Excel in Mathematics Book 10 Final_CTP (2080)_compressed (1)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 301 Vedanta Excel in Mathematics - Book 10 Unit 15 Probability 15.1 Probability -Looking back 1. Let’s discuss and answer the following questions. a) What is probability? b) What is the sample space in a random experiment? c) What is the probability of sure event? d) What is the probability of impossible event? e) What is the probability scale of any event E? 2. Let’s discuss and fill the blank space as quickly as possible. Mr. Manish tosses a coin once. a) The sample space is …..................... b) The probability of getting head is … c) The probability of getting tail is … 15.2 Definition of basic terms on probability -review Random experiments Any experiment whose outcome cannot be predicted or determined in advance is called a random experiment. For example, tossing a coin, rolling a die, etc. are random experiments. Outcomes The results of a random experiment are called outcomes. For example, while tossing a coin, the occurrence of head or tail is the outcome. Sample space The set of all possible outcomes of a random experiment is known as sample space. Usually, it is denoted by S. For example, (i) When a coin is tossed, S = {H, T} (ii) When a die is thrown, S = {1, 2, 3, 4, 5, 6} Event A subset of the sample space S related to an experiment is known as event. For example, While tossing a coin 3 times, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Let A = {HHH, HHT, HTH, THH}, A is the event in which at least two heads are obtained. Exhaustive cases The total number of all possible outcomes of a random experiment is known as exhaustive cases. For example, While tossing a coin, S = {H, T}. So, exhaustive cases = 2. While tossing two coins simultaneously, S = {HH, HT, TH, TT} So, exhaustive cases = 4. Favourable cases The outcomes in an random experiment which are desirable (or expected) to us are called favourable cases. For example, While tossing a coin, S = {H, T} Here, the favourable number of case of head is 1 and tail is also 1. The favourable number of cases of getting a ‘king’ when a card is drawn from a well shuffled pack of 52 cards is 4 Equally likely events Two or more events are said to be equally likely if the chance of occurring any one event is equal to the chance of occurring other events. For example, while throwing a die, the chance of coming up the numbers 1 to 6 is equal. So, they are equally likely events

Vedanta Excel in Mathematics - Book 10 302 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability 15.3 Mutually exclusive and Non-mutually exclusive events Mutually exclusive events Two ore more events of a sample space S are said to be mutually exclusive if the occurrence of any one event excludes the occurrence of the other events. For example, while tossing a coin, the occurrence of head excludes the occurrence of tail or vice versa. So, they are mutually exclusive events. Furthermore, let’s consider the experiment of throwing a die. Let A be the event, ‘the number obtained is even’. Then A = {2, 4, 6}. Again, let B be the event, ‘the number obtained is odd’. Then, B = {1, 3, 5}. Here, A ∩ B = φ None-mutually exclusive events If A and B are any two events on a sample space S and A ∩ B ≠ 0, the events are said to be none-mutually exclusive events. For example, Again, let's consider the experiment of throwing a die. Let A be the event 'the number obtained is even' and B be the event ' the number obtained is less than 5'. Then, A = {2, 4, 6} B = {1, 2, 3, 4} ∴ A ∩ B = {2, 4} It means, ‘an even number from 1 to 6’ and ‘a number less than 5’ can occur simultaneously in a single throw of a die. So, A and B are said to be none-mutually exclusive events. 15.4 Law of addition of mutually exclusive events If A and B are any two mutually exclusive events of a sample space S, the probability of the occurrence of either A or B is equal to the sum of the individual probabilities of the occurrence of A and B. i.e. P (A or B) = P (A) + P (B) i.e. P (A ∪ B) = P (A) + P (B) Further more, if A, B and C are three mutually exclusive events, then, P (A or B or C) = P (A ∪ B ∪ C) = P (A) + P (B) + P (C) and so on. A ∩ B = φ A B S A ∩ B ≠ φ A B S 2 4 1 3 5 A B S 6 A B S

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 303 Vedanta Excel in Mathematics - Book 10 Probability Worked-out Examples Example 1: One card is drawn at random from the numbered cards, numbered from 10 to 21. Find the probability that the card may be prime numbered or even numbered card. Solution: Let, n(S) be the exhaustive cases, n(A) be the number of events of prime numbered cards, and n(B) be the number of events of even numbered cards. Here, S = {10, 11, 12, ... 21}. So, n(S) = 12 A = {11, 13, 17, 19}. So, n(A) = 4 B = {10, 12, 14, 16, 18, 20}. So, n(B) = 6 Now, P(A) = n(A) n(S) = 4 12 = 1 3 P(B) = n(B) n(S) = 6 12 = 1 2 By using the law of addition of probabilities of mutually exclusive events, P (A or B) = P (A ∪ B) = P(A) + P(B) = 1 3 + 1 2 = 5 6 . So, the probability that the card may be prime numbered or even numbered is 5 6 . Example 2: From a pack of 52 cards, a card is drawn at random. What is the probability of getting a face card or an ace? Solution: Let n (S) be the exhaustive events, n(A) the number of events of face card and n(B) the number of events of ace. Here, n(S) = 52, n(A) = 12 and n(B) = 4 Now, by using the law of addition of probabilities of mutually exclusive events, P(A or B) = P(A ∪ B) = P(A) + P(B) = n(A) n(S) + n(B) n(S) = 12 52 + 4 52 = 16 52 = 4 13 So, the probability that the card may be a face card or an ace is 4 13 . Example 3: A bag contains 3 red, 5 blue, 7 black, and 9 white identical balls. If a ball is drawn randomly from the bag, find the probability that it may be either red, or blue or white.

Vedanta Excel in Mathematics - Book 10 304 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability Solution: Let, n(S) be the exhaustive cases, n(A) the number of cases of red balls, n(B) the number of cases of blue balls, and n(C) the number of cases of white balls. Here, n(S) = 3 + 5 + 7 + 9 = 24 n(A) = 3, n(B) = 5 and n(C) = 9 Now, by using the law of addition of probabilities of mutually exclusive events, P (A or B or C) = P (A ∪ B ∪ C) = P(A) + P(B) + P(C) = n(A) n(S) + n(B) n(S) + n(C) n(S) = 3 24 + 5 24 + 9 24 = 17 24 15.5 Law of addition of none-mutually exclusive events If A and B are any two none-mutually exclusive events of a sample space S, then the probability of occurrence of either A or B in single trial is given by P(A or B) = P(A ∪ B) = P(A) + P (B) – P(A ∩ B) Example 4: From the pack of cards, a card is drawn randomly. Find the probability of getting it red or face card. Solution: Let, n(S) be the exhaustive cases, n(A) be the number of events of red card, and n(B) be the number of events of face cards. Here, n(S) = 52, n(A) = 26, and n(B) = 12. Also, n(A ∩ B) = 6 Now, using the law of addition of probabilities of non-mutually exclusive events, P(A or B) = P (A ∪ B) = P(A) + P(B) – P(A ∩ B) = n(A) n(S) + n(B) n(S) – n(A ∩ B) n(S) = 26 52 + 12 52 – 6 52 = 26 + 12 – 6 52 = 32 52 = 8 13 So, the probability of getting the card red or face is 8 13 . Example 5: Two unbiased dice are rolled. Find the probability that the sum of the numbers on the two faces is either divisible by 3 or divisible by 4. Solution: Two dice can be thrown in 6n = 62 = 36 ways. (6 is the exhaustive cases of a die and n is the number of rolling dice.) S A B A ∩ B 3 faced card of heart + 3 faced card of diamond

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 305 Vedanta Excel in Mathematics - Book 10 Probability Let n(S) be the exhaustive cases, n(A) the number of events that the sum is divisible by 3 and n(B) the number of events that the sum is divisible by 4. Here, A = {(1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)} ∴ n(A) = 12 Also, B = {(1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)} ∴ n(B) = 9 The event A ∩ B happens when the sum is divisible by 3 as well as 4, i.e., when the sum is 12 which is (6, 6). ∴ n(A ∩ B) = 1. Now, by using the law of addition of probabilities of none-mutually exclusive events, P(A or B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = n(A) n(S) + n(B) n(S) – n(A ∩ B) n(S) = 12 36 + 9 36 – 1 36 = 20 36 = 5 9 . So, the probability that the sum of the numbers on the two faces is either divisible by 3 or by 4 is 5 9 . EXERCISE 15.1 General section 1. Define the following terms with examples. a) Sample space b) Exhaustive cases c) Mutually exclusive events d) Non-mutually exclusive events 2. a) If A and B are two mutually exclusive events, what is the probability of the occurrence of either A or B? b) If X, Y and Z are three mutually exclusive events, what is the probability of the occurrence of either X or Y or Z ? c) If A and B are two non-mutually exclusive events, what is the probability of the occurrence of either A or B? 3. a) A and B are mutually exclusive events. If n(A) = 4, n(B) = 6 and n(S) = 12, find P (A ∪ B). b) If P(A) = 2 9 , P (B) = 4 9, where A and B are mutually exclusive events, find P (A ∪ B). c) If P(A ∪ B) = 7 13 and P(A) = 4 13 , where A and B are mutually exclusive events, find P(B). d) A and B are mutually exclusive events. If P(A ∪ B) = 0.8 and P(B) = 0.3, find P(A). e) Two events Q and R are mutually exclusive with P(Q) = 3 5 and P(R) = 1 5 . Find the probabilities of the following events. (i) P(Q ∪ R) (ii) P( Q ∪ R )

Vedanta Excel in Mathematics - Book 10 306 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability 4. a) A and B are non-mutually exclusive events. If n(A) = 5, n(B) = 7, n(A ∩ B) = 2 and n(S) = 20, find P(A ∪ B). b) If P(A) = 4 25 , P(B) = 1 5 and P(A ∩ B) = 2 25 , where A and B are none-mutually exclusive events, find P(A ∪ B). c) A and B are non-mutually exclusive events. If P (A ∪ B) = 9 16 , P(A) = 3 16 and P(B) = 3 4 , find P(A ∩ B). d) If P(A ∪ B) = 0.7, P(A ∩ B) = 0.1 and P(A) = 0.35, find P(B). Creative Section 5. From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probabilities of the following events. a) It is either a king or a queen. b) It is a black ace or a red king. c) It is a king or an ace. d) It is either a non-faced card or a jack. e) It is either a black king or a red king. f) It is either a red or a club. g) It is either a spade or a heart. h) It is a queen or a black ace. i) It is neither a king nor an ace. j) It is neither a black jack nor a red queen. 6. a) When a fair die is thrown, find the probabilities of the following events. (i) Getting either 1, or 6 (ii) Getting either 1, 3, or 5 (iii) Getting an even number less than 6 or an odd number less than 5. (iv) Getting a multiple of 2 or a multiple of 3 (v) Getting either 1 or a prime number. b) Two unbiased dice are rolled. Find the probability that the sum of the numbers on the two faces is either divisible by 5 or divisible by 6. 7. a) From the number cards numbered from 1 to 30, a card is drawn at random. Find the probabilities of the following events. (i) The number is either divisible by 5 or by 7. (ii) The number is either divisible by 5 or by 9. (iii) The number is either a multiple of 7 or a multiple of 8. (iv) The number is either a prime number less than 20 or a composite number more than 20. www.geogebra.org/classroom/cptpggjg Classroom code: CPTP GGJG Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 307 Vedanta Excel in Mathematics - Book 10 Probability b) A number card is drawn randomly from the set of numbered cards, numbered from 6 to 39. Find the probability that the card may be a prime number or a cubed number. c) From the number cards, numbered from 7 to 27 a card is drawn at random. Find the probability of getting the card of a prime or an even number. 8. a) A box contains 4 yellow, 6 green, and 8 red balls. A ball is drawn at random from the box. Find the probabilities of the following events. (i) The ball is either yellow or red. (ii) The ball is either green or yellow. (iii) The ball is neither red nor green. b) A basket contains 5 yellow, 3 blue, and 2 green balls. If a ball is drawn randomly from the basket, find the probability of not getting a blue ball. 9. From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probabilities of the following events. a) It is either a black or a face card. b) It is either a spade or a queen card. c) It is either a red card or an ace. d) It is either a heart or a jack. 10. a) From the number cards numbered from 1 to 30, a card is drawn at random. Find the probabilities of the following events. (i) The number is either divisible by 4 or by 5. (ii) The number is either divisible by 5 or by 6. (iii) The number is either a multiple of 3 or a multiple of 4. (iv) The number is either a prime or an even number. b) Find the probability of getting a square number or an even number card while drawing a flash card from the set of cards numbered from 5 to 15. c) Find the probability of occurring a cube number or an odd number card while drawing a flash card from the set of cards numbered from 7 to 17. d) A natural number is chosen at random from the first 30 natural numbers. What is the probability that the number chosen is neither divisible by 4 nor by 6? e) Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is either less than 6 or divisible by 4? 15.5 Dependent and Independent events Dependent events Two or more events are said to be dependent if the occurrence of one of the events affects the occurrence of the other events. For example, while drawing a ball in two successive trials from a bag containing 2 red and 3 blue balls without a replacement, getting any one coloured ball in the first trial affects to get other coloured ball in the second trial. So, these are the dependent events. Independent events Two or more events are said to be independent if the occurrence or none-occurrence of one of the events does not affect the occurrence or none-occurrence of the other events.

