NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 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) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called sure event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 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 is 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:
(i) The outcome is not equally likely because the car starts normally only when there is some defect, the car does not start.
(ii) The outcome is not equally likely because the outcome depends on the training of the player.
(iii) The outcome in the trial of true-false question is, either true or false. Hence, the two outcomes are equally likely.
(iv) A baby can be either a boy or a girl and both the outcomes have equally likely chances.

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the bail at the beginning of a football game?
Solution:
When we toss a coin, the outcomes head and tail are equally likely. So, the result of an individual coin toss is completely 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:
We know that probability of an event cannot be less than 0 and greater than 1.
Correct option is (B).

Question 5.
If P (E) = 0.05, what is the probability of ‘not E’?
Solution:
We have, P (E) + P (not E) = 1
Given: P(E) = 0.05
P (not 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) A bag contains only lemon flavoured candies.
P (an orange flavoured candy) = 0
(ii) P (a lemon flavoured candy) = 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:
We have, P (E) + P (not E) = 1
⇒ P (E) + 0.992 = 1
⇒ P (E) = 1 – 0.992 = 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:
Number of red balls = 3
Number of black balls = 5
Total number of balls = 3 + 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 number of marbles = 5 + 8 + 4 = 17
(i) P (red marble) = \(\frac { 5 }{ 17 }\)
(ii) P (white marble) = \(\frac { 8 }{ 17 }\)
(iii) P (not a green marble) = \(\frac { 13 }{ 17 }\)

Question 10.
A piggy bank contains hundred 50 p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 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 ₹ 5 coin?
Solution:
Number of 50 p coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins = 20
Number of ₹ 5 coins = 10
Total number of coins = 180

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 (see figure). What is the probability that the fish taken out is a male fish?

Solution:
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 + 8 = 13
P (a male 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 (see figure.), 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:
(i) P (getting 8) = \(\frac { 1 }{ 8 }\)
(ii) P (an odd number) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\) ( odd numbers are 1, 3, 5, 7)
(iii) P (a number greater than 2) = \(\frac { 6 }{ 8 }\) = \(\frac { 3 }{ 4 }\)
(iv) P (a number less than 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
(ill) an odd number
Solution:
(i) Prime numbers on a die = 2, 3, 5
P (a prime number) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
(ii) Number lying between 2 and 6 = 3, 4, 5
P(a number lying between 2 and 6) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
(iii) Odd numbers = 1, 3, 5
P (an 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:
Number of cards in a well-shuffled deck = 52.
(i) P (a king of red colour) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
(ii) P (a face card) = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)
(iii) P (a red face card) = \(\frac { 6 }{ 52 }\) = \(\frac { 3 }{ 26 }\)
(iv) P (the jack of hearts) = \(\frac { 1 }{ 52 }\)
(v) P(a spade) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(vi) P (the queen of diamonds) = \(\frac { 1 }{ 52 }\)

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:
Out of 5 cards there is only one queen.
(i) P (getting queen) = \(\frac { 1 }{ 5 }\) [when queen is drawn, four cards are left]
(ii) (a) P (an ace) = \(\frac { 1 }{ 4 }\)
(b) P (a queen) = \(\frac { 0 }{ 4 }\) = 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:
Number of defective pens = 12
Number of good pens = 132
Total number of pens = 12 + 132 = 144
P (the pen is good one) = \(\frac { 132 }{ 144 }\) = \(\frac { 11 }{ 12 }\)

Question 17.
(i) A lot of 20 bulbs contain 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 number of bulbs = 20
Number of defective bulbs = 4
P (bulb drawn is defective) = \(\frac { 4 }{ 20 }\) = \(\frac { 1 }{ 5 }\)
(ii) Remaining bulbs = 19
P (bulb drawn is not defective) = \(\frac { 15 }{ 19 }\)

Question 18.
A box contains 90 discs 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:
Total numbers of discs = 90
(i) P (a two digit number) = \(\frac { 81 }{ 90 }\) = \(\frac { 9 }{ 10 }\)
(ii) Here, perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81
P (getting a perfect square number) = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)
(iii) Numbers divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
P (getting a number 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:
(i) P (getting A) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
(ii) P (getting D) = \(\frac { 1 }{ 6 }\)

Question 20.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution:

Question 21.
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 ball pens = 144
Number of defective pens = 20
Number of good pens = 144 – 20 = 124

