Mathematics Test - 35

We know that:

Number of arrangement of \(n\) different objects is given by n!

\({n} !={n} \times({n}-1) \times \ldots \times 1\)

The word VOWELS contains 6 letters, in which all the different letters occur only once.

Now, after fixing \({O}\) in first place and \({E}\) in the last place, we have only 4 letters.

\(\therefore\) Number of words \(=4 !=24\).

Here total frequency \(=120\)

\(N=120=\) even

\(\frac{N}{2}=60\)

In cumulative frequency (C.f.) we select 65 as it is closer to 60.

∴ Median is 5 because 65 is in line of 5.

Given, at any instant \(t\), for a sphere, \(r\) denotes the radius, \(S\) denotes the surface area.

Surface area \(S =4 \pi r ^2\)

Differentiating w.r.to t, we get

\( \Rightarrow \frac{ ds }{ dt }=4 \pi \cdot 2 r \frac{ dr }{ dt } \)

\( \Rightarrow \frac{ ds }{ dt }=8 \pi r \frac{ dr }{ dt }\)

\( \Rightarrow \frac{ dr }{ dt }=\frac{1}{8 \pi r } \frac{ ds }{ dt } \ldots\)

Volume of sphere \(V =\frac{4}{3} \pi r ^3\)

Differentiating w.r.to \(t\), we get

\( \Rightarrow \frac{ dV }{ dt }=\frac{4}{3} \pi \cdot 3 r ^2 \frac{ dr }{ dt } \)

\( \Rightarrow \frac{ dV }{ dt }=4 \pi r ^2 \frac{ dr }{ dt }\)

From equation (1), we have

\( \Rightarrow \frac{ dV }{ dt }=4 \pi r ^2 \frac{1}{8 \pi r } \frac{ ds }{ dt } \)

\( \Rightarrow \frac{ dV }{ dt }=\frac{ r }{2} \frac{ ds }{ dt }\)

Hence, if at any instant \(t\), for a sphere, \(r\) denotes the radius, \(S\) denotes the surface area and \(V\) denotes the volume, then \(\frac{ dV }{ dt }=\frac{1}{2} r \frac{ dS }{ dt }\)

Consider, the angle between \(\vec{a}\) and \(\vec{b}\) is \(\theta\)

Given, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)

\(\Rightarrow \vec{a}+\vec{b}=-\vec{c}\)

\(\Rightarrow|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|-\overrightarrow{\mathrm{c}}|\)

Squaring on both side, we get

\(\Rightarrow|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|-\overrightarrow{\mathrm{c}}|^{2}\)

\(\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}+2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+|\overrightarrow{\mathrm{b}}|^{2}=|-\overrightarrow{\mathrm{c}}|^{2}\)

\(\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2 \mathrm{ab} \cos \theta=|-\overrightarrow{\mathrm{c}}|^{2}\)

\(\Rightarrow|(3)|^{2}+|(5)|^{2}+2(3)(5) \cos \theta=(7)^{2}\)

\(\Rightarrow 30 \cos \theta=15\)

\(\Rightarrow \cos \theta=\frac{1}{2}\)

\(\Rightarrow \theta=\frac{\pi}{3}\)

Hence, If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0},|\vec{a}|=3,|\vec{b}|=5\) and \(|\vec{c}|=7,\) then the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{3}\)

Let, \(y=\frac{1}{3 x^{2}}\) ....(1)

We know that:

\(\frac{d x^{n}}{dx}=nx^{n-1}\)

Differentiating (1) with respect of x, we get:

\(\frac{dy}{dx} =\frac{d}{dx}\left(\frac{1}{3 x^{2}}\right)\)

\(=\frac{1}{3} \frac{d\left(x^{-2}\right)}{dx}\)

\(=\frac{1}{3} \times-2 \times x^{-2-1}\)

\(=\frac{-2}{3 x^{3}}\)

