Maths

(a)In the figure (i) given below, O is the centre of the circle and ∠PBA = 42°. Calculate the value of ∠PQB (b) In the figure (ii) given below, AB is a diameter of the circle whose centre is O. Given that ∠ECD = ∠EDC = 32°, calculate (i) ∠CEF (ii) ∠COF.

Solution: In ∆APB, ∠APB = 90° (Angle in a semi-circle) But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle) ∠A + 90° + 42°= 180° ∠A + 132° = 180° ⇒ ∠A = 180° – 132° = 48° But ∠A = ∠PQB (Angles in the...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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