CBSE Class 10 Mathematics Probability Assignment Set 02

Important Concepts

1. Theoretical probability of an event \( E \), written as \( P(E) \), is defined as:
\[ P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes of the experiment}} \]
Here, we assume that the outcomes of the experiment are equally likely.

2. The probability of a sure event (or certain event) is 1.

3. The probability of an impossible event is 0.

4. The probability of an event \( E \) is a number \( P(E) \) such that \( 0 \le P(E) \le 1 \).

5. Elementary events - An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.

6. For any event \( E \), \( P(E) + P(\overline{E}) = 1 \), where \( \overline{E} \) stands for 'not \( E \)'. Here, \( E \) and \( \overline{E} \) are called complementary events.

7. Performing experiments:
a. Tossing a coin.
b. Throwing a die.
c. Drawing a card from a deck of 52 cards.

8. Sample space - The set of all possible outcomes in an experiment is called the sample space.

Level-I
 

Question. Write a sample space of
a. Tossing a coin.
b. Throwing a die.
Answer:
a. When a coin is flipped, the sample space is \( \{H, T\} \), representing Head and Tail.
b. When a single die is rolled, the sample space is \( \{1, 2, 3, 4, 5, 6\} \).
In simple words: The sample space shows every possible result of an experiment. For a coin, it is heads or tails; for a die, it is any number from 1 to 6.

Exam Tip: Always write the sample space elements inside curly brackets to represent them as a set.

Question. Define probability (Theoretical probability of an event).
Answer: The theoretical probability of an event \( E \), written as \( P(E) \), is the ratio of the number of outcomes favorable to \( E \) to the total number of all possible outcomes of the experiment, assuming all outcomes are equally likely:
\[ P(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes}} \]
In simple words: Probability is a way of measuring how likely something is to happen, calculated by dividing the winning chances by all the possible chances.

Exam Tip: Remember that the probability of any event must always be between 0 and 1, inclusive.

Question. A card is drawn from a well-shuffled pack of 52 cards what is the probability that it is an ace?
Answer: There are 4 aces in a standard deck of 52 playing cards.
Total number of possible outcomes = 52
Number of favorable outcomes (getting an ace) = 4
Probability \( P(\text{Ace}) = \frac{4}{52} = \frac{1}{13} \).
In simple words: Since there are 4 aces in a full deck of 52 cards, the chance of pulling one out is 4 out of 52, which simplifies to 1 out of 13.

Exam Tip: Always simplify your fraction to the lowest terms to ensure you get full marks.

Question. A dice is thrown once. Find the probability of getting a number greater than 3.
Answer: When a die is rolled, the possible outcomes are \( \{1, 2, 3, 4, 5, 6\} \), so the total number of outcomes is 6.
The numbers greater than 3 are 4, 5, and 6, which gives 3 favorable outcomes.
Probability = \( \frac{3}{6} = \frac{1}{2} \).
In simple words: Rolling a number higher than 3 means getting a 4, 5, or 6. That is 3 possibilities out of 6, which is exactly half of the chances.

Exam Tip: Clearly list both the total outcomes and the favorable outcomes before writing the probability fraction.

Question. What is the probability that a number selected from the number 1,2,3................16 is prime number?
Answer: The total numbers from 1 to 16 are 16.
The prime numbers in this range are 2, 3, 5, 7, 11, and 13.
Number of favorable outcomes = 6
Probability = \( \frac{6}{16} = \frac{3}{8} \).
In simple words: Out of the 16 numbers, only 6 of them (2, 3, 5, 7, 11, and 13) are prime numbers. This gives a probability of 6 out of 16, which reduces to 3 out of 8.

Exam Tip: Note that 1 is neither prime nor composite, so do not include it in your list of prime numbers.

Question. A letter is chosen at random from the English alphabet. Find the probability that the letter chosen precedes 'g'.
Answer: There are 26 letters in the English alphabet, so the total number of outcomes is 26.
The letters that precede 'g' are a, b, c, d, e, and f.
Number of favorable outcomes = 6
Probability = \( \frac{6}{26} = \frac{3}{13} \).
In simple words: The letters that come before 'g' are a, b, c, d, e, and f, making 6 letters in total. The chance of choosing one of these from all 26 letters is 6 out of 26, or 3 out of 13.

Exam Tip: "Preceding 'g'" means the letters strictly before 'g', so do not include 'g' itself in the favorable outcomes.

Question. Find the probability of getting a red heart.
Answer: A standard deck of 52 cards has 4 suits: Hearts, Diamonds, Clubs, and Spades. All Hearts are red.
Number of hearts in the deck = 13
Total cards = 52
Probability of getting a heart = \( \frac{13}{52} = \frac{1}{4} \).
In simple words: Since one-fourth of the cards in a deck are hearts (which are red), the probability of drawing a heart is 13 out of 52, which simplifies to 1 out of 4.

Exam Tip: Hearts are always red, so the phrase "red heart" simply refers to the suit of Hearts.

Question. A coin is tossed twice. Find the probability of getting at least one head.
Answer: When a coin is tossed twice, the sample space is \( \{HH, HT, TH, TT\} \).
Total number of outcomes = 4
Outcomes with at least one head are \( HH \), \( HT \), and \( TH \).
Number of favorable outcomes = 3
Probability = \( \frac{3}{4} \).
In simple words: When you flip a coin twice, there are 4 possible combinations. Three of these combinations have at least one head, so the probability is 3 out of 4.

Exam Tip: "At least one" means you can have 1 head or 2 heads. Only the double-tail outcome (\( TT \)) is excluded.

Question. What is the probability of a sure event?
Answer: A sure event is an event that is certain to occur. Its probability is always 1 (or 100%).
In simple words: If an event is guaranteed to happen, its probability is 1.

Exam Tip: Always write the numerical value "1" when asked for the probability of a sure event.

Level-II
 

Question. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting.
i. An ace
ii. A face card
Answer:
i. There are 4 aces in a pack of 52 cards.
Probability of drawing an ace = \( \frac{4}{52} = \frac{1}{13} \).
ii. There are 12 face cards (4 Kings, 4 Queens, 4 Jacks) in a pack of 52 cards.
Probability of drawing a face card = \( \frac{12}{52} = \frac{3}{13} \).
In simple words: There are 4 aces and 12 face cards in a standard deck. The probability of getting an ace is 1 out of 13, and the probability of drawing a face card is 3 out of 13.

