CDS Maths(2) Questions and answers

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

41). If 64 identical small spheres are made out of a big of sphere of diameter 8 cm, what is the surface area of each small sphere?

A). \( \pi\) \(cm^{2} \)

B). \( 2\pi\) \(cm^{2} \)

C). \( 4\pi\) \(cm^{2} \)

D). \( 8\pi\) \(cm^{2} \)

View Answer

Correct Answer: \( 4\pi\) \(cm^{2} \)

Volume of big sphere = \( \Large \frac{4}{3} \pi R^{3} \)

= \( \Large \frac{4}{3} \pi 4^{3} = \frac{(4)^{4}}{3} \pi cm^{3} \)

Now volume of each small sphere

\( \Large \frac{(4)^{4} \pi }{3 \times (4)^{3}} = \frac{4 \pi }{3} cm^{3} \)

Now volume of small sphere = \( \Large \frac{4}{3} \pi r^{3} \)

\( \Large \frac{4}{3} \pi = \frac{4}{3} \pi r^{3} \)

Now surface area of each small sphere = \( \Large 4 \pi r^{2} = 4 \pi cm^{2} \)

42). AB and CD are 0 diameters of a circle of radius r and they are mutually perpendicular. What is the ratio of the area of the circle to the area of the triangle ACD.?

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

B). \( \pi\)

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

D). \( 2\pi\)

View Answer

Correct Answer: \( \pi\)

"

Area of circle = \( \Large \pi r^{2} \)

and area of \( \Large \vartriangle ACD = \frac{1}{2} \times (2r) \times r = r^{2}\)

Ratio of the area of the circle t the triangle ACD = \( \Large \frac{ \pi r^{2}}{r^{2}} = \pi \)

43). A circle of radius 10 cm has an equilateral triangle inscribed in it. The length of the perpendicular drawn from the centre to any side of the triangle is

A). \( 2.5\)\( \sqrt{3}\) \( cm\)

B). \( 5\)\( \sqrt{3}\) \( cm\)

C). \( 10\)\( \sqrt{3}\) \( cm\)

D). None of these

View Answer

44). The dimensions ofa field are 15 m by 12 in. A pit 8 m long, 2.5 m wide and 2 m deep is dug in one corner of the iield and the earth removed is evenly spread over the remaining area of the field. The level of the field is raised by

A). 15 cm

B). 20 cm

C). 25 cm

D). \( \Large \frac{200}{9} \) \( cm\)

View Answer

Correct Answer: 25 cm

"

Area of field = \( \Large 15 x 12 = 180m^{2} \)

Area of remaining field = 9.5 x 8 + 7 x 12

\( \Large = 76 + 84 = 160m^{2} \)

Now, Volume of earth\( \Large = (8 x 2.5 X 2)m^{3} = 40m^{3} \)

(Area of remaining field) x h = volume of earth

160 X h = 40

\( \Large h = \frac{40}{160}m = \frac{1}{4}m \)

h = 25 cm

The level of the field is raised by 25cm.

Hence option (C)is correct.

45). The sides of a right-angled triangle are equal to three consecutive numbers expressed in centimeters. What can be the area of such a triangle?

A). \( 6\) \( cm^{2}\)

B). \( 8\) \( cm^{2}\)

C). \( 10\) \( cm^{2}\)

D). \( 12\) \( cm^{2}\)

View Answer

Correct Answer: \( 6\) \( cm^{2}\)

The sides of a right-angled triangle are equal three consecutive numbers.

"

Now \( \Large (AC)^{2} = (AB)^{2} + (BC)^{2} \)

\( \Large (m + 2)^{2} = m^{2} + (m + 1)^{2} \)

\( \Large m^{2} + 4m + 4 = m^{2} + m^{2} + 2m + 1 \)

\( \Large m^{2} - 3m + m - 3 = 0 \)

\( \Large m(m-3)+ 1(m-3) = 0 \)

(m+1)(m-3)=0

m=3 [Where \( \Large m \ne = 1 \)]

Area of triangle = \( \Large \frac{1}{2} x (m + 1) x m \)

\( \Large = \frac{1}{2} x 4 x 3 = 6 cm^{2} \)

46). How many circular plates of diameter d can be taken out of a square plate of side 2d with minimum loss of material?

