CDS Maths(2) Questions and answers

Topics

General Knowledge
General Science
General English
Aptitude
General Computer Science
General Intellingence and Reasoning
Current Affairs
Exams
Elementary Mathematics
English Literature

Exams

These questions were asked in previous year CDS General Maths exam.You can practice these questions as real time exam in this link.

51). The shadow of a tower standing on a level plane is found to be 50 m longer when the Sun's elevation is \( 30^{\circ} \) than when it is \( 60^{\circ} \). What is the height of the tower?

A). \( 25\) \( m\)

B). \(25\sqrt{3} \) \( m\)

C). \( \Large \frac{25}{\sqrt{3}} \) \( m\)

D). \(30 \) \( m\)

View Answer

Correct Answer: \(25\sqrt{3} \) \( m\)

"

\( \Large tan 30 \^{\circ} = \frac{AB}{BC} \)

\( \Large \frac{1}{\sqrt{3}} = \frac{h}{50 + x} = > 50 + x = \sqrt{3}h \)

\( \Large x = \sqrt{3}h - 50 \) ... (i)

and \( \Large tan 60 \^{\circ} = \frac{AB}{BD} \)

\( \Large \sqrt{3} = \frac{h}{x} \) ... (ii)

Putting value of x from equation (i) in equation(ii) we get

\( \Large h = \sqrt{3} \times x = \sqrt{3} \times (\sqrt{3}h - 50) \)

\( \Large h = 3h - 50\sqrt{3} \)

\( \Large 2h = 50\sqrt{3} \)

\( \Large h = 25\sqrt{3}m\)

52). The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same . plane is \( 60^{\circ} \) and the angle of elevation of the top of the second tower from the foot of the first tower is \( 30^{\circ} \). The distance between the two towers is In times the height of the shorter tower. What is m equal to?

A). \( \sqrt{2}\)

B). \( \sqrt{3}\)

C). \( \Large \frac{1}{2} \)

D). \( \Large \frac{1}{3} \)

View Answer

Correct Answer: \( \sqrt{3}\)

"

Here CD -> first tower and AB -> second tower

In \( \Large \vartriangle ABC \),

\( \Large tan 30 \^{\circ} =\frac{AB}{BC} \)

\( \Large tan 30 \^{\circ} =\frac{h}{mh} \)

\( \Large \frac{1}{\sqrt{3}} = \frac{h}{mh} \)

\( \Large m = \sqrt{3} \)

53). If \( \large \cos x+ \sec x= 2\) then what is \( \large \cos^{n}x+ \sec^{n}x\) equal to, where n is a positive integer?

A). 2

B). \( 2^{n-2}\)

C). \( 2^{n-1}\)

D). \(2^{n} \)

View Answer

Correct Answer: 2

Here cos x + sec x = 2

Now for n = 2

\( \Large cos^{2}x + sec^{2}x = (cos x + sec x)^{2} - 2 \)

\( \Large (2)^{2} = 2 \)

For n = 3,

\( \Large cos^{3} x + sec^{3}x = (cos x + sec x) (cos^{2}x - 1 + sec^{2} x) \)

=(2)(2 -1) = 2

Hence it can be concluded that if cos x + sec x = 2, then \( \Large cos^{n}x + sec^{n} = 2 \), when n is positive integer.

54). What is \(sin 25^{\circ} sin 35^{\circ} sec 65^{\circ} sec 55^{\circ} \) equal to?

A). -1

B). 0

C). \( \Large \frac{1}{2} \)

D). 1

View Answer

Correct Answer: 1

\( \Large sin 25 \^{\circ} sin 35 \^{\circ} sec 66 \^{\circ} sec 55 \^{\circ}

= (sin 25 \^{\circ} sec 65 \^{\circ} ) (sin 35 \^{\circ} sec 55 \^{\circ} )\)

\( \Large = \frac{sin 25 \^{\circ} }{ cos65\^{\circ} } \times \frac{sin 35 \^{\circ}}{cos 55 \^{\circ} }\)

\( \Large =\frac{ sin(90 \^{\circ} - 65 \^{\circ} )}{cos65 \^{\circ} } \times \frac{sin(90 \^{\circ} - 55 \^{\circ})}{cos 55 \^{\circ} } \)

\( \Large \frac{cos65 \^{\circ}}{cos65 \^{\circ} } \times \frac{cos 55 \^{\circ} }{cos 55 \^{\circ} } = 1 \times 1 = 1 \)

55). The value of \( cos 25^{\circ} - sin 25^{\circ}\) is

A). positive but less than 1

B). positive but greater than 1

C). negative

D). 0

View Answer

Correct Answer: positive but less than 1

\( \Large cos 25 \^{\circ} - sin 25 \^{\circ} \)

\( \Large = cos 25 \^{\circ} - sin (90 \^{\circ} - 65 \^{\circ} ) \)

\( \Large = cos 25 \^{\circ} - cos 65 \^{\circ} \)

\( \Large = 2 sin \left(\frac{25 \^{\circ} + 65 \^{\circ} }{2}\right).sin \left( \frac{65 \^{\circ} -25 \^{\circ}}{2} \right)\)

\( \Large 2 sin 45 \^{\circ} . sin 20 \^{\circ} \)

\( \Large \sqrt{2}sin20 \^{\circ} \)

Hence value of \( \Large \sqrt{2} sin 20 \^{\circ} \) is positive but 1ess than 1. Hence option (A)is correct.

56). In a right-angled triangle ABC, right-angled at B, if \( \cos A\) = \( \Large \frac{4}{5} \). then what is \( \sin C\)equal to?