Exam Tip: Remember that face cards are Kings, Queens, and Jacks only. Aces are not considered face cards.

Question. A bag contains 5 red balls, 4 green ball and 7 white balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is (i) White (II) neither Red nor White.
Answer: Total number of balls = \( 5 + 4 + 7 = 16 \).
(i) Number of white balls = 7.
Probability of getting a white ball = \( \frac{7}{16} \).
(ii) "Neither Red nor White" means the ball must be green.
Number of green balls = 4.
Probability of getting a green ball = \( \frac{4}{16} = \frac{1}{4} \).
In simple words: With 16 total balls in the bag, the chance of picking a white one is 7 out of 16. A ball that is neither red nor white must be green, which gives a 4 out of 16 (or 1 out of 4) chance.

Exam Tip: "Neither A nor B" can be calculated directly by finding the probability of the remaining category (green), which is quicker than using complementary subtraction.

Question. Find the probability of getting 53 Friday in a leap year.
Answer: A leap year has 366 days, which consists of 52 complete weeks and 2 extra days.
These 2 extra days can be any of the following consecutive pairs:
1. Monday, Tuesday
2. Tuesday, Wednesday
3. Wednesday, Thursday
4. Thursday, Friday
5. Friday, Saturday
6. Saturday, Sunday
7. Sunday, Monday
Out of these 7 possible combinations, Friday occurs in 2 cases: (Thursday, Friday) and (Friday, Saturday).
Therefore, the probability of having 53 Fridays is \( \frac{2}{7} \).
In simple words: A leap year contains 52 full weeks plus 2 extra days. For these 2 extra days, there are 7 possible pairs of days, and 2 of those pairs contain a Friday. This gives a 2 out of 7 chance.

Exam Tip: For any non-leap year, the probability of 53 of any weekday is \( \frac{1}{7} \), whereas for a leap year, it is always \( \frac{2}{7} \).

Question. Two dice are thrown simultaneously. What is the probability that.
i. 5 will come up on at least one?
ii. 5 will come up at both dice?
Answer: When two dice are thrown together, the total number of outcomes in the sample space is \( 6 \times 6 = 36 \).
i. The outcomes where 5 comes up on at least one die are:
\( (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5) \).
There are 11 such outcomes.
Probability = \( \frac{11}{36} \).
ii. The outcome where 5 comes up on both dice is only \( (5,5) \).
Number of favorable outcomes = 1.
Probability = \( \frac{1}{36} \).
In simple words: When rolling two dice, there are 36 different possible results. Eleven of those results contain at least one 5, while only one result has a 5 on both.

Exam Tip: When listing outcomes for "at least one", be careful not to count \( (5,5) \) twice.

Question. Two coins are tossed once. Find the probability of getting.
i. Exactly one head
ii. Almost one head
Answer: When two coins are tossed, the sample space is \( \{HH, HT, TH, TT\} \). Total number of outcomes = 4.
i. Outcomes with exactly one head are \( HT \) and \( TH \).
Number of favorable outcomes = 2.
Probability = \( \frac{2}{4} = \frac{1}{2} \).
ii. Interpreting "almost one head" as "at most one head" (which means 0 or 1 head), the favorable outcomes are \( HT \), \( TH \), and \( TT \).
Number of favorable outcomes = 3.
Probability = \( \frac{3}{4} \).
In simple words: When you toss two coins, you have a 1 in 2 chance of getting exactly one head. The chance of getting no more than one head (meaning one head or none at all) is 3 out of 4.

Exam Tip: Double check whether a term like "almost" is a typographical error for "at most", and solve based on standard probability definitions which match the answer sheet.

Question. In a lottery there are 10 prizes and 25 blank. Find the probability of getting a prize.
Answer: Total number of lottery tickets = \( 10 \text{ prizes} + 25 \text{ blanks} = 35 \).
Number of favorable outcomes (winning a prize) = 10.
Probability of getting a prize = \( \frac{10}{35} = \frac{2}{7} \).
In simple words: There are 35 total tickets in the lottery. Since 10 of those tickets win a prize, your chance of winning is 10 out of 35, which simplifies to 2 out of 7.

Exam Tip: Don't forget to add the prizes and blanks together to get the total number of outcomes before writing your ratio.

Question. The king, the queen and the jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of .
i. Heart
ii. Queen
iii. Clubs
Answer: Three cards (King, Queen, and Jack of clubs) are removed from the deck.
Total number of remaining cards = \( 52 - 3 = 49 \).
i. None of the hearts were removed, so there are still 13 hearts in the deck.
Probability of getting a heart = \( \frac{13}{49} \).
ii. There were 4 queens originally, and 1 queen (of clubs) was removed, leaving 3 queens.
Probability of getting a queen = \( \frac{3}{49} \).
iii. There were 13 club cards originally. Since 3 club cards (King, Queen, Jack) were removed, there are \( 13 - 3 = 10 \) clubs left.
Probability of getting a club card = \( \frac{10}{49} \).
In simple words: After removing three club cards, the deck is down to 49 cards. There are still 13 hearts, only 3 queens remaining, and 10 clubs left, giving probabilities of 13/49, 3/49, and 10/49 respectively.

Exam Tip: When cards are removed, always adjust both the total number of cards (the denominator) and the count of specific cards (the numerator).

Level-III
 

Question. Nidhi and Nisha are two friends. What is the probability that both will have
a. Same birthday
b. Different birthday (ignore the leap year)
Answer: Let us assume there are 365 days in a year.
a. For both of them to have the same birthday, if Nidhi is born on any day, Nisha must be born on that exact same day. There is only 1 such favorable day out of 365.
Probability of the same birthday = \( \frac{1}{365} \).
b. For them to have different birthdays, Nisha must be born on any of the remaining 364 days of the year.
Probability of different birthdays = \( 1 - P(\text{same birthday}) = 1 - \frac{1}{365} = \frac{364}{365} \).
In simple words: The chance that two people share the exact same birthday is very low - just 1 out of 365. That means the chance they have different birthdays is extremely high - 364 out of 365.

Exam Tip: Note that the sum of the probabilities of these two complementary events is exactly 1.

