mensuration Model Questions & Answers, Practice Test for ibps clerk prelims 2024

Answer: (b)

We know that, sum of angles of any triangle = 180°

x + x + 4x = 180°

x = ${180°}/6$ = 30°

So, angle of triangles are 30°, 30°, 120°.

Let side of the triangles ABC are x, x and y

From cosine formula,

cos(120°) = ${AB^2 + AC^2 - BC^2}/{2.AB.AC} = {x^2 + x^2 - y^2}/{2.x.x}$

${-1}/2 = {2x^2 - y^2}/{2x^2} ⇒ y^2 = 3x^2$

y ∼ = $√3x$

Perimeter of the triangle = (x + x + y) = k(y) (given)

$(2/√3 y + y) = K.y = ∴ K = 1 + 2/√3$

Answer: (b)

OB = OA = radius

Also, ∠AOB = 60° $({360°}/6 = 60°)$

and ∠OAB = ∠OBA = 60°

So, ΔAOB is an equilateral triangle.

Then, AB = 5 cm

So, Area, x = Area of circle – Area of hexagon

= $πr^2 - {3√3 (a)^2}/2$

= ${22}/7 × (5)^2 - {3√3}/2 × (5)^2$

(∵ r = a = 5)

= 78.57 – 64.95 = 13.62 $cm^2$

Answer: (d)

Area (DOE) =${\text"Area(ABCD)"}/8$

Reqd. Ratio = $7/8$

Answer: (b)

Let height and radius of cylinder is h and r respectively.

Volume of cylinder = $πr^2h$

Volume of circular cone = $1/3 πr^2h$

∴ Required ratio = ${\text"Volume of utilised wood"}/{\text"Volume of wasted wood"}$

Volume of right circular cone

= ${\text"(Volume of right circular cylinder"}/{\text"–Volume of right circular cone)"}$

= ${1/3 π r^2 h}/{π r^2 h - 1/3 π r^2 h} ⇒ {1/3 π r^2 h}/{2/3 π r^2 h} ∼ = 1/2 = 1 : 2$

Answer: (a)

Surface area of the walls of the first room

= 2(XZ + YZ)

Surface area of the walls of the second room

= 2(4XZ + 4YZ) = 8(XZ + YZ) = 4 times the first area

⇒Cost required = 4 × 2500 = Rs.10,000.