Mix Test 11

Let the number to be added be x.

Then the numbers 5, 9, 17, 27 will be 5 + x, 9 + x, 17 + x, 27 + x.

Given that new numbers are in proportional,

i.e., 5 + x : 9 + x :: 17 + x : 27 + x.

\(\Rightarrow\) \(\frac{5+\text{x}}{9+\text{x}}=\frac{17+\text{x}}{27+\text{x}}\)

\(\Rightarrow\) (5 + x)(27 + x) = (9 + x)(17 + x)

\(\Rightarrow\) \(\text{x}^2\) + 32x + 135 = \(\text{x}^2\) + 26x + 153

\(\Rightarrow\) 32x − 26x = 153 − 135

\(\Rightarrow\) 6x = 18

\(\Rightarrow\) \(\text{x}=\frac{18}{6}=3.\)

Given AP is 4, 9, 14……...,254.

First term of AP is a = 4.

Common difference of AP is d \(=a_2-a_1=9-4=5.\)

\(n^{th}\) term of AP is \(a_n\) = 254.

\(\because\) \(a_n\) = a + (n – 1) d

\(\therefore\) a + (n – 1) d = 254

\(\Rightarrow\) 4 + (n − 1)5 = 254 (a = 4 & d = 5 and \(a_n\) = 254)

\(\Rightarrow\) 5(n − 1) = 254 − 4 = 250

\(\Rightarrow\) n − 1 = \(\frac{250}{5}=50\)

\(\Rightarrow\) n = 50 + 1 = 51.

\(10^{th}\) term from the end of the AP is 51 – (10 – 1) = 51– 9 = \(42^{th}\) term of AP.

\(\therefore\) \(a_{42}\) = a + (42 – 1) d = 4 + 41 × 5 = 4 + 205 = 209. (\(\because\) a = 4, d = 5)

Hence, the \(10^{th}\) term from the end of the given AP is 209.

\(\because\) Jack, queen and king are face cards.

After removing red face cards total number of remaining cards in the pack

n(s) = 52 – 3 × 2 = 52 – 6 = 46.

The number of cards which neither a red card nor a queen

n(E) = Total number of cards – Total number of red cards – Total number of black queen cards = 46 – 20 – 2 = 24.

\(\therefore\) Probability that the drawn card is neither a red card nor a queen \(=\frac{n(E)}{n(S)}=\frac{24}{46}=\frac{12}{23}\)

We have tan θ = \(\frac{8}{15}\)

\(\therefore\) sec θ = \(\sqrt{1+\tan^2\theta} \) (\(\because\) 1 + \(\tan^2\theta = \sec^2\theta\))

\(=\sqrt{1+\frac{64}{225}}\)

\(=\sqrt{\frac{225+64}{225}}\)

\(=\sqrt{\frac{289}{225}}\)

\(=\frac{17}{15}.\)

\(\therefore\) cos θ = \(\frac{1}{\sec\,\theta}\) = \(\frac{15}{17}.\)

Now, \(\frac{\tan\theta}{\sec\theta}\) = \(\cfrac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}}\) = sinθ.

\(\therefore\) sin θ = \(\frac{\tan\,\theta}{\sec\,\theta}\) = \(\cfrac{\frac{8}{15}}{\frac{17}{15}}\) = \(\frac{8}{17}.\)

Now, \(\frac{(2+2\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(2-2\cos\theta)}\)

\(=\cfrac{\left(2+2\times \frac{8}{17}\right) \left(1- \frac{8}{17}\right)}{\left(1+ \frac{15}{17}\right)\left(2-2\times \frac{15}{17}\right)}\)

\(=\cfrac{\left(\frac{34+16}{17}\right) \left(\frac{17-8}{17}\right)}{\left(\frac{17+15}{17}\right)\left(\frac{34-30}{17}\right)}\)

\(=\frac{50\times 9}{32\times 4}\)

\(=\frac{25\times 9}{32\times 2}\)

\(=\frac{225}{64}.\)

(\(\because\) sin θ = \(\frac{8}{17}\) & cos θ = \(\frac{15}{17}\))

SECOND METHOD:-

\(\frac{(2+2\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(2-2\cos\theta)}\)

\(=\frac{2(1+\sin\theta)(1-\sin\theta)}{2(1+\cos\theta)(1-\cos\theta)}\)

\(=\frac{1-\sin^2\,\theta}{1-\cos^2\,\theta}\) (\(\because\) \((a + b)(a - b)\) \(=a^2 - b^2\))

\(=\frac{\cos^2\theta}{\sin^2\theta}=\left(\frac{\cos\theta}{\sin\theta}\right)^2=\cot^2\theta\) (\(\because\) \(\sin^2\theta + \cos^2\theta = 1\, \& \,\cot\,\theta=\frac{\cos\theta}{\sin\theta}\))

\(=\frac{1}{\tan^2\theta}\)

\(=\frac{1}{\left(\frac{8}{15}\right)^2}\)

\(=\frac{15^2}{8^2}\)

\(=\frac{225}{64}.\)

Given equation is \(\mathrm{3x^2-(3k-2)x-(k-6)=0}\)

Sum of the roots \(=\frac{-b}{a}=\frac{-(-(3k-2))}{3}=\frac{3k-2}{3}\)

Product of roots \(=\frac{c}{a}=\frac{-(k-6)}{3}\)

According to given condition \(\frac{3k-2}{3}=\frac{-(k-6)}{3}\)