Question. A box contains 20 balls bearing number 1,2,3,4,.......20. A ball is drawn at random from the box. What is the probability that the number on the ball is.
a. An odd number
b. divisible by 2 or 3
c. Prime number
Answer: Total number of balls = 20.
a. Odd numbers from 1 to 20 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. There are 10 odd numbers.
Probability of an odd number = \( \frac{10}{20} = \frac{1}{2} \).
b. Numbers divisible by 2 or 3 from 1 to 20 are:
Divisible by 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 (10 numbers)
Divisible by 3: 3, 6, 9, 12, 15, 18 (6 numbers)
Common numbers (divisible by both 6): 6, 12, 18 (3 numbers)
Total numbers divisible by 2 or 3 = \( 10 + 6 - 3 = 13 \).
Probability = \( \frac{13}{20} \).
c. Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, and 19. There are 8 prime numbers.
Probability of a prime number = \( \frac{8}{20} = \frac{2}{5} \).
In simple words: Out of 20 numbered balls, half are odd (giving a 1/2 probability), 13 are divisible by either 2 or 3 (giving a 13/20 probability), and 8 are prime numbers (giving a 2/5 probability).

Exam Tip: When finding numbers "divisible by A or B", use the principle of inclusion-exclusion to avoid double-counting numbers that are divisible by both.

Question. A card is drawn at random from a well shuffled deck of 52 cards. What is the probability of drawing
a. King or a spade
b. A non spade
c. Either a king or a 10 of heart
Answer: Total cards = 52.
a. There are 4 Kings and 13 Spades in a deck. One card (the King of Spades) is counted in both groups.
Number of cards that are a King or a Spade = \( 4 + 13 - 1 = 16 \).
Probability = \( \frac{16}{52} = \frac{4}{13} \).
b. There are 13 spades, so the number of non-space cards = \( 52 - 13 = 39 \).
Probability of drawing a non-spade = \( \frac{39}{52} = \frac{3}{4} \).
c. There are 4 Kings in the deck. The "10 of Hearts" is a single specific card which is not a King.
Total favorable cards = \( 4 \text{ Kings} + 1 \text{ (10 of Hearts)} = 5 \).
Probability = \( \frac{5}{52} \).
In simple words: The chance of pulling a King or any spade is 4 out of 13. Since spades make up a quarter of the deck, the chance of getting any other suit is 3 out of 4. Getting either a King or the 10 of hearts has a 5 out of 52 chance.

Exam Tip: Be careful with "either/or" questions in card decks; check if there is any overlap between the categories to prevent counting any card twice.

Question. Tom was born in February 2000. What is the probability that he was born on 13th Feb?
Answer: The year 2000 was a leap year because it is divisible by 400.
In a leap year, February has 29 days.
The probability of being born on any specific date (like the 13th) in that month is \( \frac{1}{29} \).
In simple words: Because the year 2000 was a leap year, February had 29 days instead of 28. This means there is a 1 in 29 chance that Tom was born on February 13th.

Exam Tip: Always check if a given century year (like 2000) is a leap year by dividing it by 400.

Question. Are the following outcomes equally likely or not? A baby is born " It is a boy or a girl".
Answer: Yes, the outcomes of a baby being born as a boy or a girl are considered equally likely. The probability for each outcome is \( \frac{1}{2} \).
In simple words: When a baby is born, it is just as likely to be a boy as it is to be a girl, so these two outcomes are equally likely.

Exam Tip: In introductory probability questions, biological genders are assumed to have equal likelihoods of occurrence.

Question. Find the probability of getting 53 Mondays in a leap year 53 Tuesday in a non leap year.
Answer:
1. For a leap year (366 days), there are 52 complete weeks and 2 extra days. The 2 extra days can be 7 different pairs. Mondays occur in 2 of these pairs: (Sunday, Monday) and (Monday, Tuesday). Thus, the probability of 53 Mondays in a leap year is \( \frac{2}{7} \).
2. For a non-leap year (365 days), there are 52 complete weeks and 1 extra day. This single extra day can be any of the 7 days of the week. Therefore, the probability that the extra day is a Tuesday (giving 53 Tuesdays) is \( \frac{1}{7} \).
In simple words: A leap year has two extra days, giving a 2/7 chance of an extra Monday. A regular year has only one extra day, giving a 1/7 chance of an extra Tuesday.

Exam Tip: Remember the pattern: leap years have a probability of \( \frac{2}{7} \) for 53 occurrences of any weekday, while non-leap years have a probability of \( \frac{1}{7} \).

Question. A letter is selected from the letter of word MATHEMATICS. What is the probability that it is M?
Answer: The word "MATHEMATICS" contains 11 letters in total: M, A, T, H, E, M, A, T, I, C, S.
The letter 'M' appears 2 times in the word.
Probability of selecting 'M' = \( \frac{2}{11} \).
In simple words: There are 11 letters in "MATHEMATICS" and the letter 'M' appears twice. This means the probability of choosing an 'M' is 2 out of 11.

Exam Tip: Count the letters in the given word carefully to avoid any simple counting errors.

Question. In an N.C.C camp there are 20 boys and 15 girls. The best cadet is to be chosen. What is the probability that the best cadet is a girl?
Answer: Total number of cadets = \( 20 \text{ boys} + 15 \text{ girls} = 35 \).
Number of girls (favorable outcomes) = 15.
Probability that the best cadet is a girl = \( \frac{15}{35} = \frac{3}{7} \).
In simple words: There are 35 cadets in total, and 15 of them are girls. The chance of a girl being chosen as the best cadet is 15 out of 35, which simplifies to 3 out of 7.

Exam Tip: Always express the final probability as a simplified fraction.

Question. Out of 400 bulbs in a box 15 bulbs are defective one bulb is taken out at random from the box. Find the probability that the drawn bulb is not defective.
Answer: Total number of bulbs = 400.
Number of defective bulbs = 15.
Number of non-defective bulbs = \( 400 - 15 = 385 \).
Probability that the drawn bulb is not defective = \( \frac{385}{400} = \frac{77}{80} \).
In simple words: Out of 400 bulbs, 385 are working fine. The chance of picking a working bulb is 385 out of 400, which reduces to 77 out of 80.

Exam Tip: You can also calculate this as \( 1 - P(\text{defective}) = 1 - \frac{15}{400} = 1 - \frac{3}{80} = \frac{77}{80} \).