Vedanta Excel in Mathematics - Book 10 308 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability For example, in the random experiment of tossing a coin twice or more, the occurrence of any one event in the first trial does not affect the occurrence of any other event in the second trial. So, they are independent events. 15.6 Multiplication law of probability for independent events Two or more events are said to be independent if the occurrence or none-occurrence of any one event does not affect the occurrence or none-occurrence of the other events. Suppose, a bag contains 1 red ball and 2 black balls. If a ball is drawn and again it is replaced to draw another ball, the outcomes of these two trials are independent events. Study the following possible outcomes of two trials shown in the table. R B1 B2 R R R RB1 RB2 B1 B1 R B1 B1 B1 B2 B2 B2 R B2 B1 B2 B2 Here, the probability of Red ball, P(R) = 1 3 The probability of Black ball, P(B) = 2 3 Now, from the table, (i) Probability of Red balls in both trials P(R and R) = 1 9 = 1 3 × 1 3 = P(R) × P(R) (ii) Probability of Black balls in both trials P(B and B) = 4 9 = 2 3 × 2 3 = P(B) × P(B) (iii) Probability of Red at 1st and Black in 2nd trial P(R and B) = 2 9 = 1 3 × 2 3 = P(R) × P(B) (iv) Probability of Black at 1st and Red in 2nd trial P(B and R) = 2 9 = 2 3 × 1 3 = P(B) × P(R) From the above illustrations, we can verify the following probability relationships between any two independent events. (i) P(R and R) = P(R ∩ R) = P(R) × P(R) (ii) P(B and B) = P(B ∩ B) = P(B) × P(B) (iii) P(R and B) = P(R ∩ B) = P(R) × P(B) (iv) P(B and R) = P(B ∩ R) = P(B) × P(R) Thus, if A and B are any two independent events, P(A and B) = P(A ∩ B) = P(A) × P(B). It is called the multiplication law of probability for independent events.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 309 Vedanta Excel in Mathematics - Book 10 Probability Furthermore, if A, B and C are any three independent events, P(A ∩ B ∩ C) = P(A) × P(B) × P(C) Worked-out Examples Example 1: Two cards are drawn one after another with replacement from a well shuffled pack of 52 cards. Find the probability that both cards are king. Solution: There are 4 kings in a pack of 52 cards. Let K denotes the event of king. Since the card which is drawn at first is replaced to draw another card, it is an independent event. Now, by using multiplication law of probability for independent events, P(K and K) = P(K ∩ K)= P(K) × P(K) = n(K) n(S) × n(K) n(S) = 4 52 × 4 52 = 1 13 × 1 13 = 1 169 So, the required probability is 1 169 . Example 2: A bag contains 5 red balls and 3 blue balls. A ball is drawn at random and replaced. After that another ball is drawn. Find the probability that (i) both balls are blue and (ii) none of them are blue. Solution: Here, the number of possible outcomes = total number of balls n(S) = 5 + 3 = 8 balls Let B denotes the event of blue ball. Since the ball which is drawn at first is replaced to draw another ball, it is an independent event. Now, by using multiplication law of probability for independent events, (i) P(B and B) = P(B ∩ B) = P(B) × P(B) = n(B) n(S) × n(B) n(S) = 3 8 × 3 8 = 9 64 So, the probability that both of them are blue is 9 64 . (ii) P(B and B) = [1 – P(B)] [1 – P(B)] = 1 – 3 8 1 – 3 8 = 25 64 Alternatively, none of them are blue means both of them are red. So, P(R and R) = P(R) ∩ P (R) = P(R) × P(R) = 5 8 × 5 8 = 25 64 Example 3: A box contains 4 red, 3 blue and 5 white balls. A ball is drawn at random and it is replaced, then another ball is drawn. Find the probability that (i) the first is white and the second is blue,

Vedanta Excel in Mathematics - Book 10 310 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability (ii) the first is blue and the second is red, (iii) both of them are red, (iv) both of them are of the same colour, (v) both of them are not blue. Solution: Here, the number of possible outcomes = total number of balls n(S) = 4 + 3 + 5 = 12 balls Let, R denotes red, B denotes blue and W denotes white balls. Since, the ball which is drawn at first is replaced to draw another ball, it is an independent event. Now, by using multiplication law of probability for independent events, (i) P(W and B) = P(W ∩ B)= P(W) × P(B) = n(W) n(S) × n(B) n(S) = 5 12 × 3 12 = 5 48 ∴ The probability that the first is white and the second is blue is 5 48 . (ii) P(B and R) = P(B ∩ R) = P(B) × P(R) = n(B) n(S) × n(R) n(S) = 3 12 × 4 12 = 1 12 ∴ The probability that the first is blue and the second is red is 1 12 . (iii) P(R and R) = P(R ∩ R) = P(R) × P(R) = n(R) n(S) × n(R) n(S) = 4 12 × 4 12 = 1 9 ∴ The probability that both of them are red is 1 9. (iv) P(R, R or B, B or W, W) = [P(R) × P(R)] + [P(B) × P(B)] + [P(W) × P(W)] = n(R) n(S) × n(R) n(S) + n(B) n(S) × n(B) n(S) + n(W) n(S) × n(W) n(S) = 4 12 × 4 12 + 3 12 × 3 12 + 5 12 × 5 12 = 1 9 + 1 16 + 25 144 = 50 144 = 25 72 ∴ The probability that both of them are of the same colour is 25 72 (v) Here, both of them are not blue means, they may be either RR or RW or WR or WW. So, P(R, R or R, W or W, R or WW) = [P(R) × p(R)] + [P(R) × P(W)] + [P(W) × P(R)] + [P(W) × P(W)] = n(R) n(S) × n(R) n(S) + n(R) n(S) × n(W) n(S) + n(W) n(S) × n(R) n(S) + n(W) n(S) × n(W) n(S) = 4 12 × 4 12 + 4 12 × 5 12 + 5 12 × 4 12 + 5 12 × 5 12 = 1 9 + 5 36 + 5 36 + 25 144 = 9 16 ∴ The probability that both of them are not blue is 9 16 . Alternative process Here, n(S) = 12 n(B) = 3

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 311 Vedanta Excel in Mathematics - Book 10 Probability ∴ P(B) = n(B) n(S) = 3 12 = 1 4 Now, P ( B ∩ B ) = [1 – P(B)] [1 – P(B)] = 1 – 1 4 1 – 1 4 = 3 4 × 3 4 = 9 16 So, the probability that both of them are not blue is 9 16 . Example 4: If two unbiased coins are tossed simultaneously, find the probability of getting (i) two heads (ii) at least one head. Solution: Here, the sample space, S = {HH, HT, TH, TT} n(S) = 4 (i) Now, by using multiplication law of probability for independent events, P(H and H) = P(H) × P(H) = 1 2 × 1 2 = 1 4 (ii) Again, P(HH or HT or TH) = P(H) × P(H) + P(H) × P(T) + P(T) × P(H) = 1 2 × 1 2 + 1 2 × 1 2 + 1 2 × 1 2 = 1 4 + 1 4 + 1 4 = 3 4 Example 5: Find the probability of getting 3 on the die and head on the coin when a die is rolled and a coin is tossed simultaneously. Solution: Here, the sample space S1 for a die = {1, 2, 3, 4, 5, 6}. So, n(S1 ) = 6 The sample space S2 for a coin = {H, T}. So n(S2 ) = 2 Also, n(3) = 1 and n(H) = 1. It is the case of independent events. Now, by using the law of multiplication of probability for independent events, P(3 and H) = P(3) × P(H) = n(3) n(S1 ) × n(H) n(S2 ) = 1 6 × 1 2 = 1 12 So, the required probability is 1 12 . Example 6: The probability of solving a mathematical problem by two students A and B are 1 3 and 1 4 respectively. If the problem is given to the both students, find the probability of solving. Solution: Here, the problem may be solved by both the students. So, it is the case of none-mutually exclusive event. Alternative process Here, n(HH) = 1. ∴ P(HH) = n(HH) n(S) = 1 4 Alternative process Here, n(at least one head) n(A) = 3 ∴ P(A) = n(A) n(S) = 3 4 .

Vedanta Excel in Mathematics - Book 10 312 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability ∴ P (A or B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Also, the problem solving event is independent. ∴ P (A and B) = P(A ∩ B) = P(A) × P(B) = 1 3 × 1 4 = 1 12 Now, P(A ∪ B)= P(A) + P(B) – P(A ∩ B) = 1 3 + 1 4 – 1 12 = 4 + 3 – 1 12 = 6 12 = 1 2 So, the required probability is 1 2 . EXERCISE 15.2 General section 1. a) Define independent and dependent events with an example. b) If A and B are two independent events, find the probability of (A ∩ B). c) If X, Y and Z are three independent events, find P(X ∩ Y ∩ Z). d) A ball is drawn from a box and it is replaced to draw another ball. Are the probabilities of outcomes independent? Why? 2. a) A coin is tossed two times. Find the probability that (i) both the outcomes are head (ii) the first is head and the second is tail (iii) at least one tail (iv) at most one head b) What is the probability of getting 3 on the die and head on the coin when a die is rolled and a coin is tossed simultaneously? c) If a card is drawn at random from a pack of 52 cards and at the same time a marble is drawn at random from a bag containing 2 red marbles and 3 blue marbles. Find the probability of getting a blue marble and a king. d) A card is drawn randomly from a pack of 52 cards and a die is thrown once. Determine the probability of not getting a card king as well as 6 on the die. 3. a) Two cards are drawn one after another with replacement from a well-shuffled pack of 52 cards. Find the probability that both the cards are ace. b) A card is drawn from a well-shuffled pack of playing cards. If another card is drawn after replacing the first card, find. (i) the probability that both of them are red queen, (ii) the probability that both of them are not king,

