Mensuration Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 2 EXERCISES

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The following question based on Mensuration topic of quantitative aptitude

Questions : A cistern 6 m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:

(a) 53.5 $m^2$

(b) 49 $m^2$

(c) 50 $m^2$

(d) 55 $m^2$

The correct answers to the above question in:

Answer: (b)

Area of the wet surface = [2(lb + bh + lh) – lb]

= 2(bh + lh) + lb

= [2(4 × 1.25 + 6 × 1.25) +6 × 4] $m^2 = 49 m^2$.

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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers

Question : 1

The curved surface area of a right circular cone is 1.76 $m^2$ and its base diameter is 140 cm. What is the height of the cone?

a) 20 $√2$ cm

b) 10 cm

c) 10 $√2$ cm

d) 10 $√{15}$ cm

Answer: (d)

Radius of cone = ${140}/2 = 70 cm = {70}/{100}$ = 0.7m

Curved surface are = πrl = 1.76 $m^2$

${22}/7$ × 0.7 × l = 1.76

l = ${1.76 × 7}/{22 × 0.7}$ = 0.8m = 80 cm

height of cone = $√{l^2 - r^2} = √{80^2 - 70^2}$

= $√{1500} = 10√{15}$ cm

Question : 2

Let AB and AC be two rays intersecting at A. If D, E be the points lying on AB, AC respectively and P be the point such that P divides the line DE such that PD : PE = AD : AE. Then, what is the locus of the point P?

a) The perpendicular bisector of angle A

b) The angle bisector of angle A

c) The angle trisector of angle A

d) None of the above

Answer: (b)

Since, ${PD}/{PE} = {AD}/{AE} = {AP}/{AP}$

mensuration-area-and-volume-aptitude-mcq

ΔDAP and ΔAPE are similar. So, ∠1 = ∠2

AP is bisector of ∠A.

Hence, the locus of P is the bisector of angle A.

Question : 3

A horse is tethered to one corner of a rectangular grassy field 40 m by 24 m with a rope 14 m long. Over how much area of the field can it graze ?

a) 150 $m^2$

b) 154 $cm^2$

c) 308 $m^2$

d) None of these

Answer: (b)

mensuration-aptitude-mcq

Area of the shaded portion

= $1/4 × π (14)^2$

= 154 $m^2$

Question : 4

A gardener increased the area of his rectangular garden by increasing its length by 40% and decreasing its width by 20%. The area of the new garden

a) has increased by 8%.

b) has increased by 20%.

c) has increased by 12%.

d) is exactly the same as the old area.

Answer: (c)

Let initial dimensions be, l & b

∴ Final length is 1.4 l

Final breadth is 0.8 b

∴ Final area is = 1.4 l × 0.8 b

= 1.12 lb

∴ Area is increased by 12%.

Shortcut Method : + 40 – 20 + ${40 × (-20)}/{100}$

= 20 – 8 = 12%

Therefore, the area of the new garden increased by 12%

Question : 5

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 m. The height of the cylindrical portion is 11 m, while the vertex of the cone is 16 m above the ground. What is the area of the curved surface for conical portion?

a) 3432/7 sq m

b) 3434/9 sq m

c) 3431/8 sq m

d) 3234/7 sq m

Answer: (a)

Slant height of the cone = $√{5^2 + 12^2}$

= $√{2 + 144} = √{169} = 13m$

mensuration-area-and-volume-aptitude-mcq

Curved surface area for conical portion = πrl

= ${22}/7 × 12 × 13 = {3432}/7$sq m

Question : 6

The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, then what is the length of the perpendicular on the side of length 50 m from the opposite vertex?

a) 67.2 m

b) 43 m

c) 52.2 m

d) 70 m

Answer: (a)

Given, 2s = 240 ⇒ s = 120 and c = 50m, b = 78 m, a = 112m

∴ Area of triangle = $1/2$ × Base × Height

mensuration-area-and-volume-aptitude-mcq

and also, Δ = $√{s(s - a)(s - b)(s - c)}$

∴ = $√{120(120 - 112)(120 - 78)(120 - 50)}$

= $√{120 × 8 × 42 × 70} = 1680 m^2$

∵ Area of triangle

= $1/2$ × Base × Height

⇒ 1680 = $1/2$ × 50 × h

∴ h = ${2 × 1680}/{50}$ = 67.2m

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