Self Evaluation Questions
 

Question. A bag contains 5 white balls, 7 red balls and 2 blue balls. One ball is drawn at random from the bag what is probability that bulb drawn is
a. White or blue
b. Black or red
c. Not white
Answer: Total balls in the bag = \( 5 + 7 + 2 = 14 \).
a. Number of white or blue balls = \( 5 + 2 = 7 \).
Probability = \( \frac{7}{14} = \frac{1}{2} \).
b. Since there are no black balls, the number of black or red balls is just the number of red balls, which is 7.
Probability = \( \frac{7}{14} = \frac{1}{2} \).
c. Number of balls that are not white = \( 14 - 5 = 9 \).
Probability = \( \frac{9}{14} \).
In simple words: With 14 total balls, there are 7 balls that are white or blue (giving a 1/2 chance), 7 balls that are red (giving a 1/2 chance), and 9 balls that are not white (giving a 9/14 chance).

Exam Tip: Keep track of the color of the balls given in the question; since black is not listed in the bag, its count is simply zero.

Question. A child has a die whose six faces show the letters as given below.
A B C D E A
If a die is thrown once find probability of getting A and D.
Answer: The total number of outcomes on the six faces of the die is 6.
1. The letter 'A' appears on 2 faces.
Probability of getting 'A' = \( \frac{2}{6} = \frac{1}{3} \).
2. The letter 'D' appears on 1 face.
Probability of getting 'D' = \( \frac{1}{6} \).
In simple words: The die has 6 sides. Since 'A' is on 2 of those sides, the chance of rolling an 'A' is 2 out of 6, or 1/3. The letter 'D' is only on 1 side, so the chance of rolling a 'D' is 1 out of 6.

Exam Tip: When a letter appears more than once on a custom die, remember to count all its occurrences for the numerator.

Question. Find the probability of getting 53 Sundays in a leap year.
Answer: A leap year has 366 days, which contains 52 weeks and 2 extra days. These 2 extra consecutive days can be any of the 7 pairs: (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), or (Sun, Mon).
Out of these, Sunday is included in 2 pairs: (Saturday, Sunday) and (Sunday, Monday).
Probability of getting 53 Sundays = \( \frac{2}{7} \).
In simple words: Since a leap year has 2 extra days beyond the 52 full weeks, there are 2 out of 7 possible calendar combinations where those extra days include a Sunday.

Exam Tip: Writing down all 7 pairs of consecutive days helps demonstrate clear working to the examiner.

Question. From a well shuffled pack of 52 cards, a card is drawn at random. Find the probability that it is a:-
a. Spade
b. King
c. Club
d. Queen
e. A red card
f. The black king
g. The queen of diamonds.
Answer: Total cards in the pack = 52.
a. There are 13 spades. Probability = \( \frac{13}{52} = \frac{1}{4} \).
b. There are 4 kings. Probability = \( \frac{4}{52} = \frac{1}{13} \).
c. There are 13 clubs. Probability = \( \frac{13}{52} = \frac{1}{4} \).
d. There are 4 queens. Probability = \( \frac{4}{52} = \frac{1}{13} \).
e. There are 26 red cards (13 hearts and 13 diamonds). Probability = \( \frac{26}{52} = \frac{1}{2} \).
f. There are 2 black kings (King of clubs and King of spades). Probability = \( \frac{2}{52} = \frac{1}{26} \).
g. There is only 1 Queen of Diamonds. Probability = \( \frac{1}{52} \).
In simple words: In a deck of 52 cards, the probability of getting a spade or a club is 1/4; a king or a queen is 1/13; any red card is 1/2; a black king is 1/26; and the specific queen of diamonds is 1/52.

Exam Tip: Always double-check whether the question specifies a particular suit (like "diamonds") or just a general color (like "black" or "red") to choose the correct numerator.

Question. Two dice are thrown at the same time find the probability of getting.
a. Same number on both dice.
b. Different number on both dice.
Answer: Total number of outcomes when throwing two dice = 36.
a. The outcomes with the same number on both dice (doublets) are \( (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \). There are 6 such outcomes.
Probability = \( \frac{6}{36} = \frac{1}{6} \).
b. The outcomes with different numbers on both dice are all other outcomes except the 6 doublets, which is \( 36 - 6 = 30 \) outcomes.
Probability = \( \frac{30}{36} = \frac{5}{6} \).
In simple words: Out of 36 combinations, 6 have matching numbers, so the chance of rolling matching numbers is 1 out of 6. The other 30 combinations have different numbers, which gives a 5 out of 6 chance.

Exam Tip: You can find the probability of different numbers quickly using the formula \( P(E') = 1 - P(E) \).

Question. Someone is asked to take a number from 1 to 100. Find the probability that it is not a prime number.
Answer: Total numbers from 1 to 100 = 100.
There are 25 prime numbers in the range from 1 to 100.
Therefore, the number of non-prime numbers is \( 100 - 25 = 75 \).
Probability that the chosen number is not prime = \( \frac{75}{100} = \frac{3}{4} \).
In simple words: There are 25 prime numbers between 1 and 100, which means there are 75 numbers that are not prime. The chance of choosing one of these is 75 out of 100, or 3 out of 4.

Exam Tip: Note that the number 1 is not prime, so it is counted among the 75 non-prime numbers.

Question. Card marked with number 5 to 50 are placed in a box and mixed throughout. A card is drawn from the box at random. Find the probability that the number on the taken out card is
a. A Prime number less than 10
b. A number which is a perfect square
Answer: Total number of cards = \( 50 - 5 + 1 = 46 \).
a. The prime numbers less than 10 between 5 and 50 are 5 and 7. This gives 2 favorable outcomes.
Probability = \( \frac{2}{46} = \frac{1}{23} \).
b. The perfect squares in the range 5 to 50 are 9, 16, 25, 36, and 49. This gives 5 favorable outcomes.
Probability = \( \frac{5}{46} \).
In simple words: With 46 total cards from 5 to 50, only two numbers (5 and 7) are primes less than 10, giving a 1/23 chance. Five numbers (9, 16, 25, 36, and 49) are perfect squares, giving a 5/46 chance.

Exam Tip: To find the total number of items from \( A \) to \( B \), always use the formula \( B - A + 1 \).

