Geometry - Basics - Previous Year CAT/MBA Questions
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At any point of time, let x be the smaller of the two angles made by the hour hand with the minute hand on an analogue clock (in degrees). During the time interval from 2:30 p.m. to 3:00 p.m., what is the minimum possible value of x?
- (a)
45
- (b)
105
- (c)
90
- (d)
0
- (e)
75
Answer: Option C
Text Explanation :
Workspace:
ABC is a triangle with integer-valued sides AB = 1, BC >1, and CA >1. If D is the mid-point of AB, then, which of the following options is the closest to the maximum possible value of the angle ACD (in degrees)?
- (a)
15
- (b)
30
- (c)
45
- (d)
75
- (e)
60
Answer: Option A
Text Explanation :
Given, AB = 1, AC > 1 and BC > 1
If ABC was an equilateral triangle i.e if AB = AC = BC = 1, ACD would have been 30°.
As C moves away from A and B, the angle decreases, hence the angle should be less than 30 and among the given options only 15 is less than 30.
Workspace:
If area of the adjacent faces of a cuboid is given as p, q and r respectively and the volume isgiven as 'V' then the square of the volume will be
- (a)
pqr
- (b)
- (c)
- (d)
(pqr)2
Answer: Option A
Text Explanation :
Let the length, breadth and height of the cubhoid be l, b and h respectively.
Therefore of adjacent faces will be lb, bh and hl.
If lb = p, then bh = q and hl = r.
Volume (V) = lbh
V2 = (lbh)2
or, V2 = pqr
Workspace:
In the figure (not drawn to scale) given below, if AD = CD = BC and angle BCE = 81°, how much is the value of angle DBC?
- (a)
27%
- (b)
33%
- (c)
54%
- (d)
66%
Answer: Option C
Text Explanation :
Given,
AD = DC, this implies angle ACD = angle DAC = x
DC = BC, this implies angle CDB = angle CBD = 2x
∠DCB = 180 - x - 81 = 99 - x
In triangle BCD, sum of the angles is 180 degrees
99 - x + 2x + 2x = 180
3x = 81
x = 27
∠DBC = 2x = 54 degrees
The answer is option C
Workspace:
A man, from the foot of the building P, walks towards the building Q. After walking for 2 mts, he finds that buildings P and Q makean angle of elevation of 60° and 30° respectively. If building Q is 1.5 mts high, find the distance between the tops of both buildings P and Q.
- (a)
5 mts
- (b)
4 mts
- (c)
mts
- (d)
mts
Answer: Option A
Text Explanation :
The situation given in the problem can be summarised as:
By 30 - 60 - 90 theorem, we can say that P'M = 4 meters and MQ' = 3 meters
Also, triangle P'MQ' would be a right angled triangle.
Hence, P'Q' would be the hypotenuse of length = root(16 + 9) = 5 meters
Workspace:
ABC is an isosceles triangle. BDE, EFG, GHI and IJC are four equal isosceles triangles inside ABC triangle. D and F,F and H,H and J are connected by circular arcs. The angle ABC is 30 degrees and BE is 1m. What is the area of the shaded region?
- (a)
- m2
- (b)
- m2
- (c)
- m2
- (d)
- m2
Answer: Option C
Text Explanation :
Since ∠ABC = 30° and ∆DBE is an isosceles triangle, ∠BDE = 120°
Applying the cosine rule in △BDE.
BE2 = BD2 + DE2 - 2 × BD × DE cos(120)
1 = 2BD2 + BD2 [since BD = DE]
BD =
Thus, BD = DE = EF = FG = GH = HI = IJ = JC =
Area of △BDE = × BD × DE × sin(120) = × × × = 4
The total area of all smaller isosceles triangles = 4 × 4 = m2
∠DEF = ∠FGH = ∠HIJ = 120°
Thus, the total area of circular arcs = π × ()2 = m2
Now, side BC = BE + EG + GI + IC = 4m
Using the cosine rule in △ABC, we get AB = AC = .
Thus, Area of △ABC = × × × sin(120) = m2
Thus, the area of the shaded region = Area of △ABC - Area of smaller isosceles triangles - Area of the circular sections
= - - = -
Hence, the answer is option C.
Workspace:
A rod is cut into 3 equal parts. The resulting portions are then cut into 12, 18 and 32 equal parts, respectively. If each of the resulting portions have integer length, the minimum length of the rod is
- (a)
6912 units
- (b)
864 units
- (c)
288 units
- (d)
240 units
Answer: Option B
Text Explanation :
As the three portions of the rod correspond to 12, 18 and 32 equal parts, their length will be equal to LCM of 12, 18 and 32.
LCM of 12, 18 and 32 = 288
Since there are three equal parts of the rod, total length = 288 × 3 = 864
Hence, option (b).
Workspace:
In a square of side 2 meters, isosceles triangles of equal area are cut from the corners to form a regular octagon. Find the perimeter and area of the regular octagon.
- (a)
;
- (b)
;
- (c)
;
- (d)
None of these
Answer: Option D
Text Explanation :
Let the side of isosceles triangle be x.
∴ The side of octagon x.
∴ x + x + x = 2
∴ 2x + x = 2
∴ x =
∴ Side of octagon = x = =
∴ Perimeter of octagon = 8 × = units
Area of octagon = Area of square – 4 × Area of isosceles triangle
= 22 - 4 × x2
= 4 - 4 × ×
= 4 - = sq. units
Hence, option (d).
Workspace:
Let A1 be a square whose side is ‘a’ metres. Circle C1 circumscribes the square A1 such that all its vertices are on C1. Another square A2 circumscribes C1. Circle C2 circumscribes A2, and A3 circumscribes C2, and so on. If DN is the area between the square AN and the circle CN, where N is a natural number, then the ratio of the sum of all DN to D1 is:
- (a)
1
- (b)
- 1
- (c)
Infinity
- (d)
None of the above
Answer: Option C
Text Explanation :
Area of square 1 = a2
Area of circle 1 = π
∴ D1 = (π - 2)
Area of square 2 = (a)2
Area of circle 2 = πa2
∴ D2 = a2(π - 2)
Similarly,
D3 = 2a2 (π - 2), and so on.
∴ D1 + D2 + D3 + ... = a2(π - 2) [0.5 + 1 + 2 + 4 + ...]
= Infinite
Hence, option (c).
Workspace:
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