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) 28π $cm^2$
(b) 26π $cm^2$
(c) 16π $cm^2$
(d) 30π$cm^2$
The correct answers to the above question in:
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 $cm^2$
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
In the figure given below, PQ is parallel to RS and PR is parallel to QS. If ∠LPR = 35° and ∠ UST = 70°, then what is ∠MPQ equal to ?
![mensuration area and volume aptitude mcq 21 50](https://careericons.com/adminicon/bunch/images/mensuration-area-and-volume-aptitude-mcq-21-50.png)
a) 75°
b) 55°
c) 70°
d) 80°
Answer »Answer: (a)
PQ || RS
PR|| QS
∴ PQRS is a || gm
∠LPR = 35° and ∠UST = 70°
∠UST = ∠RSQ (Vertically opposite)
∠RSQ = ∠RPQ (opposite angle of 11 gm)
∠LPR + ∠RPQ + ∠MPQ = 180°
35° + 70° + ∠MPQ = 180°
∠MPQ = 180 – 105
∠MPQ = 75°
Question : 3
Consider an equilateral triangle of a side of unit length. A new equilateral triangle is formed by joining the midpoints of one, then a third equilateral triangle is formed by joining the mid-points of second. The process is continued. The perimeter of all triangles, thus formed is
a) 6 units
b) 2 units
c) 3 units
d) Infinity
Answer »Answer: (a)
Sides of equilateral triangle are follows:
3, $3/2, 3/4, 3/8$ ... so on
These sequence formed a GP serves.
So sum of GP for Infinite terms.
S = $a/{1 - r}$
Here a = 3, r = $1/2$
S = $3/{1 - 1/2}$ =6 units
Question : 4
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 : 5
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 : 6
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
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