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

Question :1

A right circular cone is cut by a plane parallel to its base in such a way that the slant heights of the original and the smaller cone thus obtained are in the ratio 2 : 1. If $V_1$ and $V_2$ are respectively the volumes of the original cone and of the new cone, then what is $V_1 : V_2$ ?

Answer: (d)

Here similar triangle

ΔABC ~ In ΔADE

mensuration-area-and-volume-aptitude-mcq

∴ $V_1/V_2 = {1/3 π × 4r^2 × √{(2l)^2 - (2r)^2}}/{1/3 π × r^2 × √{l^2 - r^2}} = 8/1$ = 8 : 1

Question :2

mensuration-area-and-volume-aptitude-mcq
In the figure given above, PQ = QS and QR = RS. If ∠SRQ = 100°, then how many degrees is ∠QPS?

Answer: (a)

ΔQRS is an isosceles triangle.

mensuration-area-and-volume-aptitude-mcq

∴ ∠QSR = ∠SQR = 40°

⇒ ∠PQS= ∠180° – ∠RQS = ∠140°

Again, Δ∠PQS is an isosceles triangle.

∴ ∠PQS = ∠QSR = ${180 - 140}/2 = {40}/2$ = 20°

Question :3

A drinking glass of height 24 cm is in the shape of frustum of a cone and ' diameters of its bottom and top circular ends are 4 cm and 18 cm respectively. If we take capacity of the glass as &ohi;x $cm^3$ , then what is the value of x ?

Answer: (b)

Frustum

r = $4/2 , R = {18}/2$ , h = 24

Volume = $π/3 h [r_1^2 + r_2^2 + r, r_2]$

π x = ${22}/{7 × 3} × 24 [(2)^2 + (9)^2 + 2 × 9] = 824 cm^3$

Question :4

Consider the following statements in respect of two chords XY and ZT of a circle intersecting at P.
I. PX . PY = PZ . PT
II. PXZ and PTY are similar triangles.
Which of the statements given above is/are correct?

Answer: (a)

When two chords of a circle are intersect internally, then they are divided in a proportion.

mensuration-area-and-volume-aptitude-mcq

i.e., PX . PY = PZ . PT

In ΔPXZ and ΔPTY,

∠ZPX = ∠YPT

(vertically opposite angles)

∠PZX = ∠PYT

(angle in same segment)

∠PXZ = ∠PTY

(angles in same segment)

∴ ΔPXZ ∼ ΔPTY

Hence, the both statements are correct.

Question :5

A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the radius of the wheel is 30 cm, the speed of the cycle is

Answer: (c)

Distance covered in one revolution ∼

= 2π × 30 = 60π cm ∼

Distance covered in one minute by 140 revolutions ∼

= 140 × 60 π cm ∼

Distance covered in one hour ∼

= 140 × 60 π × 60 cm = 15.82560 cm = 15.825 km ∼

So, option (c) is correct. ∼

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