Question. There are 20 cards numbered 1,2,3,.........20 in a box. One card is drawn. Find the probability that the number on the card is
a. A number divisible by 6
b. A number divisible by 7
Answer: Total cards in the box = 20.
a. Numbers divisible by 6 from 1 to 20 are 6, 12, and 18. This gives 3 favorable outcomes.
Probability = \( \frac{3}{20} \).
b. Numbers divisible by 7 from 1 to 20 are 7 and 14. This gives 2 favorable outcomes.
Probability = \( \frac{2}{20} = \frac{1}{10} \).
In simple words: Out of 20 cards, three are divisible by 6, which gives a 3 out of 20 chance. Two cards are divisible by 7, which gives a 2 out of 20 (or 1 in 10) chance.

Exam Tip: Always list the specific numbers to prevent missing or adding incorrect values.

Formative Assessment - III

Time: 1½ HR          Marks: 40

Section A (Each Question Carries 1 Mark)

Question. Which of the following equations has two distinct real roots?
(a) \( 2x^2 + 3\sqrt{2}x + 9/4 = 0 \)
(b) \( x^2 + x - 5 = 0 \)
(c) \( x^2 + 3x + 2\sqrt{2} = 0 \)
(d) \( 5x^2 - 3x + 1 = 0 \)
Answer: (b) \( x^2 + x - 5 = 0 \)
For a quadratic equation \( ax^2 + bx + c = 0 \) to have two distinct real roots, its discriminant \( D = b^2 - 4ac \) must be greater than zero.
Checking option (b):
Here, \( a = 1 \), \( b = 1 \), and \( c = -5 \).
\( D = (1)^2 - 4(1)(-5) = 1 + 20 = 21 \).
Since \( D > 0 \), this equation has two distinct real roots.
In simple words: We check the discriminant \( b^2 - 4ac \) for each option. Since option (b) gives a positive value (21), it has two distinct real roots.

Exam Tip: Remember: \( D > 0 \) gives distinct real roots, \( D = 0 \) gives equal real roots, and \( D < 0 \) gives no real roots.

Question. The sum of first 16 terms of the A.P. : 10,6,2, ............. is.
(a) -320
(b) 320
(c) -352
(d) -400.
Answer: (a) -320
In the given Arithmetic Progression, the first term \( a = 10 \) and the common difference \( d = 6 - 10 = -4 \).
We use the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \) for \( n = 16 \):
\( S_{16} = \frac{16}{2}[2(10) + (16-1)(-4)] \)
\( S_{16} = 8[20 + 15(-4)] \)
\( S_{16} = 8[20 - 60] = 8[-40] = -320 \).
In simple words: The terms of the pattern go down by 4 each time. When we add the first 16 terms using the AP sum formula, the final result is -320.

Exam Tip: Be careful with the negative common difference (\( d = -4 \)); multiplying negative numbers correctly is crucial for finding the correct sum.

Question. The point (-4,0),(4,0) ,(0,3) are the vertices of a .
(a) right angled triangle
(b) Isosceles triangle
(c) Equilateral triangle
(d) scalene triangle
Answer: (b) Isosceles triangle
Let the vertices be \( A(-4,0) \), \( B(4,0) \), and \( C(0,3) \). Using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
\( AB = \sqrt{(4 - (-4))^2 + (0-0)^2} = \sqrt{8^2 + 0} = 8 \)
\( BC = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)
\( AC = \sqrt{(-4-0)^2 + (0-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)
Since two sides are equal (\( BC = AC = 5 \)), it forms an Isosceles triangle.
In simple words: When we calculate the length of all three sides using the distance formula, we find that two of the sides are exactly 5 units long. A triangle with two equal sides is an isosceles triangle.

Exam Tip: If the squares of the sides satisfied Pythagoras' theorem (\( a^2 + b^2 = c^2 \)), it would be a right-angled triangle. Here \( 5^2 + 5^2 = 50 \neq 8^2 \), so it's only isosceles.

Question. A pole 6m high casts a shadow 2\(\sqrt{3}\)m long an the ground, then the sun's elevation is.
(a) \( 60^\circ \)
(b) \( 45^\circ \)
(c) \( 30^\circ \)
(d) \( 90^\circ \)
Answer: (a) \( 60^\circ \)
Let \( \theta \) represent the angle of elevation of the sun. The height of the pole is the opposite side and the length of the shadow is the adjacent side of a right-angled triangle.
\( \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \)
We know that \( \tan 60^\circ = \sqrt{3} \), so \( \theta = 60^\circ \).
In simple words: We can think of the pole and its shadow as the two legs of a right triangle. The tangent of the angle is the height divided by the shadow, which is \( \sqrt{3} \). The angle for this tangent value is \( 60^\circ \).

Exam Tip: Always write down the trigonometric ratio (\( \tan\theta = \frac{\text{height}}{\text{shadow}} \)) first before substituting values to make your calculation clear.

Question. If radii of two concentric circles are 4cm and 5cm, Then the length of each chord of one circle which is tangent to other circle is:
(a) 3cm
(b) 6cm
(c) 9cm
(d) 1cm
Answer: (b) 6cm
Let the two concentric circles have center \( O \). Let \( AB \) be the chord of the outer circle (radius \( R = 5 \) cm) which is tangent to the inner circle (radius \( r = 4 \) cm) at point \( P \).
Since the radius is perpendicular to the tangent at the point of contact, \( OP \perp AB \).
In the right-angled triangle \( OPA \):
\( OA^2 = OP^2 + AP^2 \)
\( 5^2 = 4^2 + AP^2 \implies AP^2 = 25 - 16 = 9 \implies AP = 3 \text{ cm} \).
Since the perpendicular from the center to a chord bisects the chord, the total length of the chord \( AB = 2 \times AP = 2 \times 3 = 6 \text{ cm} \).
In simple words: The line from the center to the contact point of the tangent is 4 cm, and the outer radius is 5 cm. This forms a 3-4-5 right triangle, meaning half the chord is 3 cm. Thus, the whole chord is 6 cm.

Exam Tip: Remember that the perpendicular from the center of a circle to a chord always divides the chord into two equal halves.

Section B (Each Question Carries 2 Marks)

Question. Find a relation between x and y such that the point (x,y) is equidistant from the point (3,6) and(-3,4).
Answer: Let \( P(x,y) \) be the point that is equidistant from \( A(3,6) \) and \( B(-3,4) \). Therefore, \( PA = PB \), which means \( PA^2 = PB^2 \).
Using the distance formula:
\( (x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2 \)
\( x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 \)
Subtracting \( x^2 + y^2 \) from both sides yields:
\( -6x - 12y + 45 = 6x - 8y + 25 \)
Grouping all terms to one side:
\( 12x + 4y - 20 = 0 \)
Dividing by 4 gives the required relationship:
\( 3x + y - 5 = 0 \).
In simple words: We set the distance from \( (x,y) \) to both points equal to each other. After expanding and simplifying the algebra, we get the line equation \( 3x + y - 5 = 0 \).