A). \( \Large \frac{3}{5} \)

B). \( \Large \frac{4}{5} \)

C). \( \Large \frac{3}{4} \)

D). \( \Large \frac{2}{5} \)

View Answer

Correct Answer: \( \Large \frac{4}{5} \)

"

\( \Large cos A = \frac{4}{5} = \frac{AB}{AC} \)

Now in \( \Large \vartriangle ABC, (AC)^{2} = (AB)^{2} + (BC)^{2} \)

\( \Large (5)^{2} = (4)^{2} + (BC)^{2} \)

\( \Large (BC)^{2}= 25 - 16 = 9 \)

BC = 3,

Now \( \Large sin C = \frac{AB}{AC} = \frac{4}{5} \)

57). If \( \alpha\) and \( \beta \) are complementary angles, then what is \( \large \sqrt{\cos\alpha cosec\beta - \cos \alpha \sin \beta}\) equal to?

A). \(\cos \beta \)

B). \( \cos \alpha \)

C). \( \sin \alpha\)

D). \( -\tan \beta\)

View Answer

Correct Answer: \( \sin \alpha\)

Here \( \Large \alpha \) and \( \Large \beta \) are complementary angles

\( \Large \alpha + \beta = 90 \^{\circ} \)

Now \( \Large \sqrt{cos \alpha .cosec \beta - cos \alpha sin \beta}\)

\( \Large = \sqrt{\frac{cos \alpha}{sin \beta } - cos \alpha.sin \beta }\)

\( \Large = \sqrt{\frac{cos \alpha - cos \alpha sin^{2} \beta }{sin \beta }} \)

\( \Large = \sqrt{\frac{cos \alpha(1 - sin^{2} \beta)}{sin \beta }} \)

\( \Large = \sqrt{\frac{cos \alpha .cos^{2} \beta}{sin \beta }} \)

Now \( \Large \alpha + \beta = 90 \^{\circ} \)

\( \Large \beta = 90 - \alpha \)

\( \Large cos \beta = cos(90 - \alpha) \)

\( \Large cos \beta = sin \alpha \)

Again \( \Large \beta = 90 - \alpha \)

\( \Large sin \beta = sin (90 - \alpha) \)

\( \Large sin \beta = cos \alpha \)

\( \Large cos \alpha.cosec \beta -cos \alpha sin \beta \)

\( \Large = \sqrt{\frac{cos \alpha -cos^{2} \beta}{sin \beta}}\)

\( \Large \sqrt{\frac{cos \alpha.sin^{2}\alpha}{cos \alpha}} = sin \alpha \)

58). If \( 2 \cot \theta =3\), then what is \( \Large \frac{2 \cos\theta - \sin\theta}{2 \cos\theta + \sin\theta } \) equal to ?

A). \( \Large \frac{2}{3} \)

B). \( \Large \frac{1}{3} \)

C). \( \Large \frac{1}{2} \)

D). \( \Large \frac{3}{4} \)

View Answer

Correct Answer: \( \Large \frac{1}{2} \)

Here \( \Large 2cot \theta = 3 \)

\( \Large cot \theta = \frac{3}{2} \)

\( \Large \frac{cos \theta }{sin \theta } = \frac{3}{2} \)

Now \( \Large \frac{2cos \theta -sin \theta}{2cos \theta + sin \theta } \) Dividing numerator and denominator by \( \Large sin \theta \) we get

\( \Large \frac{2\frac{cos \theta }{sin \theta } - 1}{2\frac{cos \theta }{sin \theta} + 1} = \frac{2 \left(\frac{3}{2}\right) - 1}{2 \left(\frac{3}{2}\right) + 1} = \frac{2}{4} = \frac{1}{2}\)

59). If \( \sin \theta \cos \theta\)= \( \Large \frac{1}{2} \), then what is \( \large \sin^{6}\theta + \cos^{6}\theta\) equal to?

A). 1

B). 2

C). 3

D). \( \Large \frac{1}{4} \)

View Answer

Correct Answer: \( \Large \frac{1}{4} \)

Here \( \Large sin \theta cos \theta = \frac{1}{2} \)

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

Now, \( \Large sin^{2} \theta + cos^{6} \theta = (sin 45 \^{\circ} )^{6} + (cos 45 \^{\circ})^{6}\)

=\( \Large \left(\frac{1}{\sqrt{2}}\right)^{6} + \left(\frac{1}{\sqrt{2}}\right)^{6} \)

=\( \Large \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \)

60). If \( \sec \theta\)+\( \tan \theta\)= 2, what is the value of \( \sec \theta\)?

A). \( \Large \frac{3}{2} \)

B). \(\sqrt{2}\)

C). \( \Large \frac{5}{2} \)

D). \( \Large \frac{5}{4} \)

View Answer

Correct Answer: \( \Large \frac{5}{4} \)

Here \( \Large sec \theta + tan \theta = 2 \) ... (i)

Now we know that

\( \Large sec^{2} \theta +tan^{2} \theta = 1\)

\( \Large (sec \theta + tan \theta )(sec \theta - tan \theta )= 1 \)

\( \Large (2)(sec \theta - tan \theta )= 1 \)

\( \Large (sec \theta - tan \theta )= \frac{1}{2} \) ... (ii)

Solving equation (i) and (ii) we get

\( \Large (sec \theta + tan \theta )= 2 \)

\( \Large (sec \theta - tan \theta )= \frac{1}{2} \)

---------------------------------

\( \Large 2sec \theta = \frac{5}{2} \)

\( \Large sec \theta =\frac{5}{4}\)

3	4	5	6	7	8	9

Go to :

Total Pages : 10