Mix Test 9

\(\because\) a and b are relatively prime

\(\therefore\) HCF(a, b) = 1.  (By definition of relatively prime number)

Now, we know that HCF(a, b) \(\times\) LCM(a, b) = a \(\times\) b

\(\therefore\) LCM(a, b) = a \(\times\) b

Hence, if a and b are relatively prime then their LCM is product of both numbers a and b.

The equations are x – 3y = 2 ...........(1)

And – 2x + 6y = 5. .............(2)

Now, \(\frac{a_1}{a_2}=\frac{1}{-2}=\frac{-1}{2}\)

\(\frac{b_1}{b_2}=\frac{-3}{6}=\frac{-1}{2}\)

And \(\frac{c_1}{c_2}=\frac{2}{5}\)

Hence, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\) which is condition that both lines represent by given equations are parallel.

Hence, lines (paths) representing by equations (1) and (2) are parallel to each other.

Given that two dice are thrown at same time.

Therefore, total possible outcomes = n(s) = 6 × 6 = 36.

Let event E be the event that sum of two numbers appearing on the top is more than 9.

Hence, favorable outcomes of event E = {(4, 6), (5, 5), (5, 6), (6, 5), (6, 6), (6, 4)}

Hence, total outcomes favorable to event E is n(E) = 6.

\(\therefore\) \(P(E) = \frac{Total\,outcomes\,favourable\,to\,event\,E}{Total\,possible\,outcomes}\) \(=\frac{6}{36}=\frac{1}{6}\)

Hence, probability of event E is \(\frac{1}{6}\)

We have sin θ = cos(θ - 45º), where θ is an acute angle.

\(\Rightarrow\) cos(90º - θ) = cos(θ - 45º)  (\(\because\) cos(90º - θ) = sin θ)

\(\Rightarrow\) 90º - θ = θ - 45º

\(\Rightarrow\) 2θ = 90º + 45º = 135º

\(\Rightarrow\) θ \(=\frac{135^o}{2}\) = 62.5º

Hence, the value of θ is 62.5º.

Let the required quadratic polynomial is \(\mathrm{ax^2 + bx + c}\) ..........(1)

Therefore sum of zeros is \(-\frac{b}{a}=-5\) (given)

\(\Rightarrow b = 5a\)

And product of zeros is \(\frac{c}{a}=6\)

\(\Rightarrow c = 6a\)

By putting the values of b and c in equation (1), we get the quadratic polynomial 

\(\mathrm{ax^2 + 5ac + 6a}\) \(\mathrm{=a(x^2 + 5x+6)}\) where a is any real number except zero.

Let \(a=\frac{1}{2},\) then the required polynomial is \(\frac{1}{2}(\mathrm x^2 + 5\mathrm x + 6)\).

\(\mathrm{2\left(\frac{2x-1}{x+3}\right)}\)\(\mathrm{-3\left(\frac{x+3}{2x-1}\right)=5}\), x \(\ne\) \(-3, \frac{1}{2}\) 

\(\mathrm{\Rightarrow \frac{2(2x-1)^2-3(x+3)^2}{(x+3)(2x-1)}}=5\)

\(\Rightarrow \mathrm{2(4x^2 - 4x + 1)}\)\(\mathrm{-3(x^2+6x+9)}\) \(\mathrm{= 5(x + 3)(2x - 1)}\)

\(\mathrm{\Rightarrow 8x^2 - 3x^2 - 8x}\) \(\mathrm{-18x + 2- 27}\) \(\mathrm{=10x^2 + 25 x-15}\)

\(\mathrm{\Rightarrow 5x^2 - 26x- 25}\) \(\mathrm{=10x^2 + 25x-15}\)

\(\Rightarrow \mathrm{10x ^2 - 5x^ 2 + 25x }\) \(\mathrm{+ 26x - 15 + 25 = 0}\)

\(\Rightarrow \mathrm{5x^2 + 51x + 10=0}\)

\(\Rightarrow \mathrm{5x^2 + x+ 50x+10=0}\)

\(\Rightarrow\mathrm{ 5x(5x+1)+10(5x+1)=0}\)

\(\mathrm{\Rightarrow 5(x+10)(5x+1)=0}\)

\(\Rightarrow \) x + 10 = 0 or 5x + 1 = 0

\(\Rightarrow \) x = -10 or x \(=\frac{-1}{5}\)

Hence, the solutions of given equation are x \(=\frac{-1}{5}, -10\).

Median:-

\(\because\) \(\frac{n}{2}=\frac{\sum f_i}{2}=\frac{30}{2}=15\)

In the above distribution table, the cumulative frequency 25 is nearest greatest value of \(\frac{n}{2}=15\)

Therefore, 200-250 is the median class.

Now, \(l\) = Lower limit of median class = 200,

h = Class interval = 250 – 200 =50,

cf = Cumulative frequency of the class before median class = 13

f = Frequency of median class = 12

\(\therefore\) Median \(=\frac{l+\frac{n}{2}-cf}{f}\times h\) 

\(=200+\frac{15-13}{12}\times 50\)

\(=200+\frac{2}{12}\times 50\)

\(=200+\frac{50}{6}\)

= 200 + 8.33 =208.33.

Hence, the median of daily expenditure on food is 208.33 rupees.

Let the speed of the stream is x km/hour.

The speed of the boat in still water is 4 km/hour.

The upstream speed of the boat is (4 – x) km/hour.

The downstream speed of the boat is (4 + x) km/hour.

Distance = 30 km

We know that time \(=\frac{Distance}{speed}\)

Now, time taken by the boat in going 30 km upstream \(=\frac{30}{4-\mathrm x}\) hours.

And time taken by the boat in going 30 km downstream \(=\frac{30}{4+\mathrm x}\) hours.

Now, according to given conditions \(3\left(\frac{30}{4+\mathrm x}\right)=\frac{30}{4-\mathrm x}\)

\(\Rightarrow \frac{3}{4+\mathrm x}=\frac{1}{4-\mathrm x}\)

\(\Rightarrow \) 4 + x = 12 − 3x

\(\Rightarrow \) 4x = 12 − 4 = 8

\(\Rightarrow \) x = 2.

Hence, the speed of the stream is 2 km/hour.