Given,

The objective function is ,

\(F=4 x+6 y\)

At\((0,2)\),

\(F=4 \times 0+6 \times 2=12\)

At \((3,0)\),

\(F=4 \times 3+6 \times 0=12\)

At \((6,0)\),

\(F=4 \times 6+6 \times 0=24\)

At \((0,5)\),

\(F=4 \times 0+6 \times 5=30\)

Thus, the maximum value of \(\mathrm{F}=30\)

The minimum value of \(\mathrm{F}=12\)

Andthe value of the maximum value of \(\mathrm{F}\) - minimum value of \(\mathrm{F}\) \(=30-12\)

\(=18\)

\(f(x)=2 x^{2}+\frac{1}{x}\) .... (1)

We know that: If \(f(x)=x^{n}\), then: \(f'(x)=n x^{n-1}\)

Differentiating (1) with respect to x, we get:

\(\Rightarrow f^{\prime}(x)=4 x-\frac{1}{x^{2}}\)

Putting x = 1 in above, we get:

\(f'(1)=4 \times 1-\frac{1}{1^{2}}\)

\(f'(1)=4-1=3\)

\(\therefore\) The value of \(f^{\prime}(1)\) is 3.

The odds against an event \(A\) are \(5: 3\).

Probability of not occurring the individual event \(A=\frac{5}{8}\)

Probability of occurring the individual event \(A=\frac{3}{8}\)

Again, Odds in favor of another independent event \(B\) and \(6: 5\).

Probability of occurring the individual event \(B=\frac{6}{(6+5)}\) \(=\frac{6}{11}\)

Probability of not occurring the individual event \(B=\frac{5}{(6+5)}\) \(=\frac{5}{11}\)

The chances that neither \(A\) nor \(B\) occurs is \(=\) Probability of not occurring event \(A \times\) probability of not occurring event \(B\) [As the two events are independent therefore multiplication will occur]

\(=\frac{5}{8} \times \frac{5}{11}\) \(=\frac{25}{88}\)

Let \(x\) and \(y\) respectively be the number of machines \(A\) and \(B\) which the factory owner should buy.

Now, according to the given information, the linear programming problem is:

Maximise \(Z=60 x+40 y\)

Subject to the constraints:

\(1000 x+1200 y \leq 9000\)

\(\Rightarrow 5 x+6 y \leq 45 \ldots(1)\)

\(12 x+8 y \leq 72 \)

\(\Rightarrow 3 x+2 y \leq 18 \ldots(2) \)

\(x\geq 0, y \geq 0 \ldots \text { (3) }\)

The inequalities (1), (2), (3) can be graphed as:

The shaded portion OABC is the feasible region.

The value of Z at the corner points are given in the following table.

Total number of boys in a class \(=16\)

Total number of girls in a class \(=10\)

Selecting number of boy \(=1\)

Selecting number of girls \(=2\)

Total number of students \(=(16+10)=26\)

Number of all combinations of \(n\) things, taken \(r\) at a time, is givenby \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

Number of ways to select three students\(={ }^{26} C_{3}=\frac{26 !}{23 ! 3 !}=\frac{26 \times 25 \times 24}{3 \times 2}=2600\)

Number of ways to select \(1\) boy \(={ }^{16} C_{1}=\frac{16 !}{1 ! 15 !}=16\)

Number of ways to select \(2\) girls \(={ }^{10} C_{2}=\frac{10 !}{2 ! 8 !}=\frac{10 \times 9}{2}=45\)

Probability of selecting \(1\) boy and \(2\) girls\(=\frac{\text { Number of ways to select } 1 \text { boy } \times \text { Number of ways to select } 2 \text { girls }}{\text { Number of ways to select three students }}\)

\(\therefore\) Probability of selecting \(1\) boy and \(2\) girls \(=\frac{(16 \times 45)}{2600}\)

\(=\frac{720}{2600}\)

\(=\frac{18}{65}\)