MATHEMATICS | Study Mode

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Question 21
A

2 < x < 3

B

-1 < x < 5

C

x < 1

D

x < 5

Explanation

 Solve for x in |x-2|<3

| x − 2 | < 3 

x − 2 < ± 3

Set up the positive portion of the ± solution.

x − 2 < 3

Move all terms not containing x to the right side

x < 5

Set up the negative portion of the ± solution.

x − 2 < − 3

Move all terms not containing x to the right side 

Add 2 to both sides 

x < − 3 + 2 

x < − 1

The solution to the equation includes both the positive and negative portions of the solution. 

− 1 < x < 5 (B) 


Correct option: B
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Question 22
A

14877

B

77814

C

81477

D

77148

Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: A
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Question 23
A

2-32-352

B

232-3-52

C

232-352

D

2-32-3-52

Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: A
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Question 24
MATHEMATICS Q 24 image
A

40

B

45

C

15

D

30

Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: A
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Question 25
MATHEMATICS Q 25 image
A

100

B

70

C

130

D

110

Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: C
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Question 26
A

42

B

66

C

12

D

30

Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: D
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Question 27
A

576

B

720

C

336

D

420

Explanation

Explanation not provided, please join discuss, your contributions are welcomed


Correct option: C
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Question 28
A

\(8\sqrt{2} cm\)

B

\(8\sqrt{3} cm\)

C

4 cm

D

8 cm

Explanation

Length of chord = \(2r \sin (\frac{\theta}{2})\)

= \(2 \times 8 \times \sin (\frac{90}{2})\)

= \(16 \times \frac{\sqrt{2}}{2}\)

= \(8\sqrt{2} cm\)


Correct option: A
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Question 29
A

16π cm2

B

32π cm2

C

4π cm2

D

8πcm2

Explanation

Diameter = 4\(\sqrt{3}\) cm

radius = 2\(\sqrt{3}\) cm

Area of major sector = \(\frac{\theta}{360} \times \pi r^{2}\)

\(\theta = 360 - 120 = 240°\)

= \(\frac{240}{360} \times \pi \times 12\)

= \(8\pi cm^{2}\)


Correct option: D
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Question 30
A

pair of parallel lines to PQ

B

perpendicular line to PQ

C

circle centre P

D

circle centre Q.

Explanation

The locus of points at a fixed distance from the point P is a circle with the given P at its centre.

 

The locus of points at a fixed distance from the point Q is a circle with the given point Q at its centre

 

The locus of points equidistant from two points P and Q i.e line PQ is the perpendicular bisector of the segment determined by the points

 

Hence, The locus of a point which is equidistant from the line PQ forms a perpendicular line to PQ


Correct option: B

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