Exam Tip: When working with equidistant points, squaring both sides of the distance equation (\( PA^2 = PB^2 \)) avoids having to deal with square roots.

Question. Check whether the following equation is a quadratic equation.
\( x^2-4x+6=0 \)
Answer: The given equation is \( x^2 - 4x + 6 = 0 \).
A standard quadratic equation is of the form \( ax^2 + bx + c = 0 \) where \( a \neq 0 \) and the highest exponent of the variable \( x \) is 2.
Comparing the given equation, we have \( a = 1 \), which is non-zero, and the degree of the equation is 2. Therefore, yes, this is a quadratic equation.
In simple words: A quadratic equation must have \( x^2 \) as its highest power. Since this equation has \( x^2 \) and no higher powers, it is indeed a quadratic equation.

Exam Tip: Make sure the coefficient of \( x^2 \) is not zero after fully expanding any given equation.

Question. The angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower is 30\(^0\). Find the height of the tower.
Answer: Let \( h \) be the height of the tower in meters. The point on the ground is 30 m from the foot, forming a right triangle with an angle of elevation of \( 30^\circ \).
Using the tangent ratio:
\( \tan 30^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{30} \)
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\( \frac{1}{\sqrt{3}} = \frac{h}{30} \)
\( h = \frac{30}{\sqrt{3}} = 10\sqrt{3} \text{ m} \).
In simple words: We can represent this with a right triangle where the opposite side is the tower's height and the adjacent side is 30 meters. Using the formula for tangent, the height is \( 10\sqrt{3} \) meters.

Exam Tip: Rationalize the denominator by multiplying both top and bottom by \( \sqrt{3} \) to write the final answer in its standard form.

Question. Find the sum of the following A.P.:
-37,-33,-29,...................... to 12 terms.
Answer: For the given Arithmetic Progression: -37, -33, -29, ...
First term \( a = -37 \)
Common difference \( d = -33 - (-37) = 4 \)
Number of terms \( n = 12 \)
Using the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( S_{12} = \frac{12}{2}[2(-37) + (12-1)4] \)
\( S_{12} = 6[-74 + 11(4)] \)
\( S_{12} = 6[-74 + 44] = 6[-30] = -180 \).
In simple words: The terms of this progression increase by 4 each time. When we add the first 12 terms together using the formula, the final sum is -180.

Exam Tip: Be careful with negative values during calculations. Double check that the common difference \( d \) is positive (\( +4 \)), not negative, since the numbers are increasing.

Question. The length of a tangent from a point A at distance 5cm from the center of the circle is 4cm. Find the radius of the circle.
Answer: Let \( O \) be the center of the circle and \( P \) be the point where the tangent from point \( A \) meets the circle. The radius \( OP \) is perpendicular to the tangent line \( AP \) at the point of contact, making triangle \( OPA \) a right-angled triangle.
Using Pythagoras' theorem:
\( OA^2 = OP^2 + AP^2 \)
Given that the distance from the center is \( OA = 5 \text{ cm} \) and the tangent length is \( AP = 4 \text{ cm} \):
\( 5^2 = OP^2 + 4^2 \)
\( 25 = OP^2 + 16 \)
\( OP^2 = 25 - 16 = 9 \)
\( OP = 3 \text{ cm} \).
Thus, the radius of the circle is 3 cm.
In simple words: The radius, tangent, and line to the center form a right triangle. Since the hypotenuse is 5 cm and one side is 4 cm, the missing side (the radius) must be 3 cm.

Exam Tip: The radius is a physical length, so it must always be a positive value (3 cm), even if algebra technically yields \( \pm 3 \).

Question. Find the distance between the points [ -8/5,2] and [2/5,2].
Answer: Let the two points be \( A\left(-\frac{8}{5}, 2\right) \) and \( B\left(\frac{2}{5}, 2\right) \).
Using the distance formula:
\( d = \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2} \)
\( d = \sqrt{\left(\frac{2}{5} - \left(-\frac{8}{5}\right)\right)^2 + \left(2 - 2\right)^2} \)
\( d = \sqrt{\left(\frac{10}{5}\right)^2 + 0^2} \)
\( d = \sqrt{2^2} = 2 \text{ units} \).
In simple words: Since both points lie on the same horizontal line (both have a y-coordinate of 2), we can find the distance by just finding the difference between their x-coordinates, which is 2.

Exam Tip: When coordinates share a common value (like the y-value here), you can double-check your distance formula calculation by simply subtracting the other coordinates.

Section C (Each Question Carries 3 Marks)
 

Question. If (1,2),(4,y),(x,6) and (3,5) are the vertices of a || gm taken in order, find x and y
Answer: Let the vertices of the parallelogram be \( A(1,2) \), \( B(4,y) \), \( C(x,6) \), and \( D(3,5) \).
Since the diagonals of a parallelogram bisect each other, the midpoint of diagonal \( AC \) is the same as the midpoint of diagonal \( BD \).
Midpoint of \( AC = \left(\frac{1+x}{2}, \frac{2+6}{2}\right) = \left(\frac{1+x}{2}, 4\right) \)
Midpoint of \( BD = \left(\frac{4+3}{2}, \frac{y+5}{2}\right) = \left(\frac{7}{2}, \frac{y+5}{2}\right) \)
Equating the x-coordinates:
\( \frac{1+x}{2} = \frac{7}{2} \implies 1 + x = 7 \implies x = 6 \)
Equating the y-coordinates:
\( \frac{y+5}{2} = 4 \implies y + 5 = 8 \implies y = 3 \).
Therefore, \( x = 6 \) and \( y = 3 \).
In simple words: In a parallelogram, the diagonals cut each other exactly in half. By finding the middle point of both diagonals and setting them equal, we solve to find \( x = 6 \) and \( y = 3 \).

Exam Tip: Using the midpoint property of diagonals is much faster and simpler than using the distance formula for opposite sides when dealing with parallelograms.

