Mathematics Test

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Mathematics Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Consider the following circles
I. \(x^{2}+y^{2}+4 x-6 y-12=0\)
II. \(x^{2}+y^{2}-12 x-14 y+60=0\)
III. \(x^{2}+y^{2}-10 x+8 y+18=0\)
Which of the above circle has equal area?

A
I, II

B
I, III

C
II, III

D
I, II, III

Solution

I. Centre \(=(-2,3)\)
Radius \(=\sqrt{(-2)^{2}+(3)^{2}+12}\)
\(=\sqrt{25}=5\)
\(\therefore\) Area \(=25 \pi\)
II. Centre \(=(6,7)\)
Radius \(=\sqrt{6^{2}+7^{2}-60}=\sqrt{25}=5\)
\(\therefore\) Area \(=25 \pi\)
III. Centre \(=(5,-4)\)
Radius \(=\sqrt{5^{2}+(-4)^{2}-18}=\sqrt{23}\)
\(\therefore\) Area \(=23 \pi\)
So I \(\&\) II have equal area
Question 2
1 / -0

What is the value of \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x} ?\)

A
0

B
1/2

C
1

D
-1/2

Solution

\(\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}=\left(\frac{0}{0}\right)\) form
Using L' Hospital rule
\(\lim _{x \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+x}}}{1}=\lim _{x \rightarrow 0} \frac{1}{\sqrt[3]{1+x}}=\frac{1}{2}\)
Question 3
1 / -0

The area of a rectangle will be maximum for the given perimeter, when rectangle is a

A
Parallelogram

B
trapezium

C
square

D
None of these

Solution

We know that, perimeter of a rectangle
\(S=2(x+y)\) where \(x \& y\) are adjacent sides
\(\Rightarrow y=\frac{s-2 x}{2}\)
Now area of rectangle \(A=x y=\frac{x}{2}(S-2 x)=\frac{1}{2}\left(S x-2 x^{2}\right)\)
On differentiating w.r.t x of A, we get

\(\frac{d A}{d x}=\frac{1}{2}(S-4 x)=0\)
\(\Rightarrow x=\frac{s}{4}\) and \(y=\frac{s}{4}\)
Again \(\frac{d^{2} A}{d x^{2}}=-v e\)
Hence the area of rectangle will be maximum when rectangle is a square
Question 4
1 / -0

What is the area bounded by the lines x = 0, y = 0 and x + y + 2 = 0 ?

A
1/2 sq unit

B
1 sq unit

C
2 sq units

D
4 sq units

Solution
Question 5
1 / -0

Kamal & Monika appeared for an interview for 2 vacancies. The probability of Kamal’s selection is 1/3 & that of Monika’s selection is 1/5. Find the probability that only one of them will be selected.

A
2/5

B
1/5

C
5/9

D
2/3

Solution

Required probability \(=\frac{1}{3} \times\left(1-\frac{1}{5}\right)+\frac{1}{5} \times\left(1-\frac{1}{3}\right)\)
\(=\frac{4}{15}+\frac{2}{15}=\frac{6}{15}=\frac{2}{5}\)
Question 6
1 / -0

If two vertices of an equilateral triangle be (0,0), (3, √3). Find the third vertex.

A
(3, √3)

B
(0, 2√3)

C
(1, 2)

D
(√2, √3 )

Solution

\(O(0,0)\) and \(A(3, \sqrt{3})\) be the given points and Let \(B(x, y)\) be the third vertex of equilateral \(\Delta O A B\). Then \(O A=O B=A B\) \(\Rightarrow \mathrm{OA}^{2}=\mathrm{OB}^{2}=\mathrm{AB}^{2}\)
We have, \(O A^{2}=(3-0)^{2}+(\sqrt{3}-0)^{2}=12\)
\(O B^{2}=x^{2}+y^{2}\)
And, \(A B^{2}=(x-3)^{2}+(y-\sqrt{3})^{2}\)
\(\Rightarrow A B^{2}=x^{2}+y^{2}-6 x-2 \sqrt{3} y+12\)
\(\therefore O A^{2}=O B^{2}=A B^{2}\)
\(\Rightarrow \mathrm{OA}^{2}=\mathrm{OB}^{2}\) and \(\mathrm{OB}^{2}=\mathrm{AB}^{2}\)
\(\Rightarrow x^{2}+y^{2}=12\)
and \(x^{2}+y^{2}=x^{2}+y^{2}-6 x-2 \sqrt{3} y+12\)

\(\Rightarrow x^{2}+y^{2}=12\) and \(6 x+2 \sqrt{3} y=12\)
\(\Rightarrow x^{2}+y^{2}=12 \& 3 x+\sqrt{3} y=6\)
\(\Rightarrow x^{2}+\left(\frac{6-3 x}{\sqrt{3}}\right)^{2}=12 \quad\left[\begin{array}{c}\because 3 x+\sqrt{3} y=6 \\ \therefore y=\frac{6-3 x}{\sqrt{3}}\end{array}\right]\)
\(\Rightarrow 3 x^{2}+(6-3 x)^{2}=36\)
\(\Rightarrow 12 x^{2}-36 x=0\)
\(\Rightarrow x=0,3\)
\(\mathrm{x}=0 \Rightarrow \sqrt{3} y=6 \Rightarrow y=\frac{6}{\sqrt{3}}=2 \sqrt{3}[\text {Putting } x=0 \text { in } 3 x+\sqrt{3} y=6]\)
And \(x=3 \Rightarrow 9+\sqrt{3} y=6 \Rightarrow y=\frac{6-9}{\sqrt{3}}=-\sqrt{3}[\text { Putting } x=3 \text { in } 3 x+\sqrt{3} y=6]\)
Hence the coordinates of the third vertex B are \((0,2 \sqrt{3})\) or, \((3,-\sqrt{3})\)
Question 7
1 / -0

