Selina

### Prove that:

Solution: To Prove: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$ Formula Used: $\cos 2 A=2 \cos ^{2} A-1$ Proof: $\text { LHS }=\cos ^{-1}\left(2 x^{2}-1\right) \ldots(1)$ Let $x=\cos A \ldots$...

### Find the equation of a circle passing through the origin and intercepting lengths a and b on the axes.

Answer:           AD = b units and AE = a units. D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre. The general equation of a circle: (x - h)2 + (y - k)2 = r2...

### Using elementary row transformations, find the inverse of each of the following matrices:

Solution: We have $A=\left(\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...

### Why does the acetylation of —NH2 group of aniline reduce its activating effect?

Solution: The acetylation of —NH2 group of aniline reduces its activating effect because the lone pair of electrons on the nitrogen of acetanilide interacts with oxygen atom due to resonance.

### What is Hinsberg reagent?

Solution: Hinsberg's reagent is benzene sulphonyl chloride, also known as $C_6H_5SOCl$. To distinguish between primary, secondary, and tertiary amines, Hinsberg's reagent is used.

### What is the product when C6H5CH2NH2 reacts with HNO2?

Solution: The main product is $C_6H_5CH_2OH$ When $C_6H_5CH_2NH_2$ reacts with HNO2, it produces an unstable diazonium salt, which is then converted to alcohol. When $C_6H_5CH_2NH_2$ reacts with...

### What is the role of HNO3 in the nitrating mixture used for nitration of benzene?

Solution: The nitration of organic compounds is done with a nitration mixture, which is a 1:1 solution of HNO3 and H2SO4. In the nitration of benzene, it acts as a base and provides electrophile.

### Solve the system of equations by using the method of cross multiplication:

Solution: The given equations are: $\begin{array}{l} 6 x-5 y-16=0\dots \dots(i) \\ 7 x-13 y+10=0\dots \dots(ii) \end{array}$ Here $a_{1}=6, b_{1}=-5, c_{1}=-16, a_{2}=7, b_{2}=-13$ and $c_{2}=10$ On...

### If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66o and ∠ABC = 80o. Calculate: (i) ∠DBC, (ii) ∠IBC,

Solution: Join $DB\text{ }and\text{ }DC,\text{ }IB\text{ }and\text{ }IC$ Given, if $\angle BAC\text{ }=\text{ }{{66}^{o~}}and~\angle ABC\text{ }=~{{80}^{o}}$ $I$ is the incentre of the...

### In the given figure, ABC is a triangle in which ∠BAC = 30o. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose center is O.

Solution: Firstly, join $OB\text{ }and\text{ }OC$ Proof: $\angle BOC\text{ }=\text{ }2\angle BAC\text{ }=\text{ }2\text{ }x\text{ }{{30}^{o}}~=\text{ }{{60}^{o}}$ Now, in $\vartriangle OBC$...

### In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30degree, find: angle APB

Solution: In the given fig, $O$is the centre of the circle and, $CA\text{ }and\text{ }CB$are the tangents to the circle from $C$Also, $\angle ACO\text{ }=\text{ }{{30}^{o}}$ $P$ is any...

### In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30degree, find: (i) angle BCO (ii) angle AOB

Solution: In the given fig, $O$is the centre of the circle and, $CA\text{ }and\text{ }CB$are the tangents to the circle from $C$Also, $\angle ACO\text{ }=\text{ }{{30}^{o}}$ $P$ is any...

### In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20degree, find angle AOD.

Solution: Join $OB$ In $\vartriangle OBC$ we have $BC\text{ }=\text{ }OD\text{ }=\text{ }OB$[Radii of the same circle] $\angle BOC\text{ }=\angle BCO\text{ }=\text{ }{{20}^{o}}$ And ext....

### Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = 90degree – ½ ∠A

Join $ED,\text{ }EF\text{ }and\text{ }DF$ And $BF,\text{ }FA,\text{ }AE\text{ }and\text{ }EC$ $\angle EBF\text{ }=\angle ECF\text{ }=\angle EDF\text{ }\ldots ..\text{ }\left( i \right)$ [Angle...

### Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Join $AD$ And $AB$ is the diameter. We have $\angle ADB\text{ }=\text{ }90{}^\text{o}~$[Angle in a semi-circle] But,  $\angle ADB\text{ }+~\angle ADC\text{ }=\text{ }180{}^\text{o}~$[Linear...

