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

Answer: (a)

Surface area of sphere, $S_1 = 4πr^2$

If radius is 2r, then surface area of sphere,

$S_2 =4π(2r)^2 = 16πr^2$

∴ $S_2 = 4S_1$

Hence, it increases four times.

Answer: (b)

In ΔABD

cos 60° = ${2^2 + 5^2 - BD^2}/{2.2.5}$

$1/2 = {4 + 25 - BD^2}/{20}$

$BD^2$ = 29 – 10

BD = $√{19}$

In ΔBCD

cos 120° = ${3^2 + CD^2 - BD^2}/{2.3.CD}$

- $1/2 = {9 + CD^2 - ⊥ 9}/{6 CD}$

$CD^2$ – 10 = –3 CD

$CD^2$ + 3 CD = x

Let CD = x

$x^2$ + 3x – 10 = 0

(x + 5) (x – 2) = 0

x ≠ –5

x = 2

CD = 2 cm

So, option (a) is correct.

Answer: (c)

Let side of a cube be a unit.

∴ Volume of cube, V = $a^3$

and total surface area of cube, S = 6 $(a)^2$

V = $a^3$

Squaring on both sides

$V^2 = a^6 ⇒ V^3 = (a^2)^3$

$V^2 = (S/6)^3 ⇒ 216V^2 = S^3$

Answer: (a)

Given, AC = $√3$ r

AB = $√3/2$ r

In ΔOAB,

$(OB)^2 = (OA)^2 – (AB)^2$

= $r^2 - (√3/2 r)^2$

= $r^2 - 3/4 r^2 = r^2/4$

⇒ OB = $r/2$

Answer: (a)

According to theorem:– the sum of any two sides of a triangle is greater than twice the median drawn to the third side.

(AB + AC) > 2 AD

(3 + 5) > 2AD

AD < 4

Thus, AD is always less than 4 cm.