Mix Test 8

Given that HCF of two numbers is 27 and their LCM is 162.

One number is 81.

Let the other number is a.

We know that for two positive integer a & b, HCF (a, b) \(\times\) LCM (a,b) = a \(\times\) b.

Therefore, HCF (81, a) \(\times\) LCM (81, a) = 81 \(\times\) a

\(\Rightarrow a = \frac{HCF(81, a)\times LCM(81,a)}{81}\)

\(\Rightarrow a =\frac{27\times 162}{81}=27\times 2 = 54\)

Hence, the other number is 54.

Given AP is 7, 11, 15, …., 139.

First term of AP is a = 7.

And common difference of AP is d \(=a_2-a_1\) = 11 – 7 = 4.

Let \(n^{th}\) term of AP is 139

\(\therefore\) a + (n – 1) d = 139. (\(\because\) \(n^{th}\) term of AP is given by \(a_n= a+(n-1)d\))

\(\Rightarrow\) 7 + (n – 1)4 = 139 (\(\because\) a = 7 & d = 4)

\(\Rightarrow\) 4(n – 1) = 139 – 7 = 132

\(\Rightarrow\) n – 1 \(=\frac{132}{4}=33\)

\(\Rightarrow\) n = 33 + 1 = 34.

Hence, total term in given AP is 34.

Given system of equations are \(\mathrm{\frac{3}{x+y}+\frac{2}{x-y}=2}\) and \(\mathrm{\frac{9}{x+y}-\frac{4}{x-y}=1}\)

Given that u \(=\frac{1}{\mathrm x+y}\) and v \(=\frac{1}{\mathrm x-y}\)

Then, the system of equations becomes

3u + 2v = 2.  ...........(1)

And 9u – 4v = 1. ..........(2)

Given system of equations are \(\frac{3}{\mathrm x+y}+\frac{2}{\mathrm x-y}=2\) and \(\frac{9}{\mathrm x+y}-\frac{4}{\mathrm x-y}=1\)

Given that \(u = \frac{1}{\mathrm x+y}\) and \(v = \frac{1}{\mathrm x-y}\)

Then, the system of equations becomes

3u + 2v = 2. ............(1)

And 9u – 4v = 1. ...........(2)

Multiplying equation (1) by 2, we get

6u + 4v = 4. ............(3)

By adding equations (2) and (3), we get (9u – 4v) + (6u + 4v) = 1 + 4

\(\Rightarrow\) 9u + 6u = 5

\(\Rightarrow\) 15u = 5 \(\Rightarrow\) \(u = \frac{5}{15}=\frac{1}{3}\)

Now, putting \(u = \frac{1}{3}\) in equation (1), we get \(3\times \frac{1}{3}+2v = 2\)

\(\Rightarrow\) 1 + 2v = 2

\(\Rightarrow\) 2v = 2 – 1 = 1

\(\Rightarrow\) v \(=\frac{1}{2}\)

Hence, the solution is \(u =\frac{1}{3}\) and \(v = \frac{1}{2}\)

Given system of equations are \(\frac{3}{\mathrm x+y}+\frac{2}{\mathrm x-y}=2\) and \(\frac{9}{\mathrm x+y}-\frac{4}{\mathrm x-y}=1\)

Given that \(u = \frac{1}{\mathrm x+y}\) and \(v = \frac{1}{\mathrm x-y}\)

Then, the system of equations becomes

3u + 2v = 2. ..........(1)

And 9u – 4v = 1. ..........(2)

Multiplying equation (1) by 2, we get

6u + 4v = 4.  ..............(3)

By adding equations (2) and (3), we get (9u – 4v) + (6u + 4v) = 1 + 4

\(\Rightarrow\) 9u + 6u = 5 \(\Rightarrow\) 15 u = 5

\(\Rightarrow\) \(u = \frac{5}{15}=\frac{1}{3}\)

Now, putting \(u=\frac{1}{3}\) in equation (1), we get \(3\times \frac{1}{3}+2v = 2\)

\(\Rightarrow\) 1 + 2v = 2 \(\Rightarrow\) 2v = 2 – 1 = 1

\(\Rightarrow\) \(v = \frac{1}{2}\)

Now, \(u = \frac{1}{3}\) but \(u = \frac{1}{\mathrm x+y}\)

Therefore, \(\frac{1}{\mathrm x+y}=\frac{1}{3}\) \(\Rightarrow\) x + y = 3. ..........(4)

And \(v = \frac{1}{2}\) but \(v = \frac{1}{\mathrm x-y}\)

Therefore, \(\frac{1}{\mathrm x-y}=\frac{1}{2}\) \(\Rightarrow\) x – y = 2. ......(5)

After substituting the value of u and v, the original system of equations becomes

x + y = 3 and x – y = 2.

