mensuration Model Questions & Answers, Practice Test for ssc mts tier 1 2024

Answer: (a)

Volume of frustum

= ${π H}/3 (R^2 + r^2 + Rr)$

= ${π}/3H{(√{Q/π})^2 + (√{P/π})^2 + √{Q/π} √{P/π}}$

= ${πH}/3 (Q/π + P/π + {√{PQ}}/π)$

= $H/2 (P + Q + √{PQ})$

Answer: (a)

Answer: (a)

∵ ∠PCT + ∠PCB = π

∠PCB = π – (π – b°) = b°

In ΔBPC,

∠PCB + ∠BPC + ∠PBC = π

⇒ ∠PBC = π – ∠PCB – ∠BPC

= π – b° – a°

∵∠ABE + ∠EBC = π (linear pair)

⇒ ∠ABE = π – ∠PBC = π– (π – b° – a°)

= a° + b° (∵ ∠PBC = ∠EBC)

Now, Sum of two interior angles = Exterior angle

∴ in ΔABE, ∠EAB + ∠ABE = ∠BES ⇒ c° + b° + a° = x°

∴ x° = a° + b° + c°

Answer: (d)

Diagonal of a cube of side 'a'

= $√3$. a = l

∴ a = $l/√3$

total surface area of cube = 6 × $a^2$

= $6 × (l/√3)^2 = 2l^2$

Answer: (a)

Let two cocentric circular rings with centre O and

Radius of large ring = $r_2$

Radius of smaller ring = $r_1$

Area of circular both

= Area of larger ring – Area of smaller ring ...(1)

Given circumference of larger ring = 3 × circumference of smaller ring

⇒ 2π$r_2 = 3 × 2πr_1$

⇒ $r_2 = 3r_1$

Also given

Area of circular both = n (area of smaller ring) ...(2)

Comparing (1) & (2) we have

Area of larger ring – Area of smaller ring

= n(Area of smaller ring)

⇒ $πr_2^2 - π1_1^2 = n(πr_1^2)$

⇒ $πr_2^2 (n + 1) πr_1^2$

⇒ $(n + 1)r_1^2 = r_2^2$

⇒ $(n + 1)r_1^2 = (3r_1)^2$

(∵ $r_2 = 31_1$)

⇒ n + 1 = 9

⇒ n = 8

∴ Option (c) is correct.