The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5,6} Number of possible outcomes of a dice = 6 (i) Prime numbers on a dice are 2, 3, and 5. Total prime numbers on a dice = 3 Probability of getting a prime number = 3/6 = 1/2 (ii) Numbers lying between 2 and 6 = 3, 4, 5 Total numbers lyingRead more
The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5,6}
Number of possible outcomes of a dice = 6
(i) Prime numbers on a dice are 2, 3, and 5.
Total prime numbers on a dice = 3
Probability of getting a prime number = 3/6 = 1/2
(ii) Numbers lying between 2 and 6 = 3, 4, 5
Total numbers lying between 2 and 6 = 3
Probability of getting a number lying between 2 and 6 = 3/6 = 1/2
(iii) Odd numbers on a dice = 1, 3, and 5
Total odd numbers on a dice = 3
Probability of getting an odd number = 3/6 = 1/2
Total number of cards in a well-shuffled deck = 52 (i) Total number of kings of red colour = 2 P(getting a number greater than 2) = (Number of favourable outcomes)/(Number of total possible outcomes) = 2/52 = 1/26 (ii) Total number of face cards = 12 p(getting a face card) = (Number of favourable ouRead more
Total number of cards in a well-shuffled deck = 52
(i) Total number of kings of red colour = 2
P(getting a number greater than 2) = (Number of favourable outcomes)/(Number of total possible outcomes) = 2/52 = 1/26
(ii) Total number of face cards = 12
p(getting a face card) = (Number of favourable outcomes)/(Number of total possible outcomes) = 12/52 = 3/13
(iii) Total number of red cards = 6
P(getting a red face card) = (Number of favourable outcomes)/(Number of total possible outcomes) = 6/52 = 3/26
(iv) Total number of jack of hearts = 1
P(getting a jack of hearts) = (Number of favourable outcomes)/(Number of total possible outcomes) = 1/52
(v) Total number of spade cards = 13
P(getting a Spade card) = (Number of favourable outcomes)/Number of total possible outcomes) = 13/52 = 1/4
(vi) Total number of queen of diamonds = 1
P(getting a queen of Diamonds) = (Number of favorable outcomes)/(Number of total possible outcomes) = 1/52
(i) Total number of cards = 5 Total number of queens = 1 P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 1/5 (ii) When the queen is drawn and put aside, the total number of remaining cards will be 4. (a) Total number of aces = 1 P(getting Ace) = (Number ofRead more
(i) Total number of cards = 5
Total number of queens = 1
P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 1/5
(ii) When the queen is drawn and put aside, the total number of remaining cards will be 4.
(a) Total number of aces = 1
P(getting Ace) = (Number of favourable outcomes) = 1/4
(b) As queen is already drawn, therefore, the number of queens will be 0.
P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 0/4 = 0
Total number of pens = 12 + 132 = 144 Total number of good pens = 132 P(getting a good pen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 132/144 = 11/12
Total number of pens = 12 + 132 = 144
Total number of good pens = 132
P(getting a good pen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 132/144 = 11/12
(i) Total number ofbulbs = 20 Total number of defective bulbs =4 P(getting a defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 4/20 = 1/5 (ii) Remaining total number of bulbs = 19 Remaining total number of non-defective bulbs = 16 - 1 = 15 P(getting a not defectRead more
(i) Total number ofbulbs = 20
Total number of defective bulbs =4
P(getting a defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 4/20 = 1/5
(ii) Remaining total number of bulbs = 19
Remaining total number of non-defective bulbs = 16 – 1 = 15
P(getting a not defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 15/19
Total number of discs = 90 (i) Total number of two-digit numbers between 1 and 90 81 P (getting a two-digit number) = 81/90 = 9/10 (ii) Perfect squares between 1 and 90 are 1,4,9, 16, 25, 36, 49, 64, and 81. Therefore, total number of perfect squares between 1 and 90 is 9. P (getting a perfect squarRead more
Total number of discs = 90
(i) Total number of two-digit numbers between 1 and 90 81
P (getting a two-digit number) = 81/90 = 9/10
(ii) Perfect squares between 1 and 90 are 1,4,9, 16, 25, 36, 49, 64, and 81.
Therefore, total number of perfect squares between 1 and 90 is 9.
P (getting a perfect square) = 9/90 = 1/10
(iii) Numbers that are between 1 and 90 and divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 55, 60, 65, 70, 75, 80, 85, and 90.