Given system of equations are \(\frac{3}{\mathrm x+y}+\frac{2}{\mathrm x-y}=2\) and \(\frac{9}{\mathrm x+y}-\frac{4}{\mathrm x-y}=1\)

Given that \(u = \frac{1}{\mathrm x+y}\) and \(v = \frac{1}{\mathrm x-y}\)

Then, the system of equations becomes

3u + 2v = 2. ...........(1)

And 9u – 4v = 1. ..........(2)

Multiplying equation (1) by 2, we get

6u + 4v = 4. .............(3)

By adding equations (2) and (3), we get (9u – 4v) + (6u + 4v) = 1 + 4

\(\Rightarrow 9u + 6u = 5\)

\(\Rightarrow 15 u = 5\) \(\Rightarrow u = \frac{5}{15}=\frac{1}{3}\)

Now, puting \(u=\frac{1}{3}\) in equation (1), we get \(3\times \frac{1}{3}+2v=2\)

\(\Rightarrow\) 1 + 2v = 2

\(\Rightarrow\) 2v = 2 – 1 = 1

\(\Rightarrow\) v \(=\frac{1}{2}\)

Now, \(u = \frac{1}{3}\) but \(u = \frac{1}{\mathrm x+y}\)

Therefore, \(\frac{1}{\mathrm x+y}=\frac{1}{3}\) \(\Rightarrow \mathrm x+y = 3\) ..........(4)

And v \(=\frac{1}{2}\) but v \(=\frac{1}{\mathrm x-y}\)

Therefore, \(\frac{1}{\mathrm x+y}=\frac{1}{2}\) \(\Rightarrow \mathrm x -y = 2\) ..........(5)

By adding equations (4) and (5), we get (x + y) + (x – y) = 3 + 2

\(\Rightarrow\) 2x = 5 \(\Rightarrow \mathrm x =\frac{5}{2}\) 

Now, putting \(\mathrm x = \frac{5}{2}\) in equation (4), we get \(\frac{5}{2}+y = 3\)

\(\Rightarrow y = 3-\frac{5}{2}=\frac{1}{2}\)

Hence, the solution of given system of equations is \(\mathrm x =\frac{5}{2}\) and \(y = \frac{1}{2}\).

Given system of equations are \(\frac{3}{\mathrm x+y}+\frac{2}{\mathrm x-y}=2\) and \(\frac{9}{\mathrm x+y}-\frac{4}{\mathrm x-y}=1\).

Given that \(u = \frac{1}{\mathrm x+y}\) and \(v = \frac{1}{\mathrm x-y}\)

Then, the system of equations becomes

3u + 2v = 2.  ............(1)

And 9u – 4v = 1. ..........(2)

Multiplying equation (1) by 2, we get

6u + 4v = 4.  ..........(3)

By adding equations (2) and (3), we get (9u – 4v) + (6u + 4v) = 1 + 4

\(\Rightarrow\) 9u + 6u = 5 \(\Rightarrow\) 15 u = 5

\(\Rightarrow\) \(u = \frac{5}{15}=\frac{1}{3}\)

Now, putting \(u = \frac{1}{3}\) in equation (1), we get \(3\times \frac{1}{3}+2v = 2\)

\(\Rightarrow \) 1 + 2v = 2 \(\Rightarrow \) 2v = 2 – 1 = 1

\(\Rightarrow \) \(v = \frac{1}{2}\)

Now, \(u = \frac{1}{3}\) but \(u = \frac{1}{\mathrm x+y}\)

Therefore, \(\frac{1}{\mathrm x+y}=\frac{1}{3}\) \(\Rightarrow \mathrm x+y = 3\)  ............(4)

And \(v=\frac{1}{2}\) but \(v = \frac{1}{\mathrm x-y}\) 

Therefore, \(\frac{1}{\mathrm x-y}\) \(=\frac{1}{2}\) \(\Rightarrow \) x – y = 2. ........(5)

After substituting the value of u and v, the original system of equations converts in modified system of equations.

