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) 12 times
(b) 5 times
(c) 6 times
(d) 17 times
The correct answers to the above question in:
Answer: (d)
Let the length, breadth and height of the cuboid be x, 2x and 3x, respectively.
Therefore, volume = x × 2x × 3x = $6x^3$
New length, breadth and height = 2x, 6x and 9x, respectively.
New volume = $108x^3$
Thus, increase in volume = $(108 – 6)x^3 = 102x^3$
${\text"Increase in volume"}/{\text"Original volume"} = {102x^3}/{6x^3}$ = 17
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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers
Question : 1
Two circles touch each other internally. Their radii are 4 cm and 6 cm. What is the length of the longest chord of the outer circle which is outside the inner circle?
a) 6 $√3$ cm
b) 4 $√2$ cm
c) 4 $√3$ cm
d) 8 $√2$ cm
Answer »Answer: (d)
Let O is centre of big circle, and O' is centre of smaller circle. Both are touch internally each other.
OA = 6 cm O'A = 4cm
Here PR is longest chord of big circle
P.S = ${PR}/2$
OS = AS – OA
= 8 – 6 = 2 cm.
In ΔPSO
$(PS)^2 + (OS)^2 = (OP)^2$
⇒ $(PS)^2 + (2)^2 = (6)^2$
PS = $√{36 - 4} = √{32} = 4 √2$
Now,
PS = ${PR}/2$
PR = 2 × PS
= 2 × $4 √2 = 8 √2$
Question : 2
Let LMNP be a parallelogram and NR be perpendicular to LP. If the area of the parallelogram is six times the area of ΔRNP and RP = 6 cm what is LR equal to?
a) 9 cm
b) 15 cm
c) 12 cm
d) 8 cm
Answer »Answer: (c)
Area of parallelogram = 6 × Area of ΔNPR
∴ NR × PL= 6 × $1/2$ × NR × PR
⇒ PL = 3PR (here, PL = PR + RL)
⇒ PR + RL= 3PR
⇒ RL= 2PL = 2 × 6 = 12 cm
Question : 3
In the figure given, ∠B = 38°, AC = BC and AD = CD. What is ∠D equal to?
a) 38°
b) 26°
c) 28°
d) 52°
Answer »Answer: (c)
Given, AC = BC
In ΔABC, ∠ABC = ∠CAB
∠ABC = ∠CAB = 38° (∴ AC = BC)
∠ACB = 180° – (∠ABC + ∠CAB)
= 180° – (38° + 38°) = 180° – 76° = 104°
In ΔACD, ∠ACD = 180° – 104° = 76°
and ∠ACD = ∠CAD = 76°
(∵ CD = AD)
∴ ∠ADC = 180° – (∠ACD + ∠CAD)
= 180° – (76° + 76°) = 28°
Question : 4
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 : 5
In the figure given below, ABC is a triangle. BC is parallel to AE. If BC = AC, then what is the value of ∠CAE?
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 : 6
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)
GET Mensuration PRACTICE TEST EXERCISES
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