Therefore, total numbers divisible by 5 = 18
Probability of getting a number divisible by 5 = 18/90 = 1/5
(i) False, Let A = 30° and B = 60° Therefore, LHS = sin(A + B) = sin(30° + 60°) = sin 90° 1 and RHS = sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = (1+√3)/2 ≠ 1 Hence, sin (A + B) ≠ sin A + sin B (ii) True, As we know that sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2 and sin 90° = 1Read more
(i) False,
Let A = 30° and B = 60°
Therefore, LHS = sin(A + B) = sin(30° + 60°) = sin 90° 1 and
RHS = sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = (1+√3)/2 ≠ 1
Hence, sin (A + B) ≠ sin A + sin B
(ii) True,
As we know that sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2 and sin 90° = 1
Hence, for the increasing values of θ, sin θ is also increasing.
(iii) False,
As we know that cos 0° = 1, cos 30° = √3/2, cos 45° = 1√2, cos 60° = 1/2 and cos 90° = 0
Hence, for the increasing values of θ, cos θ is decreasing.
(iv) False,
∵ cos 30° = √3/2, but sin 30° = 1/2.
(v) True,
∵ tan 0° = 0, we know that cot 0° = 1/tan 0° = 1/0, which is not defined.
See the explanation video of the above solution here✌
A die is thrown once. Find the probability of getting
The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5,6} Number of possible outcomes of a dice = 6 (i) Prime numbers on a dice are 2, 3, and 5. Total prime numbers on a dice = 3 Probability of getting a prime number = 3/6 = 1/2 (ii) Numbers lying between 2 and 6 = 3, 4, 5 Total numbers lyingRead more
The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5,6}
See lessNumber of possible outcomes of a dice = 6
(i) Prime numbers on a dice are 2, 3, and 5.
Total prime numbers on a dice = 3
Probability of getting a prime number = 3/6 = 1/2
(ii) Numbers lying between 2 and 6 = 3, 4, 5
Total numbers lying between 2 and 6 = 3
Probability of getting a number lying between 2 and 6 = 3/6 = 1/2
(iii) Odd numbers on a dice = 1, 3, and 5
Total odd numbers on a dice = 3
Probability of getting an odd number = 3/6 = 1/2
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds.
Total number of cards in a well-shuffled deck = 52 (i) Total number of kings of red colour = 2 P(getting a number greater than 2) = (Number of favourable outcomes)/(Number of total possible outcomes) = 2/52 = 1/26 (ii) Total number of face cards = 12 p(getting a face card) = (Number of favourable ouRead more
Total number of cards in a well-shuffled deck = 52
See less(i) Total number of kings of red colour = 2
P(getting a number greater than 2) = (Number of favourable outcomes)/(Number of total possible outcomes) = 2/52 = 1/26
(ii) Total number of face cards = 12
p(getting a face card) = (Number of favourable outcomes)/(Number of total possible outcomes) = 12/52 = 3/13
(iii) Total number of red cards = 6
P(getting a red face card) = (Number of favourable outcomes)/(Number of total possible outcomes) = 6/52 = 3/26
(iv) Total number of jack of hearts = 1
P(getting a jack of hearts) = (Number of favourable outcomes)/(Number of total possible outcomes) = 1/52
(v) Total number of spade cards = 13
P(getting a Spade card) = (Number of favourable outcomes)/Number of total possible outcomes) = 13/52 = 1/4
(vi) Total number of queen of diamonds = 1
P(getting a queen of Diamonds) = (Number of favorable outcomes)/(Number of total possible outcomes) = 1/52
Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
(i) Total number of cards = 5 Total number of queens = 1 P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 1/5 (ii) When the queen is drawn and put aside, the total number of remaining cards will be 4. (a) Total number of aces = 1 P(getting Ace) = (Number ofRead more
(i) Total number of cards = 5
See lessTotal number of queens = 1
P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 1/5
(ii) When the queen is drawn and put aside, the total number of remaining cards will be 4.
(a) Total number of aces = 1
P(getting Ace) = (Number of favourable outcomes) = 1/4
(b) As queen is already drawn, therefore, the number of queens will be 0.
P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 0/4 = 0
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Total number of pens = 12 + 132 = 144 Total number of good pens = 132 P(getting a good pen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 132/144 = 11/12
Total number of pens = 12 + 132 = 144
See lessTotal number of good pens = 132
P(getting a good pen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 132/144 = 11/12
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
(i) Total number ofbulbs = 20 Total number of defective bulbs =4 P(getting a defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 4/20 = 1/5 (ii) Remaining total number of bulbs = 19 Remaining total number of non-defective bulbs = 16 - 1 = 15 P(getting a not defectRead more
(i) Total number ofbulbs = 20
See lessTotal number of defective bulbs =4
P(getting a defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 4/20 = 1/5
(ii) Remaining total number of bulbs = 19
Remaining total number of non-defective bulbs = 16 – 1 = 15
P(getting a not defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 15/19
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Total number of discs = 90 (i) Total number of two-digit numbers between 1 and 90 81 P (getting a two-digit number) = 81/90 = 9/10 (ii) Perfect squares between 1 and 90 are 1,4,9, 16, 25, 36, 49, 64, and 81. Therefore, total number of perfect squares between 1 and 90 is 9. P (getting a perfect squarRead more
Total number of discs = 90
See less(i) Total number of two-digit numbers between 1 and 90 81
P (getting a two-digit number) = 81/90 = 9/10
(ii) Perfect squares between 1 and 90 are 1,4,9, 16, 25, 36, 49, 64, and 81.
