NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-10/

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 10
Chapter Name Circles
Exercise Ex 10.1
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
There can be infinitely many tangents to a circle.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ………… point(s).
(ii) A line intersecting a circle in two points is called a ………… .
(iii) A circle can have parallel tangents at the most ………… .
(iv) The common point of a tangent to a circle and the circle is called ……….. .
Solution:
(i) One
(ii) Secant
(iii) Two
(iv) Point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) \( \sqrt{199} \) cm
Solution:
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 1
Radius of the circle = 5 cm
OQ = 12 cm
∠OPQ = 90°
[The tangent to a circle is perpendicular to the radius through the point of contact]
PQ2 = OQ2 – OP2 [By Pythagoras theorem]
PQ2 = 122 – 52 = 144 – 25 = 199
PQ = \( \sqrt{199} \) cm.
Hence correct option is (d).

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 2
A line m is parallel to the line n and a line l which is secant is parallel to the given line.

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-8-ex-8-4/

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 8
Chapter Name Introduction to Trigonometry
Exercise Ex 8.4
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
From trigonometric identity, cosec² A – cot² A = 1, we get
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 1

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
Since sin² A + cos² A = 1, therefore
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 2

Question 3.
Evaluate:
(i)
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 3
(ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 4
(ii) sin 25° cos 65° + cos 25° sin 65° = sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° cos 25°
= sin² 25° + cos² 25° = 1

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A =
(A) 1
(B) 9
(C) 8
(D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ ) =
(A) 0
(B) 1
(C) 2
(D) -1

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\)
(A) sec² A
(B) -1
(C) cot² A
(D) tan² A
Solution:
(i) 9 sec² A – 9 tan² A = 9(sec² A – tan² A) = 9 x 1 = 9
Correct option is (B)
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 5

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 6
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 7
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 8

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 9

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NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1

NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-15/

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 15
Chapter Name Probability
Exercise Ex 15.1
Number of Questions Solved 25
Category NCERT Solutions

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
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 1

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
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 2

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?
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 3
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?
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 4
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:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 5
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?
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 6
Solution:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 7

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
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 8

Question 22.
Two dice, one blue and one grey, are thrown at the same time. Now
(i) Complete the following table:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 9

(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 13.3

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-13-ex-13-3/

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 13
Chapter Name Surface Areas and Volumes
Exercise Ex 13.3
Number of Questions Solved 9
Category NCERT Solutions

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

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

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Given: radius of metallic sphere = 4.2 cm
∴ Volume = \(\frac { 4 }{ 3 }\)π(4.2)³ …. (i)
∵ Sphere is melted and recast into a cylinder of radius 6 cm and height h.
∴ Volume of the cylinder =πr²h = π(6)² x h … (ii)
According to question,
Volume of the cylinder = Volume of the sphere
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 1

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radius of 1st metallic sphere = 6 cm
∴ Volume of 1st metallic sphere = \(\frac { 4 }{ 3 }\)π(6)³ cm³
Radius of 2nd metallic sphere = 8 cm
∴ Volume of 2nd metallic sphere = \(\frac { 4 }{ 3 }\)π(8)³ cm³
Radius of 3rd metallic sphere = 10 cm
∴ Volume of 3rd metallic sphere = \(\frac { 4 }{ 3 }\)π(10)³ cm³
Volume of all three metallic spheres = \(\frac { 4 }{ 3 }\)π(6³+8³+10³) cm³
∵ 3 spheres are melted and recast into a new metallic sphere of radius r.
∴ Volume of new metallic sphere = \(\frac { 4 }{ 3 }\)πr³
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 2

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Given: diameter of the well = 7 m Radius = \(\frac { 7 }{ 2 }\)m
and depth of the well = 20 m
Volume of the earth taken out from the well = πr²
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 3

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Given: diameter of the well = 3 m
⇒ Radius = \(\frac { 3 }{ 2 }\)m
Depth of the well = 14 m
Volume of the earth taken out from the well = πr²h
= π(\(\frac { 3 }{ 2 }\))² x 14 = \(\frac { π×9×14 }{ 4 }\) = \(\frac { 63 }{ 2 }\)πm³
∵ Earth taken out from the well evenly spread to form an embankment having height h and width of embankment around the well is 4 m.
∴ External radius (R) = radius of well + width of the embankment
= \(\frac { 3 }{ 2 }\)m + 4m = \(\frac { 11 }{ 2 }\)m
Internal radius = \(\frac { 3 }{ 2 }\)m = radius of well
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 4

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 5
Let total number of ice cream cones are n.
∴ All ice cream cones are filled from ice cream in the cylinder.
Total volume of n number of ice cream cones = Volume of ice cream in the cylinder
n x π x 54 = π(36)15
⇒ n x 54 = 36 x 15
⇒ n = \(\frac { 36 × 15 }{ 54 }\) = 10

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
Solution:
Given: diameter of each coin = 1.75 cm ⇒ radius = \(\frac { 1.75 }{ 2 }\)cm
and thickness of each coin = 2 mm
Let n number of coins are melted to form a cuboid.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 6

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Given: radius of the cylindrical bucket = 18 cm
and height = 32 cm
∴ Volume of the cylindrical bucket = π²h = π(18)² x 32 cm³
Let radius of the conical heap = r cm
Given: height of the conical heap = 24 cm
∴ Volume of the conical heap = \(\frac { 1 }{ 3 }\)π(r²) 24cm³
According to question,
Volume of the cylinderical bucket = Volume of the conical heap
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 7

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Given: width of canal = 6m, depth = 1.5 m
Rate of flowing water – 10 km/h
Volume of the water flowing in 30 minutes = \(\frac { 6×1.5×30×10 }{ 60 }\)km³
= \(\frac { 6×1.5×10×1000×30 }{ 10×60 }\)km³ = 45000 m³
We require water for standing up to height = 8 cm = \(\frac { 8 }{ 100 }\) m
Let the required area he A
∴ Volume of water required = A(\(\frac { 8 }{ 100 }\))m³
According to question. 45000 = \(\frac { A×8 }{ 100 }\)
⇒ \(\frac { 45000×100 }{ 8 }\) = A ⇒ A = 562500 m²

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his Held, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Given: diameter of the pipe = 20 cm ⇒ radius of the pipe = 10 cm
Water flowing from the pipe at rate = 3 km
Let it filled the tank in ‘t’ hours.
Volume of the water flowing in ‘t’ hours.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 8

NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2

NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-15-ex-15-2/

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 15
Chapter Name Probability
Exercise Ex 15.2
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 1

Question 2.
A die is numbered in such a way that its faces show the number 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 2

What is the probability that the total score is at least 6?
(i) even
(ii) 6
(iii) at least 6
Solution:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 3

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is doubles that of a red ball, determine the number of blue balls in the bag.
Solution:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 4

Question 4.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 5

Question 5.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue balls in the jar.
Solution:
NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 6

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