\(\therefore\) The modified system of equations is x + y = 3 and x – y = 2.

Now, straight lines are x + y = 3 and x – y = 2.

\(\Rightarrow \) y = –x + 3 and y = x – 2

Hence, the slope of line x + y = 3 is \(m_1\) = –1. (By comparing y = mx + c with equations of line)

Now, \(m_1m_2=-1\times 1 = -1\) which is condition that both lines are perpendicular.

\(\because\) Room shape is cuboid and dimension of room is \(12m\times 9m\times 8 m\)

\(\therefore\) \(l\) = 12m, b = 9m & h = 8m

Height of longest pole is \(\sqrt{l^2+b^2+h^2}\) 

\(=\sqrt{12^2+9^2+8^2}\)

\(=\sqrt{144+81+64}\)

\(=\sqrt{289}\) = 17 m

Hence, the length of the longest pole that can be kept in a room of dimension \(12m\times 9m\times 8m\) is 17m.

In a ΔABC, ∠C = 3∠B = 2(∠A + ∠B).

\(\therefore\) 3∠B = 2(∠A + ∠B)

\(\Rightarrow\) 3∠B = 2∠A + 2∠B

\(\Rightarrow\) 3∠B – 2∠B = 2∠A

\(\Rightarrow\) ∠B = 2∠A

And ∠C = 3∠B = 6∠A       (\(\because\) ∠B = 2∠A)

We know that sum of angles in a triangle is 180°.

\(\therefore\) ∠A + ∠B + ∠C = 180º

\(\Rightarrow\) ∠A + 2∠A + 6∠A = 180º

\(\Rightarrow\) 9∠A = 180º

\(\Rightarrow\) ∠A = \(\frac{180^o}{9}=20^o\)

Now, ∠B = 2∠A = 2 \(\times\) 20º = 40º

And ∠C = 6∠A = 6 \(\times\) 20º = 120º

∠A − ∠B + ∠C = 20º − 40º + 120º = 140º − 40º = 100º

We have to find the value of \(\frac{1}{\sec\theta-\tan\theta}-\frac{1}{\cos\theta}\)

Now, \(\frac{1}{\sec\theta - \tan\theta}+\frac{1}{\sec\theta + \tan\theta}\) 

\(=\frac{(\sec\theta + \tan\theta)+(\sec\theta - \tan\theta)}{(\sec\theta - \tan\theta)(\sec\theta + \tan\theta)}\) 

\(=\frac{2\sec\theta}{\sec^2\theta - \tan^2\theta}\)  \((\because (a+b)(a-b) = a^2-b^2)\)

\(=2\sec\theta = \frac{2}{\cos\theta}\) \(\left(\because \sec^2\theta -\tan^2\theta = 1 \,\&\, \sec\theta = \frac{1}{\cos\theta}\right) \)

Hence, \(\frac{1}{\sec\theta - \tan\theta}+\frac{1}{\sec\theta + \tan\theta}\) \(=\frac{2}{\cos\theta}=\frac{1}{\cos\theta}+\frac{1}{\cos\theta}\)

\(\Rightarrow \frac{1}{\sec\theta -\tan\theta}-\frac{1}{\cos\theta}\) \(=\frac{1}{\cos\theta}-\frac{1}{\sec\theta + \tan\theta}\)

Ascending order of observations is 29, 32, 48, 50, x, x+2, 72, 78, 84, 95.

Total number of observations is 10 (even).

Therefore, the median of the data is average of \(5^{th}\) & \(6^{th}\) observations.

\(\Big(\because \frac{n}{2}=5, \frac{n}{2}+1=6\) and if n is even then median \(=\frac{\frac{n}{2}th \,observation+\left(\frac{n}{2}+1\right)th\,observation}{2}\Big)\)

And \(5^{th}\) & \(6^{th}\) observations are x & x + 2 respectively.

Therefore, median \(=\frac{\mathrm x+\mathrm x+2}{2}=\frac{2\mathrm x+2}{2}\) = x + 1.

But given that median of the data is 63.

Therefore, x + 1 = 63

\(\Rightarrow\) x = 63 - 1 = 62

Ascending order of observations is 29, 32, 48, 50, x, x+2, 72, 78, 84, 95.

