Previous Year OMET Questions for Geometry

1. XAT 2020 QADI | Geometry - Mensuration

A rectangular swimming pool is 50 meters long and 25 meters wide. Its depth is always the same along its width but linearly increases along its length from 1 meter at one end to 4 meters at the other end. How much water (in cubic meters) is needed to completely fill the pool?

A.
3750
B.
2500
C.
1250
D.
1874
E.
3125

2. XAT 2020 QADI | Geometry - Mensuration

Six drums are used to store water. Five drums are of equal capacity, while the sixth drum has double the capacity of each of these five drums. On one morning, three drums are found half full, two are found two-thirds full and one is found completely full. It is attempted to transfer all the water to the smaller drums. How many smaller drums are adequate to store the water?

A.
Four but not three
B.
Three or four, depending on which drum had how much water initially
C.
Five but not four
D.
Five may be inadequate, depending on which drum had how much water initially
E.
Three but not two

3. XAT 2019 QADI | Geometry - Mensuration

A gold ingot in the shape of a cylinder is melted and the resulting molten metal molded into a few identical conical ingots. If the height of each cone is half the height of the original cylinder and the area of the circular base of each cone is one ﬁfth that of the circular base of the cylinder, then how many conical ingots can be made?

A.
40
B.
30
C.
60
D.
20
E.
10

4. IIFT 2019 QA | Geometry - Mensuration

A square of length 1 m is inside a square of length 2 m and four quarter circles are joined as shown in the figure. The value of y – x is given by,

A.
(8 - π)/10
B.
(4 - π)/5
C.
(2π - 1)/8
D.
(π - 3)/4

5. IIFT 2019 QA | Geometry - Mensuration

Consider a cuboidal underground tunnel of length 500 m whose cross-section is given in the figure. If 1 m³ of concrete costs 1000 rupees, find the amount of money needed to build the tunnel.

A.
8(4 - π) × 10⁶ rupees
B.
64(4 - π) × 10⁶ rupees
C.
16(4 - π) × 106 rupees
D.
32(4 - π) × 106 rupees

6. XAT 2018 QADI | Geometry - Mensuration

It takes 2 liters to paint the surface of a solid sphere. If this solid sphere is sliced into 4 identical pieces, how many liters will be required to paint all the surfaces of these 4 pieces.

A.
2.2 liters
B.
2.5 liters
C.
3 liters
D.
4 liters
E.
None of the above

7. XAT 2018 QADI | Geometry - Mensuration

A cone of radius 4 cm with a slant height of 12 cm was sliced horizontally, resulting into a smaller cone (upper portion) and a frustum (lower portion). If the ratio of the curved surface area of the upper smaller cone and the lower frustum is 1:2, what will be the slant height of the frustum?

A.
12 - √3
B.
12 - 2√3
C.
12 - 3√3
D.
12 - 4√3
E.
None of the above

8. IIFT 2018 QA | Geometry - Mensuration

A bucket contains 200 cc of liquid. A solid ball is dropped in the bucket resulting in the rise of liquid level to 1.3 times of its original level. If the radius of the base of the bucket is 3 cm and the radius of the surface of the liquid level is 1 cm more than the radius of the base of the bucket before the ball is dropped. Find the volume of the solid metal ball.

A.
68cc
B.
80cc
C.
92cc
D.
Can’t be determined

Answer: Option B

Explanation :

The radius of the bucket is 3 cm and the radius of the original water level is (3 + 1) = 4 cm

Original volume = 200 cc

Let the original height of water = h cm

Using the formula for volume of frustum of cone:

$200 = \frac{h}{3} π [4^{2} + 3^{2} + (4) (3)]$

∴ h = 600/37π

Now, the height of the liquid increases by 0.3 times itself i.e. it becomes 1.3h

In a frustum of cone, ratio of radii is the same as ratio of heights (applying the concepts of similarity).

