SOLVED PAPER 2024 (CDS) (I) (Mathematics)

Total Questions: 100

91. (Questions 91-93) Consider the following for the next three (03) items that follow :

Consider two identical semicircles and one circle inscribed in a rectangle of length 10 cm as shown in the figure given below..

(Take, π = 3.14 and √2 = 1.4)

What is the area of triangle EOF?

Correct Answer: (c) 12.5 cm²
Solution:

92. What is the area of trapezium AEFB?

Correct Answer: (a) 30 cm²
Solution:

93. What is the area of the shaded region?

Correct Answer: (b) 14.25 cm²
Solution:

94. (Questions 94 and 95) Consider the following for the next to (02) items that follow :

Consider a circle of area 9pi square unit and an equitateral triangle ABC as shown in the figure given below.

What is the length of the side of the triangle ABC?

Correct Answer: (b) 4√3 unit
Solution:

95. What is the area of the shaded region?

Correct Answer: (a) 3(π + √3) cm²
Solution:

96. (Questions 96-98) Consider the following for the next three (03) items that follows:

Two circles with centres at O₁ and O₂ touching each other are placed inside a rectangle of sides 9 cm and 8 cm as shown in the figure given below.


What is the sum of the areas of the two circles ?

Correct Answer: (a) 17π cm²
Solution:

97. Which one of the following is correct in respect of angle 0?

Correct Answer: (c) 45° < θ < 60°
Solution:

98. What is the area of the shaded region ?

Correct Answer: (d) (240 - 12π - πθ)/24 unit²
Solution:

99. (Questions 99 and 100) Consider the following for the next two (02) items that follow :

Let ABCD be the diameter of a circle of radius 6 cm. The lengths AB, BC and CD are equal. Semi- circles are drawn with AB and BD as diameters as shown in the figure given below.


What is the ratio of the area of the shaded region to that of the non-shaded region?

Correct Answer: (a) 2 : 7
Solution:

100. What is the perimeter of the shaded region?

Correct Answer: (d) 12π cm
Solution:Perimeter of shaded region,
= Perimeter of AED + Perimeter of AFB + Perimeter of BGD
= π(6) + π(2) + π(4)
= 6π + 2π + 4π = 12π