Mensuration Model Questions Set 2 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) 324 $m^3$
(b) 512 $m^3$
(c) 480 $m^3$
(d) 256 $m^3$
The correct answers to the above question in:
Answer: (b)
Let side of cubical water tank be 'x' meter.
Capacity of tank = $x^3$
According to question
⇒ $x^3 – 128 = (x – 2).x^2$
⇒ $x^3 – 128 = x^3 – 2x^2$
⇒ $2x^2$ = 128
⇒ $x^2$ = 64
⇒ x = 8 metre
Capacity of tank $(8)^3 = 512 m^3$
So, option (b) is correct.
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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers
Question : 1
Consider the following statements : Two triangles are said to be congruent, if
- Three angles of one triangle are equal to the corresponding three angles of the other triangle.
- Three sides of one triangle are equal to the corresponding three sides of the other triangle.
- Two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle.
- Two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.
a) 1, 2 and 4
b) 1, 2 and 3
c) 1, 3 and 4
d) 2, 3 and 4
Answer »Answer: (d)
Question : 2
In the figure given below, ABC is a triangle. BC is parallel to AE. If BC = AC, then what is the value of ∠CAE?
![mensuration area and volume aptitude mcq 21 38](https://careericons.com/adminicon/bunch/images/mensuration-area-and-volume-aptitude-mcq-21-38.png)
a) 40°
b) 20°
c) 30°
d) 50°
Answer »Answer: (d)
Given that, BC || AE
∠CBA + ∠EAB = 180°
⇒ ∠EAB =180° – 65° = 115°
∵ BC= AC
Hence, ΔABC is an isosceles triangle
∠CBA= ∠CAB = 65°
Now, ∠EAB = ∠EAC + ∠CAB
⇒ 115° = x + 65° ⇒ x = 50°
Question : 3
ABCD is a quadrilateral such that BC = BA and CD > AD. Which one of the following is correct?
a) ∠BAD > ∠BCD
b) ∠BAD = ∠BCD
c) ∠BAD < ∠BCD
d) None of these
Answer »Answer: (a)
Construction : In quadrilateral ABCD, form A to C.
Now, in ΔABC
∵ AB = BC ...(Given)
∴ ∠BAC = ∠BCA
(angles opposite to equal side)
In ΔADC,
∵ CD > AD
∴ ∠DAC> ∠DCA
(since in a triangle, angle opposite to greater side is bigger than the angle opposite to smaller side)
On adding eqs. (i) and (ii), we get
∠BAD > ∠BCD
Question : 4
The length of a minute hand of a wall clock is 9 cm. What is the area swept (in $cm^2$) by the minute hand in 20 min? (take π = 3.14)
a) 67.74
b) 88.78
c) 84.78
d) 57.78
Answer »Answer: (c)
The angle made by the minute hand in 20 min = 120°
∴ The area swept by the minute hand in 20 min
= $θ/{360°} × π r^2 = ∼ {120°}/{360°} × 3.14 × 9 × 9 = 84.78 cm^2$
Question : 5
In the figure given below, SPT is a tangent to the circle at P and O is the centre of the circle. If ∠QPT = α, then what is ∠POQ equal to?
![mensuration area and volume aptitude mcq 24 112](https://careericons.com/adminicon/bunch/images/mensuration-area-and-volume-aptitude-mcq-24-112.png)
a) 90° – α
b) α
c) 2α
d) 180° – 2α
Answer »Answer: (c)
Question : 6
Let the incircle to a ΔABC touch BC, AC and AB respectively at the points X, Y and Z.
- Statement I
- If AB > BC, then AB + AZ < BC + XC
- Statement II
- AZ = AY
a) Statement I is correct and Statement II is incorrect
b) Statement I and II are correct and Statements II is the correct explanation of Statement I
c) Statement I and II are correct and Statement II is not the correct explanation of Statement I
d) Statement I is incorrect and Statement II is correct
Answer »Answer: (d)
In ΔAOZ and ΔAOY,
AO = OA [common]
∠OAZ = ∠OAY [Since, OA bisect ∠A]
and ∠AZO = ∠AYO [each 90°]
∴ ΔAZO ≅ ΔAYO
So, AZ = AY [by CPCT]
Similarly, CX = CY and BX = BZ
Now, AB > BC
∴ AZ + ZB > BX + XC
AZ > XC [∵ BX = ZB]
If AB > BC, then AB + AZ > BC + XC
So, Statement I is incorrect and Statement II is correct.
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