∴ New radius = 4.3 cm

∴ Volume of ball = new volume – original volume

$= \frac{1.3 h}{3} π [4 . 3^{2} + 3^{2} + (4.3) (3)] - 200$

$= \frac{1.3 h}{3} \times π \times \frac{600}{37 π} \times 40.4 - 200$

= 283.9 - 200 = 83.9 cc

The closest value in the options is 80 cc.

Hence, option 2.

9. IIFT 2018 QA | Geometry - Mensuration

In the given figure, PA = QB and PRQ is the arc of the circle, centre of which is O such that angle POQ = 90⁰. If AB = 25√2 cm and the perpendicular distance of AB from centre O is 30cm. Find the area if the shaded region?

A.
625√2 sq.cm
B.
$625 (\frac{1}{2} + \frac{π}{4})$
C.
$750 \sqrt{2} - 625 (\frac{1}{2} + \frac{π}{4})$
D.
None

Answer: Option C

Explanation :

PA = QB and angles A and B are right angles.

Hence, PABQ is a rectangle.

∴ Required area

= A(rectangle PABQ) – A(segment O-PRQ)

= A(rectangle PABQ) – [A(sector O-PRQ) – A(∆OPQ)]

= A(rectangle PABQ) – A(sector O-PRQ) + A(∆OPQ)

Now, PQ = AB = 25√2 cm and ∠POQ = 90°

∴ Radius of sector = PO = OQ = PQ/√2 = 25

Now, A(∆OPQ) = (1/2) × OP × OQ
= (1/2) × 25 × 25 = 625/2 sq.cm

A(sector O-PRQ) = (90/360) × π × (OP)²
= (π/4) × (25)² = (625π/4)

Now, height of ∆OPQ = 25/√2

Distance from O to AB is 30 cm

∴ PA = QB = 30 – (25/√2) cm

∴ A(rectangle PABQ) = PA × AB

= [30 – (25/√2)] × (25√2) = 750√2 – 625 sq.cm

∴ Required area = $(750 \sqrt{2} - 625)$ - $\frac{625 π}{4}$ + $\frac{625}{2}$

= 750√2 - 625/2 - 625π/4

= 750√2 - 625 $(\frac{1}{2} + \frac{π}{4})$

Hence, option 3.

10. IIFT 2018 QA | Geometry - Mensuration

Let S₁ be a square of side 4 cm. Circle C₁ circumscribes the square S₁ such that all its corners are on C₁. Another square S₂ circumscribes the circle C₁. Circle C₂ circumscribes the square S₂, and square S₃ circumscribes circle C₂, & so on. If A_N is the area between the square S_N and the circle C_N, where N is the natural number, then the ratio of sum of all A_N to A₁ is

A.
1
B.
$\frac{π}{2} - 1$
C.
Can’t be determined
D.
None of the above

11. IIFT 2018 QA | Geometry - Mensuration

A metallic solid is made up of a solid cylindrical base with a solid cone on its top. The radius of the base of the cone is 5cm. and the ratio of the height of the cylinder and the cone is 3:2. A cylindrical hole is drilled through the solid with height equal to 2/3rd of the height of solid. What should be the radius (in cm) of the hole so that the volume of the hole is 1/3rd of the volume of the metallic solid after drilling?

A.
$\sqrt{\frac{45}{8}}$
B.
$\sqrt{\frac{35}{8}}$
C.
$\sqrt{\frac{65}{8}}$
D.
$\sqrt{\frac{55}{8}}$

Answer: Option D

Explanation :

Volume of original solid = volume of cylinder + volume of cone

Radius of original cylinder = radius of cone = 5 cm

Let height of the original cylinder be 3x cm. Hence, height of cone = 2x cm

Volume of solid = $π (5^{2}) (3 x) + \frac{π}{3} (5^{2}) (2 x)$

$= 25 π \times \frac{11 x}{3} = \frac{275 π x}{3}$

Let the radius of the cylindrical hole be r cm.

