Mathematics Test-18

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Mathematics Test-18

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Question 1
4 / -1

In a group of 600 students, there are 575 students who can speak Marathi and 300 can speak Hindi. What is the number of students who can speak Marathi only?

A
275

B
300

C
325

D
350

Solution

Total number of students \(=600\)

Let \(\mathrm{M}=\) Set of students speaking Marathi

Let \(\mathrm{H}=\) Set of students speaking Hindi

So, \(\mathrm{n}(\mathrm{M})=575, \mathrm{n}(\mathrm{H})=300\) and \(\mathrm{n}(\mathrm{M} \cup \mathrm{H})=600\)

We know that: \(\mathrm{n}(\mathrm{M} \cup \mathrm{H})=\mathrm{n}(\mathrm{M})+\mathrm{n}(\mathrm{H})-\mathrm{n}(\mathrm{M} \cap \mathrm{H})\)

\(\Rightarrow600=575+300-\mathrm{n}(\mathrm{M} \cap \mathrm{H})\)

\(\Rightarrow \mathrm{n}(\mathrm{M} \cap \mathrm{H})=875-600\)

\(=275\)

So, there are 275 common students among 600 who can speak both Marathi and Hindi.

Now, students who speak Marathi only = \(\mathrm{n}(\mathrm{M})-\mathrm{n}(\mathrm{M} \cap \mathrm{H})\)

\(=575-275\)

\(=300\)
Question 2
4 / -1

Directions For Questions

Direction: In the following question two equation numbered I and II are given. Solve the equation and answer the question.

...view full instructions

I. \(3 x^{2}-2 \sqrt{21} x+7=0\)

II. \(3 y^{2}+\sqrt{3 y}-2=0\)

A
\(x>y\)

B

C
\(x \geq y\)

D
\(x \leq y\)

Solution
Question 3
4 / -1

Directions For Questions

Direction: In the following question two equation numbered I and II are given. Solve the equation and answer the question.

...view full instructions

I. \(10 x^{2}-59 x+45=0\)

II. \(9 y^{2}-100 y+100=0\)

A
\(x>y\)

B

C
\(x \geq y\)

D
\(x=y\) or relationship between \(x\) and \(y\) cannot be established.

Solution
Question 4
4 / -1

Directions For Questions

Direction: In the following question two equation numbered I and II are given. Solve the equation and answer the question.

...view full instructions

I. \(2 x^{2}+11 x+12=0\)

II. \(y^{2}-y-2=0\)

A
\(x>y\)

B

C
\(x \geq y\)

D
\(x \leq y\)

Solution
Question 5
4 / -1

\(\left|\begin{array}{ccc}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|\) then :

A
\(\cos \alpha+\cos \beta+\cos \gamma=0\)

B
\(\cos \alpha \cdot \cos \beta \cdot \cos \gamma=0\)

C
\(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\)

D
\(\sum \cos \alpha \cos \beta=0\)

Solution

\(\left|\begin{array}{ccc}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|\)

\(\Rightarrow 1-\cos ^2 \gamma-\cos \alpha(\cos \alpha-\cos \beta \cos \gamma)+\cos \beta(\cos \alpha \cos \gamma-\)

\(\cos \beta)=-\cos \alpha(-\cos \beta \cos \gamma)+\cos \beta(\cos \alpha \cos \gamma)\)

\(\Rightarrow 1-\cos ^2 \gamma-\cos ^2 \alpha+\cos \alpha \cos \beta \cos \gamma+\cos \alpha \cos \beta \cos \gamma-\cos ^2 \beta=\)

\(\cos \alpha \cos \beta \cos \gamma+\cos \alpha \cos \beta \cos \gamma\)
Question 6
4 / -1

There are 20 points in a plane, how many triangles can be formed by these points if 5 are colinear?

A
1130

B
550

C
1129

D
1140

Solution

Given:

Number of points in plane n = 20.

Number of colinear points m = 5.

Number of triangles from by joining n points of which m are colinear \(={ }^n C_3 -\)\({ }^m C_3 \)

Therefore the number of triangles = \({ }^ {20}C_3 -{ }^5 C_3\)

\(=\frac{20!}{(20-3)!.{3!}}-\frac{5!}{(5-3)!.3!}\)

= 1140-10

= 1130
Question 7
4 / -1

A
75

B
110

C
395

D
225

Solution
Question 8
4 / -1

A pair of dice is thrown and the numbers appearing have sum greater than or equal to \(10\). The probability of getting a sum of \(10\) is:

A
\(\frac{2}{26}\)

B
\(\frac{1}{4}\)

C
\(\frac{1}{12}\)

D
\(\frac{5}{36}\)

Solution

Total no. of outcomes when two dice are thrown are:-

\((1,1)(1,2)(1,3)(1,4)(1,5)(1,6)\)

\((2,1)(2,2)(2,3)(2,4)(2,5)(2,6)\)

\((3,1)(3,2)(3,3)(3,4)(3,5)(3,6)\)

\((4,1)(4,2)(4,3)(4,4)(4,5)(4,6)\)

\((5,1)(5,2)(5,3)(5,4)(5,5)(5,6)\)

\((6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\)

So, \(n(S)=36\)

Sum of \(10\) in pair of die can be:

\((4,6),(5,5),(6,4)\)

Outcome of sum of \(10\) is \(n(E)=3\)

\(\therefore\) Probability of getting a total of \(10\) is \(=\frac{3}{36}=\frac{1}{12}\)
Question 9
4 / -1

What is the curve which passes through the point \((1,1)\) and whose slope is \(\frac{2 y}{x}\)?

A
Circle

B
Parabola

C
Ellipse

D
Hyperbola

Solution

Let the curve be \(y=f(x)\)
\(\therefore\) Slope of the tangent drawn at any point on the curve is \(\frac{d f(x)}{d x}=f^{\prime}(x)\)
Given that slope at any point on the curve is \(\frac{2 y}{x}\),
\(\Rightarrow \frac{d y}{d x}=\frac{2 y}{x}\)
\(\Rightarrow \int \frac{1}{y} d y=\int \frac{2}{x} d x\)
\(\Rightarrow \ln y=2 \ln x+c\)
Where \(\mathrm{c}\) is the integration constant,
Given that the curve passes through the point \((1,1),\)
\(\Rightarrow \mathrm{c}=0\)
\(\therefore \mathrm{y}=\mathrm{x}^{2}\) is the equation of the curve which is a parabola.
Question 10
4 / -1

A line makes an angle \(\alpha,\beta, \gamma\) with the \(X, Y, Z\) axes. Then \(\sin ^2 \alpha+\sin ^2 \beta+\) \(\sin ^2 \gamma\) is equal to:

A
\(1\)

B
\(2\)

C
\(\frac{3}{2}\)

D
\(4\)

Solution

For a vector,

\(\cos ^2(\alpha)+\cos ^2(\beta)+\cos ^2(\gamma)=1\)

\(\Rightarrow 1-\sin ^2(\alpha)+1-\sin ^2(\beta)+1-\sin ^2(\gamma)=1\)

\(\Rightarrow \sin ^2(\alpha)+\sin ^2(\beta)+\sin ^2(\gamma)=3-1\)

\(\Rightarrow \sin ^2(\alpha)+\sin ^2(\beta)+\sin ^2(\gamma)=2\)