### Prove that, of any two chords of a circle, the greater chord is nearer to the center.

Given: $A$ circle with center $O$ and radius $r$ $AB\text{ }and\text{ }CD$are two chords such that $AB\text{ }>\text{ }CD$ Also, $OM\bot AB\text{ }and\text{ }ON\bot CD$ Required to...

### Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.

Let $DE$be the tangent to the circle at $P$ And, $DE\text{ }||\text{ }QR$[Given] $\angle EPR\text{ }=\angle PRQ$[Alternate angles are equal] $\angle DPQ\text{ }=\angle PQR$[Alternate...

### One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting: (v) a diamond or a spade

(v) Number of favorable outcomes for a diamond or a spade = 13 + 13 = 26 So, number of favorable outcomes = 26 Hence, P(getting a diamond or a spade) = 26/52 = 1/2

### One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting: (iii) the jack or the queen of the hearts (iv) a diamond

(iii) Favorable outcomes for jack or queen of hearts = 1 jack + 1 queen So, the number of favorable outcomes = 2 Hence, P(jack or queen of hearts) = 2/52 = 1/26 (iv) Number of favorable outcomes for...

### One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting: (i) a queen of red color (ii) a black face card

Solution: We have, Total possible outcomes = 52 (i) Number queens of red color = 2 Number of favorable outcomes = 2 Hence, P(queen of red color) = 2/52 (ii) Number of black cards = 26 Number of...

### A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.If the outcomes are equally likely, find the probability that the pointer will point at: (v) a number less than or equal to 9 (vi) a number between 3 and 11

(v) Favorable outcomes for a number less than or equal to 9 are 1, 2, 3, 4, 5, 6, 7, 8, 9 So, number of favorable outcomes = 9 Hence, P(the pointer will be at a number less than or equal to 9) =...

### A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.If the outcomes are equally likely, find the probability that the pointer will point at: (iii) a prime number (iv) a number greater than 8

(iii) Favorable outcomes for a prime number are 2, 3, 5, 7, 11 So, number of favorable outcomes = 5 Hence, P(the pointer will be at a prime number) = 5/12 (iv) Favorable outcomes for a number...

### A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.If the outcomes are equally likely, find the probability that the pointer will point at: (i) 6 (ii) an even number

Solution: We have, Total number of possible outcomes = 12 (i) Number of favorable outcomes for 6 = 1 Hence, P(the pointer will point at 6) = 1/12 (ii) Favorable outcomes for an even number are 2, 4,...

### A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin: (i) will be a Re 1 coin? (ii) will not be a Rs 2 coin? (iii) will neither be a Rs 5 coin nor be a Re 1 coin?

(iii) Number of favourable outcomes for neither Re 1 nor Rs 5 coins = Number of favourable outcomes for Rs 2 coins = 50 = n(E) Hence, probability (neither Re 1 nor Rs 5 coin) = n(E)/ n(S) = 50/100 =...

### A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin: (i) will be a Re 1 coin? (ii) will not be a Rs 2 coin?

Solution: We have, Total number of coins = 20 + 50 + 30 = 100 So, the total possible outcomes = 100 = n(S) (i) Number of favourable outcomes for Re 1 coins = 30 = n(E) Probability (Re 1 coin) =...

### A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be: (iii) white or green?

(iii) Number of favorable outcomes for white or green ball = 16 + 8 = 24 = n(E) Hence, probability for drawing a white or green ball = n(E)/ n(S) = 24/34 = 12/17

### A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be: (i) not red? (ii) neither red nor green?

Solution: Total number of possible outcomes = 10 + 16 + 8 = 34 balls So, n(S) = 34 (i) Favorable outcomes for not a red ball = favorable outcomes for white or green ball So, number of favorable...

### The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday?

Solution: We know that, P(do not have the same birthday) + P(have same birthday) = 1 0.897 + P(have same birthday) = 1 Thus, P(have same birthday) = 1 – 0.897 P(have same birthday) =...

### A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is: (i) black (ii) red

Solution: We have, Total possible outcomes = number of red balls. (i) Number of favourable outcomes for black balls = 0 Hence, P(black ball) = 0 (ii) Number of favourable outcomes for red balls =...

### If P(E) = 0.59; find P(not E)

Solution: We know that, P(E) + P(not E) = 1 So, 0.59 + P(not E) = 1 Hence, P(not E) = 1 – 0.59 = 0.41

### Which of the following cannot be the probability of an event? (iii) 37% (iv) -2.4

(iii) As 0 ≤ 37 % = (37/100) ≤ 1 Thus, 37 % can be a probability of an event. (iv) As -2.4 < 0 Thus, -2.4 cannot be a probability of an event.