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 313 Vedanta Excel in Mathematics - Book 10 Probability (iii) the probability that one is jack and other is ace, (iv) the probability that one is not a king and other is ace, (v) the probability that both of them are of the same colour. 4. a) A bag contains 6 red balls and 4 white balls. A ball is drawn at random and it is replaced to draw another ball. (i) Find the probability that both of them are red. (ii) Find the probability that both of them are white. (iii) Find the probability that the first is red and the second is white. (iv) Find the probability that none of them are red. (v) Find the probability that none of them are white. b) A box contains 5 black, 7 blue and 4 yellow balls. A ball is drawn at random and it is replaced, then another ball is drawn. Find the probability that (i) the first is blue and the second is black, (ii) both of them are yellow, (iii) both of them are of the same colour, (iv) the first is black and the second is yellow. (v) both of them are not black. c) A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same colour. 5. a) The probability that Bishwant gets scholarship is 0.9 and Sunayana will get is 0.8. What is the probability that at least one of them gets the scholarship? [Hint: P(B or S) = P(B) + P(S) – P(B ∩ S)] b) The probability of solving a mathematical problem by two students A and B are 1 2 and 1 5 respectively. If the problem is given to both the students, find the probability of solving the problems. 15.7 Probability tree diagram The probabilities of all possible outcomes of a series of trials of random experiment can be shown by a diagram called probability tree diagram. Every branch of a tree diagram shows the probability of respective event in a particular trial. From a tree diagram, the probability of a particular event of a random experiment can be obtained as the product of the probabilities of the event in different trials along the path of the event. Let’s consider an example of tossing of a coin 3 times and calculate the probabilities of the events. www.geogebra.org/classroom/snw4y23d Classroom code: SNW4 Y23D Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 314 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability P(HHH) = 1 2 × 1 2 × 1 2 = 1 8 P(HHT) = 1 2 × 1 2 × 1 2 = 1 8 P(HTH) = 1 2 × 1 2 × 1 2 = 1 8 P(HTT) = 1 2 × 1 2 × 1 2 = 1 8 P(THH) = 1 2 × 1 2 × 1 2 = 1 8 P(THT) = 1 2 × 1 2 × 1 2 = 1 8 P(TTH) = 1 2 × 1 2 × 1 2 = 1 8 P(TTT) = 1 2 × 1 2 × 1 2 = 1 8 H T H T H T H T H T H T H T P(H1 ) = 1 2 P(T1 ) = 1 2 P(T2 ) = 1 2 P(T2 ) = 1 2 P(T3 ) = 1 2 P(T3 ) = 1 2 P(T3 ) = 1 2 P(T3 ) = 1 2 P(H2 ) = 1 2 P(H2 ) = 1 2 P(H3 ) = 1 2 P(H3 ) = 1 2 P(H3 ) = 1 2 P(H3 ) = 1 2 Toss 1 Toss 2 Toss 3 From the tree diagram (i) The probability of getting 3 heads = P(HHH) = 1 2 × 1 2 × 1 2 = 1 8 (ii) The probability of getting 2 heads and 1 tail = P(HHT or HTH or THH) = P(HHT) + P(HTH) + P(THH) = 1 8 + 1 8 + 1 8 = 3 8 (iii) The probability of getting at least 2 tails = P(HTT or THT or TTH or TTT) = P(HTT) + P(THT) + P(TTH) + P(TTT) = 1 8 + 1 8 + 1 8 + 1 8 = 4 8 = 1 2 (iv) The probability of getting at most 1 head = P(HTT or THT or TTH or TTT) = P(HTT) + P(THT) + P(TTH) + P(TTT) = 1 8 + 1 8 + 1 8 + 1 8 = 4 8 = 1 2 (v) The probability of getting exactly 1 tail = P(HHT, HTH, THH) = P(HHT) + P(HTH) + P(THH) = 1 8 + 1 8 + 1 8 = 3 8 15.8 Multiplication law of probability of dependent events Two or more events are said to be dependent if the occurrence of any one event affects the occurrence of the other events. Suppose, a box contains 5 red and 3 black balls. Let a ball be drawn at random and not replaced.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 315 Vedanta Excel in Mathematics - Book 10 Probability Here, in the first trial, P(R) = 5 8 and P(B) = 3 8 Let, the outcome in the first trial be red ball. Then, in the second trial, P(R) = 4 7 and P(B) = 3 7 Let, the outcome in the first trial be black ball. Then, in the second trial, P(R) = 5 7 and P(B) = 2 7 Now, let’s denote the occurrence of red ball by R1 , or the occurrence of white ball by W1 in the first trial; the occurrence of red ball by R2 or the occurrence of white ball by W2 in the second trial. Then, from the above illustrations, P(R2 /R1 ) [read as ‘probability of R2 given that R1 has occurred] = 4 7 P(B2 /B1 ) [read as ‘probability of B2 given that B1 has occurred] = 2 7 P(R2 /B1 ) [read as ‘probability of R2 given that B1 has occurred] = 5 7 P(B2 /R1 ) [read as ‘probability of B2 given that R1 has occurred] = 3 7 Now, the multiplication law of probability in the case of dependent events can be stated as following. P(R1 and R2 ) = P(R1 ∩ R2 ) = 5 8 × 4 7 = P(R1 ) × P(R2 /R1 ) P(B1 and B2 ) = P(B1 ∩ B2 ) = 3 8 × 2 7 = P(B1 ) × P(B2 /B1 ) P(R1 and B2 ) = P(R1 ∩ B2 ) = 5 8 × 3 7 = P(R1 ) × P(B2 /R1 ) P(B1 and R2 ) = P(B1 ∩ R2 ) = 3 8 × 5 7 = P(B1 ) × P(R2 /B1 ) Study the following tree diagram to illustrate these probabilities. 5 red 3 black 4 red 3 black 5 red 2 black P(R1 ) = 5 8 P(B1 ) = 3 8 P(R2 /R1 ) = 4 7 P(B2 /R1 ) = 3 7 P(R2 /B1 ) = 5 7 P(B2 /B1 ) = 2 7 RR → P(RR) = 5 8 × 4 7 = 5 14 RB → P(RB) = 5 8 × 3 7 = 15 56 BR → P(BR) = 3 8 × 3 7 = 15 56 BB → P(BB) = 3 8 × 3 7 = 3 28 5 red 3 black 4 red 3 black 5 red 2 black

Vedanta Excel in Mathematics - Book 10 316 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability Worked-out Examples Example 1: Two children were given birth by a couple at the interval of four years. Illustrate the probabilities of son or daughter by drawing a tree diagram and calculate the probability that both are girls. Solution: Let S be the event of son and D be the event of daughter. Probability tree diagram: P(S1 ) = 1 2 P(D1 ) = 1 2 P(D2 ) = 1 2 P(S2 ) = 1 2 P(D2 ) = 1 2 P(S2 ) = 1 2 S D S D S D P(SS) = 1 2 × 1 2 = 1 4 P(SD) = 1 2 × 1 2 = 1 4 P(DS) = 1 2 × 1 2 = 1 4 P(DD) = 1 2 × 1 2 = 1 4 From the tree diagram, The probability of both are girls= P(DD) = P(D) × P(D) = 1 2 × 1 2 = 1 4 So, the probability of both are girls is 1 4 . Example 2: There were three black and five white balls in a box. Two balls were drawn at random one by one. Find the probability that both balls are of the same colour. (no replacement) Solution: Let B1 or W1 denote the occurrence of black or white ball in the first trial and let B2 or W2 denote the occurrence of black or white ball in the second trial. Here, the number of possible outcomes = total number of balls = 3 + 5 = 8 balls. Since the ball which is drawn at first is not replaced to draw another ball, it is a dependent event. Now, by using multiplication law of probability for a dependent event, P(B1 B2 ) = P(B1 ) × P(B2 /B1 ) = 3 8 × 2 7 = 3 28 Also, P(W1 ,W2 ) = P(W1 ) × P(W2 /W1 ) = 5 8 × 4 7 = 5 14 Again, the probability of both balls are same colour is P(B1 B2 or W1 W2 ) = P(B1 B2 ) + P(W1 W2 ) = 3 28 + 5 14 = 3 + 10 28 = 13 28 So, the probability that both balls are of the same colour is 13 28.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 317 Vedanta Excel in Mathematics - Book 10 Probability Example 3: There are 5 black balls and 4 white balls in a bag. Two balls are drawn randomly one after the other. Show the probability in a tree diagram. Also find the probability of (i) getting both white balls and (ii) getting none of white balls. Solution: Let W1 and B1 denote the occurrence of white or black ball in the first trial and let W2 and B2 denote the occurrence of white or black ball in the second trial. Here, the number of possible outcomes = total number of balls = 5 + 4 = 9 balls Clearly, the balls are drawn one after the other indicates that there is no replacement. So, it is a dependent event. Probability tree diagram 5 black 4 white 4 black 4 white 5 black 3 white P(B1 ) = 5 9 P(W1 ) = 4 9 P(B2 /B1 ) = 4 8 P(W2 /B1 ) = 4 8 P(B2 /W1 ) = 5 8 P(W2 /W1 ) = 3 8 P(B1 B2 ) = 5 9 × 4 8 = 5 18 P(B1 W2 ) = 5 9 × 4 8 = 5 18 P(W1 B2 ) = 4 9 × 5 8 = 5 18 P(W1 W2 ) = 4 9 × 3 8 = 1 6 (i) From the tree diagram, the probability of getting two white balls = P(W1 W2 ) = 1 6 (ii) Also, probability of getting non of white balls P( W1 ∩ W2 ) = [1 – P(W1 )] [1 – P (W2 / B1 )] = 1 – 4 9 1 – 4 8 = 5 9 × 4 8 = 5 18 Example 4: Two cards are drawn from a well-shuffled deck of 52 cards without replacing the first card. By drawing a tree-diagram, find the probability that both card are jacks. Solution: Let J1 or J1 denote the occurrence of jack or the card except jack in the first trial and J2 or J2 denotes the occurrence of jack or the card except jack in the second trial. Here, the number of possible outcomes = 52. Since the card which is drawn at first is not replaced, it is a dependent event.

Vedanta Excel in Mathematics - Book 10 318 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability Probability tree diagram: P(J1 J2 ) = 4 52 × 3 51 = 1 221 P(J1, J1 ) = 4 52 × 48 51 = 16 221 P(J1 J2 ) = 48 52 × 4 51 = 16 221 P(J1 J2 ) = 48 52 × 47 51 = 188 221 P(J1 ) = 4 52 P(J2 /J1 ) = 3 51 4J 48J1 3J 48J1 4J 47J1 P(J1 ) = 48 52 P(J2 /J1 ) = 48 51 P(J2 /J1 ) = 4 51 P(J2 /J1 ) = 47 51 From the tree diagram, The probability that both cards are jack = P(J1 J2 ) = 1 221 Example 5: There is one red, one blue and one white ball in a bag. A ball is drawn randomly and not replaced, then another ball is drawn. Write the sample space of the experiment using a tree diagram and find the probability of getting at least one red ball. Solution: Let R, B and W denote the red, blue and white ball. Here, the number of possible outcomes = 1 + 1 + 1 = 3 When the first ball is drawn, it is not replaced to draw another ball. So, it is a dependent event. Probability tree diagram P(RB) = 1 3 × 1 2 = 1 6 P(RW) = 1 3 × 1 2 = 1 6 P(BR) = 1 3 × 1 2 = 1 6 P(BW) = 1 3 × 1 2 = 1 6 P(WR) = 1 3 × 1 2 = 1 6 P(WB) = 1 3 × 1 2 = 1 6 1R 1W 1B 1W 1R 1B P(R) = 1 3 P(B) = 1 3 P(W) = 1 3 P(B) = 1 2 P(R) = 1 2 P(W) = 1 2 P(R) = 1 2 P(W) = 1 2 P(B) = 1 2 1R 1B 1W From the tree diagram, The sample space, S = {RB, RW, BR, BW, WR, WB} Also, the probability of getting at least one red ball = P(RB or RW or BR or WR) = P(RB) + P(RW) + P(BR) + P(WR) = 1 6 + 1 6 + 1 6 + 1 6 = 4 6 = 2 3

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 319 Vedanta Excel in Mathematics - Book 10 Probability Alternative process The sample space, n(S) = 6 The number of events where there is at least one red ball, n(A) = 4 ∴ Probability of getting at least one red ball = n(A) n(S) = 4 6 = 2 3 EXERCISE 15.3 General section 1. a) A coin is tossed two times in succession, write down the sample space by drawing the tree diagram. b) A fair coin is tossed three times. Draw a probability tree diagram to show all the possible outcomes and determine, (i) the probability of getting 3 heads (ii) the probability of getting 1 head and 2 tails (iii) the probability of getting at least 2 heads (iv) the probability of getting at most 1 tail (v) the probability of getting exactly 1 head 2. a) Two children were born from a married couple. Find the probability of having at least one son by drawing a tree diagram. b) Two children were given birth by a couple at the interval of five years. Illustrate the probabilities of son or daughter by drawing a tree diagram and calculate the probability of both are son. c) Two children were born in a family. By drawing a tree diagram, find the probability of having no daughter. d) Three children were born in a family. By drawing a tree diagram, find the probability of having at least a girl. e) Three children were born in a family. By drawing a tree diagram, find the following probabilities. (i) at least two daughters (ii) all of them are boys (iii) at least a boy. 3. a) A bag contains 3 red and 5 white balls. A ball is drawn at random and replaced, then another ball is drawn. Draw a probability tree diagram and find the probability that both of them are not of the same colour. b) A bag contains 6 black and 4 white marbles. A marble is drawn at random from the bag and it is replaced to draw another marble. By using a tree diagram, find the probability of getting (i) both of them of the same colour. (ii) Both of them of the different colours. c) Two cards are drawn from a well-shuffled deck of 52 cards. Find the probability that both cards are king (the first card is replaced) by showing a tree diagram.