Therefore, total number of perfect squares between 1 and 90 is 9.
P (getting a perfect square) = 9/90 = 1/10
(iii) Numbers that are between 1 and 90 and divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 55, 60, 65, 70, 75, 80, 85, and 90.
Therefore, total numbers divisible by 5 = 18
Probability of getting a number divisible by 5 = 18/90 = 1/5
State whether the following are true or false. Justify your answer.
(i) False, Let A = 30° and B = 60° Therefore, LHS = sin(A + B) = sin(30° + 60°) = sin 90° 1 and RHS = sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = (1+√3)/2 ≠ 1 Hence, sin (A + B) ≠ sin A + sin B (ii) True, As we know that sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2 and sin 90° = 1Read more
(i) False,
Let A = 30° and B = 60°
Therefore, LHS = sin(A + B) = sin(30° + 60°) = sin 90° 1 and
RHS = sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = (1+√3)/2 ≠ 1
Hence, sin (A + B) ≠ sin A + sin B
(ii) True,
As we know that sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2 and sin 90° = 1
Hence, for the increasing values of θ, sin θ is also increasing.
(iii) False,
As we know that cos 0° = 1, cos 30° = √3/2, cos 45° = 1√2, cos 60° = 1/2 and cos 90° = 0
Hence, for the increasing values of θ, cos θ is decreasing.
(iv) False,
∵ cos 30° = √3/2, but sin 30° = 1/2.
(v) True,
∵ tan 0° = 0, we know that cot 0° = 1/tan 0° = 1/0, which is not defined.
See the explanation video of the above solution here✌
See lessEvaluate the following trigonometry questions?
(i) (sin 18°)/(cos 72°) = (sin 18°)/(cos 72°) = cos (90° - 18°))/(cos 72°) [∵ cos (90° - θ) = sin θ] = (cos 72°)/(cos 72°) = 1 (ii) (tan 26°)/(cot 64°) = (tan 26°)/(cot 64°) = (cot (90° - 26°))/(cot 64°) [∵ cot (90° - θ) = tan θ] = (cot 64°)/(cot 64°) = 1 (iii) cot 48° – sin 42° = cot 48° – sin 42°Read more
(i) (sin 18°)/(cos 72°)
= (sin 18°)/(cos 72°) = cos (90° – 18°))/(cos 72°) [∵ cos (90° – θ) = sin θ]
= (cos 72°)/(cos 72°) = 1
(ii) (tan 26°)/(cot 64°)
= (tan 26°)/(cot 64°) = (cot (90° – 26°))/(cot 64°) [∵ cot (90° – θ) = tan θ]
= (cot 64°)/(cot 64°) = 1
(iii) cot 48° – sin 42°
= cot 48° – sin 42° = cos 48° – cos (90° – 42°) [∵ cos (90° – θ) = sin θ]
(iv) cosec 31° – sec 59°
cosec 31° – sec 59° = cosec 31° – cosec (90° – 59°) [∵ cosec (90° – θ) = sec θ]
= cosec 31° – cosec 31° = 0
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See lessFind the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 6x² – 3 – 7x.
6x² - 3 - 7x = 6x² - 7x - 3 = 6x² - 9x + 2x -3 = 3x(2x - 3) + 1(2x -3) = (3x +1)(2x -3) The value of 6x² - 7x - 3 is zero if 3x + 1 = 0 or 2x -3 = 0. ⇒ x = -1/3 or x = 3/2, Therefore, the zeroes of 6x² - 7x - 3 are -1/3 and 3/2. Now, Sum of zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6 = -(7)/6 = -(CofficiRead more
6x² – 3 – 7x
= 6x² – 7x – 3
= 6x² – 9x + 2x -3
= 3x(2x – 3) + 1(2x -3)
= (3x +1)(2x -3)
The value of 6x² – 7x – 3 is zero if 3x + 1 = 0 or 2x -3 = 0.
⇒ x = -1/3 or x = 3/2,
Therefore, the zeroes of 6x² – 7x – 3 are -1/3 and 3/2.
Now, Sum of zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6 = -(7)/6 = -(Cofficient of x)/Cofficient of x²
Product of zeroes = (-1/3) x 3/2 = -1/2 = (-3)/6 = Constant term/Cofficient of x².
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