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) 25 cm
(b) 15 cm
(c) 20 cm
(d) ${200}/9$ cm
The correct answers to the above question in:
Answer: (a)
Volume of pit = l b h = 8 × 2.5 × 2 = 40 $m^3$.
Let the label of the earth spread over remaining area = h.
Volume of the earth spread = Volume of a pit
⇒ [(12 × 15) – (8 × 2.5)] × h = 40
∴ h = ${40}/{180 - 20} = {40}/{160} = 1/4$m = 25 cm
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
ABCD is a quadrilateral with AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm and angle ABC is a right-angle.
What is the area of quadrilateral ABCD ?
a) 312 $cm^2$
b) 300 $cm^2$
c) 306 $cm^2$
d) 316 $cm^2$
Answer »Answer: (c)
In right triangle ABC,
AC = $√{(AB)^2 + (BC)^2} = √{(40)^2 + (9)^2}$ = 41 cm
Area of ΔABC = $1/2 × 9 × 40 = 180 cm^2$
∵ Area of quadrilateral ABCD
= 126 + 180 = 306 $cm^2$
Question : 2
A line segment AB is the diameter of a circle with centre at O having radius 6.5 cm. Point P is in the plane of the circle such that AP = x and BP = y. In which one of the following cases the point P does not lie on the circle ?
a) x = 5 cm and y = 12 cm
b) x = 6.5 cm and y = 6.5 cm
c) x = 12 cm and y = 5 cm
d) x = 0 cm and y = 13 cm
Answer »Answer: (b)
P does not lie on the circle at x = 6.5cm and
y = 6.5 cm because, as <APB = 90°
then, $AB^2 ∼ ∼ = AP^2 ∼ ∼ + BP^2$
$(13)^2 ≠ (6.5)^2 ∼ ∼ + (6.5)^2$
Hence, point P does not lies on the circle ∼
Question : 3
In the figure given above, ABCD is a trapezium. EF is parallel to AD and BG. ∠y is equal to
![mensuration area and volume aptitude mcq 23 38](https://careericons.com/adminicon/bunch/images/mensuration-area-and-volume-aptitude-mcq-23-38.png)
a) 60°
b) 30°
c) 45°
d) 65°
Answer »Answer: (a)
From figure.
BC || EF || AD
∴ x° = z° = 50° (corresponding interior angle
∴ θ + z° = 180° (linear pair)
∴ ∼ θ = 180° – 50° = 130°
In quadrilateral
AQFD, x° + y° + 120° + θ = 360°
50° + y° + 120° + 130° = 360°
y = 360° – 300° = 60°
Question : 4
In the figure given above, LM is parallel to QR. If LM divides the ΔPQR such that area of trapezium LMRQ is two times the area of ΔPLM, then what is ${PL}/{PO}$ equal to?
a) $1/2$
b) $1/√2$
c) $1/√3$
d) $1/3$
Answer »Answer: (c)
In the given figure.
ar MRQL = 2 ar ΔPLM
Let area of ΔPLM be x, then
∴ the area of trapezium = 2x
∴ ar ΔPQR = 2x + x = 3x
Here it is clear from the given figure that ΔPQR ∼ ΔPLM
∴ ${\text"ar ΔPQR"}/{\text"ar ΔPLM"} = {3x}/x$
${PL^2}/{PQ^2} = 1/3 ∴ = {PL}/{PQ} = 1/√3$
Question : 5
Two straight lines AB and AC include an angle. A circle is drawn in this angle which touches both these lines. One more circle is drawn which touches both these lines as well as the previous circle. If the area of the bigger circle is 9 times the area of the smaller circle, then what must be the angle A?
a) 75°
b) 45°
c) 60°
d) 90°
Answer »Answer: (c)
Let the radius of the bigger circle be a and radius of the smaller circle be b.
Then the angle made by direct common tangents when two circles of radius a and b touch externally is given by θ = $2sin^{- 1}({a - b}/{a + b})$
We are given that area of the bigger circle = 9 area of the smaller circle
⇒ $πa^2 = 9πb^2 ⇒ a^2 = 9b^2 ⇒ a = 3b$
Let us consider ∠BAC = θ
Thus,
θ = $2 sin^{-1} ({a - b}/{a + b}) = 2 sin^{-1} ({3b - b}/{3b + b}) = 2sin^{-1} ({2b}/{4b})$
= $2 sin^{-1} (1/2) = 2 sin^{-1}$ (sin 30°) = 2 × 30° = 60°
Question : 6
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
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