Also, height of cylindrical hole = (2/3)(5x) = (10x/3) cm

∴ Volume of hole = $π (r^{2}) \frac{10 x}{3}$

Hence, as per the given condition:

$\frac{π (r^{2}) \frac{10 x}{3}}{\frac{275 x}{3} - π (r^{2}) \frac{10 x}{3}} = \frac{1}{3}$

$∴ \frac{10 r^{2}}{275 - 10 r^{2}} = \frac{1}{3}$

∴ 275 – 10r² = 30r²
∴ 40r² = 245

$∴ r^{2} = \frac{245}{40} = \frac{55}{8}$

$∴ r = \sqrt{(\frac{55}{8})}$

Hence, option 4.

12. XAT 2017 QADI | Geometry - Mensuration

The Volume of a pyramid with a square base is 200 cubic cm. The height of the pyramid is 13 cm. What will be the length of the slant edges (i.e. the distance between the apex and any other vertex), rounded to the nearest integer?

A.
12 cm
B.
13 cm
C.
14 cm
D.
15 cm
E.
16 cm

13. XAT 2017 QADI | Geometry - Mensuration

ABCD is a rectangle. P, Q and R are the midpoint of BC, CD and DA. The point S lies on the line QR in such a way that SR: QS = 1:3. The ratio of the area of triangle APS to area of rectangle ABCD is

A.
36/128
B.
39/128
C.
44/128
D.
48/128
E.
64/128

Answer: Option A

Explanation :

Let the length of the rectangle be ‘l’ and breadth be ‘b’.

The given figure is.

Let us draw a line MS parallel to DC

∆RMS ~ ∆RDQ

∴ RM : MD = RS : SQ = 1 : 3

Let the length of the rectangle (i.e., AB = CD) ABCD be 8l and width be 8b (i.e., AD = BC).

∴ AR = 4b and RM = b

Also, MS = l and SL = 7l.

Also, PL : LC = 1 : 3 [∵ RS : SQ = 1 : 3]

Area (∆ARS) = ½ × AR × SM = 2lb

Area (∆RDQ) = ½ × RD × DQ = 8lb

Area (∆ABP) = ½ × AB × BP = 16lb

Area (PSQC) = Area (∆PSL) + Area(∆SLQC)
= ½ × SL × PL + ½ × (SL + QC) × LC
= 3.5lb + 16.5lb
= 20lb

Area of ∆ASP = Area of rectangle ABCD – [Area (∆ARS) + Area (∆RDQ) + Area (∆ABP) + Area (PSQC)]
= 64lb – [2lb + 8lb + 16lb + 20lb]
= 64lb - 46lb = 18lb

∴ $\frac{A r e a (∆ A S P)}{A r e a (A B C D)}$ = $\frac{18 l b}{64 l b}$ = $\frac{36}{128}$

Hence, option (a).

14. IIFT 2017 QA | Geometry - Mensuration

A rectangular plank √10 m wide, is placed symmetrically along the diagonal of a square of side 10 m as shown in the figure. The area of the plank is:

A.
10(√20 − 1) sq.m
B.
10(√5 − 1) sq.m
C.
10√20 − 1 sq.m
D.
None of the above

15. IIFT 2017 QA | Geometry - Mensuration

A pest control person uses a particular machine for his job. It moves along the circumference of a circular hall of radius 49 meters in 148 minutes to finish the pest control. How many minutes more will it take him to move along the perimeter of a hexagon of side 54 meters?

A.
7.69 minutes
B.
14.36 minutes
C.
14.00 minutes
D.
4.28 minutes

16. XAT 2016 QADI | Geometry - Mensuration

Akhtar plans to cover a rectangular floor of dimensions 9.5 meters and 11.5 meters using tiles. Two types of square shaped tiles are available in the market. A tile with side 1 meter costs Rs. 100 and a tile with side 0.5 meters costs Rs. 30. The tiles can be cut if required. What will be the minimum cost of covering the entire floor with tiles?

A.
10930
B.
10900
C.
11000
D.
10950
E.
10430

17. XAT 2016 QADI | Geometry - Mensuration

A square piece of paper is folded three times along its diagonal to get an isosceles triangle whose equal sides are 10 cm. What is the area of the unfolder original piece of paper?

