
x = \(-\frac{\sqrt{3}}{2}\) or x = \(-\frac{\sqrt{3}}{2}\)
Thus, the roots of the given quadratic equation are \(-\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\)

4. 2x² + x + 4 = 0
Dividing throughout by 2, we get

But, the square of any real number cannot be negative.
Hence, the real roots of the given quadratic equation do not exist.

Question 2.
Find the roots of the quadratic equations given in Q.1 by applying the quadratic formula.
1. 2x² – 7x + 3 = 0
Here, a = 2, b = -7 and c = 3.
Then, b² – 4ac = (-7)² – 4(2)(3) = 25
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{7 \pm \sqrt{25}}{2(2)}\)
∴ x = \(\frac{7 \pm 5}{4}\)
∴ x = 3 or x = \(\frac{1}{2}\)
Thus, the roots of the given quadratic equation are 3 and \(\frac{1}{2}\).

2. 2x² + x – 4 = 0
Here, a = 2, b = 1 and c = -4.
Then, b² – 4ac = (1)² – 4(2)(-4) = 33
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-1 \pm \sqrt{33}}{2(2)}\)
∴ x = \(\frac{-1 \pm \sqrt{33}}{4}\)
Thus, the roots of the given quadratic equation are \(\frac{-1 \pm \sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)

3. 4x² + 4\(\sqrt{3}\)x + 3 = 0
Here, a = 4, b = 4\(\sqrt{3}\) and c = 3.
Then, b² – 4ac = (4\(\sqrt{3}\))² – 4(4)(3) = 0
Then, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-4 \sqrt{3} \pm \sqrt{0}}{2(4)}\)
∴ x = \(-\frac{\sqrt{3}}{2}\) or x = \(-\frac{\sqrt{3}}{2}\)
Thus, the roots of the given quadratic equation are \(-\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\).

4. 2x² + x + 4 = 0
Here, a = 2, b = 1 and c = 4.
Then, b² – 4ac = (1)² – 4 (2)(4) = -31 < 0
Since b² – 4ac < 0 for the given quadratic equation, the real roots of the given quadratic equation do not exits.

Question 3.
Find the roots of the following equations:
1. x – \(\frac{1}{x}\) = 3, x ≠ 0
2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\); x ≠ -4, 7
Solution:
1. x – \(\frac{1}{x}\) = 3, x ≠ 0
∴ x² – 1 = 3x
∴ x² – 3x – 1 = 0
Here, a = 1, b = -3 and c = -1.
Then, b² – 4ac = (-3)² – 4(1)(-1) = 13
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
x = \(\frac{3 \pm \sqrt{13}}{2(1)}\)
x = \(\frac{3 \pm \sqrt{13}}{2}\)
Thus, the roots of the given equation are \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\).

2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\); x ≠ -4, 7
∴ \(\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}\)
∴ \(\frac{-11}{x^2-3 x-28}=\frac{11}{30}\)
∴ \(\frac{-1}{x^2-3 x-28}=\frac{1}{30}\)
∴ -30 = x² – 3x – 28
∴ x² – 3x + 2 = 0
Here, a = 1, b = -3 and c = 2.
Then, b² – 4ac = (-3)² – 4(1)(2) = 1
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{3 \pm \sqrt{1}}{2(1)}\)
∴ x = \(\frac{3 \pm 1}{2}\)
∴ x = 2 or x = 1
Thus, the roots of the given equation are 2 and 1.
Note: Here, the method of factorisation would turn out to be more easier.

Question 4.
The sum of the reciprocals of Rehman’s ages (in years), 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:
Let the present age of Rehman be x years.
So, his age 3 years ago was (x – 3) years and his age 5 years hence will be (x + 5) years.
The sum of the reciprocals of these two ages (in years) is given to be \(\frac{1}{3}\).
∴ \(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)
∴ \(\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}\)
∴ 3(2x + 2) = (x – 3)(x + 5)
∴ 6x + 6 = x² + 2x – 15
∴ x² – 4x – 21 = 0
∴ x² – 7x + 3x – 21 = 0
∴ x(x – 7) + 3(x – 7) = 0
∴ (x – 7)(x + 3) = 0
∴ x – 7 = 0 or x + 3 = 0
∴ x = 7 or x = -3
Now, since x represents the present age of Rehman, it cannot be negative, i,e., x ≠ -3.
∴ x = 7
Thus, the present age of Rehman is 7 years.

