Mix Test 5

Given number is \(\frac{11}{2^3\times 5}=\frac{11}{4}\times \frac{1}{2\times 5}\)

\(=2.75\times \frac{1}{10}=0.275\)

The decimal representation of given number is 0.275.

Therefore, the decimal representation of \(\frac{11}{2^3\times 5}\) will terminate after 3 decimal places.

Given quadratic equation is \(\mathrm{2x^2-4x+3 = 0}\)

By comparing given quadratic equation with standard quadratic equation \(\mathrm{ax^2+bx+c=0}\), we get

a = 2, b = -4 and c = 3

Now, the discriminant of given quadratic equation is \(D = b^2-4ac\)\(=(-4)^2-4\times 2\times3\)

\(=16-24=-8<0\)  

Hence, both roots are imaginary i.e., both roots are not real

(Because, if \(D=b^2-4ac < 0,\) then its roots are imaginary means no real roots.)

Hence, the quadratic equation \(\mathrm{2x^2-4x+3=0}\) has no real roots.

Since, tangent line to a circle is a line touches the circle at exactly one point, never intersecting the circle.

Therefore, there is no tangent to a circle passing through a point lying inside the circle.

Hence, zero tangents are possible to a circle passing through a point lying inside the circle.

Girish can currently run that distance in 45 sec.

i.e., the first term of formed A.P. is a = 45 sec.

With each day of practice, it takes him 2 sec. less

i.e., the common difference of formed A.P. is d = −2 sec.

Now, the first term and common difference of formed A.P. is a = 45 sec and d = –2 sec.

Therefore, the second term of formed A.P. is a + d = 45 - 2 = 43 sec.

The third term of formed A.P. is a + 2d = 45 + 2(−2) = 45 − 4 = 41 sec.

Therefore, the A.P. for the above given situation is of the form 45, 43, 41, … … …

Girish can currently run that distance in 45 sec.

i.e., the first term of formed A.P. is a = 45 sec.

With each day of practice, it takes him 2 sec. less.

i.e., the common difference of formed A.P. is d = −2 sec.

His goal is complete 200 m run in 30 sec.

\(\because\) Girish’s goal is complete 200 m run chase in 30 sec

Let Girish achieve his goal in n days.

Therefore, the \(n^{th}\) term of the A.P. for the above situation is 30 sec.

We know that the \(n^{th}\) term of the A.P. is a + (n − 1)d.

Therefore, a + (n - 1)d = 30

\(\Rightarrow 45+(n-1)(-2)=30\) (By putting a = 45 sec. and d = -2sec.)

\(\Rightarrow -2(n-1)\) \(=30-45 = -15\)

\(\Rightarrow n -1 = \frac{-15}{-2}=7.5\)

\(\Rightarrow n = 7.5 + 1=8.5\)

Since, number of days is always a positive integer, therefore, the minimum number of days he needs to practice till his goal is accomplished in 9 days.

Since, first term of A.P. formed by above given situation is 45 which is an odd number and the common difference of A.P. is d = −2 means the common difference is an even number. Therefore, all terms of this A.P. is an odd number. This sequence is a decreasing sequence of odd numbers starting with 45.

Since, 30 is an even number which never appears in this sequence.

Girish can currently run that distance in 45 sec.

i.e., the first term of formed A.P. is a = 45 sec.

With each day of practice, it takes him 2 sec. less.

i.e., the common difference of formed A.P. is d = -2sec.

We know the sum of first n terms of A.P. is given by \(S_n = \frac{n}{2}[2a+(n-1)d]\)

Now, the sum of first 5 terms of formed A.P. is \(Ss = \frac{5}{2}[2a+(5-1)d]\) \(=\frac{5}{2}[2\times 45+(5-1)(-2)]\)

\(=\frac{5}{2}[90-2\times 4]\) \(=\frac{5}{2}[90-8]\) \(=\frac{5}{2}\times 82\) \(=5\times 41 = 205\)

Hence, the sum of first 5 terms of A.P formed by above given situation is 205.

The sequence is a, a + d, a + 2d,.......

The first term of given sequence is a,

The second term of given sequence is a + d,

The third of given sequence is a + 2d

In similar ways, the \(n^{th}\) term of the sequence is a + (n - 1)d.