Total number of observations is 10 (even).

Therefore, the median of the data is average of \(5^{th}\) & \(6^{th}\) observations.

\(\Big(\because \frac{n}{2}=5, \frac{n}{2}+1=6\) and if n is even then median \(=\frac{\frac{n}{2}th \,observation+\left(\frac{n}{2}+1\right)th\,observation}{2}\Big)\)

And \(5^{th}\) & \(6^{th}\) observations are x & x + 2 respectively.

Therefore, median \(=\frac{\mathrm x+\mathrm x+2}{2}=\frac{2\mathrm x+2}{2}\) = x + 1.

But given that median of the data is 63.

Therefore, x + 1 = 63

\(\Rightarrow\) x = 63 − 1 = 62.

\(\therefore\) x + 2 = 62 + 2 = 64.

Hence, the ascending order of the observations is as follows: 29, 32, 48, 50, 62, 64, 72, 78, 84, 95.

(By putting x = 62)

Ascending order of observations is 29, 32, 48, 50, x, x+2, 72, 78, 84, 95.

Total number of observations is 10 (even).

Therefore, the median of the data is average of \(5^{th}\) & \(6^{th}\) observations.

\(\Big(\because \frac{n}{2}=5, \frac{n}{2}+1=6\) and if n is even then median \(=\frac{\frac{n}{2}th\,observation+\left(\frac{n}{2}+1\right)th\,observation}{2}\Big)\)

And \(5^{th}\) & \(6^{th}\) observations are x & x + 2 respectively.

Therefore, median \(=\frac{\mathrm x+\mathrm x+2}{2}\mathrm{=\frac{2x+2}{2}=x+1}\)

But given that median of the data is 63.

Therefore, x + 1 = 63

\(\Rightarrow\) x = 63 − 1 = 62.

\(\therefore\) x + 2 = 62 + 2 = 64.

Hence, the ascending order of the observations is as follows: 29, 32, 48, 50, 62, 64, 72, 78, 84, 95. (By putting x = 62)

\(\because\) Our observations are non-repeating.

\(\therefore\) All observations have same frequency 1.

There are total 10 observations.

Therefore, sum of all frequency is \(\sum^{10}_{i=1} f_i = \sum^{10}_{i=1}1= 10\).

And \(\sum f_i\mathrm x_=\sum1\,\mathrm x_i\) (\(\because\) All frequencies = 1)

\(=\sum \mathrm x_i \) = 29 + 32 + 48 +50 + 62 + 64 + 72 + 78 + 84 + 95 = 614.

Hence, the mean of observations \(=\frac{\sum f_i \mathrm x_i}{\sum f_i}=\frac{614}{10}=61.4\)

Hence, the mean of observations is 61.4.

Ascending order of observations is 29, 32, 48, 50, x, x+2, 72, 78, 84, 95.

Total number of observations is 10 (even).

Therefore, the median of the data is average of \(5^{th}\) & \(6^{th}\) observations.

\(\Big(\because \frac{n}{2} = 5, \frac{n}{2}+ 1 = 6\) and if n is even then median \(=\frac{\frac{n}{2}th\,observation+\left(\frac{n}{2}+1\right)th\,observation}{2}\Big)\)

And \(5^{th}\) & \(6^{th}\) observations are x & x + 2 respectively.

Therefore, median \(=\frac{\mathrm x+\mathrm x+2}{2}=\frac{2\mathrm x+2}{2}\) = x + 1

But given that median of the data is 63.

Therefore, x + 1 = 63

\(\Rightarrow\) x = 63 − 1 = 62.

\(\therefore\) x + 2 = 62 + 2 = 64.

Hence, the ascending order of the observations is as follows: 29, 32, 48, 50, 62, 64, 72, 78, 84, 95. (By putting x = 62)

\(\because\) Our observations are non-repeating.

\(\therefore\) All observations have same frequency 1.

There are total 10 observations.

Therefore, sum of all frequency is \(\sum^{10}_{i=1}f_i = \sum^{10}_{i=1}1=10\)

And \(\sum f_i \mathrm x_i = \sum 1\mathrm x_i \) (\(\because \) All frequencies = 1)

\(=\sum \mathrm x_i \) = 29 + 32 + 48 +50 + 62 + 64 + 72 + 78 + 84 + 95 = 614.