Question 22.
Two dice, one blue and one grey, are thrown at the same time. Now
(i) Complete the following table:

(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:
(i) Total number of possible outcomes = 36
(1, 2) and (2, 1) are the favourable events of getting the sum 3.
P(sum 3) = \(\frac { 2 }{ 36 }\) = \(\frac { 1 }{ 18 }\)
(1, 3) , (2, 2) and (3, 1) are the favourable events of getting the sum 4.
P(sum 4) = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)
(1, 4) , (2, 3), (3, 2) and (4, 1) are the favourable events of getting the sum 5.
P(sum 5) = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)
(1, 5) , (2, 4), (3, 3), (4, 2) and (5, 1) are the favourable events of getting the sum 6.
P (sum 6) = \(\frac { 5 }{ 36 }\)
(1, 6) , (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) are the favourable events of getting the sum 7.
P(sum 7) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
(3, 6) , (4, 5), (5, 4) and (6, 3) are the favourable events of getting the sum 9.
P(sum 9) = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)
(4, 6) , (5, 5) and (6, 4) are the favourable events of getting the sum 10.
P(sum 10 = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)
(5,6) and (6,5) are the favourable events of getting the sum 11.
P(sum 11) = \(\frac { 2 }{ 36 }\) = \(\frac { 1 }{ 18 }\)
(ii) No, because the outcomes as 11 different sum are not equally likely.

Question 23.
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:
Possible outcomes are
HHH, TTT, HHT, HTH, THH, TTH, THT, HTT = 8
P (win the game) = \(\frac { 2 }{ 8 }\) = \(\frac { 1 }{ 4 }\)
P (lose the game) = 1 – \(\frac { 1 }{ 4 }\) = \(\frac { 3 }{ 4 }\)

Question 24.
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:
Total outcomes = 36
Number of outcomes in favour of 5 is (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6) = 11
(i) P (5 will not come up either time) = \(\frac { 25 }{ 36 }\)
(ii) P (5 will come up at least once) = \(\frac { 11 }{ 36 }\)

Question 25.
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) Argument is incorrect.
The possible outcomes are (HH), (HT), (TH), (TT)
P(HH) = \(\frac { 1 }{ 4 }\)
P(TT) = \(\frac { 1 }{ 4 }\)
P(HT or TH) = \(\frac { 2 }{ 4 }\) = \(\frac { 1 }{ 2 }\)
(ii) Argument is correct.
Possible outcomes = 1, 2, 3, 4, 5, 6
Odd numbers are = 1, 3, 5
P (an odd number) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

We hope the NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1, drop a comment below and we will get back to you at the earliest.

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

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

13.1 Class 10 Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:
Volume of one cube = 64 cm³
Let edge of one cube = a
Volume of the cube = (edge)³
a³ = 64 ⇒ a = 4 cm
Similarly, edge of the another cube = 4 cm.
Now, both cubes are joined together and a cuboid is formed as shown in the figure.
Now, length of the cuboid (l) = 8 cm
breadth of the cuboid (b) = 4 cm
height of the cuboid (h) = 4 cm
Surface area of the cuboid so formed = 2 (lb + bh + hl)
= 2(8 x 4 + 4 x 4 + 4 x 8)
= 2(32 + 16 + 32) = 160 cm²

NCERT Class 10 Exercise 13.1 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:
Given: diameter of the hemisphere = 14 cm
Radius = \(\frac { 14 }{ 2 }\) = 7 cm
Curved surface area of the hemisphere = 2πr² = 2 x \(\frac { 22 }{ 7 }\) x 7 x 7 cm²
= 14 x 22 cm² = 308 cm²
Here, total height of the vessel = 13 cm
Height of the cylinder = Total height – Height of the hemisphere = 13 cm – 7 cm = 6 cm
and radius of the cylinder = radius of the hemisphere = 7 cm
Inner surface area of the cylinder = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 7 x 6
= 2 x 22 x 6 = 264 cm²
Inner surface area of the vessel = Inner surface area of the cylinder + curved surface area of the hemisphere
= 264 cm² + 308 cm² = 572 cm²

Ex 13.1 Class 10 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:

Surface Area And Volume Class 10 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:

Class 10 Maths Chapter 13 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:

Chapter 13 Class 10 Maths Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

Class 10 Maths Chapter 13 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 ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:

Exercise 13.1 Class 10 NCERT Solutions 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:

Class 10 Maths Chapter 13 Solutions Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in 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:

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

Exercise 8.1 Class 10 Question 1.
In ∆ABC right angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:

Class 10 Ex 8.1 Question 2.
In given figure, find tan P – cot R.
Solution:

Trigonometry Class 10 Question 3.
If sin A = \(\frac { 3 }{ 4 }\) , calculate cos A and tan A.
Solution:

sin A = \(\frac { 3 }{ 4 }\)
sin A = \(\frac { BC }{ AC }\) = sin A = \(\frac { 3 }{ 4 }\)
Let BC = 3k and AC = 4k

Ex 8.1 Class 10 Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:

Class 10 Trigonometry Ex 8.1 Question 5.
Given sec θ = \(\frac { 13 }{ 12 }\) , calculate all other trigonometric ratios.
Solution:

Ex 8.1 Class 10 Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:

Ex 8.1 Class 10 Question 7.
If cot θ = \(\frac { 7 }{ 8 }\), evaluate:
(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right)}\)
(ii) cot²θ
Solution:

Trigonometry NCERT Class 10 Ex 8.1 Question 8.
If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.
Solution:

Ex 8.1 Class 10 Maths Solutions Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac { 1 }{ \surd 3 }\), find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:

Class 10 Ex 8.1 Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:

NCERT Solutions For Class 10 Maths Chapter 8 Question 11.
State whether the following statements are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac { 12 }{ 5 }\) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = \(\frac { 4 }{ 3 }\) for some angle.
Solution:
(i) tan 60° = √3 , Since √3 > 1. (False)
(ii) sec A is always ≥ 1. (True)
(iii) cos A is the abbreviation for cosine A. (False)
(iv) cot without ∠A is meaningless. (False)
(v) sin θ can never be greater than 1.
∴ sin θ = \(\frac { P }{ H }\) , hypotenuse is always greater than other two sides. (False)

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Solution:

Question 2.
Find the distance between the points (0, 0) and (36, 15).
Solution:
Let points be A (0, 0) and B (36, 15)
The distance between two points is

Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let points be A (1, 5), B (2, 3) and C (-2, -11)

AB + BC ≠ AC
Hence, the given points are not collinear.

Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let points be A(5, -2), B (6, 4) and C (7, -2)

Here, AB = BC
ΔABC is an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Solution:
Points A (3, 4), B (6, 7), C (9, 4) and D (6, 1)

Here, AB = BC = CD = DA and AC = BD
ABCD is a square.
Hence, Champa is correct.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let points be A (-1, -2), B (1, 0), C (-1, 2) and D (-3, 0)

Here, AC = BD, AB = BC = CD = AD
Hence, the quadrilateral ABCD is a square.
(ii) Let points be A (-3, 5), B (3, 1), C (0, 3) and D (-1, -4)

The given points do not form any quadrilateral.
(iii) Let points be A(4, 5), B (7, 6), C (4, 3) and D (1, 2)

Here, AB = CD, BC = AD
and AC ≠ BD
The quadrilateral ABCD is a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
Let points be A (2, -5) and B (-2, 9)
Let P (x, 0) be the point on x-axis.
AP = BP

Question 8.
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
Solution:
Points P (2, -3), Q (10, y) and PQ = 10 units
The distance between two points is

Question 9.
If Q (0, 1) is equidistant from P (5, -3), and R (x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Points are P (5, -3) and R (x, 6)
Point Q (0, 1) is equidistant from points P (5, -3) and R (x, 6).

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Solution:
Points A(3, 6) and B(-3, 4) are equidistant from point P(x, y)

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) 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.
(iii) 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.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Given:
a₁ = ₹ 15
a₂ = ₹ 15 + ₹ 8 = ₹ 23
a₃ = ₹ 23 + ₹ 8 = ₹ 31
List of fares is ₹ 15, ₹ 23, ₹ 31
and a₂ – a₁ = ₹ 23 – ₹ 15 = ₹ 8
a₃ – a₂ = ₹ 31 – ₹ 23 = ₹ 8
Here, a₂ – a₁ = a₃ – a₂
Thus, the list of fares forms an AP.