### Which of the following cannot be the probability of an event? (i) 3/7 (ii) 0.82

Solution: We know that probability of an event E is 0 ≤ P(E) ≤ 1 (i) As 0 ≤ 3/7 ≤ 1 Thus, 3/7 can be a probability of an event. (ii) As 0 ≤ 0.82 ≤ 1 Thus, 0.82 can be a probability of an...

### Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:(iii) less than or equal to 12

(iii) All the outcomes are favourable to the event E = ‘sum of two numbers ≤ 12’. Thus, P(E) = n(E)/ n(S) = 36/36 = 1

### Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is: (i) 8 (ii) 13

Solution: We have, the number of possible outcomes = 6 × 6 = 36. (i) The outcomes favourable to the event ‘the sum of the two numbers is 8’ = E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} The number...

### In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that: (i) it is acceptable to a trader who accepts only a good shirt? (ii) it is acceptable to a trader who rejects only a shirt with major defects?

Solution: We have, Total number of shirts = 50 Total number of elementary events = 50 = n(S) (i) As, trader accepts only good shirts and number of good shirts = 44 Event of accepting good shirts =...

### In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds?

Solution: Total result = 0 sec to 40 sec Total possible outcomes = 40 So, n(S) = 40 Favourable results = 0 sec to 15 sec Favourable outcomes = 15 So, n(E) = 15 Hence, the probability that the music...

### All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting:(iii) a black card

(iii) Number of black cards left = 23 cards (13 club + 10 spade) Event of drawing a black card = E = 23 So, n(E) = 23 Hence, probability of drawing a black card = n(E)/ n(S) = 23/49

### All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting: (i) a black face card (ii) a queen

Solution: We have, Total number of cards = 52 If 3 face cards of spades are removed Then, the remaining cards = 52 – 3 = 49 = number of possible outcomes So, n(S) = 49 (i) Number of black face cards...

### A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is: (i) white (ii) neither red nor white.

Solution: We have, Total number of balls in the box = 7 + 8 + 5 = 20 balls Total possible outcomes = 20 = n(S) (i) Event of drawing a white ball = E = number of white balls = 5 So, n(E) = 5 Hence,...

### A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets: (i) at least one head? (ii) at most one head?

Solution: We know that, When two coins are tossed simultaneously, the possible outcomes are {(H, H), (H, T), (T, H), (T, T)} So, n(S) = 4 (i) The outcomes favourable to the event E, ‘at least one...

### A and B are friends. Ignoring the leap year, find the probability that both friends will have: (i) different birthdays? (ii) the same birthday?

Solution: Out of the two friends, A’s birthday can be any day of the year. Now, B’s birthday can also be any day of 365 days in the year. We assume that these 365 outcomes are equally likely. So,...

### In a match between A and B: (i) the probability of winning of A is 0.83. What is the probability of winning of B? (ii) the probability of losing the match is 0.49 for B. What is the probability of winning of A?

Solution: (i) We know that, The probability of winning of A + Probability of losing of A = 1 And, Probability of losing of A = Probability of winning of B Therefore, Probability of winning of A +...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:(v) a card with number less than 8 (vi) a card with number between 2 and 9

(v) Numbers less than 8 = { 2, 3, 4, 5, 6, 7} Event of drawing a card with number less than 8 = E = {6H cards, 6D cards, 6S cards, 6C cards} So, n(E) = 24 Thus, probability of drawing a card with...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is: (iii) a queen of black card (iv) a card with number 5 or 6

(iii) Event of drawing a queen of black colour = {Q(spade), Q(club)} = E So, n(E) = 2 Thus, probability of drawing a queen of black colour = n(E)/ n(S) = 2/52 = 1/26 (iv) Event of drawing a card...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is: (i) a face card (ii) not a face card

Solution: We have, the total number of possible outcomes = 52 So, n(S) = 52 (i)  No. of face cards in a deck of 52 cards = 12 (4 kings, 4 queens and 4 jacks) Event of drawing a face cards = E = (4...

### A dice is thrown once. What is the probability of getting a number: (i) greater than 2? (ii) less than or equal to 2?

Solution: The number of possible outcomes when dice is thrown = {1, 2, 3, 4, 5, 6} So, n(S) = 6 (i) Event of getting a number greater than 2 = E = {3, 4, 5, 6} So, n(E) = 4 Thus, probability of...