Vedanta Excel in Mathematics - Book 10 320 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability 4. In the following tree diagram B1 and W1 denotes black or white ball drawn in the first trial and B2 and W2 denotes black or white ball drawn in the second trial. (i) How many balls are there altogether? How many of them are black and white? (ii) What do you mean by P(B2 /B1 ), P (W2 /B1 ), P (B2 /W1 ) and P (W2 /W1 )? (iii) Name the outcomes of the events A, B, C, and D and write the sample space of this experiment. (iv) Are the probabilities of the events dependent or independent? (v) Find the probabilities of the events A, B, C, and D. 5. a) 8 black and 5 white pens of the same size are kept in a bag. Two pens are drawn one after the other without replacement. Show the probabilities of all possible outcomes in a tree diagram. b) A bag contains 3 red and 4 blue balls of the same size. Two balls are drawn one after another without replacement. Show all the probabilities in a tree diagram. 6. a) A bag contains 7 black and 5 white balls. Two balls are drawn randomly one by one without replacing the first one. By drawing a tree diagram find the probability that (i) the first is black and the second is white. (ii) both of them are not black. (iii) both of them are of the different colour. (iv) both of them are of the same colour. b) There are 3 red, 5 blue and 2 yellow identical pencils in a box. Two pencils are drawn at random and without replacing the first one. By drawing a tree diagram find the probabilities of following events. (i) both of them are of blue colour. (ii) both of them are not red colour. (iii) at least one is yellow. (iv) none of them are blue and yellow. (v) all of them are of the same colour. 7. a) A card is drawn from a packet of numbered cards numbered from 1 to 20. If the second card is also drawn without replacing the first one, find the probability of each of the following events, (i) Both of them are of the multiples of 5. (ii) None of them are of multiples of 3. (iii) Show these probabilities in a tree diagram. B W B A B C D W B W P(B1 ) = 5 9 P(W1 ) = 4 9 P(B2 /B1 ) = 1 2 P(W2 /B1 ) = 1 2 P(B2 /W1 ) = 5 8 P(W2 /W1 ) = 3 8

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 321 Vedanta Excel in Mathematics - Book 10 Probability b) From a well-shuffled pack of 52 cards two cards are drawn at random without replacing the first one. (i) Find the probability that both of them are king. (ii) Find the probability that both of them are faced cards. (iii) Find the probability that both of them are black none-faced cards. (iv) Find the probability that the first is ace and the second is queen of heart. c) From a packet of cards, three cards are drawn in succession at random without replacement. Find the probability that the cards so drawn are no face cards. Objective Questions Tick the correct alternatives. 1. The probability scale of any event E is (A) 0 < P (E) < 1 (B) 1 < P (E) < 0 (C) 0 ≤ P (E) ≤ 1 (D) 1 ≤ P (E) ≤ 0 2. For any event E, P ( E ) is equal to (A) - P (E) (B) 1 – P (E) (C) P (E) – 1 (D) P (E) 3. If A and B are mutually exclusive events then P(A or B) is ... (A) P(A) + P(B) (B) P(A) – P(B) (c) P(A) × P(B) (D) P(A + B) 4. Three unbiased coins are tossed simultaneously. What is the probability of getting at most two heads? (A) 1 8 (B) 3 8 (C) 7 8 (D) 1 2 5. A box contains the lottery tickets numbered from 3 to 32. If a ticket is drawn at random, what is the probability of the ticket bearing square or cube number? (A) 1/5 (B) 1/10 (C) 1/15 (D) 2/15 6. A dice is rolled once. What is the probability that the digit turned off is a prime number or composite number? (A) 1 3 (B) 1 2 (C) 1 6 (D) 5 6 7. An ace of diamond is lost from a deck of 52 playing cards and a card is drawn at random. What is the probability of getting black faced card or ace? (A) 10 51 (B) 3 17 (C) 4 13 (D) 16 51 8. If A and B are two mutually exclusive events then which of the following relations is NOT correct? (A) Probability of solving the problem = P (A) + P (B) – P (A∩B) (B) Probability of solving the problem by both= P (A) × P (B)] (C) Probability of solving the problem by at least one of them = 1 – [P ( A ) × P (B )] (d) Probability of solving the problem by at least one of them = 1 – [P (A) × P (B)] 9. Two children were born in the interval of five years. What is the probability of having the children of alternate sexes? (A) 0.25 (B) 0.5 (C) 0.75 (D) 0.2 10. The probability of solving a mathematical problem by two students A and B are 0.2 and 0.25 respectively, then the probability of solving the problem is (A) 0.2 (B) 0.25 (C) 0.4 (D) 0.45

Vedanta Excel in Mathematics - Book 10 322 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Probability Assessment –VI 1. The table given below shows the marks obtained by the students of class 10 of a school in a terminal examination. Marks obtained 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Number of students 4 8 11 14 10 3 Based on this data, answer the following questions. (a) Construct the cumulative frequency table. (b) Find the class in which the median lies. (c) Compute the median mark. (d) If a student Sunita says that the median and mode lie on the same class, is she correct? Give reason. 2. The monthly household expenditure (in Rs thousands) of a community is shown the table given below. Expenditure (in 1000) 18 - 24 24 - 30 30 - 36 36 - 42 42 - 48 Number of family 4 8 11 14 10 Answer the following questions. (a) Write the class in which the mode lies. (b) Write the formula to find the mode of a continuous data. (c) Find the monthly expenditure of maximum number of families. (d) Find the maximum expenditure of minimum 25% of the families. 3. From a well-shuffled pack of 52 playing cards, a card is drawn at random. (a) What is the probability that the card so drawn is a king? (b) Find probability of getting an ace or a faced card. (c) Find the probability of getting a queen with heart or a jack with diamond. (d) If Hari says that the events of getting a faced card and a black card are mutually exclusive events, is he right? Give reason. 4. A bag contains 6 black and 7 blue identical marbles. Two marbles are drawn from the bag one after another. (a) If the first marble is not replaced in the bag and another marble is drawn, find the probability of getting a blue marble. (b) If the first marble is replaced in the bag and another marble is drawn, find the probability of getting a blue marble. (c) Are these events independent events? Give reason. 5. A coin is flipped and a die is rolled together. (a) Write the sample space. (b) Find the probability that the head occurring on coin and an even numbered face occurring on die. (c) Show the events of the above experiment on a tree diagram. (d) Are the events of coin and die independent events? Give reason.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 323 Vedanta Excel in Mathematics - Book 10 Unit 16 Trigonometry 16.1 Trigonometric ratios of an acute angle of a right-angled triangle - review In the given right-angled triangle ABC, right angled at B. ∠BAC = q is an acute angled and taken as angle of reference. Here, side AC is opposite to the right angled and it is hypotenuse (h). Side BC is opposite to the angle of reference (q) and it is perpendicular (p). Side BC is adjacent to the angle of reference (q) and it is base (b). Now, we can make six different ratios of the various pairs of the sides of a right-angled triangle. These ratios are given below: (i) perpendicular hypotenuse = p h which is called sine of θ or sinθ . (ii) base hypotenuse = b h which is called cosine of θ or cosθ. (iii) perpendicular base = p b Which is called tangent of θ or tanθ. (iv) hypotenuse perpendicular = h p Which is called cosecant of θ or cosecθ. (v) hypotenuse base = h b which is called secant of θ or secθ. (vi) base perpendicular = b p which is called cotangent of θ or cotθ. Here, sinθ, cosθ, and tanθ are the main ratios and cosecθ, secθ, and cotθ are inverse ratios of the main ratios respectively. 16.2 Values of trigonometric ratios of some standard angles The angles like 0°, 30°, 45°, 60°, and 90° are commonly known as the standard angles. The values of trigonometric ratios of these standard angles can also be obtained geometrically without using the table. The table given below shows the value of trigonometric ratios of these standard angles perpendicular(p) A base (b) B C hypotenuse (h) θ www.geogebra.org/classroom/pdc3x9un Classroom code: PDC3 X9UN Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 324 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Trigonometry Trigonometric ratios Angles 0° 30° 45° 60° 90° sin 0 1 2 1 2 3 2 1 cos 1 3 2 1 2 1 2 0 tan 0 1 3 1 3 ∞ 16.3 Height and distance One of the important applications of trigonometry is to find the heights of objects like, tree, building, tower, etc. and distance between objects. In such cases, we have to solve right angled triangles. For example, in the figure, AB is the height of a tree, BC is the distance between the tree and the observer at C. Here, if θ and BC are known, we can calculate the height AB or if θ and AB are known, we can calculate the distance between the tree and the observer by using a trigonometric ratio. 16.4 Angle of elevation and angle of depression In the adjoining figure, an observer is looking at the top of a tower. Here, OA is called the line of sight and OB is the horizontal line. ∠BOA is the angle between the upward line of sight and the horizontal line. ∠BOA is called the angle of elevation. Thus, the angle made by upward line of sight with horizontal line is known as angle of elevation. In this figure, ∠BOA is the angle between the downward line of sight and the horizontal line. ∠BOA is called the angle of depression. C B Distance Height A θ Line of sight Horizontal Line of sight Angle of elevation Angle of depression A B O

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 325 Vedanta Excel in Mathematics - Book 10 Trigonometry Thus, the angle made by downward line of sight with horizontal line is known as angle of depression. In the adjoining figure, the horizontal lines AC and BO are parallel to each other. The angle of depression CAO and the angle of elevation BOA are alternate angles between parallel lines. ∴ ∠ CAO = ∠ BOA i.e., Angle of depression= Angle of elevation. The instrument which is used to measure the angle of elevation or the angle of depression is called Theodolite. Worked-out Examples Example 1: In the figure given alongside, AB = height of a boy, CD = height of a tree, BD = distance between the boy and the tree, ∠CAE = angle made by the line of sight with horizontal line. (i) What is the ∠CAE called? (ii) What are the lengths of AE and ED? (iii) Find the height of the tree. (iv) What change will be in ∠CAE when the boy moves towards the tree? Give reason. Solution: Here, AB = height of a boy = 5 ft, CD = height of a tree BD = distance between the boy and the tree = 50 ft, ∠CAE = angle made by the line of sight with horizontal line = 45o (i) ∠CAE is the angle of elevation of the top of the tree. (ii) AE = BD = 50 ft and ED = AB = 5 ft. (iii) In right angled triangle ACE; tan 45o = p b = CE AE or, 1 = CE 50 ft or, CE = 50 ft. Now, CD = CE + ED = 50 ft. + 5 ft. = 55 ft. So, the height of the tree is 55 ft. (iv) ∠CAE becomes larger when the boy moves towards the tree because the distance between the boy and the tree decreases. Angle of depression Angle of elevation O A B C A 45° B C D E 50 ft. 5 ft.

