# Geometry - Mensuration - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Geometry - Mensuration. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**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

Answer: Option E

**Explanation** :

Volume of the tank = Volume of the cuboid + Volume of the triangular prism

The volume of cuboid will be 50 × 25 × 1 = 1250

The volume of triangular prism = Area of triangle × width = 1/2 × 3 × 50 × 25 = 1875

∴ Total volume = 1875 + 1250 = 3125.

Hence, option (e).

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**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

Answer: Option C

**Explanation** :

Let the volume of 5 smaller drums be V

∴ Volume of the bigger drum = 2V

On one morning, three drums are found half full, two are found two-thirds full and one is found completely full.

Case 1: We will get the maximum water when the bigger drum is completely full

∴ Amount of water = 3 × V/2 + 2 × 2V/3 + 1 × 2V = 29V/6 = 4.83V

∴ Number of smaller drums required to store 29V/6 water = 5

Case 2: We will get the minimum water when the bigger drum is only half full

∴ Amount of water = 1 × 2V/2 + 2 × V/2 + 2 × 2V/3 + 1 × V = 13V/3 = 4.133V

∴ Number of smaller drums required to store 13V/3 water = 5

∴ In both cases we need at least 5 smaller drums to store the water.

Hence, option (c).

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**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

Answer: Option B

**Explanation** :

Volume of a cylinder = π × (base area) × height

Volume of a cylinder = 1/3 × π × (base area) × height

Volume of a cone with same base and height as that of a cylinder is 1/3rd the volume of the cylinder.

If height of the cone is half that of the cylinder, volume will also become half. Also, if the base area is 1/5th the volume will also be 1/5th.

∴ Volume of the given cone = 1/3 × 1/2 × 1/5 = 1/30^{th} the volume of the original cylinder.

∴ 30 such cones can be made from the original cylinder.

**Alternately,**

Let A and h be the base area and height of the cylinder.

Base area of the cone = A/5 and height = h/2

Volume of a cylinder (V) = πAh

Volume of a cylinder = 1/3 × π × A/5 × h/2 = πAh/30 = V/30

∴ Volume of the given cone is 1/30^{th} the volume of the cylinder.

⇒ 30 such cones can be made from the original cylinder.

Hence, option (b).

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**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

Answer: Option D

**Explanation** :

Area of the shaded region = (Area of the bigger square) − (4 × Area of the quarter of radius = 1) − 4x.

= 22 − [4 × π × 1^{2}/4)] − 4x ...(1)

Also, area of the same shaded region = (Area of the smaller square) − 4y.

= 12 − 4y ...(2)

Equating (1) and (2), we get,

4 -π – 4x = 1 – 4y

⇒ (y − x) = (π − 3)/4

Hence option (d).

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**IIFT 2019 QA | Geometry - Mensuration**

Consider a cuboidal underground tunnel of length 500 m whose cross-section is given in the figure. If 1 m^{3} of concrete costs 1000 rupees, find the amount of money needed to build the tunnel.

- A.
8(4 - π) × 10

^{6}rupees - B.
64(4 - π) × 10

^{6}rupees - C.
16(4 - π) × 106 rupees

- D.
32(4 - π) × 106 rupees

Answer: Option C

**Explanation** :

Total volume of Tunnel = 8 × 16 × 500 = 64000 m^{3}

Radius of the given semicircle = 8 m

Volume of the semi-circle = (π × 8 × 8)/2 × 500 = 32π × 500 = 16000π m^{3}

∴ Volume of the required cross-section = 64000 - 16000π = 16000(4 − π) m^{3}

Total amount of money needed = 16000(4 − π) × 10^{3} = 16(4 − π) × 10^{6} rupees

Hence, option (c).

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**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

Answer: Option D

**Explanation** :

Total outer surface area of a sphere of radius r = 4πr^{2}

When the sphere is cut into 4 identical pieces, each piece will expose πr^{2}/2 + πr^{2}/2 = πr^{2} area more.

∴ Now, 4πr^{2} additional area needs to be painted which will require 2 more liters of paint.

∴ 4 litres of paint will be needed.

Hence, option (d).

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**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

Answer: Option D

**Explanation** :

Curved surface area of the original larger cone = π × Radius × Slant height = 48π

Therefore, the curved surface of the smaller cone = 1/3 × 48π = 16π

Now, the radius and slant of height of the smaller cone would be reduced in equal proportions from the larger cone.

Therefore, slant height of the frustum would be 1/√3 time the corresponding values of the larger cone.

⇒ Slant height of the smaller cone = 1/√3 × 12 = 4√3

∴ Slant height of the frustum = 12 - 4√3

Hence, option (d).

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**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:

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

∴ 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

$=\frac{1.3h}{3}\pi [4.{3}^{2}+{3}^{2}+(4.3\left)\right(3\left)\right]-200\phantom{\rule{0ex}{0ex}}$

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

= 283.9 - 200 = 83.9 cc

The closest value in the options is 80 cc.

Hence, option 2.

