Mensuration Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 2 EXERCISES
The following question based on Mensuration topic of quantitative aptitude
(a) 20 $√2$ cm
(b) 10 cm
(c) 10 $√2$ cm
(d) 10 $√{15}$ cm
The correct answers to the above question in:
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
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 »Answer: (b)
Since, ${PD}/{PE} = {AD}/{AE} = {AP}/{AP}$
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 : 2
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 »Answer: (b)
Area of the shaded portion
= $1/4 × π (14)^2$
= 154 $m^2$
Question : 3
Consider a circle C of radius 6 cm with centre at O. What is the difference in the area of the circle C and the area of the sector of C subtending an angle of 80º at O?
a) 28π $cm^2$
b) 26π $cm^2$
c) 16π $cm^2$
d) 30π$cm^2$
Answer »Answer: (a)
Radius of circle, r = 6 cm
∴ Area of circle = $πr^2 = π × 6^2 = 36π cm^2$
and Area of sector subtending an angle of 80° at O
= ${π r^2 θ}/{360°} = {π × 6^2 × 80°}/{360°} = 8 π cm^2$
∴ Required difference = 36π – 8π = 28π $cm^2$
Question : 4
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 $m^2$
Answer »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] $m^2 = 49 m^2$.
Question : 5
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 »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 : 6
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 »Answer: (a)
Slant height of the cone = $√{5^2 + 12^2}$
= $√{2 + 144} = √{169} = 13m$
Curved surface area for conical portion = πrl
= ${22}/7 × 12 × 13 = {3432}/7$sq m
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