Mix Test 6

The probability of an impossible event is 0. (Because it cannot occur in any situation).

Total number of cards in the deck = 52.

Total number of king card in the deck = 4.

The probability that the drawn card is a king

\(=\frac{Total\,number\,of\,king\,cards}{Total\,number\,of\,cards\, in\,deck}\) \(=\frac{4}{52}=\frac{1}{13}\)

Total number of cards in the deck = 52.

Total number of cards which shown 8 in the deck is 4 in which 2 are black and 2 are red suited.

Therefore total number of red 8 cards in the deck = 2

Now, the probability that the draw card is a red eight

\(=\frac{Total\,number\,of\,red\,8\,cards}{Total\,number\,of\,cards\,in\,deck}\) \(=\frac{2}{52}=\frac{1}{26}\)

Total number of cards in the deck = 52.

The face cards are jack, queen and king cards, each of them have 4 quantities in the deck.

Therefore, total number of face cards in the deck = \(4\times 3=12\)

Hence, the probability that the drawn card is a face card

\(=\frac{Total\,number\,of\,face\,cards}{Total\,number\,of\,cards\,in\,deck}\) \(=\frac{12}{52}=\frac{3}{13}\)

Total number of cards in the deck = 52.

Total queen cards is 4 in the deck in which 2 are black suited and 2 are red suited.

Therefore, total number of queen cards of black suit = 2.

The probability of getting a queen of black suit

\(= \frac{total\,number\,of\,queen\,cards\,in\,black\,suit}{total\,number\,of\,cards\,in\,the\,deck}\) \(=\frac{2}{52}=\frac{1}{26}\)

Total number of cards in the deck = 52.

Total number of ace cards in the deck = 4

Total number of spade cards in the deck = 13.

Hence, total number of cards which are either an ace or a card of spades = 4 + 13 – 1 = 16.

(\(\because\) One ace card is spade card)

Therefore, the probability that the drawn card is a card of spades or an ace card

\(=\frac{Total\,number\,of\,cards\,which\,are\,either\,an\,ace\,or\,a\,card\,of\,spades}{Total\,number\,of\,cards\,in\,the\,deck}\)

\(=\frac{16}{52}=\frac{4}{13}\)

Given that \(a\cos\theta + b\sin\theta = m\) ...........(1)

And \(a\sin\theta -b\cos\theta =n\) ..........(2)

Now, squaring equation (2) and (3), we get

\(m^2 = (a\cos\theta +b\sin\theta)^2\) \(=a^2\cos^2\theta + b^2\sin^2\theta + 2ab \,\sin\theta\cos\theta\) ........(3)

And \(n^2 = (a\sin\theta - b\cos\theta)^2\) \(=a^2\sin^2\theta + b^2\cos^2\theta - 2ab\sin\theta \cos\theta\) ......(4)

Now, adding equations (3) & (4), we get

\(m^2 + n^2 \) \(=a^2 (\cos^2+\sin^2\theta)\) \(+b^2(\sin^2\theta + \cos^2\theta)\)\(+2ab\sin\theta \cos\theta\) \(-2ab\sin\theta \cos\theta\)

\(\Rightarrow m^2 + n^2 = a^2+b^2\)  \((\because \sin^2\theta + \cos^2\theta =1)\)

Let the bigger number is x and smaller number is y

Given that difference between both numbers is 26.

\(\therefore\) x – y = 26.  ............(1)

Given that one number is three times the other

Therefore, x = 3y. ..........(2)

Now, putting x = 3y in equation (1), we get 3y – y = 26

\(\Rightarrow 2y = 26 \) \(\Rightarrow y = \frac{26}{2}=13\)

Now, putting y = 13 in equation (2), we get x = 3 \(\times\) 13 = 39.

Hence, the numbers are 13 and 39.

Formula for general term is \(T_n = a+(n-1)d\)

(\(\because\) \(n^{th}\) term of AP is given by \(T_n = a+(n-1)d\))

Formula for general term is \(T_n = a+(n-1)d\) .........(1)

\(T_7 = a+6d\) (By putting n = 7 in equation (1))

Formula for general term is \(T_n = a+(n-1)d\) ........(1)

\(T_7=a+8d\) (By putting n = 9 in equation (1))

Let the first term of the AP is a and common difference of the AP is d.

Given that \(7^{th}\) term of AP = \(\frac{1}{9}\) and \(9^{th}\) term of AP = \(\frac{1}{7}\)

(\(\because\) \(n^{th}\) term of AP whose first term is a and common difference is d is given by a + (n –1)d)

\(\therefore\) a + 6d = \(\frac{1}{9}\)  ..............(1)

And a + 8d = \(\frac{1}{7}\) .............(2)

By subtracting equation (1) from equation (2),

we get (a + 8d) – (a + 6d) \(=\frac{1}{7}-\frac{1}{9}\)

\(\Rightarrow\) \(2d=\frac{9-7}{63}=\frac{2}{63}\)

\(\Rightarrow\) \(d=\frac{1}{63}\)

Let the first term of the AP is a and common difference of the AP is d.

Given that \(7^{th}\) term of AP = \(\frac{1}{9}\) and \(9^{th}\) term of AP = \(\frac{1}{7}\)

(\(\because\) \(n^{th}\) term of AP whose first term is a and common difference is d is given by a + (n –1)d)

\(\therefore\) a + 6d = \(\frac{1}{9}\)  .................(1)

And a + 8d = \(\frac{1}{7}\) ..............(2)

By subtracting equation (1) from equation (2),

we get (a + 8d) – (a + 6d)  \(=\frac{1}{7}-\frac{1}{9}\)

\(\Rightarrow 2d = \frac{9-7}{63}=\frac{2}{63}\)

\(\Rightarrow d=\frac{1}{63}\)

By putting d = \(\frac{1}{63}\) in equation (1), we get \(a=\frac{1}{9}-\frac{6}{63}\) \(=\frac{7-6}{63}=\frac{1}{63}\)

Therefore, the first term of \(AP = a = \frac{1}{63}\)

We know that quadratic equation \(\mathrm{ax^2 + bx + c=0}\) has equal roots if \(D = b^2 -4ac = 0\)

Therefore, given quadratic equation \(\mathrm {x^2 - 4kx + k = 0}\) has equal roots if D = 0.

\(\Rightarrow (-4k)^2-4\times 1\times k=0\) (\(\because\) In equation \(\mathrm{x^2 - 4kx + k = 0}\), a = 1, b = -4k & c = k)

\(\Rightarrow 16k^2 - 4k = 0\)

\(\Rightarrow 4k(4k - 1) = 0\)

\(\Rightarrow k = 0\,or\, 4k-1 = 0\)

\(\Rightarrow k = 0 \,or\, k = \frac{1}{4}\)

Hence, for k = 1 and k = \(\frac{1}{4},\) the quadric equation \(\mathrm{x^2 - 4kx + a=0}\) has equal roots.

Total number of observations are n, where n is an even number.

Then the median is calculated by the average of \(\frac{n}{2}th\) and \(\left(\frac{n}{2}+1\right)th\) observations.

Hence, median \(=\frac{\frac{n}{2}th\,observation + \left(\frac{n}{2}+1\right)th\,observation}{2}\)

Write 148 and 185 in the prime factorization, we get \(148 =2\times 2\times 37\) and \(185 = 5\times 37\)

Therefore, \(\frac{148}{185}=\frac{2\times 2\times 37}{5\times 37}=\frac{4}{5}\)

Hence, the simplest form of \(\frac{148}{185}\) is \(\frac{4}{5}\).