### A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is: (iii) not yellow (iv) neither yellow nor red

(iii) Probability of not drawing a yellow ball = 1 – Probability of drawing a yellow ball Thus, probability of not drawing a yellow ball = 1 – 1/8 = (8 – 1)/ 8 = 7/8 (iv) Neither yellow ball nor red...

### A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is: (i) yellow (ii) red

Solution: The total number of balls in the bag = 3 + 4 + 1 = 8 balls So, the number of possible outcomes = 8 = n(S) (i) Event of drawing a yellow ball = {Y} So, n(E) = 1 Thus, probability of drawing...

### If two coins are tossed once, what is the probability of getting: (iii) both heads or both tails.

(iii) E = event of getting both heads or both tails = {HH, TT} n(E) = 2 Hence, probability of getting both heads or both tails = n(E)/ n(S) = 2/4 = ½

### If two coins are tossed once, what is the probability of getting: (i) both heads. (ii) at least one head.

Solution: We know that, when two coins are tossed together possible number of outcomes = {HH, TH, HT, TT} So, n(S) = 4 (i) E = event of getting both heads = {HH} n(E) = 1 Hence, probability of...

### A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.

Solution: In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6} So, n(S) = 6 For two dice, n(S) = 6 x 6 = 36 Favorable cases where the sum is 10 or more with 5 on 1st die = {(5, 5), (5,...

### A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?

Solution: We know that, Number of pages in the book = 85 Number of possible outcomes = n(S) = 85 Out of 85 pages, pages that sum up to 8 = {8, 17, 26, 35, 44, 53, 62, 71, 80} So, pages that sum up...

### A die is thrown once. Find the probability of getting a number: (iii) less than 8 (iv) greater than 6

(iii) On a dice, numbers less than 8 = {1, 2, 3, 4, 5, 6} So, n(E) = 6 Hence, probability of getting a number less than 8 = n(E)/ n(S) = 6/6 = 1 (iv) On a dice, numbers greater than 6 = 0 So, n(E) =...

### A die is thrown once. Find the probability of getting a number: (i) less than 3 (ii) greater than or equal to 4

Solution: We know that, In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6} So, n(S) = 6 (i) On a dice, numbers less than 3 = {1, 2} So, n(E) = 2 Hence, probability of getting a number...

### From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: (iii) 3 and 5 (iv) 3 or 5

(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15} So, favorable number of events = n(E) = 1 Hence, probability of selecting a card with a multiple of 3 and...

### From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: (i) 3 (ii) 5

Solution: We know that, there are 25 cards from which one card is drawn. So, the total number of elementary events = n(S) = 25 (i) From numbers 1 to 25, there are 8 numbers which are multiple of 3...

### Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: (v) less than 48

(v) From numbers 1 to 100, there are 47 numbers which are less than 48 i.e. {1, 2, ……….., 46, 47} So, favorable number of events = n(E) = 47 Hence, probability of selecting a card with a number less...

### Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: (iii) between 40 and 60 (iv) greater than 85

(iii) From numbers 1 to 100, there are 19 numbers which are between 40 and 60 i.e. {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59} So, favorable number of events = n(E)...

### Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: (i) a multiple of 5 (ii) a multiple of 6

Solution: We kwon that, there are 100 cards from which one card is drawn. Total number of elementary events = n(S) = 100 (i) From numbers 1 to 100, there are 20 numbers which are multiple of 5...

### Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:(iii) an even number and a multiple of 3 (iv) an even number or a multiple of 3

(iii) From numbers 2 to 10, there is one number which is an even number as well as multiple of 3 i.e. 6 So, favorable number of events = n(E) = 1 Hence, probability of selecting a card with a number...

### Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be: (i) an even number (ii) a multiple of 3

Solution: We know that, there are totally 9 cards from which one card is drawn. Total number of elementary events = n(S) = 9 (i) From numbers 2 to 10, there are 5 even numbers i.e. 2, 4, 6, 8, 10...

### In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of: (i) winning of Geeta (ii) not winning of Ritu

Solution: (i) Winning of Geeta is a complementary event to winning of Ritu Thus, P(winning of Ritu) + P(winning of Geeta) = 1 P(winning of Geeta) = 1 – P(winning of Ritu) P(winning of Geeta) =...

### (i) If A and B are two complementary events then what is the relation between P(A) and P(B)? (ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?

