aptitude mensuration mcq pdf Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024

Answer: (b)

Let ∠XOB = θ

In ΔOXB,

∠XOB + ∠OBX + ∠OXB = 180°

⇒ θ + 45° + ∠OXB = 180°

⇒ ∠OXB= 180° – 45° – θ

⇒ ∠OXB= 135° – θ

Here, ∠OXA + ∠OXB = 180°

⇒ ∠OXA + 135° – θ = 180°

⇒ ∠OXA= 45° + θ

In ΔOAX,

AO = AX

∴ ∠OXA= ∠AOX = 45° + θ

Since, ∠AOX+ ∠XOB = 90°

⇒ 45 + θ + θ ∼ = 90°

⇒ 2θ = 45°

∴ θ = 22.5°

Answer: (b)

Diameter of a cone = 7 cm

∴ Radius of cone = $7/2$ cm

Slant height of a right circular cone (l) = 10 cm

∴ Lateral surface area of a cone = π r l

= ${22}/7 × 7/2 × 10 = 11 × 10 = cm^2$

Answer: (a)

In ΔADC, ∠ADB = 180° – ∠ADC = 180° – 60° = 120°

∠DAC = 180° – 60° – 30° = 90°

Again, in right angled ΔDAC,

tan 60° = ${AC}/{DA} ⇒ √3 = a/x ⇒ x = a/√3$

Answer: (a)

Curved surface area of the well = 2 πrh

= $2 × {22}/7 × 2 × 14 = 176 m^2$

∴ Expense of getting per square metre plastered = Rs.25

∴ Expense of 176 $m^2$ = 176 × 25

= Rs.4400

Answer: (c)

From the given figure. We know that sum of all angles is 180° because of straight lines.

∠LOQ +∠QOP + ∠POM = 180°

∴ (x° + 20°) + 50° + (x° – 10°) = 180°

⇒ 2x° + 60° = 180°

⇒ 2x° = 120°

x° = 60°