Question. The angle of elevation of the top of a building from the foot of the tower is 30\(^0\) and the angle of elevation of the top of tower from the foot of the building is 600 .If the tower is 50 meter high find the height of the building.
Answer: Let \( AB \) represent the building of height \( h \) and \( CD \) represent the tower of height 50 m. Let the distance between their bases on the ground be \( BD = x \).
In right-angled triangle \( CDB \) (looking at the tower from the foot of the building):
\( \tan 60^\circ = \frac{CD}{BD} = \frac{50}{x} \)
\( \sqrt{3} = \frac{50}{x} \implies x = \frac{50}{\sqrt{3}} \quad \text{--- (Equation 1)} \)
In right-angled triangle \( ABD \) (looking at the building from the foot of the tower):
\( \tan 30^\circ = \frac{AB}{BD} = \frac{h}{x} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{x} \implies x = h\sqrt{3} \quad \text{--- (Equation 2)} \)
Equating the two equations for \( x \):
\( h\sqrt{3} = \frac{50}{\sqrt{3}} \)
\( h = \frac{50}{3} = 16\frac{2}{3} \text{ m} \).
Thus, the height of the building is \( 16\frac{2}{3} \) m.
In simple words: We use two right triangles sharing the same baseline. First, we use the tower's height and the \( 60^\circ \) angle to find the distance between them. Then, we use that distance and the \( 30^\circ \) angle to find the building's height, which is \( 16\frac{2}{3} \) meters.

Exam Tip: Clearly define which variable represents what object (tower vs. building) in a diagram to avoid swapping the elevation angles during calculations.

Question. If triangle ABC is drown to circumscribe a circle of radius 4cm, such that the segment BD And DC into which BC is divided by the point of contact D are of the length 8cm and 6cm respectively. Find the sides AB and AC (fig.)
Answer: Let the circle touch the sides \( BC \), \( CA \), and \( AB \) of \( \triangle ABC \) at points \( D \), \( E \), and \( F \) respectively. Let \( O \) be the center of the circle of radius 4 cm.
Tangents from the same external point to a circle are equal in length:
\( BD = BF = 8 \text{ cm} \)
\( CD = CE = 6 \text{ cm} \)
Let \( AF = AE = x \text{ cm} \).
Thus, the sides of the triangle are:
\( a = BC = 6 + 8 = 14 \text{ cm} \)
\( b = AC = x + 6 \text{ cm} \)
\( c = AB = x + 8 \text{ cm} \)
The semi-perimeter \( s \) of \( \triangle ABC \) is:
\( s = \frac{a+b+c}{2} = \frac{14 + (x+6) + (x+8)}{2} = x + 14 \)
Using Heron's Formula, the area of \( \triangle ABC \) is:
\( \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \)
\( \text{Area} = \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)} \quad \text{--- (Equation 1)} \)
Alternatively, the area of \( \triangle ABC \) is the sum of the areas of \( \triangle OBC \), \( \triangle OCA \), and \( \triangle OAB \):
\( \text{Area} = \frac{1}{2} \times \text{radius} \times (a + b + c) = \text{radius} \times s \)
\( \text{Area} = 4 \times (x + 14) \quad \text{--- (Equation 2)} \)
Equating both areas:
\[ \sqrt{48x(x+14)} = 4(x+14) \]
Squaring both sides:
\[ 48x(x+14) = 16(x+14)^2 \]
Since \( x + 14 \neq 0 \), we divide both sides by \( 16(x+14) \):
\[ 3x = x + 14 \implies 2x = 14 \implies x = 7 \]
Now, we calculate the lengths of the sides:
\( AB = x + 8 = 7 + 8 = 15 \text{ cm} \)
\( AC = x + 6 = 7 + 6 = 13 \text{ cm} \).
In simple words: We find the lengths of the tangents from each corner of the triangle. By calculating the total area of the triangle in two different ways—using Heron's formula and by summing the three smaller triangles—we can solve for \( x = 7 \). This gives the side lengths of 15 cm and 13 cm.

Exam Tip: Remember that the area of a circumscribing triangle is always equal to its semi-perimeter multiplied by the radius of the in-circle (\( A = r \cdot s \)).

Question. Find the value of k for the following quadratic equation, so that they have two equal roots.
\( kx(x-2)+6=0 \)
Answer: First, let's expand the given equation to the standard quadratic form \( ax^2 + bx + c = 0 \):
\( kx(x-2) + 6 = 0 \implies kx^2 - 2kx + 6 = 0 \)
Here, the coefficients are \( a = k \), \( b = -2k \), and \( c = 6 \).
For the quadratic equation to have two equal roots, the discriminant must be zero (\( b^2 - 4ac = 0 \)):
\( (-2k)^2 - 4(k)(6) = 0 \)
\( 4k^2 - 24k = 0 \)
\( 4k(k - 6) = 0 \)
This gives two possible values: \( k = 0 \) or \( k = 6 \).
However, if \( k = 0 \), the coefficient of \( x^2 \) becomes zero, which means the equation is no longer quadratic. Therefore, \( k = 6 \).
In simple words: We expand the equation and set its discriminant to zero to find the conditions for equal roots. This gives us two potential values, but since \( k \) cannot be 0 (otherwise the \( x^2 \) term disappears), our final answer is \( k = 6 \).

Exam Tip: Always check if your calculated values of \( k \) make the leading coefficient zero. Since a quadratic equation must have a non-zero leading coefficient, \( k = 0 \) is rejected.

Question. Which term of the A.P. 3,15,27,39,.....................will be 132 more than its 54\(^{th}\) term.
Answer: For the given Arithmetic Progression: 3, 15, 27, 39, ...
First term \( a = 3 \)
Common difference \( d = 15 - 3 = 12 \).
Let the required term be the \( n^{\text{th}} \) term \( a_n \). According to the problem:
\( a_n = a_{54} + 132 \)
Using the general term formula \( a_k = a + (k-1)d \):
\( a + (n-1)d = a + (54-1)d + 132 \)
\( (n-1)d = 53d + 132 \)
Substituting \( d = 12 \):
\( (n-1)(12) = 53(12) + 132 \)
Dividing the entire equation by 12:
\( n - 1 = 53 + 11 \)
\( n - 1 = 64 \implies n = 65 \).
Thus, the \( 65^{\text{th}} \) term is 132 more than the \( 54^{\text{th}} \) term.
In simple words: Instead of calculating the actual value of the 54th term, we can write an equation based on the term formula. Dividing by the common difference of 12 quickly tells us that the 65th term is the one we are looking for.