Vedanta Excel in Mathematics - Book 10 326 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Trigonometry Example 2: In the given figure, AB = height of a house, BC = distance between the house and a car, ∠DAC = angle made by the line of sight with horizontal line. (i) What is the ∠DAC called? (ii) What is the measure of ∠ACB? (iii) How far is the car from the basement of the house? (iv) If the car was moved a few meter back, what change will be in angle of depression of the car as observed from the roof of the house? Give reason. Solution: Here, AB = height of a house = 25 m, BC = distance between the house and a car, ∠DAC = angle made by the line of sight with horizontal line = 30o (i) ∠DAC is the angle of depression of the car when it is observed from the roof of the house. (ii) ∠ACB = ∠DAC = 30o [Being alternate angles within parallel lines] (iii) In right angled triangle ABC; tan 30o = p b = AB BC or, 1 3 = 25 BC or, BC = 25 3 m So, distance between the house and a car is 25 3 m. (iv) When the car was moved back, the angle of depression as observed from the roof of the house would decrease because the distance between the house and the car increases. Example 3: A man observes the top of a tree 30 3 m high and finds the elevation of 30°. Find the distance between the man and the foot of the tree. Solutions: Let AB be the height of the tree and BC be the distance between the foot of the tree and the man. Here, AB = 30 3 m Angle of elevation, ∠ BCA = 30°. In rt. ∠ ed D ABC, tan 30° = p b = AB BC = 30 3 BC or, 1 3 = 30 3 BC or, BC = 30 3 × 3 = 30 × 3 = 90 m Hence, the required distance is 90 m. C A B 30 3 m 30° A C B 30° D

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 327 Vedanta Excel in Mathematics - Book 10 Trigonometry Example 4: The diameter of a circular pond is 90 m and a pole is fixed at the centre of the pond. The height of the pole is 48 m and the pond is 3 m deep. Find the angle of elevation of the top of the pole from a point in the circumference of the pond. Solutions: Let BC be the height of the pole above the surface of water. AB be the radius of the circular pond. Here, diameter of the pond = 90 m Then, radius of the pond = 90 2 m = 45 m Height of the pond above the surface of water, BC = 48 m – 3 m = 45 m In right-angled triangle ABC, tan q = p b = BC AB = 45 45 = 1 or, tan q = tan45° ∴ q = 45° Hence, the required angle of elevation is 45°. Example 5: A girl, 1.2 m tall, is flying a kite. When the length of the string of the kite is 180 m, it makes an angle of 30° with the horizontal line. At what height is the kite from the ground? Solution: Let CK be the height of the kite from the ground, AB be the height of the girl and AK be the length of the string. Here, AB = 1.2 m AK = 180 m Angle of elevation DAK = 30°. In rectangle ABCD, AB = DC = 1.2 m. In rt. ∠ ed ∆ ADK, sin30° = p h = KD AK or, 1 2 = KD 180 or, KD = 90 m Now, CK = KD + DC = 90 m + 1.2 m = 91.2 m So, the kite is at a height of 91.2 m from the ground. A q B 3 m 45 m 45 m C 30° 180 m 1.2 m A K D C B

Vedanta Excel in Mathematics - Book 10 328 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Trigonometry Example 6: A woman is 1.6 m tall and the length of her shadow in the sun is 1.6 3 m. Find the altitude of the sun. Solution: Let AB be the height of the woman and BC be the length of her shadow. Let θ be the altitude of the sun. Here, AB = 1.6 m BC = 1.6 3 m In rt. ∠ ed ∆ ABC, tanθ = p b = AB BC or, tanθ = 1.6 1.6 3 or, tanθ = 1 3 or, tanθ = tan30° ∴ θ = 30° So, the required altitude of the sum is 30°. Example 7: A boy 3 m tall is 72 m away from a tower 25 3 m high. Find the angle of elevation of the top of the tower from his eyes. Solutions: Let AB be the height of the boy and CD be the height of the tower. Let BD be the distance between the boy and the tower. Here, BD = AE = 72 m AB = 3 m and CD = 25 3 ∴ CE = CD – ED = 25 3 – AB = 25 3 – 3 = 24 3 m Now, in rt. ∠ ed D AEC, tan A = p b = CE AE or, tan A = 24 3 72 tan A = 3 3 = 3 3 × 3 = 1 3 A = tan–1 1 3 = 30° So, the required angle of elevation is 30°. A B 1.6 m C θ 1.6 3 m = ? A B 72 m C D E 25 3 m 3 m

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 329 Vedanta Excel in Mathematics - Book 10 Trigonometry Example 8: Two vertical poles are fixed 60 m apart. The angle of depression of the top of the first pole from the top of the second pole which is 150 m high is 30°. Find the height of the first pole. Solutions: Let AB be the height of the first (shorter) pole, CD be the height of the second (taller) pole and BD be the distance between them. Here, BD = AE = 60 m CD = 150 m The angle of depression = angle of elevation, ∠ CAE = 30°. Now, in rt. ∠ ed D CEA, tan 30° = p b = CE AE = CE 60 or, 1 3 = CE 60 or, CE = 60 3 = 60 3 × 3 3 = 20 3 = 20 × 1.732 = 34.64 m Again, in rectangle ABDE, AB = DE = CD – CE = 150 m – 34.64 m = 115.36 m So, the required height of the the first pole is 115.36 m Example 9: The top of a tree which is broken, by the wind makes an angle of 60° with the ground at a distance 3 3 m from the foot of the tree. Find the height of the tree before it was broken. Solutions: Let AC be the height of the tree before it was broken, BD be the broken part of the tree and CD be the distance between the foot of the tree and the point on the ground at which the top of the tree touched. Here, CD = 3 3 m The angle of elevation, ∠ CDB = 60° Now, in rt. ∠ed D BCD, tan 60° = p b = BC CD = BC 3 3 or, 3 = BC 3 3 or, BC = 9 m Also, cos 60° = b h = 3 3 BD A 30° 30° B 60 m C D E 150 m A B C D 60° 3 3 m

Vedanta Excel in Mathematics - Book 10 330 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Trigonometry or, 1 2 = 3 3 BD or, BD = 6 3 = 6 × 1.732 = 10.39 m Then, the height of the tree before broken = AC = AB + BC = BD + 9 m = 10.39 m + 9 m = 19.39 m So, the required height of the tree was 19.39 m. Example 10: Two men are on the opposite side of a tower of 30 m high. They observed the angle of elevation of the top of the tower and found to be 30° and 60° respectively. Find the distance between them. Solution: Let AB be the height of the tower and CD be the distance between the two men. Here, AB = 30 m Angles of elevation BCA = 30° and ∠ BDA = 60°. In rt. ∠ed ∆ ABC, tan30° = p b = AB BC or, 1 3 = 30 BC or, BC = 30 3 = 30 × 1.732 = 52.05 m Again, in rt. ∠ ed ∆ ABD, tan60° = p b = AB BD or, 3 = 30 BD or, BD = 30 3 = 30 3 × 3 3 = 30 3 3 = 10 3 = 17.32 m Now, CD = BC + BD = 52.05 m + 17.32 m = 69.37 m So, the required distance between the men is 69.37 m. EXERCISE 16.1 General section 1. a) To which trigonometric ratio is the ratio of perpendicular and the base associated in a right angled triangle? b) Which is the trigonometric ratio represented by the ratio of perpendicular and the hypotenuse in a right angled triangle? 2. a) Define angle of elevation. b) Write the definition of angle of depression. 30 m 30° 60° B A C D

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 331 Vedanta Excel in Mathematics - Book 10 Trigonometry 3. Find the unknown lengths of the sides of the following figures: a) b) c) d) e) f) g) h) 4. Find the unknown sizes of angles in the following figures: a) b c) d) C A B x 45 m 15 3 m C A B20 m x 40 m R P S Q x 50 m 25 2 m C D A E B2 m 32 m x 10 3 m Creative section 5. a) In the given figure, AB = height of a tower, C = a point of observation, AC = length of line of sight, BC = distance between the point of observation and the tower, ∠ACB = angle made by the line of sight with horizontal ground. (i) What is the ∠ACB called? (ii) Find the distance between the point of observation and the tower. (iii) What change will be in ∠ACB when the point of observation moves towards the tower? Give reason. (iv) What change will be in ∠ACB when the height of tower is decreased? Give reason. b) In the given figure, AB = height of a tree, A = the position of an eagle, C = position of a snake, BC = distance between the snake and the tree, ∠DAC = angle made by the line of sight with horizontal ground when the eagle looks down the snake. (i) What is the ∠DAC called? (ii) What is the measure of ∠ACB? C A 20 m B y m x m 45° B C A x m y m 60° 40 3 m R P Q y m 150 m 30° x m F E G 30° y m 80 3 m x m C D A E B 1.5m 121.5 m x m 60° A B E C D 2 m x m 60° 30° 60 3 m R P Q S T 45° 80 m x m 200 m x m A B C D 60° 10 3 m A C B 30° 30 m A D C B 30°

Vedanta Excel in Mathematics - Book 10 332 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Trigonometry (iii) How far is the snake from the foot of the tree? (iv) If the snake moves towards the tree, what change will be in the angle of depression of the snake? Give reason. 6. a) In the figure given alongside, AB = height of a girl, CD = height of a house and BD = distance between the girl and the house. (i) Write the name of angle of elevation of the top of the house. (ii) What are the lengths of AE and ED? (iii) Find the height of the house. (iv) What change will be in the angle of elevation of the roof of the house when the height of the house is increased? Give reason. b) In the figure given alongside, AB = height of a house, CD = height of a tower, and BD = distance between the house and the tower. (i) Write the name of angle of elevation of the top of the tower as observed from the roof of the house. (ii) What is the length of CE? (iii) Find the distance between the house and the tower. 7. a) The angle of elevation of the top of a tree observed from a point 60 m away from its foot is 45°. Find the height of the tree. b) The angle of depression at a point on the ground from the top of a building is 60°. If the point is 36 m far from the foot of the building, find the height of the building. c) From the top of a tower of height 120 m the angle of depression of an object lying in front of the tower is observed and found to be 60°. Find the distance between the object and the foot of the tower. d) A tower on the bank of a river is of 20 m high and the angle of elevation of the top of the tower from the opposite bank is 30°. Find the breadth of the river. e) When the sun's altitude is 45°, the length of the shadow of a vertical pole is 12.5 m. Find the height of the pole. f) Find the length of the shadow cost by a tree of height 15 m at a time of the sun's attitude to be 30°. g) A man was flying a kite. The string was 300 m long and an angle of 45° was formed with horizon. What was the height of the kite above the horizon? h) A ladder 7.5 m long is resting against a wall. If the ladder makes an angle of 60° with the ground, find (i) the height where the ladder is resting on the wall. (ii) The distance between the feet of the ladder and the wall. 8. a) A circular pond has a pole standing vertically at its centre. The top of the pole is 30 m above the water surface and the angle of elevation of it from a point on the circumference is 60°. Find the length of radius of the pond. b) The diameter of a circular pond is 120 m. The angle of elevation of the top of the pillar situated in the middle of the pond observed from the edge of the pond is found to be 60°. Find the height of the pillar above the surface of water. A 45° B C D E 30 ft. 3 ft. 30 m 60° 72 m A E B D C

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 333 Vedanta Excel in Mathematics - Book 10 Trigonometry c) The circumference of a circular meadow is 220 m and a pillar is situated at the centre of the meadow. If the angle of elevation of the top of the pillar from a point at the circumference of the meadow is found to be 30°, find the height of the pillar. 9. a) An observer finds the angle of elevation of the top of a pole to be 60°. If the height of the pole and the observer are 25.3 m and 1.3 m respectively, find the distance between the observer and the pole. b) A woman 1.6 m tall, observes the angle of elevation of the top of a house and it is found to be 60°. If the distance of the woman from the foot of the house is 100 m, find the height of the house. c) A man observes the top of a pole of 52 m height situated in front of him and finds the angle of elevation to be 30°. If the distance between the man and the pole is 86m, find the height of the man. d) An observer observed from the top of a house of height 8 m, the angled of elevation of the top of the temple is 60°. If the distance between the temple and the house is 15 3 m, find the height of the temple. e) From the top of a hill 51.6 m high, the angle of depression of the top of a tree of height 6.6 m situated in front of the hill, was observed and found to be 30°. Find the distance between the hill and the tree. 10. a) On a windy day, a girl of height 1.1 m was flying her kite. When the length of the string of the kite was 33 metres, it makes an angle of 30° with horizon. At what height was the kite above the ground? b) On the roof of a house 9 m high, a 1.5 m tall man was flying a kite and the kite was at a height of 58 m above the ground. If the string of the kite makes an angle of 30° with the horizon, calculate the length of the string of the kite. 11. a) From the top of a building 20 m high, a 1.7 m tall man observes the elevation of the top of a tower and finds it 45°. If the distance between the building and the tower is 50 m, find the height of the tower. b) A man of height 1.8 m observes the angle of elevation of the top of a building of height 107 m from the roof of a house and it is found to be 30°. If the distance between the house and the building is 25 3 m, find the height of the house. 12. a) A man is 2m tall and the length of his shadow in the sun is 2 3 m. Find the altitude of the sun. b) A woman observes the top of a tower of 80 3 m height from a point on the ground 240 m far from the foot of the tower. Find the angle of elevation. c) A man 1.75 m tall is 50 m away from a tower 51.75 m high. Find the angle of elevation of the top of the tower from his eyes. 13. a) The upper part of a straight tree broken by the wind makes an angle of 45° with the plane surface at a point 9 m from the foot of the tree. Find the height of the tree before it was broken. b) A tree of 14 m height is broken by the wind so that its top touches the ground and makes an angle of 60° with the ground. Find the length of the broken part of the tree. c) The top of a tree broken by the wind makes an angle of 30° with the ground at a distance of 4 3 m from the foot of the tree. Determine the height of the tree before it was broken. 14. a) Two men are on the opposite side of a tower 40 m high on the same horizontal line. They observed the angle of elevation of the top of the tower and found to be 45° and 30°. Find the distance between them. b) From a lighthouse 100 3 m high the angles of depression of two ships on its opposite sides are observed to be 30° and 60° respectively. Find the distance between the ships if the bases of the ships and the lighthouse are on the same horizontal line.

