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) 2p = q
(b) P = q
(c) p = 2q
(d) 2p = 3
The correct answers to the above question in:
Answer: (b)
p = 4π$r^2$
q = 2πr.h = 2πr. 2r
= 4π $r^2$
Hence, P = q.
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, then what is the length of the perpendicular on the side of length 50 m from the opposite vertex?
a) 67.2 m
b) 43 m
c) 52.2 m
d) 70 m
Answer »Answer: (a)
Given, 2s = 240 ⇒ s = 120 and c = 50m, b = 78 m, a = 112m
∴ Area of triangle = $1/2$ × Base × Height
and also, Δ = $√{s(s - a)(s - b)(s - c)}$
∴ = $√{120(120 - 112)(120 - 78)(120 - 50)}$
= $√{120 × 8 × 42 × 70} = 1680 m^2$
∵ Area of triangle
= $1/2$ × Base × Height
⇒ 1680 = $1/2$ × 50 × h
∴ h = ${2 × 1680}/{50}$ = 67.2m
Question : 2
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
Question : 3
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 : 4
Statement I Let the side DE of a ΔDEF be divided at S. so that ${DS}/{DE} = 1/√2$. If a line through S parallel to EF meets DF at T, then the area of ΔDEF is twice the area of the ΔDST.
Statement II The areas of the similar triangles are proportional to the squares on the corresponding sides. Which one of the following is correct in respect of the above statements?
a) Statement I is true but Statement II is false
b) Both Statements I and II are true and Statement II is the correct explanation of Statement I
c) Both Statements I and II are true but Statement II is not the correct explanation of Statement I
d) Statement II is true but statement I is false
Answer »Answer: (b)
Given that, ${DS}/{DE} = 1/√2$
DE = $√2$ × DS
When two triangles ΔDTS and ΔDEF are similar then their ratio of area is equal to square of corresponding sides.
⇒ ${ΔDST}/{ΔDEF} (1/√2)^2 = 1/2 ⇒ ΔDEF = 2ΔDST$
So, both Statements I and II are true and Statement II is the correct explanation of Statement I.
Question : 5
The hypotenuse of a right-angled triangle is 10 cm and its area is 24 cm2 . If the shorter side is halved and the longer side is doubled, the new hypotenuse becomes
a) $√265$ cm
b) $√245$ cm
c) $√255$ cm
d) $√275$ cm
Answer »Answer: (a)
Let shorter and longer side of right angle triangle are x and y cm respectively.
Then, $x^2 + y^2 = (10)^2 ⇒ x^2 + y^2$ = 100 ...(i)
and Area = $1/2 xy = 24 ⇒ x = {48}/y$
Plug. in ∼ x = ${48}/y$ into equation (i), we get
$({48}/y)^2 + y^2 = 100$
$(48)^2 + y^4 = 100 y^2$
$y^4 – 100y^2 + (48)^2$ = 0
On solving, we get y = 6 or 8
∴ x = 6, y = 8
when, x becomes half and y becomes double then,
x' = 3, y' = 16
Hypotenuse 2 2 = $√{(x)^2 + (y')^2}$
= $√{3^2 + (16)^2} = √{265}$ cm
Question : 6
The base of an isosceles triangle is 300 unit and each for its equal sides is 170 units. Then the area of the triangle is
a) 12000 square units
b) 9600 square units
c) 10000 square units
d) None of the above
Answer »Answer: (a)
Let ABC be an isosceles triangle.
Area = $1/2$ × AD × BC
= $1/2 × √{(170)^2 - (150)^2} × 300$
= $1/2 × √{28900 - 22500} × 300$
(∵ ΔADC is a right angled triangle then by pythagoras theorem, we find AD)
= 150 × $√{6400}$ = 150 × 80 = 12000 units.
∴ Option (a) is correct.
GET Mensuration PRACTICE TEST EXERCISES
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