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of those marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics be x.
Then, her marks in English = 30 – x, as her total marks in Mathematics and English is 30.
Had she scored 2 marks more in Mathematics and 3 marks less in English, her score in Mathematics would be x + 2 and in English would be 30 – x – 3 = 27 – x.
∴ (x + 2) (27 – x) = 210
∴ 27x – x² + 54 – 2x = 210
∴ -x² + 25x + 54 – 210 = 0
∴ -x² + 25x – 156 = 0
∴ x² – 25x + 156 = 0
∴ x² – 13x – 12x + 156 = 0
∴ x(x – 13) – 12(x – 13) = 0
∴ (x – 13)(x – 12) = 0
∴ x – 13 = 0 or x – 12 = 0
∴ x = 13 or x = 12
Then, 30 – x = 30 – 13 = 17 or
30 – x = 30 – 12 = 18
Thus, Shefall’s marks in Mathematics and in English are 13 and 17 respectively or 12 and 18 respectively.

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of the rectangular field be x m.
Then, it diagonal is (x + 60) m and the longer side is (x + 30) m.
In a rectangle, all the angles are right angles.
Hence, by Pythagoras theorem,
(Shorter side)² + (Longer side)² = (Diagonal)²
∴ x² + (x + 30)² = (x + 60)²
∴ x² + x² + 60x + 900 = x² + 120x + 3600
∴ x² – 60x – 2700 = 0
Here, a = 1, b = -60 and c = -2700.
Then, b² – 4ac = (-60)² – 4(1)(-2700)
= 3600 + 10800
= 14400
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{60 \pm \sqrt{14400}}{2(1)}\)
= \(\frac{60 \pm 120}{2}\)
∴ x = \(\frac{60+120}{2}\) or x = \(\frac{60-120}{2}\)
∴ x = 90 or x = -30
Since x denotes the shorter side of the rectangular field, x cannot be negative.
∴ x = 90
Then, x + 30 = 90 + 30 = 120
Thus, the shorter side (breadth) of the rectangular field is 90 m and the longer side (length) of the rectangular field is 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the smaller number be x.
Then, the larger number = \(\frac{x^2}{8}\)
Now, the difference of their squares is 180.
∴ \(\left(\frac{x^2}{8}\right)^2-(x)^2=180\)
∴ \(\frac{x^4}{64}-x^2=180\)
∴ x⁴ – 64x² – 11520 = 0
Let x² = y
∴ x⁴ = y²
∴ y² – 64y – 11520 = 0
Here, a = 1, b = -64 and c = -11520.
Then, b² – 4ac = (-64)² – 4(1)(-11520)
= 4096 + 46080
= 50176
Now, y = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{64 \pm \sqrt{50176}}{2(1)}\)
= \(\frac{64 \pm 224}{2}\)
∴ y = \(\frac{64+224}{2}\) or y = \(\frac{64-224}{2}\)
∴ y = 144 or y = -80
∴ x² = 144 or x² = -80
But, x² = -80 is not possible.
∴ x² = 144
∴ x = 12 or x = -12
Then, \(\frac{x^2}{8}=\frac{144}{8}=18\)
Thus, the required numbers are 12 and 18 or -12 and 18.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let the usual uniform speed of the train be x km/h.
∴ Time taken to travel 360 km at the speed of x km/h = \(\frac{\text { distance }}{\text { speed }}\) = \(\frac{360}{x}\) hours.
If the speed had been 5 km/h more, the new speed would be (x + 5) km/h and the time taken to travel 360 km at this increased speed would be \(\frac{360}{x+5}\) hours.
Now, New time = Usual time – 1
∴ \(\frac{360}{x+5}=\frac{360}{x}-1\)
∴ 360x = 360x + 1800 – x(x + 5) (Multiplying by x(x + 5))
∴ 0 = 1800 – x² – 5x
∴ x² + 5x – 1800 = 0
∴ x² + 45x – 40x – 1800 = 0
∴ x(x + 45) – 40(x + 45) = 0
∴ (x + 45) (x – 40) = 0
∴ x + 45 = 0 or x – 40 = 0
∴ x = -45 or x = 40
As, x is the speed (in km/h) of the train, x = -45 is not possible.
∴ x = 40
Thus, the usual uniform speed of the train is 40 km/h.