Exam Tip: Avoid calculating the large numerical value of the \( 54^{\text{th}} \) term directly; algebraic cancellation makes the problem much simpler and less prone to arithmetic mistakes.

Section D (Each Question Carries 4 Marks)
 

Question. Two water taps together can fill a tank in 9 (3/8) hours. The tap of longer 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.
Answer: Let the time taken by the smaller tap to fill the tank separately be \( x \) hours.
Then, the time taken by the tap of longer diameter is \( (x - 10) \) hours.
The work done by the smaller tap in 1 hour = \( \frac{1}{x} \)
The work done by the larger tap in 1 hour = \( \frac{1}{x-10} \)
Together, they fill the tank in \( 9\frac{3}{8} = \frac{75}{8} \) hours. Therefore, the combined work done in 1 hour is \( \frac{8}{75} \).
According to the problem:
\[ \frac{1}{x} + \frac{1}{x-10} = \frac{8}{75} \]
\[ \frac{x - 10 + x}{x(x - 10)} = \frac{8}{75} \]
\[ \frac{2x - 10}{x^2 - 10x} = \frac{8}{75} \]
Cross-multiplying:
\[ 75(2x - 10) = 8(x^2 - 10x) \]
\[ 150x - 750 = 8x^2 - 80x \]
\[ 8x^2 - 230x + 750 = 0 \]
Dividing the entire equation by 2:
\[ 4x^2 - 115x + 375 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{115 \pm \sqrt{(-115)^2 - 4(4)(375)}}{8} \]
\[ x = \frac{115 \pm \sqrt{13225 - 6000}}{8} \]
\[ x = \frac{115 \pm \sqrt{7225}}{8} = \frac{115 \pm 85}{8} \]
This yields two solutions:
1. \( x = \frac{115 + 85}{8} = \frac{200}{8} = 25 \)
2. \( x = \frac{115 - 85}{8} = \frac{30}{8} = 3.75 \)
If \( x = 3.75 \), the time taken by the larger tap would be \( 3.75 - 10 = -6.25 \) hours, which is impossible. Thus, we select \( x = 25 \).
So, the smaller tap takes 25 hours and the larger tap takes 15 hours to fill the tank separately.
In simple words: We write equations representing the rate at which each tap fills the tank per hour. Combining these rates gives us a quadratic equation, which we solve to find that the smaller tap takes 25 hours and the larger tap takes 15 hours.

Exam Tip: When a quadratic equation yields two positive values, always test both to see if they make physical sense in the context of the problem.

Question. If the angle of elevation of a cloud from a point h meters above a lake is \( \alpha \) and the depression of its reflection in the lake \( \beta \), prove that the height of the cloud is \( \frac{h(\tan \beta + \tan \alpha)}{\tan \beta - \tan \alpha} \)
Answer: Let \( XY \) represent the surface of the lake. Let \( P \) be the observation point which is \( h \) meters above the lake surface, so \( PY = h \).
Let \( C \) be the cloud at height \( H \) above the lake, and \( C' \) be its reflection inside the lake. Since the reflection's depth equals the cloud's height above the water, the depth of \( C' \) below the lake surface is also \( H \).
Draw a horizontal line \( PA \) meeting the vertical line \( CC' \) at \( A \). Let the horizontal distance \( PA = x \).
The height of the cloud above \( PA \) is \( CA = H - h \).
The depth of the reflection \( C' \) below \( PA \) is \( AC' = H + h \).
In right-angled triangle \( PAC \):
\( \tan \alpha = \frac{CA}{PA} = \frac{H - h}{x} \implies x = \frac{H - h}{\tan \alpha} \quad \text{--- (Equation 1)} \)
In right-angled triangle \( PAC' \):
\( \tan \beta = \frac{AC'}{PA} = \frac{H + h}{x} \implies x = \frac{H + h}{\tan \beta} \quad \text{--- (Equation 2)} \)
Equating both expressions for \( x \):
\[ \frac{H - h}{\tan \alpha} = \frac{H + h}{\tan \beta} \]
Cross-multiplying:
\[ (H - h)\tan \beta = (H + h)\tan \alpha \]
\[ H\tan \beta - h\tan \beta = H\tan \alpha + h\tan \alpha \]
\[ H\tan \beta - H\tan \alpha = h\tan \beta + h\tan \alpha \]
\[ H(\tan \beta - \tan \alpha) = h(\tan \beta + \tan \alpha) \]
\[ H = \frac{h(\tan \beta + \tan \alpha)}{\tan \beta - \tan \alpha} \]
This completes the proof.
In simple words: We use trigonometry on two triangles - one looking up at the cloud and one looking down at its reflection. By setting their horizontal distances equal, we solve for the height of the cloud \( H \) and prove the required expression.

Exam Tip: In reflection problems, always remember that the height of the object above the reflecting surface (the lake) is exactly equal to the depth of the reflection below it.

Question. (OR). The length h of tangents drown from an external point to a circle are equal, prove it.
Answer: Let us prove that the lengths of tangents drawn from an external point to a circle are equal.
Given: A circle with center \( O \), an external point \( P \), and two tangents \( PQ \) and \( PR \) touching the circle at \( Q \) and \( R \).
To Prove: \( PQ = PR \).
Construction: Join \( OQ \), \( OR \), and \( OP \).
Proof:
We know that the radius is perpendicular to the tangent at the point of contact. Therefore:
\( \angle OQP = \angle ORP = 90^\circ \)
Now, compare the right-angled triangles \( \triangle OQP \) and \( \triangle ORP \):
1. \( OQ = OR \) (Radii of the same circle)
2. \( OP = OP \) (Common side/Hypotenuse)
3. \( \angle OQP = \angle ORP = 90^\circ \)
By the RHS (Right angle - Hypotenuse - Side) congruence criterion:
\( \triangle OQP \cong \triangle ORP \)
By CPCT (Corresponding Parts of Congruent Triangles):
\( PQ = PR \).
This completes the proof.
In simple words: We draw lines from the center of the circle to the points where the tangents touch it, forming two right-angled triangles. Since these two triangles share the same hypotenuse and have equal radii as legs, they are congruent, meaning the tangent lengths must be equal.
Exam Tip: Mentioning the "RHS congruence criterion" and "CPCT" specifically is highly important to score full marks on this standard geometry proof.