Hence, the mean of observations \(=\frac{\sum f_i \mathrm x_i}{\sum f_i }= \frac{614}{10}=61.4\)

Hence, the mean of observations is 61.4.

We know that mode = 3 median – 2 mean.

Therefore, mode of given data \(= 3\times 63 - 2\times 61.4\)

\(=189 - 122.8 = 66.2\)

(\(\because\) mean = 61.4 and median = 63)

\(\because\) 29 is smallest observation in given data, we are taking class-width = 6 and starts class interval from 29.

It means our first class interval is 29 – 35.

Greatest observation in given data is 95. It means we stop making class interval if 95 present in any class interval.

\(\therefore\) Our class intervals are 29 – 35, 35 – 41, 41 – 47, 47 – 53, 53 – 59, 59 – 65, 65 – 71, 71 – 77, 77 – 83, 83 – 89, 89 – 95, 95 – 101.

\(\therefore\) Total class intervals = 12.

Let the radius of the circle is r.

Therefore, the area of the circle is \(\pi r^2\)

After decreasing 30% radius of the circle

The actual radius of the circle is \(r-\frac{30}{100}r\) \(=r-0.3r\) \(=0.7r\)

The area of the circle after decreasing  \(=\pi(0.7r)^2\) \(=0.49\pi r^2\).

Decreased area = Area of circle before decreasing – Area of circle after decreasing

\(=\pi r^2 - 0.49\pi r^2\)

\(=0.51\pi r^2 \) = 51% \(\times\) Area of circle before decreasing.

Hence, 51% area is decreased if the radius of a circle is decreased by 30%.

Since, \(\pi\) = 3.141592653………. which is non-terminating and non-repeating decimal number.

Hence, \(\pi\) is an irrational number.

Total number of cars in the deck = 52.

Total king cards are 4 in which 2 are black suited and 2 are red suited.

Therefore, total number of king card which is red suited = 2.

Probability of getting a king of red suit 

\(=\frac{Total\,red\,suited\,king\,card}{Total\,number\,of\,cards\,in\,the\,deck}\)

\(=\frac{2}{52}=\frac{1}{26}\)

Total number of cars in the deck = 52

Total king cards are 4 in which 2 are black suited and 2 are red suited.

Therefore, total number of king card which is red suited = 2.

Probability of getting a king of red suit \(=\frac{Total\,red\,suited\,king\,card}{Total\,number\,of\,cards\,in\,the\,deck}\) \(=\frac{2}{52}=\frac{1}{26}\)

The probability of not getting a king of red suit = 1 – probability of getting a king of red suit

\(=1-\frac{1}{26}=\frac{25}{26}\)

Total number of cars in the deck = 52.

Total number of face cards = \(3\times 4 = 12\) (\(\because\) Jack, Queen & king cards are face cards)

Therefore, the probability of getting a face and \(=\frac{Total\,number\,of\,face\,card}{Total\,number\,of\,cards\,in\,the\,deck}\) \(=\frac{12}{52}=\frac{3}{13}\)

Total number of cars in the deck = 52.

Total number of red face card = 3 \(\times\) 2 = 6.

Therefore, the probability of getting a red face card\(=\frac{Total\,number\,of\,red\,face\,cards}{Total\,number\,of\,cards\,in\,deck}\) \(=\frac{6}{52}=\frac{3}{26}\)

Total number of cars in the deck = 52.

Total number of spade cards = 13.

Therefore, the probability of getting a spade card \(=\frac{Total\,number\,of\,spade\,cards}{Total\,number\,of\,cards\,in\,the\,deck}\) \(=\frac{13}{52}=\frac{1}{4}\)

Three coins are tossed simultaneously

Therefore, the outcomes are {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Hence, total outcomes = n(S) = 8.

Let event E be the event of getting exactly two heads.

Outcomes which are favorable to event E are {HHT, HTH, THH}.

Total outcomes which is favorable to event E = n(E) = 3.

Now, probability of getting exactly two heads \(=\frac{Total\,outcomes\,favourable\,to\,event\,E}{Total\,outcomes\,in\,experiment}\) \(=\frac{n(E)}{n(S)}=\frac{3}{8}\)

Hence, the event of getting exactly two heads = \(\frac{3}{8}\)