Vedanta Excel in Mathematics - Book 10 334 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Trigonometry Project Work and Activity Section 15. a) Make the groups of your friends. Find the heights of the basketball poles, roof of your school buildings, height of trees, etc. inside your school premises. b) An inclinometer or clinometer is an instrument used for measuring angles of slopes, elevation and depression. Use this instrument from your math lab to measure the height and distance of poles, trees, and so on. Objective Questions Tick the correct alternative. 1. The angle of elevation is made by (A) upward line of sight with horizontal line (B) upward line of sight with vertical line (C) downward line of sight with horizontal line (D) downward line of sight with vertical line 2. The angle of depression is made by (A) upward line of sight with horizontal line (B) upward line of sight with vertical line (c) downward line of sight with horizontal line (D) downward line of sight with vertical line 3. A tree of the height 18 3m is situated on the edge of a river. If the angle of elevation of the tree observed from the opposite edge of the river is found to be 60°, what is the breadth of the river? (A) 3m (B) 9 3m (C) 18 m (D) 18 ft 4. The angle of elevation of the top of the tree from a point on the ground, which is 24m away from the foot of the tree, is 30°. What is the height of the tree? (A) 8 3m (B) 24 3m (C) 24 m (D) 12m 5. A circus artist is climbing a 30m long rope, which is tightly stretched and tied to the top of a vertical pole from a point on the ground. If the angle made by the rope with ground is 600 , then what is the height of the pole? (A) 3m (B) 15 3m (C) 30 3m (D) 60 3m 6. What is the altitude of the sun when the length of shadow of a vertical pole is equal to its height? (A) 300 (B) 450 (C) 600 (D) none of these 7. A 5 ft tall person observed the top of a tower of 55 ft high from a point 50 ft away from the bottom of the tower on the horizontal level. Then, the angle of elevation of the tower is (A) 450 (B) 300 (C) 600 (D) 750 8. If a 3 ft tall child observes the angle of elevation of the top of a building 47 ft high which is in front of him and finds to be 450 , the distance between the child and the building is... (A) 44 ft (B) 54 ft (C) 44 3ft (D) 44 m 9. If the angle of depression of a the top of a temple as observed from the roof of a house 30 ft high is found to be 300 and the distance between the temple and the house is 10 3 ft, what is the height of the temple? (A) 10 ft (B) 20 ft (C) 30 ft (D) 60 ft 10. A tree 18m high is broken by the wind so that its top touches the ground. If the length of broken part of the tree is 12m, the angle made by the top of the tree with the ground is... (A) 450 (B) 300 (C) 600 (D) 750

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 335 Vedanta Excel in Mathematics - Book 10 Trigonometry Assessment –VII 1. A house and a tree are situated on a horizontal plain. The height of the house is 10 m and the height of the tree is 25 m. The angle of elevation of the top of a tree from the roof of the house is 30o . Based on this information, answer the following questions. (a) Sketch the suitable figure. (b) Find the distance between the house and the tree. (c) If the height of the tree increases slowly, does the angle of elevation increase? Give reason. 2. The height of the newly constructed school building is 35 m. The height of a student standing on the play ground of the school is 1 m. The distance between the school building and the student is 34 m. (a) Show the above information in a diagram. (b) Find the angle of elevation of the top of the building as observed by the student. (c) Does the angle of elevation increase or decrease as the student moves towards the school building? Give reason. 3. A house is situated in front of a tower 80 m high. The distance between the house and the tower is 60 3 m and a man at the top of tower observes the roof of the house and finds the angle of 30o . (a) Name the angle made by the downwards line of sight with the horizon when the man observes the roof of the house. (b) Write the relation between the angles made by the line of sight with horizon when the top of the tower is observed from the roof of the house and angle made by the line of sight with horizon when the roof of the house is observed from the top of the tower. (c) Find the height of the house. 4. A pine tree is broken by the wind so that its top touches the ground and makes an angle of 30o with the ground. The broken part of the tree is 18 m more than the remaining part of the tree. (a) Represent the given information in the diagram. (b) Find the length of broken part of the tree. (c) Find the height of the tree before it is broken. (d) Find the distance on the ground from the foot of the tree at which the top of the tree touches the ground.

Vedanta Excel in Mathematics - Book 10 336 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revision and Practice Time Revision and Practice Time Set 1. In a survey of 2,000 tourists who visited Nepal, it was found that 250 visited neither Pokhara nor Janakpur, 1,125 visited Pokhara and 750 visited Janakpur. a) Write the above information using set notations. b) How many tourists visited either Pokhara or Janakpur? c) By using Venn-diagram, find the number of tourists who visited Pokhara only. d) How many tourists visited at most one of these places? 2. In a class of 75 students, 30 students like cricket but not badminton, 20 like badminton but not cricket and 10 students like both games. a) How many students like cricket? b) How many students like at least one of these games? c) How many students don’t like both games? Represent the data in a Venn-diagram. d) What is the ratio of number of students who like badminton to the cricket? 3. In an examination, 40 % of students passed in Mathematics only, 30 % passed in Science only and 10 % of the students were failed in both the subjects. a) What percent of the students passed in either of these subjects? b) What percent of the students passed in only one subjects? c) Find the percent of the students who passed in both subjects by drawing in a Venndiagram. d) What percent of the students passed in at most one of these subjects? 4. In a survey of 500 farmers of Pakhribas, Dhankuta, it was found that 75 % were farming crops, 20 % were farming crops as well as vegetables and 10 % of them were not involved in farming. a) How many farmers were involved in crops farming? b) How many farmers were involving in at least one these farming? c) By the help of a Venn-diagram, find the number of farmers who were involving in vegetable farming? d) How many of them were not involving in farming vegetables? 5. A survey was conducted among 200 Nepali people about their tours in foreign countries. It was found that 30% preferred to visit Thailand, 20% of them preferred to visit India and 60% preferred to visit other countries. a) How many people preferred at least one of these two countries? b) How many people preferred Thailand only? c) Illustrate the information in a Venn-diagram and find the number of people who preferred only one of these two countries. d) Find the ratio of number of people who preferred Thailand only to India only.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 337 Vedanta Excel in Mathematics - Book 10 Revision and Practice Time 6. In a group of 80 music lovers, 40 people liked modern songs and 30 % liked modern songs but not folk songs. a) How many people liked modern songs only? b) How many people liked both types of songs? c) If 30 people liked folk songs only, how many people did not like both types of songs? d) If each people liked at least one type of the songs, how many people would like folk songs only? 7. In an examination, 80 % examinees passed in English, 60 % in Mathematics, 50 % passed in both the subjects, and 4 examinees failed in both subjects. a) Draw a Venn-diagram to represent the above information. b) Find the number of number of examinees participated in the survey. c) Find the number of examinees who passed only one subject. d) Find the number of students who failed Mathematics. 8. In a survey of a community, it was found that 60% of people liked folk songs, 50% liked modern songs, and 10% of people did not like both types of songs. a) Illustrate the above information in a Venn-diagram. b) What percentage of people liked either folk songs or modern songs? c) If 150 people liked both types of songs, how many people were surveyed? d) How many people liked modern songs only? 9. In a group of people, 110 people like nuclear family but not joint family and 40 like joint family but not nuclear family. a) How many people like only one type of family? b) If the number of people who like nuclear family is twice the number of people who like joint family, find the number of people who like both types of family? c) Show the above information in a Venn-diagram. 10. Out of 140 students, 50 passed in English and 20 passed in both Nepali and English. The number of students who passed in Nepali is twice the number of students who passed in English. Using a Venn-diagram, find the number of students who passed in Nepali only and who didn’t pass in both subjects. 11. In a survey, 100 people liked winter but not summer and 30 liked summer but not winter. If the number of people who liked summer is one-third of the number of people who liked winter, then by using Venn diagram, find the number of people who liked: a) both seasons b) summer season (c) winter season d) only one season 12. Among 300 students of a school, 32 students took part in quiz contest, and 20 took part in debate competition. Also, the number of students who took part only in debate competition is half of the number of students who took part only in quiz contest. a) Find the number of students who took part in both the programs.

Vedanta Excel in Mathematics - Book 10 338 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revision and Practice Time b) Find the number of students who did not take part in both competitions. c) Draw a Venn-diagram to represent the above information. 13. In a survey of people, 120 drink tea and 90 do not drink tea. 150 drink coffee and 75 drink coffee but not tea. a) How many people drink tea? b) How many people are participated in the survey? c) How many people do not drink both drinks? Find it by using a Venn-diagram. d) How many people only one the drinks? 14. In a group of 95 students, the ratio of students who like Mathematics and Science is 4 : 5. If 10 of them like both the subjects and 15 of them like none of the subjects, then by drawing a Venn-diagram, find the number of students who like: a) only Mathematics b) only Science c) only one subject. d) at most one subject. 15. 40 students in a class like Mathematics or Science or both. Out of them, 14 like both the subjects. The ratio of number of students who like Mathematics to those who like Science is 4: 5. How many students like: a) Mathematics? b) only Science? c) only one subject d) Draw a Venn-diagram to show the above information. 16. In an examination, 60 examinees passed at least one subject Nepali or Social studies. The ratio of the number of examinees who passed only Nepali and only Social studies is 3 : 2 and 15 passed both the subjects. a) How many students passed in Nepali? b) How many students passed in Social studies? c) How many students passed only one subject? d) Represent the above information in a Venn-diagram. 17. In a class, the ratio of the number of students who passed Maths but not Science and those who passed Science but not Maths is 3: 5. Also, the ratio of the number who passed both the subjects and those who failed both the subjects is 2: 1. If 80 students passed only one subject, and 100 students passed at least one subject, find a) the total number of students b) Draw a Venn-diagram to show all results. 18. In the group of some students, it was found that the ratio of the number of football-lover and basketball-lover is 9: 10. 20 % of them liked both games, 25 % liked only football and 45 did not like both the games. a) Find the total number of students in the group. b) Illustrate the above information in a Venn-diagram. (Hint: n(F) = 9x, n(B) = 10x, 9x = 45 %, x = 5 %, 19. During the lockdown period by COVID-19, most of the schools of a municipality used zoom or google meet to run their regular classes virtually. Among 60 schools, 5 schools

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 339 Vedanta Excel in Mathematics - Book 10 Revision and Practice Time used google meet but not zoom. If the number of schools that used zoom only was twice the number of schools that used both the platforms and thrice the number of schools that used none of these platforms, find the number of schools that used both the platforms by using Venn-diagram. 20. Of the 84 farmers who attended a meeting at a Ruler Municipality, 35 volunteered to supervise the use of harmful pesticides in vegetable farming, and 11 volunteered both in supervising and in counseling not to use harmful pesticides. If the number of farmers who volunteered for counseling was 1.5 times the number of farmers who neither volunteered in supervision nor in counseling, how many of the farmers volunteered for counseling? 21. Out of 200 students of a college, it is found that 60 students are in Mathematics class and 90 students are in Physics class. Find the number of students who are neither in Mathematics class nor in Physics class in the following cases. a) When two classes run at the same hour b) When two classes run at the different hours and 30 students enroll in both the courses. 22. In a class of 60 students, 40 students play cricket and 30 students play volleyball. a) Find the greatest number of students who play either cricket or volleyball. b) Find the least number of students who play both the games. 23. Of 300 students of a school, it was found that 20 students use the exercise books of neither brand A nor brand B. Also, the number of students who use both the brands is twice the number of students who use brand A only and 30 fewer than the number of students who use brand B only. a) Find the number of students who use the exercise books of both the brands. b) Find the number of students who use the exercise books of only one of these brands. c) Show the result in a Venn-diagram. 24. In a survey of 60 people, it was found that they love playing either cricket or football or both or none. If the ratio of the number people who love playing only one to both to none of these games is 3:2:1 and the ratio of the number of people who love playing cricket only to football only is 3:2, find: a) the number of people who love playing cricket. b) The number of people who love playing football. c) Represent the above data in a Venn-diagram. 25. A survey was conducted in a group of people regarding the use of mask and sanitizer for prevention from the flu. Half of people said that they use only mask and 10 don’t use mask at all. Also, 40% of the people said that they use sanitizer and 6 use none of them. a) Show the above information in a Venn-diagram. b) How many people were surveyed? c) How many people use both mask and sanitizer? d) How many people use the mask?

