mensuration area volumes Model Questions & Answers, Practice Test for ibps so prelims 2023

Answer: (b)

OA = OB = AB

(given)

So, ΔAOB is an equilateral triangle.

⇒ ∠AOB = 60°

We know that, the angle subtended at the centre of circle is twice the angle subtended at circumference of the circle.

∴ ∠AOB = 2 ∠APB

⇒ ∠APB = ${60°}/2$ = 30°

Answer: (a)

AB = 2a i.e., AL = LM = MN = NB = $a/2$

Let OC = r

In ΔOML, OM= a – r and OL = OP + PL = r + $a/2$

∴ $(a/2)^2 + (a - r)^2 = (r + a/2)^2$

⇒ $a^2/4 + a^2 - 2ar + r^2 = r^2 + a^2/4 + ar$

⇒ $a^2 = 3ar ⇒ r = a/3$

So, option is correct.

Answer: (c)

Water running per hour in the river

= 2.5 × 45 × 3600 $m^3$.

Water running per minutes in the river

= ${2.5 × 45 × 3600}/{60} m^3$

= 2.5 × 45 × 60

= 6750 $m^3$

So, option (c) is correct

Answer: (a)

Let BC is shortest side in right angle triangle ΔABC,

such that AB : AC = 4 : 5.

∴ BC = $√{(AC)^2 - (AB)^2}$

= $√{(5x)^2 - (4x)^2}$ = 3x

ABC

Perimeter = AB + BC + AC = 4x + 3x + 5x = 12x

According to the question

12x = k(3x)

∴ k = ${12x}/{3x}$ = 4

Answer: (d)

Given ∠BAC = y°

and ∠BPC = x°

According to triangle property, the angle subtended by the bisector of an internal angle and an external angle is half of the third angle.

∴ y° = 2x° or x = $y/2$

Alternate Method:—

In ΔABC, y° + ∠ABC + ∠ACB = 180° ... (i)

In ΔBPC, ∠BPC + ∠PBC + ∠PCB = 180°

x° + ${∠ABC}/2 + ({∠ACD}/2 + ∠ACB)$ = 180° ... (ii)

Comparing equation (i) and (ii)

x° = ${y°}/2$