Given that tan(A + B) = \(\sqrt 3\) and tan(A – B) = \(\frac{1}{\sqrt 3}\) where 0º < A + B < 90º ; A > B.

Since, tan(A + B) = \(\sqrt 3\) = tan 60º and 0º < A + B < 90º

Therefore, A + B = 60º .........(1)

And tan(A – B) = \(\frac{1}{\sqrt 3}\) = tan 30º

Therefore, A – B = 30º  ............(2)

By adding equations (1) and (2), we get A + B + (A – B) = 60º + 30º

\(\Rightarrow\) 2A = 90º \(\Rightarrow\) A = 45º

Now, putting the value of A = 45º in equation (1), we get

45º + B = 60º \(\Rightarrow\) 60º – 45º = 15º

Hence, A = 45º and B = 15º

Given that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

Therefore, the distance of point (x, y) from (7, 1) = The distance of point (x, y) from (3, 5).

\(\Rightarrow \sqrt{(7-\mathrm x)^2+(1-y)^2}\) \(\mathrm{=\sqrt{(3-x)^2+(5-y)^2}}\)  (By distance formula between two points)

\(\Rightarrow (7-\mathrm x)^2+(1-y)^2\) \(=(3-\mathrm x)^2+(5-y)^2\) (By squaring both sides)

\(\Rightarrow \mathrm{x^2-14x+49+y^2-2y+1}\) \(=\mathrm x^2 -6\mathrm x + 9 +y^2-10y + 25\) (\(\because\) \((a-b)^2\) \(=a^2-2ab+b^2\))

\(\Rightarrow -14\mathrm x - 2 y+50\) \(=-6\mathrm x -10y + 34\) (By cancelation of \(\mathrm x^2+y^2\) from both sides)

\(\Rightarrow \mathrm{14x -6x + 2y-10y}\) \(=50-34\)

\(\Rightarrow 8\mathrm x - 8y = 16\)

\(\Rightarrow \mathrm x-y = 2\) . (Dividing by 8 from both sides )

Hence, the relation between x and y is x − y = 2 such that the point (x, y) is equidistant from the points (7,1) and (3,5).

Let the quadratic polynomial is \(\mathrm{p(x) = ax^2 + bx + c}\)

Therefore, the sum of zeros of polynomial is \(\frac{-b}{a}\) and the product of zeros of polynomial is \(\frac{c}{a}\).

(Because, sum and product of roots of quadratic equation \(\mathrm {ax^2 + bx + c=0}\) are \(\frac{-b}{a}\) and \(\frac{c}{a}\))

Given that the sum and product of zeros of quadratic polynomial are 0 and \(\sqrt 5\), respectively.

Therefore, \(\frac{-b}{a}= 0\) and \(\frac{c}{a}=\sqrt 5\)

\(\Rightarrow b = 0\) and \(c= \sqrt 5a\)

Therefore, the quadratic polynomial is \(\mathrm{p(x) = ax^2 + \sqrt 5a},\) where a is any real number.

Assuming a = 2, then the quadratic polynomial is \(\mathrm{p(x) =2x^2+2\sqrt 5}\)

We know that Mode \(=l + \frac{f_1-f_0}{2f_1-f_0-f_2}\times h\)

Where, \(n = \sum f_i = 100\). Therefore, \(\frac{n}{2}=\frac{100}{2}=50\)

In the above distribution table the highest frequency is 8 and interval of highest frequency is 3 − 5.

Therefore, modal class = Interval with highest frequency = 3 − 5.

Now, \(l\) = Lower limit of modal class = 3.

h = Class-interval = 5 − 3 = 2.

\(f_1\) = Frequency of modal class = 8.

\(f_0\) = Frequency of the class before modal class = 7.

\(f_2\) = Frequency of the class after modal class = 2.

Therefore, the mode \(=l +\frac{f_1 -f_0}{2f_1 -f_0-f_2}\times h\) \(=3+\frac{8-7}{2\times 8-7-2}\times 2\)

\(=3+\frac{1}{16-9}\times 2=3+\frac{2}{7}\)

= 3 + 0.286 = 3.286.

Hence, the mode of given data is 3.286.