Vedanta Excel in Mathematics - Book 10 340 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revision and Practice Time 26. 100 employees in an office were asked about their preference for tea and coffee. It was observed that for every 3 people who preferred tea, there were 2 people who preferred coffee and there was a person who preferred both the drinks. The number of people who drink neither of them is same as those who drink both. a) How many people preferred both the drinks? b) How many people preferred only one drink? 27. In an examination, 40 % of the examinees passed in Maths, 45 % passed in Science and 55 % passed in Health. If 10 % of them passed in Maths and Science, 20 % in Science and Health and 15 % in Health and Maths and every student passed at least one subject. a) What percent of examinees passed in all the three subjects? b) Represent the above information in a Venn-diagram. c) What percent of examinees passed in only one subject? d) What percent of examinees passed in only two subjects? 28. In a class of 175 students, 100 are studying Maths, 70 are studying physics, 46 studying chemistry, 30 studying Maths and Physics, 28 studying maths and chemistry, 23 studying physics and chemistry and 18 studying all the three subjects. Find by using a Venn-diagram, a) how many students are studying at least one these subjects? b) How many students are studying only one subject? c) How many students are studying only two subjects? d) How many students are studying none of these three subjects? 29. In an examination out of 270 examinees, 85 succeeded in English, 110 succeeded in Japanese, 15 in English and Japanese, 25 in Japanese and Korean and 20 in Korean and English. Likewise, 10 examinees succeeded in all three languages and 20 in none of these languages. a) Draw a Venn- diagram to show the result. b) How many examinees were succeeded in at most one of these languages? c) How many examinees were succeeded in at most two of these languages? d) How many examinees were succeeded in at least two of these languages? Compound interest 1. Ganesh deposits Rs. 8,00,000 in a bank at the rate of 10% per annum for 2 years. a) Calculate his interest if the bank pays simple interest. b) Calculate his interest if the bank pays yearly compound interest. c) Calculate the interest if the bank pays half-yearly compound interest. d) By what percent is the yearly compound interest more than the simple interest of the same sum of money for the same interval of time at the same rate of interest? 2. Himani tooks a loan of Rs. 5,00,000 on January 1, 2022 till June 30, 2023 from a bank at the rate of 10% p.a. a) How much interest should she pay if it will be compounding annually?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 341 Vedanta Excel in Mathematics - Book 10 Revision and Practice Time b) How much interest should she pay if it will be compounding semi-annually? c) How much interest should she pay if it will be compounding quarterly? d) By what percent is the interest compounded quarterly more than the interest compounded annually? 3. Bina borrowed a sum of money from a bank at the rate of 10% simple interest for 2 years and he lent it to Rabina at the same rate of annual compound interest for the same duration of time. Mina also, lent it to Sabina at the same rate of semi-annual compound interest for the same duration of time. Sabina paid Rs. 194481 to Mina and cleared her debt. a) Find the sum borrowed by Bina. b) Find the profit made by Bina. c) Fina the profit made by Rabina. d) Who made more profit and by how much? 4. Mahesh borrows a certain sum of money from a bank at 9% p.a. simple interest and invests the same sum to Rakesh at 10% p.a. compound interest compounded annually. Mahesh makes a profit of Rs 7,550 after 3 years. a) What is the sum that Mahesh borrowed? b) Calculate the interest that Mahesh has to pay. c) Calculate the interest that Rakesh has to pay. d) By what percent does Rakesh pay more interest than Mahesh? 5. The difference between the compound interest compounded half yearly and yearly on a sum of money for 2 years at the rate of 20% p.a. is Rs 289.20. a) Find the sum. b) Find the compound interest compounded annually. c) Find the compound interest compounded semi-annually. d) By what percent is the interest compounded annually is less than the interest compounded semi-annually? 6. Mr. Gurung lent altogether Rs 6,000 to Ram and Rahim for 2 years. Ram agreed to pay simple interest at 10% p.a. and Rahim agrees to pay compound interest at the rate of 8 % p.a. If Rahim paid Rs 50 more than Ram as the interest, find: a) How much did Mr. Gurung lend to Ram? b) How much did Mr. Gurung lend to Rahim? c) How much interest did Ram pay to Mr. Gurung? d) How much amount did Rahim pay to clear his debt? 7. The compound interest on a certain sum of money in 2 years and 4 years are Rs 4,200 and Rs 9,282 respectively. a) Find the rate of interest. b) Find the sum. c) Calculate the compound interest of the sum at the same rate for 3 years.

Vedanta Excel in Mathematics - Book 10 342 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revision and Practice Time 8. The compound interest of a certain sum of money compounded semi-annually in 1 year and 2 years are Rs 16,400 and Rs 34,481 respectively. a) Find the rate of interest. b) Find the sum. c) Calculate the compound interest of the sum at the same rate for 1.5 years. 9. The compound interest on a certain sum for 2 years at 10% p.a. is Rs 420. a) What is the principal? b) What would be the simple interest on the same sum at the same rate for the same time? c) By what percent is the simple interest less than the compound interest? 10. Bikash borrows Rs. 50,000 for 2 years at the rate of 20% compounded per annum. a) What will be the compound interest compounded annually? b) At what time will it fetch the same interest on the same sum at the same rate of simple interest? 11. Mr. Magar borrowed Rs 3,00,000 from a commercial bank at the rate of 10% p.a. compounded half yearly for 2 years. After one year, the bank changed its policy to pay the interest compounded quarterly at the same rate. a) How much interest should he have to pay in the first year? b) How much interest should he have to pay in the second year? c) How much less interest would he have to pay if the bank would take interest compounded annually for 2 years? 12. A person took a loan of Rs. 40,000 for 2 years at the rate of 10% annual compound interest. To reduce the interest and the loan partly, he/she paid Rs. 25,000 at the end of the first year. a) Write the formula to calculate the compound amount. b) How much should he/she has to pay at the end of second year to clear his/her debt? c) Find the total interest paid by his/her in two years. d) If he/she paid the loan only at the end of second year, how much less or more interest should have to be paid? 13. Mr. Mandal has a son Binesh of 14 years old and a daughter Dhaniya of 16 years. He divides Rs 66,300 between his son and daughter and deposits in their own accounts in a bank at 10% per annum compound interest. If both children receive equal amounts at the age of 18 years, a) What is the share of each of them? b) What amount will each receive at the age of 18 years? 14. Riya borrowed a certain sum of money from a bank and paid it back in 2 installments. The rate of the compound interest was 4% p.a. and she paid back Rs 6760 annually. a) What sum did she borrow? b) By what percent is the principal of the 1 year more than that of the 2 years?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 343 Vedanta Excel in Mathematics - Book 10 Revision and Practice Time 15. Sunil borrowed a certain sum from a bank at 5% per annum compound interest. He cleared the loan by paying Rs. 31,500 at the end of the first year and Rs. 22,050 at the end of the second year. a) Find the sum that he borrowed. b) How much more amount would he need if he cleared the loan only at the end of the second year? Population growth and compound depreciation 1. The present population of a town is 66,550. The annual population growth rate is 10% p.a. a) What was the population of the town 3 years ago? b) What will be the population of the town after 1 year? 2. A year ago, the population of a metropolitan was 1,10,000. The population of the place increase every year by 3%. a) What is the present population of the place? b) What will be the population of the municipality after 1 year? 3. The population of a village increases every year by 5%. At the end of two years, the total population of the village was 10,000 after 1,025 of them migrating to other places. Find the population of the village in the beginning. 4. In the beginning of 2078 B.S., the population of a village was 6,000 and the rate of growth is 4% p.a. a) If in the beginning of 2079 B.S., 260 people migrated there from different places, what was the population in the beginning of 2079 B.S.? b) What will be the population of the town in the beginning of 2080 B.S.? c) If in the beginning of 2079 B.S., 240 people migrated to other places, what will the population in the beginning of 2081 B.S.? 5. In the beginning of 2077 B.S. the population of a rural municipality was 15,625 and it was 17,576 at the end of 2079 B. S. a) What was the annual growth rate of population of the place? b) What is the population in the beginning of 2079 B.S.? c) If 424 people migrated there at the end of 2079 B.S. and then annual growth rate of population was 5%, what will be the population of the place in the beginning of 2082 B.S.? 6. Janak bought a new bike for Rs 4,00,000. He used it for 2 years and sold to his friend Pradip at 10% per year compound depreciation. a) How much did Pradip pay for the bike? b) If Pradip used it for 3 years and sold to a recondition house for Rs. 1,65,888; find the annual rate of compound depreciation? 7. At a certain rate of annual compound depreciation, the value of a computer decreases to Rs. 38,400 in 2 years and Rs. 30,720 in 3 years.

Vedanta Excel in Mathematics - Book 10 344 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revision and Practice Time a) Find the rate of depreciation. b) Find the original cost of the computer. c) In how many years, will its value become Rs. 24,576? 8. A businessman imports watches from Japan and sells in Nepal. He imports each watch for JPY 5,000 with custom duty charge. He makes 25% profit and sells each watch with 13% value added tax. (JPY 10 = NPR 9.50) (a) How much Nepali rupee does he pay for each watch? (b) If Mrs. Bajracharya buys a watch with him, how much should she pay for it? (c) If Mrs. Bajracharya uses the watch for 2 years and sells it at the rate of 20% p.a. compound depreciation, at what cost will she sell it? (d) If Mrs. Bajracharya sells it at the rate of 10% p.a. compound depreciation., how much more or less amount would she receive? Money exchange 1. The buying and selling rates of USD 1 in a bank are NPR 127.80 and NPR 128.10 respectively. a) Which rate is used by the bank when a person needs US dollar? b) How many USD can be exchanged for NPR 3,20,250 on the day? c) How many US dollars should it buy and sell to make a profit of NPR 9,900 on the day? d) Rubina needs USD 4,000 for the higher study in America and the bank charges 1% as commission, how much Nepali rupees does she require? 2. Mr. Sanjiv and Mrs. Suwani bought some EURO (€) for NPR 5,50,000 at the exchange rate of EURO (€) 1 = NPR 125 to visit a few European countries. Unfortunately, because of Visa problem, they cancelled the trip. Within a week Nepali rupee is devaluated by 3%. They again exchanged their EURO to Nepali rupee after a week. a) How many EURO was bought? b) What is the new exchange rate after devaluation in Nepali rupee? c) How much did they gain or lose? d) After devaluation in Nepali rupee, if the bank took 2.5% commission, how much rupees would he gain or lose? 3. Mr. Sherpa has a flight to Japan for the business purpose. He has some US dollars and he wants to exchange US $ 1,850 into Japanese Yen. The exchange rates fixed by NRB are 1 US dollar = NPR 124.75 and 10 Japanese Yen = NPR 9.25 a) How much Japanese Yen can he exchange with US $ 1,850? b) If he exchanges the dollar on the day when the Nepali rupee is revaluated by 5% relative to Dollar and Yen, how much Yen can he exchange with US $ 1,850? c) If the bank charges 1% commission after revaluation of Nepali rupee, how much Japanese Yen can he exchange with US $ 1,850? 4. Mr. Gurung bought 10 Tola gold in Hong Kong at KHD$ 4,000 per Tola at the rate of HKD$ 1 = NPR 16.50 and bought to Nepal.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 345 Vedanta Excel in Mathematics - Book 10 Revision and Practice Time a) What is the rate of Gold per Tola in rupees? b) Find the cost of gold in NPR if 25% custom duty was charged. c) Find the selling price of gold in NPR if 20% profit was made. d) Find the selling price of the gold with 13% VAT in Nepal. Mensuration 1. Given figure is a square based pyramid. The slant height of the pyramid is 13 cm and the side of the base 10 cm. a) Find its lateral surface area. b) Find its total surface area. c) Find its vertical height? d) Find the volume of the pyramid. 2. The solid given alongside is a pyramid with square base, where the length of the side is 16 cm and height is 6 cm. a) Find its volume. b) Find its slant height. c) Find the area of its triangular faces. d) By what percent is the total surface area more than its lateral surface area? 3. The given solid is a square-based pyramid of side 14 cm and the length of the edge of its triangular face is 25 cm. a) What is the length of its slant height? b) Find its total surface area. c) Find its vertical height. d) Find its volume. 4. The length of the base of a square-based pyramid is 18 cm and its volume is 1,296 cm3 . Find the total surface area of the pyramid. 5. The length of a side of the base of the pyramid having a square base is 12 cm and the total surface area of the pyramid is 384 cm2 , find the volume of the pyramid. 6. The total surface area of a square-based pyramid is 896 cm2 and its slant height is 25 cm. Find the volume of the pyramid. 7. The length of the side of a square-based pyramid is 12 cm. If the lateral surface area of the pyramid is 240 cm2 , find its total surface area and volume. 8. The height and length of side of a square based pyramid are in the ratio 2:3. If the area of the triangular surfaces of the pyramid is 1500 cm2 , find its volume. 13 cm O 10 cm 6 cm 16 cm 14 cm 25 cm