A). 8

B). 6

C). 4

D). 2

View Answer

Correct Answer: 4

Area of square plate \( \Large = 2d x 2d = 4d^{2} \)

Now Area of circular plate = \( \Large \pi \left(\frac{d}{2}\right)^{2} = \frac{ \pi d^{2}}{4} \)

\( \Large \frac{22}{7 \times 4}d^{2} = \frac{11}{14}d^{2} \)

Number of circular plate taken out

\( \Large =14 \times \frac{4d^{2}}{11d^{2}} = \frac{14 \times 4}{11} \)

\( \Large \frac{56}{11} = 5.04 \)

4Circular plates of diameter d can be taken out of a square plate of side 2d with minimum loss of material.

47). The volume of a hollow cube is \( 216\)\( x^{2}\). The surface area of the largest sphere which can be enclosed in it is

A). \( 18\) \( \pi\)\(x^{2} \)

B). \( 27\) \( \pi\)\(x^{2} \)

C). \( 36\) \( \pi\)\(x^{2} \)

D). \( 72\) \( \pi\)\(x^{2} \)

View Answer

Correct Answer: \( 36\) \( \pi\)\(x^{2} \)

Volume of hollow cube \( \Large = 216x^{3} \)

a = 6x

where a is side of cube

Now radius of sphere enclosed in cube \( \Large =\frac{6x}{2} = 3x\)

Now surface area of largest sphere enclosed in cube = \( \Large 4 \pi (3x)^{2} = 36 \pi x^{2}\)

48). What is the diameter of the largest circle lying on the surface of a sphere of surface area 616 square cm?

A). 14 cm

B). 10.5 cm

C). 7cm

D). 3.5 cm

View Answer

Correct Answer: 14 cm

Surface area of sphere \( \Large = 4 \pi r^{2} \)

\( \Large 616 = 4 \times \frac{22}{7} \times r^{2} \)

\( \Large r^{2} = \frac{616 x 7}{88} \)

r = 7 cm

Diameter of largest circle lying on the surface sphere = 2(7) = 14 cm

49). What is the total area of three equilateral triangles inscribed in a semicircle of radius 2 cm?

A). \( 12\) \( cm^{2}\)

B). \( \Large \frac{3 \sqrt{3}}{4} \) \(cm^{2} \)

C). \( \Large \frac{9\sqrt{3}}{4} \) \(cm \)

D). \( 3\)\( \sqrt{3}\) \( cm^{2}\)

View Answer

Correct Answer: \( 3\)\( \sqrt{3}\) \( cm^{2}\)

"

Area of three equilateral triangle = Area of trapezium ABCD

\( \Large = \frac{1}{2} \times BM \times (BC + AD) \)

Now in \( \Large \vartriangle BMO \)

\( \Large tan 60 = \frac{BM}{OM} \)

\( \Large \sqrt{3} = \frac{BM}{1} \)

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

Area of three equilateral triangle

\( \Large =\frac{1}{2} \times BM \times (BC + AD) \)

\( \Large = \frac{1}{2} \times \sqrt{3} \times (2 + 4) \)

\( \Large =\frac{1}{2} \times \sqrt{3} \times 6 = 3\sqrt{3}cm^{2} \)

50). In a trapezium, the two non-parallel sides are equal in length, each being of 5 units. The parallel sides are at a distance of 3 units apart. If the smaller side-of the parallel sides is of length 2 units, then the sum of the diagonals of the trapezium is

A). \(10 \)\(\sqrt{5} \) \( units\)

B). \(6 \)\(\sqrt{5} \) \( units\)

C). \(5 \)\(\sqrt{5} \) \( units\)

D). \(3 \)\(\sqrt{5} \) \( units\)

View Answer

Correct Answer: \(6 \)\(\sqrt{5} \) \( units\)

"

Now in \( \Large \vartriangle AFD \)

\( \Large (DF)^{2} = (AD)^{2} - (AF)^{2} \)

25 - 9 = 16

DF = 4 = EC

In \( \Large \vartriangle BED, \)

\( \Large (BD)^{2} = (DE)^{2} + (BE)^{2}\)

\( \Large = (6)^{2} + (3)^{2} \)

\( \Large (BD)^{2} = 36 + 9 = 45 \)

\( \Large BD = 3\sqrt{5}\) units

Now in \( \Large \vartriangle AFC \);

\( \Large (AC)^{2} = (AF)^{2} + (FC)^{2} \)

\( \Large = (3)^{2} + (6)^{2} = 9 + 36 = 45 \)

Sum of diagonals of trapezium = AC + BD = \( \Large 3\sqrt{5} + 3\sqrt{5} \)

\( \Large = 6\sqrt{5} units. \)

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