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) 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
The correct answers to the above question in:
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.
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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 : 3
The medians of ΔABC intersect at G. Which one of the following is correct?
a) Three times the area of ΔAGB is equal to the area of ΔABC
b) Five times the area of ΔAGB is equal to four times the area of ΔABC
c) Four times the area of ΔAGB is equal to three times the area of ΔABC
d) None of the above
Answer »Answer: (a)
Suppose ΔABC is an equilateral triangle.
A median divides an equilateral triangle into the three equal area of triangles.
ΔAGB = ar ${(ΔABC)}/3$ = ar BGC = ar ΔAGC
∴ ar ΔAGB = $1/3$ ΔABC.
Question : 4
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 »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 : 5
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 : 6
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%
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