A.
400 sq.cm.
B.
800 sq.cm.
C.
800√2 sq.cm.
D.
1600 sq.cm.
E.
Insufficient data to answer

18. IIFT 2016 QA | Geometry - Mensuration

A right circular cylinder has a height of 15 and a radius of 7. A rectangular solid with a height of 12 and a square base, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. Liquid is then poured into the cylinder such that it reaches the rim. The volume of the liquid is

A.
147(2π - 8)
B.
180(π - 5)
C.
49(5π - 24)
D.
49(15π - 8)

19. IIFT 2015 QA | Geometry - Mensuration

If a right circular cylinder of height 14 is inscribed in a sphere of radius 8, then the volume of the cylinder is:

A.
110
B.
220
C.
440
D.
660

20. XAT 2015 QA | Geometry - Mensuration

The figure below has been obtained by folding a rectangle. The total area of the figure (as visible) is 144 square meters. Had the rectangle not been folded, the current overlapping part would have been a square. What would have been the total area of the original unfolded rectangle?

A.
128 square meters
B.
154 square meters
C.
162 square meters
D.
172 square meters
E.
None of the above

Answer: Option C

Explanation :

Had the rectangle not been folded, the overlapping part would have been a square of side 6.

While unfolding, the increase in area

= Area of the triangle = $\frac{1}{2} \times 6 \times 6 = 18$

Area of the given figure = 144 square meters
∴ Area of unfolded rectangle = 144 + 18 = 162 square meters
Hence, option 3.

21. XAT 2015 QA | Geometry - Mensuration

A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?

A.
9%
B.
16%
C.
25%
D.
50%
E.
None of the above

22. XAT 2015 QA | Geometry - Mensuration

A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7√2 ft. and cost of construction is Rs. 100 per sq. ft. Find the lowest possible cost to construct 50% of the total road.

A.
Rs. 70,400
B.
Rs. 125,400
C.
Rs. 140,800
D.
Rs. 235,400
E.
None of the above

Answer: Option B

Explanation :

Radius of inner circle = ½ × diagonal of the square = 25√2 cm
Radius of outer circle = Radius of inner circle + width of the road = 32 cm
50% of the area of the road

$= \frac{1}{2} \times π \times [{(32 \sqrt{2})}^{2} - {(25 \sqrt{2})}^{2}]$

= 1254 cm²

Cost = 1254 ×100 = Rs. 125,400

Hence, option 2.

23. XAT 2012 QA | Geometry - Mensuration

Ram, a farmer, managed to grow shaped-watermelons inside glass cases of different shapes. The shapes he used were: a perfect cube, hemi-spherical, cuboid, cylindrical along with the normal spherical shaped watermelons. Thickness of the skin was same for all the shapes. Each of the glass cases was so designed that the total volume and the weight of the all the water-melons would be equal irrespective of the shape.

A customer wants to but water-melons for making juice, for which the skin of the water-melon has to be peeled off, and therefore is a waste. Which shape should the customer buy?

A.
Cube
B.
Hemi-sphere
C.
Cuboid
D.
Cylinder
E.
Normal spherical

24. XAT 2012 QA | Geometry - Mensuration

A spherical metal of radius 10 cm is molten and made into 1000 smaller spheres of equal sizes. In this process the surface area of the metal is increased by:

A.
1000 times
B.
100 times
C.
10 times
D.
No change
E.
None of the above

25. XAT 2012 QA | Geometry - Mensuration

Suresh, who runs a bakery, uses a conical shaped equipment to write decorative labels (e.g., Happy Birthday etc.) using cream. The height of this equipment is 7 cm and the diameter of the base is 5 mm. A full charge of the equipment will write 330 words on an average. How many words can be written using three fifth of a litre of cream?

A.
45090
B.
45100
C.
46000
D.
43200
E.
None of the above

Geometry - Mensuration - Previous Year CAT/MBA Questions

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