If \(A+i B=\frac{4+2 i}{1-2 i},\) where \(i=\sqrt{-1},\) then what is the value of \(A ?\)

A
–8

B
0

C
4

D
8

Solution

Given \(A+i B=\frac{4+2 i}{1-2 i} \times \frac{1+2 i}{i+2 i}\)
\(=\frac{4+10 i-4}{1+4}=\frac{10 i}{5}=2 i\)
\(\Rightarrow A+i B=0+2 i\)
\(\Rightarrow A=0 \& B=2\)
Question 8
1 / -0

If \(y=\log \left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right),\) then \(\frac{d y}{d x}\) is equal to

A

\(\frac{1}{\sqrt{x(1+x)}}\)

B

\(\frac{1}{\sqrt{x(1-x)}}\)

C

\(\frac{\sqrt{x}}{1+x}\)

D

None of these

Solution

Given \(y=\log \left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)\)
\(y=\log (1+\sqrt{x})-\log (1-\sqrt{x})\)
\(\frac{d y}{d x}=\frac{1}{1+\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}-\frac{1}{(1-\sqrt{x})}\left(-\frac{1}{2 \sqrt{x}}\right)\)
\(=\frac{1}{2 \sqrt{x}}\left[\frac{1}{1+\sqrt{x}}+\frac{1}{1-\sqrt{x}}\right]\)
\(=\frac{1}{2 \sqrt{x}}\left[\frac{1-\sqrt{x}+1+\sqrt{x}}{1-x}\right]\)
\(=\frac{1}{2 \sqrt{x}} \cdot \frac{2}{(1-x)}=\frac{1}{\sqrt{x}(1-x)}\)
Question 9
1 / -0

The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle.

A
39.2 cm.

B
2.56 cm.

C
3.92 cm.

D
25.6 cm.

Solution

Let the radius of the circle be \(r \mathrm{cm}\). Then,
Diameter \(=2 r \mathrm{cm} \&\) circumference \(=2 \pi r \mathrm{cm}\)
It is given that the circumference exceeds
the diameter by \(16.8 \mathrm{cm}\)
\(\therefore\) Circumference \(=\) Diameter +16.8 \(\Rightarrow 2 \pi r=2 r+16.8\)

\(\Rightarrow 2 \times \frac{22}{7} \times r=2 r+16.8\)
\(\Rightarrow 44 \mathrm{r}=14 \mathrm{r}+16.8 \times 7\)
\(\Rightarrow 44 \mathrm{r}-14 \mathrm{r}=117.6\)
\(\Rightarrow 30 \mathrm{r}=117.6\)
\(\Rightarrow \mathrm{r}=\frac{117.6}{30}=3.92 \mathrm{cm}\)
Question 10
1 / -0

The barrel of a fountain pen , cylindrical in shape , is 7cm long and 5mm in diameter. A full barrelof ink in the pen will be used up on writing 330 words on an average . How many words would useup a bottle of ink containing one fifth of a litre?

A
36000

B
48000

C
24500

D
41625

Solution

We have,
\[
\begin{array}{l}
\text { Volume of a barrel }=\left(\frac{22}{7} \times 0.25 \times 0.25 \times 7\right) \mathrm{cm}^{3}=1.375 \mathrm{cm}^{3} \\
\text { Volume of ink in the bottle }=\frac{1}{5} \text { litre }=\frac{1000}{5} \mathrm{cm}^{3}=200 \mathrm{cm}^{3}
\end{array}
\]
\(\therefore\) Total number of barrels that can be filled from the given volume of ink \(=\frac{200}{1.375}\)
So, required numbers of words \(=\frac{200}{1.375} \times 330=48000\)

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Mathematics Test - 27

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Mathematics Test - 27

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Question 1

I, II

I, III

II, III

I, II, III

Solution

Question 2

0

1/2

1

-1/2

Solution

Question 3

Parallelogram

trapezium

square

None of these

Solution

Question 4

1/2 sq unit

1 sq unit

2 sq units

4 sq units

Solution

Question 5

2/5

1/5

5/9

2/3

Solution

Question 6

(3, √3)

(0, 2√3)

(1, 2)

(√2, √3 )

Solution

Question 7

–8

0

4

8

Solution

Question 8

\(\frac{1}{\sqrt{x(1+x)}}\)

\(\frac{1}{\sqrt{x(1-x)}}\)

\(\frac{\sqrt{x}}{1+x}\)

None of these

Solution

Question 9

39.2 cm.

2.56 cm.

3.92 cm.

25.6 cm.

Solution

Question 10

36000

48000

24500

41625

Solution

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