\(\Rightarrow \) 3k - 2 + k - 6 = 0

\(\Rightarrow \) 4k = 8

\(\Rightarrow \) k = 2

Given that \(\alpha\) and \(\beta \) are zeros of the polynomial \(\mathrm{x^2-x-12}\)

The sum of zeros \(=\alpha + \beta = \frac{-b}{a}= \frac{-(-1)}{1}=1\)

The product of zeros \(=\alpha \beta = \frac{c}{a}=\frac{-12}{1}=-12\)

Now, \((\alpha - \beta)^2\) \(=(\alpha + \beta)^2-4\alpha \beta\) \(=1^2 - 4\times -12 \) \(=1 + 48 = 49\)

\(\Rightarrow \alpha - \beta=\pm 7\)

Case I :- \(\alpha - \beta = 7\) and \(\alpha + \beta =1\)

\(\Rightarrow 2\alpha = 8\)

\(\Rightarrow \alpha = 4\)

Then \(4+\alpha = 1\) \(\Rightarrow\) \(\beta = 1- 4 = -3\)

i.e., \(\alpha = 4 \) and \(\beta = -3\)

Case II :- \(\alpha - \beta = -7\) and \(\alpha + \beta = 1\)

\(\Rightarrow 2\alpha = -6\)

\(\Rightarrow \alpha = -3\)

Then \(-3+\beta = 1\) \(\Rightarrow \beta = 1 + 3 = 4\)

i.e., \(\alpha = - 3\) and \(\beta = 4\)

Hence, –3 and 4 are zeros of given polynomial \(\mathrm{x^2-x-12}\)

We want to find a quadratic equation whose zeros are \(2\alpha\) and \(2\beta\)

i.e., quadratic equation is \((\mathrm x - 2\alpha)(\mathrm x - 2\beta)=0\)

\(\Rightarrow \) (x - 8)(x + 6) = 0  (\(\because\) either \(\alpha = -3\) & \(\beta = 4\) or \(\alpha = 4\) and \(\beta = -3\))

\(\Rightarrow \) \(\mathrm{x^2-2x-48=0}\)

\(\therefore\) Required quadratic equation is \(\mathrm{x^2-2x - 48=0}\)

Given that sum of frequencies = 181.

\(\therefore\) x + 15 + 18 + 30 + 50 + 48 + x = 181

\(\Rightarrow \) 2x + 161 = 181

\(\Rightarrow \) 2x = 20

\(\Rightarrow \) x = 10

Hence, the value of x is 10.

In the given distribution table the highest frequency is 50 and the interval age of highest frequency is 13 – 15.

\(\therefore\) Modal class = Interval with highest frequency = 13 – 15.

The distribution table is given below:

\(\because\) Modal class is 13-15.

\(\therefore\) \(l\) = lower limit of modal class = 13

h = class interval = 15 – 13 =2

\(\mathrm f_1\) = frequency of modal class = 50

\(\mathrm f_0\) = frequency of class before modal class = 30

\(\mathrm f_2\) = frequency of class after modal class = 48

Therefore, mode \(=l +\frac{f_1-f_0}{2f_1-f_0-f_2}\times h\)

\(=13+\frac{50-30}{2\times 50-30-48}\times 2\) \(=13+\frac{20}{100-78}\times2 \) \(=13+\frac{20}{11}\) = 13 + 1.818 = 14.818.

Hence, the mode of given distribution is 14.818.

Mean: -

We get \(\sum f_i = 181\) & \(\sum f_i \mathrm x_i = 2368\)

\(\therefore\) Mean \(=\frac{\sum f_i \mathrm x_i}{\sum f_i}=\frac{2368}{181}=13.083\)

Hence, mean of given distribution is 13.083.

Given system of equations are

\(\frac{35}{\mathrm x+y}+\frac{14}{\mathrm x-y}=19;\)

And \(\frac{14}{\mathrm x+y}+\frac{35}{\mathrm x-y}=37\)

Put \(\frac{1}{\mathrm x+y}=X\) and \(\frac{1}{\mathrm x-y}=Y\)

The system of equation change to

35 X + 14Y = 19  ..............(1)

And 14X + 35Y = 37 ............(2)

Multiplying equation (1) by 2 and equation (2) by 5,

We get 70X + 28Y = 38  ...........(3)

and 70X + 175Y = 185 ............(4)

Now subtracting equation (3) from equation (4), we get (70X + 175Y) – (70X + 28Y) = 185 – 38

\(\Rightarrow\) 147Y = 147

\(\Rightarrow\) \(Y = \frac{147}{147}=1\)

Now, putting Y= 1 in equation (1), we get 35X + 14 = 19

\(\Rightarrow\) 35X = 19 − 14 = 5

\(\Rightarrow\) \(X = \frac{5}{35}=\frac{1}{7}\)

\(\therefore \) X = \(\frac{1}{7}\) & Y = 1

\(\Rightarrow\) x + y = 7 & x - y = 1 (\(\because\) x = \(\frac{1}{\mathrm x+y}\) & Y = \(\frac{1}{\mathrm x-y}\))

\(\Rightarrow\) (x + y) + (x - y) = 7 + 1 = 8

\(\Rightarrow\) 2x = 8

\(\Rightarrow\) x = 4

Now, putting x = 4 in x + y = 7, we get 4 + y = 7

\(\Rightarrow\) y = 7 - 4 = 3

\(\therefore \) x = 4 & y = 3 is solution of given system of equations.