Vedanta Excel in Mathematics - Book 10 346 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revision and Practice Time b) In the adjoining combined solid, the uppermost part is the square-based pyramid of side 1.2 m and the lowermost part is the square-based cuboid. Calculate the total surface area and volume of the solid. d) The adjoining solid is the combination of a square-based pyramid and a square-based prism. If the total surface area of the solid is 960 cm2 , find the volume of the solid. e) If the volume of the combined solid made up of the squarebased pyramid and a prism is 228 cm3 , find the total surface area of the solid. 10. 40 tourists are coming from Switzerland to visit Mt. Everest. They planned to stay at Everest base camp for 4 days. For this purpose, they ordered some square-based pyramid tents in Nepal. A tent can hold 8 people and each person has 6 ft × 3 ft space on the ground with 48 cu. ft of air to breathe. a) Find the side length of base of each tent. b) Find the height of each tent. c) Find the surface area of required tents. d) Find the total cost of all tents at the rate of Rs.400 per sq. ft. 11. Find the total surface area and the volume of the following solids. a) b) c) 12. A gate has two cylindrical pillars with a hemispherical top in each pillar. The height of each pillar is 13 ft. and the height of cylindrical part in each pillar is 12.3 ft. find the cost of colouring the surface of the pillars at Rs 25 per sq. ft. 13. There are 10 cylindrical pillars which are surmounted by the hemispherical ends of same radius all around in the ground of a temple. The radius of each pillar is 9 inch and total height 15ft 9 inch. If 1ft 9 inch of each pillar is inside the ground, calculate the cost of painting the pillars at Rs 15 per sq. ft. (12 inches = 1 ft.) 1.2m 1.2m 1.3m 50cm 12cm 12cm 20cm 8cm 6cm 6cm 5cm 5cm 40cm 25cm 14cm 31 cm 14cm 15cm 7cm 24cm

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 347 Vedanta Excel in Mathematics - Book 10 Revision and Practice Time 14. A vertical iron pole consisting of a cylindrical portion 3.7 m high and of base diameter 12 cm is surmounted by a cone 8 cm high. If 20 cm of its lower part is inside the earth, find the total cost of painting its surfaces at 15 paisa per sq. cm. (Use S=3.14) 17. A circus tent of diameter 70m is composed up of cylindrical part of height 9 m and conical part of height 12m above it. a) How much canvas is used in building the tent? b) If the width of the canvas is 2m, calculate the total cost of canvas used at Rs 25 per meter. 18. An exhibition tent is in the form of a cylinder surmounted by a cone. The diameter of base is 48m, height of the tent is 19 m and the height of the cylindrical part is 12 m. a) If 10% extra canvas is used for folding and stitching, how much canvas is required to make the tent? b) If the tent is 2.2m wide, find the cost of the canvas at Rs 65 per meter. Sequence and series 1. There are 5 arithmetic means between 5 and 20. Find the common difference and the means. 2. In an A.P., there are some arithmetic means are inserted between 7 and – 8. If the third mean is – 2, find: a) common difference b) number of means c) the remaining means 3. 7 arithmetic means are inserted between a and b. If the second and last means are 39 and 9 respectively, find: a) the first term b) the last term c) the remaining means 4. Ritima decides to save an extra sum of money in each month for a few months and saves for 5 months between the first and the last months of saving. If she deposits Rs. 1,000 in the first month and Rs. 1,500 in the second mean month, find: a) How much money will she save in each month? b) How much money will she save in the last month of saving? 5. The first term of an A.P. having 20 terms is 10 and its 20th term is 45. a) Find the sum of the series. b) Find the common difference. c) Find the sum of first 10 terms of the series. 6. Given arithmetic series is 25 + 20 + 15 + 10 + ………………. a) Find the sum of the first 10 terms of the series. b) If the sum of the first n terms of the series is 60, find the possible values of n. c) Explain the double values of n. 7. Krishna wishes to buy a refrigerator. He can buy it by paying Rs. 50,000 cash or by giving it in 10 installments as Rs. 10,000 in the first month, Rs. 9,000 in the second month, Rs. 8,000 in the third month and so on. a) If he pays the money in the installment basis, how much total amount will he pay? b) How much extra amount that he has to pay than the cost?

Vedanta Excel in Mathematics - Book 10 348 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revision and Practice Time 8. A man saves Rs. 200 in the first month of his job. Thereafter, he saves Rs. 10 more in each succeeding months. a) How much total amount will he save till 6 months? b) How long will he take to save a total of Rs. 2,450? 9. A contract on construction of a bridge specifies a penalty for delay of completion beyond a certain date as follows Rs. 2,000 for the first day, Rs. 2,500 for the second day, Rs. 3,000 for the third day etc. The penalty for each succeeding day being Rs. 500 more than for the preceding day, how much money the contractor has to pay as penalty if he has delayed the work by 10 days? 10. A club consists of members whose ages are in an A.P., the common difference being 3 months. If the youngest member of the club is 7 years and the sum of the ages of all the members is 250 years, a) How many members are there in the club? b) What is the age of eldest member in the club? 11. There are 4 geometric means between 2 and 486. Find the common ratio and the means. In a G.P., there are some geometric means are inserted between 13 and 81. If the third mean is 9, find: a) the common ratio b) number of means c) the remaining means 12. In a G.P., some geometric means are inserted between 54 and 2. If the first mean is three times the last mean, find: (i) common ratio (ii) the numbers of means (iii) the means 13. Find the two numbers whose arithmetic mean is 20 and geometric mean is 16. 14. Given geometric series is – 16 + 32 – 64 + …………………………. a) The sum of the terms of the series is 336, how many terms are there? b) What is the last term of the series? 15. A man borrows Rs 76,500 without interest and repays the loan in 8 installments, each installment being double the preceding one. Find: a) the first installment b) the last installment Quadratic equations 1. Solve these equations by factorization method: a) x – 3 x + 3 + x + 3 x – 3 = 66 7 b) x + 3 x – 2 – 1 – x x = 41 4 c) x – 10 x – 11 + x – 12 x – 13 = 31 3 d) 1 x + 1 + 2 x + 2 = 4 x + 4 e) x – 1 2x + 1 + 2x +1 x – 1 = 2 f) 2 2x – 1 x + 3 – 3 x + 3 2x – 1 = 5 g) p px – 1 + q qx – 1 = p + q h) 1 a + b + x = 1 a + 1 b + 1 x 2. The present ages of father and son are 37 years and 8 years respectively. (a) What were their ages before x years? (b) How many years ago, the product of their ages was 96? (c) How many years later, the father will be twice as old as his son? 3. The present ages of mother and daughter are 40 years and 16 years respectively. (a) What will be their ages after x years? (b) How many years hence, the product of their ages will be 756? (c) How many years ago, the mother was thrice as old as her daughter?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 349 Vedanta Excel in Mathematics - Book 10 Revision and Practice Time 4. Mr. Thapa has two sons. The sum of the present ages of his sons is 15 years and the product of his sons’ ages is 54. (a) Make the equations according to the given conditions. (b) Find their present ages. (c) What will the sum of the ages of his two sons be when the younger son becomes as old as the elder son at present? 5. The present age of Susmita is equal to the square of the age of son daughter Ayush after one year. If the age of son in 10 years hence will be 1 year less than the age of his mother in 10 years ago, answer the following questions. (a) Make the equations as the the given conditions. (b) Find their present ages. (c) If Susmita’s husband Kedar is 5 years older than her, find the difference between ages of the Kedar and his son Ayush? 6. The product of present ages of mother and son is 400. If they both live on till the son becomes as old as mother at present, the sum of their ages will be 110, answer the following questions. (a) Make the equations for the given conditions. (b) Find their present ages. (c) How many years ago was the age of the mother equal to the age of the son after the same number of years? 7. The product of present ages of father and son is 1500. When the father’s age was equal to the present age of the son, the sum of their ages was 40, answer the following questions. (a) Find their present ages. (b) How many years ago was the age of the mother twice the age of the son after the same number of years? 8. The area of a rectangular ground 1200 square metres and its perimeter is 140 metres. (a) Find the length and breadth of the ground. (b) By what percentage should the length of the ground be decreased so that it reduces to a square? 9. The students studying in Class 10 of a school organized a picnic with a total budget of Rs 66,000. They decided to collect an equal amount for the picnic. But, 2 students could not attend the picnic and each of the participating students paid Rs 55 more. (a) How many students were participated in the picnic? (b) Calculate the amount paid by each of the participated students. 10. A bus travels a journey of 360 km at a uniform speed. If the speed of the bus had been 5 km/hr more, it would have been taken 1 hr less for the same journey. (a) Find the usual speed of the bus. (b) If the speed of the bus is reduced by 4 km per hour, how much more time will it take to travel the same journey than the usual time? 11. In a two-digit number, the product of digits is 24. If 18 is subtracted from the number, the digits are reversed. Find the number.

Vedanta Excel in Mathematics - Book 10 350 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Revision and Practice Time 12. The product of digits in a two digit number is 21. The number formed by interchanging the place of digits is 36 more than the original number. Find the number. 13. In a positive number of two digits, the ten’s place digit exceeds the unit’s place digit by 5. If the product of digits is 36, find the number. 14. A number between 10 and 100 is equal to four times the sum of the digits. If the product of digits is 18, find the number. Simplification of Rational Expressions Simplify the following rational expressions. 1. 1 + 1 p 1 + 1 p – 1 1 – 2 p + 1 2. 1 – 2 m 1 – 1 m – 2 1 + 3 m – 3 3. 2ab a2 – b2 – a – b a + b + a + b a – b 4. 2x + 5 x + 1 – 3x – 2 x – 1 + x2 + 1 x2 – 1 5. x2 + y2 xy – x2 y(x + y) – y2 x(x + y) 6. x + 1 2x3 – 4x2 + x – 1 2x3 + 4x2 – 1 x2 – 4 7. ax2 + b 2x – 1 + ax2 – b 2x + 1 + 4ax3 1 – 4x2 8. x + 1 x – 1 + x – 1 x + 1 – 4x x2 + 1 9. 2 (x – 2) (x – 3) + 2 (x – 1) (3 – x) + 1 (1 – x) (2 – x) 10. 1 (x – 2) (x – 3) + 2 (x – 1) (3 – x) + 3 (1 – x) (2 – x) 11. x (x + 3) (x – 1) + x – 1 (x + 3) (2 – x) – x – 3 (2 – x) (x – 1) 12. 1 (a – b) (a – c) + 1 (b – a) (b – c) + 1 (c – a) (c – b) 13. 1 x2 – 5x + 6 + 1 x2 – 3x + 2 – 1 x2 – 4x + 3 14. 1 a2 – 5a + 6 – 2 a2 – 4a + 3 + 3 a2 – 3a + 2 15. 2 a2 – 2a – 24 + 2 a2 – 3a – 18 – 4 a2 + 7a + 12 16. x – 3 x2 – x – 6 + 2x – 1 2x2 + 5x – 3 – 2x + 5 x2 + 5x + 6 17. x – 3 x2 – 7x + 12 + 2(x – 1) x2 – 4x + 3 – 3(x – 4) x2 – 5x + 4 18. a2 + ab + b2 c2 – c(a + b) + ab + b2 + bc + c2 a2 – a(b + c) + bc + c2 + ca + a2 b2 – b(c + a) + ca 19. 4a2 – (3b – 4c) 2 (4c + 2a) 2 – 9b2 + 9b2 – (4c – 2a) 2 (2a + 3b) 2 – 16c2 + 16c2 – (2a – 3b) 2 (3b + 4c) 2 – 4a2 20. x + y (x + y)2 – z2 + y – z x2 – (y – z)2 – z + x (z + x)2 – y2