Solution: (i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1. P(A) + P(B) = 1 (ii) P(A) = 0.46 Let P(B) be...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:(v) be a face card of red colour

(v) There are 26 red cards in a deck, and 6 of these cards are face cards (2 kings, 2 queens and 2 jacks). The number of favourable outcomes for the event of drawing a face card of red color = 6...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will: (i) be a black card. (ii) not be a red card. (iii) be a red card. (iv) be a face card.

(iii) Number of red cards in a deck = 26 The number of favourable outcomes for the event of drawing a red card = 26 Then, probability of drawing a red card = 26/52 = ½   (iv) There are 52 cards...

### From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will: (i) be a black card. (ii) not be a red card.

Solution: We know that, Total number of cards = 52 So, the total number of outcomes = 52 There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are...

### In a single throw of a die, find the probability that the number: (iii) will be an odd number.

(iii) If E = event of getting an odd number = {1, 3, 5} So, n(E) = 3 Then, probability of a getting an odd number = n(E)/ n(s) = 3/6 = ½

### In a single throw of a die, find the probability that the number: (i) will be an even number. (ii) will not be an even number.

Solution: Here, the sample space = {1, 2, 3, 4, 5, 6} n(s) = 6 (i) If E = event of getting an even number = {2, 4, 6} n(E) = 3 Then, probability of a getting an even number = n(E)/ n(s) = 3/6 = ½...

### In a single throw of a die, find the probability of getting a number: (iii) not greater than 4.

(iii) E = event of getting a number not greater than 4 = {1, 2, 3, 4} So, n (E) = 4 Then, probability of getting a number not greater than 4 = n(E)/ n(s) = 4/6 = 2/3

### In a single throw of a die, find the probability of getting a number: (i) greater than 4. (ii) less than or equal to 4.

Solution: Here, the sample space = {1, 2, 3, 4, 5, 6} So, n (s) = 6 (i) If E = event of getting a number greater than 4 = {5, 6} So, n (E) = 2 Then, probability of getting a number greater than 4 =...

### . A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:(v) not a black ball

(v) There are 3 + 2 = 5 balls which are not black So, the number of favourable outcomes = 5 Thus, P(getting a white ball) = 5/10 = ½

### A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is: (iii) a white ball. (iv) not a red ball.

(iii) There are 3 white balls So, the number of favourable outcomes = 3 Thus, P(getting a white ball) = 3/10 = 3/10 (iv) There are 3 + 5 = 8 balls which are not red So, the number of favourable...

### A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is: (i) a black ball. (ii) a red ball

Solution: Total number of balls = 3 + 5 + 2 = 10 So, the total number of possible outcomes = 10 (i) There are 5 black balls So, the number of favourable outcomes = 5 Thus, P(getting a black ball) = ...

### A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is: (i) a black ball. (ii) a red ball

Solution: Total number of balls = 3 + 5 + 2 = 10 So, the total number of possible outcomes = 10 (i) There are 5 black balls So, the number of favourable outcomes = 5 Thus, P(getting a black ball) = ...

### 1. A coin is tossed once. Find the probability of: (i) getting a tail (ii) not getting a tail

Solution: Here, the sample space = {H, T} i.e. n(S) = 2 (i) If A = Event of getting a tail = {T} Then, n(A) = 1 Hence, the probability of getting a tail = n(A)/ n(S) = 1/2 (ii) Not getting a tail As...

### From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is drilled out. Find: (i) the surface area of the remaining solid (ii)the volume of remaining solid

Dimensions of rectangular solid l = 42 cm, b = 30 cm and h = 20 cm Conical cavity’s diameter = 14 cm its radius = 7 cm Depth (height) = 24 cm (i) Total surface area of cuboid...

### The volume of one sphere is 27 times that of another sphere. Calculate the ratio of their: (i) radii (ii) surface areas

The volume of first sphere = 27 x volume of second sphere Let the radius of the first sphere = r1 and, radius of second sphere = r2 (i) Then, according to the question we have...

### The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold?

The inner radius of the pipe = 2.1 cm Length of the pipe = 12 m = 1200 cm Volume of the pipe \[\begin{array}{*{35}{l}} =\text{ }\pi {{r}^{2}}h\text{ }=\text{ }22/7\text{ x }{{2.1}^{2}}~x\text{...

### The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find: (i) the volume (ii) the total surface area.

Since, a circular cylinder whose Height, h = 20 cm and base radius, r = 7 cm (i) Volume of cylinder \[\begin{array}{*{35}{l}} =\text{ }\pi {{r}^{2}}h\text{ }=\text{ }22/7\text{ }x\text{...

