SOLVED PAPER 2023 (CDS) (I) (Elementary Mathematics)

Total Questions: 100

91. (Question 91 and 92) Consider the following for the next three questions that follow. In the triangle ABC, AB = 6cm BC = 8cm and AC = 10cm The perpendicular dropped from B meets the side AC at D. A circle of radius BD (with centre B) cuts AB and BC at P and Q, respectively as shown in the figure.

What is the radius of the circle?

Correct Answer: (b) 4.8 cm
Solution:Area of ΔABC,
(1÷2) × AB × BC = (1÷2) × AC × BD
⇒ 6 × 8 = 10 × r
⇒ r = 48 ÷ 10 = 4.8cm

92. What is the length of QC?

Correct Answer: (d) 3.2 cm
Solution:QC = BC - r
= 8 - 4.8 = 3.2cm

93. If angle ∠ABD = theta then what is sin theta equal to?

Correct Answer: (c) 0.6
Solution:If ABC is a right-angle triangle,
(AB)² = AD × AC
(6)² = AD × 10
AD = 3.6 ÷ 10 = 3.6cm
sin θ = P ÷ H = 3.6 ÷ 6 = 0.6cm

94. (Question 94-96) Consider the following for the next three questions that follow.

In the figure given below, a circle is inscribed in a square PQRS.
A rectangle at the corner P that measures 4 cm x 2 cm and a square at the corner R are drawn.

What is the area of the circle?

Correct Answer: (a) 100π sq cm
Solution:
Let OAPB is a square and OA and
OB are the radius of the circle
∴ OA = OB = r
r² = (r - 2)² + (r - 4)²
r² = r² + 4 - 4r + r² + 16 - 8r
r² = 2r² + 20 - 12r
r² - 12r + 20 = 0
r² - 10r - 2r + 20 = 0
r(r - 10) - 2(r - 10) = 0
(r - 2)(r - 10) = 0
r = 10 (∵ r ≠ 2)
∴ Area of the circle = πr²
= 100π sqcm

95. What is the area of the smaller square?

Correct Answer: (d) 50(3 - 2√2) cm²
Solution:

96. What is the area of the shaded region?

Correct Answer: (b) (92 - 25π) cm²
Solution:

97. What is the ratio of the area of the circle to the area of the rectangle?

Correct Answer: (b) π ÷ √3
Solution:

98. What is the area of triangle AEC ?

Correct Answer: (a) r² ÷ √3
Solution:

99. What is the relation between x and y ?

Correct Answer: (c) x = 3y
Solution:Given, OB = BQ
∠OQB= ∠QOB = yº
By exterior angle theorem,
∠OBA = 2yº
Also, OA = OB = radius
∠OBA = ∠OAB = 2yº
In ΔOAB , ∠AOB = 180º - 4yº .....(i)
POQ is a straight line
⇒ x + 180º - 4y + yº = 180º
⇒ x = 3y

100. If y = 15 then what is angle ACB equal to ?

Correct Answer: (d) 60°
Solution:∵ ∠AOB = 180 - 4y
[from Equation (i)]
= 180 - 4 × 15 [ y = 15º , given]
= 180 - 60 = 120º
Hence, ∠ACB = (∠AOB÷2)
= 120 ÷ 2 = 60º