Mathematics Test-15

The given word: WINDOW

Consonants: W N D W

Vowels: 10

Total ways \(=\frac{6 !}{2 !}=360\)

ways when all the vowels always come together \(=\frac{5 ! \times 2 !}{2 !}=120\)

Ways when all the vowels never come together \(=360-120=240\)

Concept:

The equation of line is y = mx + c, where m is slope.

Given equation of line is x - y = 4

\(\Rightarrow y=x-4\) which is equal to \(y=m x+c\), where \(m\) is slope of the line

\(\Rightarrow m =1\)

\(\Rightarrow m =\tan \theta=1\)

\(\Rightarrow \theta=45^{\circ}\)

So,the equation of line \(x-y=4\) makes angle \(\theta_{1}=45^{\circ}\) with \(X\)-axis.

The required line makes an angle \(\theta_{2}=45^{\circ}\) with Y-axis. and makes an angle \(45^{\circ}\) with \(X\)-axis also.The line passing through the point \((0,1)\).

Slope of the required line is \(m =\tan \theta=1\)

Equation of line is,

y = mx+c

⇒ y = x + c.

Since, the line passing through the point (0,1) ,put x = 0 and y = 1 to find c.

⇒ 1 = 0 + c

⇒ c = 1.

Equation of line is y = x + 1

Hence, the equation of line passing through (0, 1) and making an angle with the y-axis equal to the inclination of the line x - y = 4 with x-axis is y = x + 1.

As we know that, Total number of card (outcomes) \(=52\)

Now, Total number of black face card (favourable outcomes) \(= 6\)

We know that,\(probability\) =\(\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

The probability of drawing a black face card\(=\frac{6}{52}\)

\(=\frac{3}{26}\)

Given equation dy = (1 + y²) dx

On solving, we get-

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

Integrating both sides

\(\int \frac{d y}{\left(1+y^{2}\right)}=\int d x\)

\(\tan ^{-1}(y)=x+c\)

\(y=\tan (x+c)\)

Fundamental Principle of Multiplication:

Let us suppose there are two tasks \(\mathrm{A}\) and \(\mathrm{B}\) such that the task \(\mathrm{A}\) can be done in \(m\) different ways following which the second task \(\mathrm{B}\) can be done in \(\mathrm{n}\) different ways. Then the number of ways to complete the task \(\mathbf{A}\) and \(\mathrm{B}\) in succession respectively is given by: \(\mathrm{m} \times \mathrm{n}\) ways.

Fundamental Principle of Addition:

Let us suppose there are two tasks \(A\) and \(B\) such that the task \(A\) can be done in \(\mathrm{m}\) different ways and task \(B\) can be completed in \(n\) ways. Then the number of ways to complete either of the two tasks is given by: \((m+n)\) ways.

Here, we have to find how many 3 -digit numbers can be formed without using the digits \(0,2,3,4,5\) and 6 .

i.e we have to find how many 3 -digit numbers can be formed using the digits \(1,7,8,9\).

Clearly, repetition of digits is allowed.

The number of ways to fill unit's digit \(=4\)

The number of ways to fill ten's digit \(=4\)

The number of ways to fill hundred's digit \(=4\)

\(\therefore\) Total number of required numbers \(=4 \times 4 \times 4=64\)

Let \(A B\) be the rod making an angle \(\theta\) with \(O X\) and \(P(x, y)\) be the point on it such that \(A P=3 ~cm\).

Then, \(PB=AB-AP=(12-3) cm=9 ~cm[AB=12 ~cm]\) From \(P\), draw \(PQ \perp OY\) and \(PR \perp OX\).

In \(\Delta PBQ, \cos \theta=\frac{P Q}{P B}=\frac{x}{9}\)

In \(\Delta PRA\),\(~\sin \theta=\frac{P R}{P A}=\frac{y}{3}\)

Since, \(\sin ^{2} \theta+\cos ^{2} \theta=1\)

\(\left(\frac{y}{3}\right)^{2}+\left(\frac{x}{9}\right)^{2}=1\)

Or, \(\frac{x^{2}}{81}+\frac{y^{2}}{9}=1\)

Thus, the equation of the locus of point \(P\) on the rod is \(\frac{x^{2}}{81}+\frac{y^{2}}{9}=1\)

Multiplying the equation throughout by \(5\), we get

\(25 x^2-30 x-10=0\)

This is the same as,

\((5 x)^2-2 \times(5 x) \times 3+3^2-3^2-10 =0\)

\(\Rightarrow (5 x-3)^2-9-10 =0\)

\(\Rightarrow (5 x-3)^2-19 =0\)

\(\Rightarrow (5 x-3)^2 =19\)

\(\Rightarrow 5 x-3 =\pm \sqrt{19}\)

\(\Rightarrow 5 x =3 \pm \sqrt{19}\)

So, \(x=\frac{3 \pm \sqrt{19}}{5}\)

Therefore, the roots are \(\frac{3+\sqrt{19}}{5}\) and \(\frac{3-\sqrt{19}}{5}\).

It can be seen that the student have scored \(3\) out of \(5\) times more than \(80\) marks.

\(\therefore\) probability that the student has scored more than \(80=\)\(\frac {\text{number of occurrence of event}}{\text {Total number of trials}}\)

\(=\frac{3}{5}\)

\(=0.6\)