(iii) Given:
a₁ = ₹ 150, a₂ = ₹ 200, a₃ = ₹ 250
a₂ – a₁ = ₹ 200 – ₹ 150 = ₹ 50
and a₃ – a₂ = ₹ 250 – ₹ 200 = ₹ 50
Here, a₃ – a₂ = a₂ – a₁
Thus, the list forms an AP.
(iv) Given: a₁ = ₹ 10000
a₂ = ₹ 10000 + ₹ 10000 x \(\frac { 8 }{ 100 }\) = ₹ 10000 + ₹ 800 = ₹ 10800
a₃ = ₹ 10800 + ₹ 10800 x \(\frac { 8 }{ 100 }\) = ₹ 10800 + ₹ 864 = ₹ 11664
a₂ – a₁ = ₹ 10800 – ₹ 10000 = ₹ 800
a₃ – a₂ = ₹ 11664 – ₹ 10800 = ₹ 864
a₃ – a₂ ≠ a₂ – a₁
Thus, it is not 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:
(i) a = 10, d = 10
(n) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = \(\frac { 1 }{ 2 }\)
(v) a = -1.25, d = -0.25
Solution:
(i) Given: a = 10, d = 10
a₁ = 10,
a₂ = 10 + 10 = 20
a₃ = 20 + 10 = 30
a₄ = 30 + 10 = 40
Thus, the first four terms of the AP are 10, 20, 30, 40.

(ii) Given: a = – 2, d = 0
The first four terms of the AP are -2, -2, -2, -2.

(iii) a₁ = 4, d = -3
a₂ = a₁ + d = 4 – 3 = 1
a₃ = a₂ + d = 1 – 3 = -2
a₄ = a₃ + d = -2 – 3 = -5
Thus, the first four terms of the AP are 4, 1, -2, -5.

(iv)

(v) a₁ = -1.25, d = -0.25
a₂ = a₁ + d = -1.25 – 0.25 = -1.50
a₃ = a₂ + d = -1.50 – 0.25 = -1.75
a₄ = a₃ + d = -1.75 – 0.25 = -2
Thus, the first four terms of the AP are -1.25, -1.50, -1.75, -2.

Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, ……
(ii) -5, -1, 3, 7, ……
(iii) \(\frac { 1 }{ 3 }\) , \(\frac { 5 }{ 3 }\) , \(\frac { 9 }{ 3 }\), \(\frac { 13 }{ 3 }\) , ……..
(iv) 0.6, 1.7, 2.8, 3.9, …….
Solution:
(i) a₁ = 3, a₂ = 1
d = a₂ – a₁ = 1 – 3 = -2
where, a₁ = first term and d = common difference
a₁ = 3, d = -2
(ii) a₁ = -5, a₂ = -1
d = a₂ – a₁ = -1 – (-5) = -1 + 5 = 4
So, first term a₁ = -5 and common difference d = 4
(iii) a₁ = \(\frac { 1 }{ 3 }\), a₂ = \(\frac { 5 }{ 3 }\)
d = \(\frac { 5 }{ 3 }\) – \(\frac { 1 }{ 3 }\) = \(\frac { 4 }{ 3 }\)
So, first term a₁ = \(\frac { 1 }{ 3 }\) and common difference d = \(\frac { 4 }{ 3 }\)
a₁= 0.6, a₂ = 1.7
d = a₂ – a₁ = 1.7 – 0.6 = 1.1
So, first term a₁ = 0.6 and common difference d = 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.
(i) 2, 4, 8, 16, …….
(ii) 2, \(\frac { 5 }{ 2 }\) , 3, \(\frac { 7 }{ 2 }\) , …….
(iii) -1.2, -3.2, -5.2, -7.2, ……
(iv) -10, -6, -2,2, …..
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …..
(vi) 0.2, 0.22, 0.222, 0.2222, ……
(vii) 0, -4, -8, -12, …..
(viii) \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , …….
(ix) 1, 3, 9, 27, …….
(x) a, 2a, 3a, 4a, …….
(xi) a, a2, a3, a4, …….
(xii) √2, √8, √18, √32, …..
(xiii) √3, √6, √9, √12, …..
(xiv) 12, 32, 52, 72, ……
(xv) 12, 52, 72, 73, ……
Solution:
(i) 2, 4, 8, 16, ……
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₂ – a₁ ≠ a₃ – a₂
Thus, the given sequence is not an AP.

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