Question 9.
Two water taps together can fill a tank in \(9 \frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the time taken by the tap with smaller diameter to fill the tank be x hours.
Then, the time taken by the tap with larger diameter = (x – 10) hours.
So, the part of the tank filled in one hour by the tap with smaller diameter = \(\frac{1}{x}\) and by the tap with larger diameter = \(\left(\frac{1}{x-10}\right)\)
So, the part of tank filled in one hour by both the taps together = \(\frac{1}{x}+\frac{1}{x-10}\)
Both the taps together can fill the tank in \(9 \frac{3}{8}\) hours, i.e., \(\frac{75}{8}\) hours.
∴ The part of tank filled in one hour by both the tank together = \(\frac{8}{75}\)
∴ \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\).
∴ 75(x – 10) + 75x = 8x (x – 10) (Multiplying by 75x(x – 10))
∴ 75x – 750 + 75x = 8x² – 80x
∴ 8x² – 230x + 750 = 0
Here, a = 8, b = -230 and c = 750.
∴ b² – 4ac = (-230)² – 4 (8)(750)
= 52900 – 24000
= 28900
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{230 \pm \sqrt{28900}}{2(8)}\)
∴ x = \(\frac{230 \pm 170}{16}\)
∴ x = \(\frac{400}{16}\) or x = \(\frac{60}{16}\)
∴ x = 25 or x = 3.75
But, x ≠ 3.75, because for x = 3.75, x – 10 < 0.
∴ x = 25 and x – 10 = 15
Thus, the time taken by the tap with smaller diameter is 25 hours and that by the tap. with larger diameter is 15 hours.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of the passenger train be x km/h.
Then, the average speed of the express train is (x + 11) km/h.
∴ Time taken by passenger train to cover 132 km = \(\frac{132}{x}\) hours.
∴ Time taken by express train to cover 132 km = \(\frac{132}{x+11}\) hours.
Time taken by express train = Time taken by passenger train – 1
\(\frac{132}{x+11}=\frac{132}{x}-1\)
∴ 132x = 132(x + 11) – x(x + 11) (Multiplying by x(x + 11))
∴ 132x = 132x + 1452 – x² – 11x
∴ x² + 11x – 1452 = 0
Here, a = 1, b = 11 and c = -1452.
∴ b² – 4ac = (11)² – 4(1)(-1452)
= 121 + 5808 = 5929
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-11 \pm \sqrt{5929}}{2(1)}\)
∴ x = \(\frac{-11 \pm 77}{2}\)
x = 33 or x = -44
x = -44 is inadmissible as x represents the speed of the passenger train.
∴ x = 33 and x + 11 = 44
Thus, the speed of the passenger train is 33 km/h and the speed of the express train is 44 km/h.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the side of the smaller square be x m.
Then, the perimeter of the smaller square = 4x m
and the area of the smaller square = x² m².
From the given, the perimeter of the bigger square = (4x + 24) m.
∴ Side of the bigger square = \(\frac{4 x+24}{4}\) = (x + 6) m and hence, the area of the bigger square = (x + 6)² m².
∴ x² + (x + 6)² = 468
∴ x² + x² + 12x + 36 – 468 = 0
∴ 2x² + 12x – 432 = 0
∴ x² + 6x – 216 = 0
∴ x² + 18x – 12x – 216 = 0
∴ x(x + 18) – 12(x + 18) = 0
∴ (x + 18) (x – 12) = 0
∴ x + 18 = 0 or x – 12 = 0
∴ x = -18 or x = 12
Here, x = -18 is not possible as x represents the side of a square.
∴ x = 12 and x + 6 = 18
Thus, the side of the smaller square is 12 m and the side of the bigger square is 18 m.