### Construct a circle, inscribing an equilateral triangle with side 5.6 cm.

Steps for Construction: i) Draw a line segment BC = 5.6 cm ii) With centres B and C, draw two arcs of 5.6 cm radius each which intersect each other at A. iii) And, join AB and AC. iv) Draw angle...

### Construct an equilateral triangle ABC with side 6 cm. Draw a circle circumscribing the triangle ABC.

Steps for construction: i) Draw a line segment BC = 6 cm ii) Draw two arcs of radius 6 cm with centres B and C, which intersect each other at A. iii) Then, join AC and AB. iv) Draw perpendicular...

### i) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm. ii) Find its incentre and mark it I.iii) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle.

Steps for construction: i) Draw a line segment BC = 6 cm. ii) Draw an arc with centre B and radius 8 cm. iii) Draw another arc with centre C and radius 5 cm which intersects the first arc at A. iv)...

### The bisectors of angles A and B of a scalene triangle ABC meet at O. iii) What is the relation between angle ACO and angle BCO?

iii) OC is the bisector of angle C Thus, ∠ACO = ∠BCO

### The bisectors of angles A and B of a scalene triangle ABC meet at O. i) What is the point O called? ii) OR and OQ are drawn perpendiculars to AB and CA respectively. What is the relation between OR and OQ?

i) O is called the incentre of the incircle of ΔABC. ii) OR and OQ are the radii of the incircle and OR = OQ.

### Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O.(iii) Does the perpendicular bisector of BC pass through O?

iii) Yes, the perpendicular bisector of BC will pass through O.

### Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O. (i) What do you call the point O? (ii) What is the relation between the distances OA, OB and OC?

i) O is called the circumcentre of circumcircle of ΔABC. ii) OA, OB and OC are the radii of the circumcircle.

### Using ruler and compasses only construct a triangle ABC in which BC = 4 cm, ∠ACB = 45o and perpendicular from A on BC is 2.5 cm. Draw a circle circumscribing the triangle ABC.

Steps for Construction: i) Draw a line segment BC = 4 cm. ii) At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm. iii) Draw another perpendicular line EY from E. iv) From C, draw a...

### Using ruler and compasses only, (i) Construct a triangle ABC with the following data: Base AB = 6 cm, BC = 6.2 cm and ∠CAB = 60o (ii) In the same diagram, draw a circle which passes through the points A, B and C and mark its centre as O. (iii) Draw a perpendicular from O to AB which meets AB in D. (iv) Prove that AD = BD

Steps for construction: i) Draw a line segment AB = 6 cm ii) Draw a ray at A, making an angle of 60o with BC. iii) With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C. iv)...

### Using ruler and compasses only, draw an equilateral triangle of side 5 cm. Draw its inscribed circle. Measure the radius of the circle.

Steps for Construction: i) Draw a line segment BC = 5 cm ii) With centres as B and C, draw two arcs of radius 5 cm each which intersect each other at A. iii) Then, join AB and AC. iv) Draw angle...

### Using ruler and compasses only. (i) Construct triangle ABC, having given BC = 7 cm, AB-AC = 1 cm and ∠ABC = 45o. (ii) Inscribe a circle in the ΔABC constructed in (i) above.

Steps for construction: A. Construction of triangle: a) Draw a line segment BC = 7 cm b) At B, draw a ray BX making an angle of 45o and cut off BE = AB – AC = 1 cm c) Join EC and draw the...

### Using ruler and compasses only, draw an equilateral triangle of side 4.5 cm and draw its circumscribed circle. Measure the radius of the circle.

Steps for construction: i) Draw a line segment BC = 4.5 cm ii) With centres B and C, draw two arcs of radius 4.5 cm which intersect each other at A. iii) And join AC and AB. iv) Draw the...

### Draw a circle of radius 4.5 cm. Draw two tangents to this circle so that the angle between the tangents is 60º.

Steps for Construction: i)  draw a circle with centre O, with radius BC = 4.5 cm ii) Draw arcs making an angle of 180º – 60º = 120º at O such that ∠AOB = 120º iii) At A and B, draw two rays making...

### Draw a circle of radius 5 cm. Draw two tangents to this circle so that the angle between the tangents is 45º.

Steps for construction: i)  draw a circle with centre O,with radius BC = 5 cm ii) Draw arcs making an angle of 180º – 45º = 135º at O such that ∠AOB = 135º iii) At A and B, draw two rays making an...