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**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.
$625\left(\frac{1}{2}+\frac{\pi}{4}\right)$

- C.
$750\sqrt{2}-625\left(\frac{1}{2}+\frac{\pi}{4}\right)$

- 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 = $(750\sqrt{2}-625)$ - $\frac{625\mathrm{\pi}}{4}$ + $\frac{625}{2}$

= 750√2 - 625/2 - 625π/4

= 750√2 - 625$\left(\frac{1}{2}+\frac{\mathrm{\pi}}{4}\right)$

Hence, option 3.

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**IIFT 2018 QA | Geometry - Mensuration**

Let S_{1} be a square of side 4 cm. Circle C_{1} circumscribes the square S_{1} such that all its corners are on C_{1}. Another square S_{2} circumscribes the circle C_{1}. Circle C_{2} circumscribes the square S_{2}, and square S_{3} circumscribes circle C_{2}, & 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 A_{1} is

- A.
1

- B.
$\frac{\pi}{2}-1$

- C.
Can’t be determined

- D.
None of the above

Answer: Option C

**Explanation** :

Here, the area of each square and each circle will keep on increasing. Hence, each A_{n} will keep on increasing.

Hence, if the value of n is not known, the ratio will not be found.

Here, the value of n is unknown. Hence, the value of the ratio cannot be determined.

Hence, option 3.

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**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 = $\pi \left({5}^{2}\right)\left(3x\right)+\frac{\pi}{3}\left({5}^{2}\right)\left(2x\right)$

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

Let the radius of the cylindrical hole be r cm.

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

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

Hence, as per the given condition:

$\frac{\pi \left({r}^{2}\right){\displaystyle \frac{10x}{3}}}{{\displaystyle \frac{275x}{3}}-\pi \left({r}^{2}\right){\displaystyle \frac{10x}{3}}}=\frac{1}{3}$

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

∴ 275 – 10*r*^{2} = 30*r*^{2}

∴ 40*r*^{2} = 245

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

$\therefore r=\sqrt{\left(\frac{55}{8}\right)}$

Hence, option 4.

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**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

Answer: Option C

**Explanation** :

Let the side of the square base AB = BC = CD = DA = ‘a’.

Height OB = 13.

Volume of a square base pyramid = a^{2} × h/3 = 200 cm^{3}

⇒ a^{2} = 600/13

Diagonal AC = a√2

⇒ AB = a/√2

In right ∆OAB, AO^{2} = AB^{2} + OB^{2}

⇒ AO^{2} = (a/√2)^{2} + 13^{2}

⇒ AO^{2} = a^{2}/2 + 169

⇒ AO^{2} = 300/13 + 169 = 2497/13

⇒ AO ≈ 13.86

Thus, the length of the slant slope (when rounded to the nearest integer) = 14 cm

Hence, option (c).

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**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

∴ $\frac{Area(\u2206ASP)}{Area\left(ABCD\right)}$ = $\frac{18lb}{64lb}$ = $\frac{36}{128}$

Hence, option (a).

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**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

Answer: Option A

**Explanation** :

Consider the figure given in the question.

The two small triangles formed in the region between the vertex of the square and the width of the plank (top left corner) will be congruent to each other. Also, because they are congruent and because the diagonal of the square is the angle bisector of the angle at the vertex, each of these two triangles will be a 45-45-90 triangle.

∴ Height of small triangle = width of small triangle/2 = √10/2

By symmetry, this will also be the height of the other small triangle (bottom right corner).

Diagonal of the square = 10√2 m

∴ Length of plank = 10√2 − (√10/2) − (√10/2) = 10√2 − √10 = √10(√20 − 1) m

Width of plank = √10 m

∴ Area of plank = √10 × √10(√20 − 1) = 10(√20 − 1) sq.m

Hence, option 1.

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**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

Answer: Option A

**Explanation** :

Circumference of circular hall = 2 × (22/7) × (49) = 308 m

∴ Speed of machine = (308/148) = (77/37) m/min

Perimeter of hexagon = 54 × 6 = 324 m

∴ Time taken to move along hexagon = (324) / (77/37)

= (324 × 37) / 77 = 155.69 minutes

∴ Extra time required = 155.69 − 148 = 7.69 minutes

Hence, option 1.

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**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

Answer: Option A

**Explanation** :

Consider the diagram below:

Area of a 1 × 1 tile = 1 cm^{2}

Total area of A = 9 × 11 = 99 cm^{2}

∴ Number of 1 × 1 tiles required to cover A = 99/1 = 99 tiles.

Total Area of B + C = 9 × 0.5 + 11 × 0.5 = 20 × 0.5 = 10 cm^{2}

∴ Number of 1 × 1 tiles required to cover B and C = 10/1 = 10 tiles.

Now the remaining area i.e., D = 0.5 × 0.5 = 0.25 cm^{2}

It would be better to but a 0.5 × 0.5 tile to cover D.

∴ The minimum cost of covering entire floor with tiles

= (99 + 10) × 100 + 1 × 30 = 10930

Hence, option (a).

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**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

Answer: Option A

**Explanation** :

When a square piece of paper is folded along its diagonal we will get a right isosceles triangle whose area becomes half.

Hence, after 3 such folds we get a right isosceles triangle, whose area is 1/8th the area of the original square.

Area of the folded triangle = ½ × 10 × 10 = 50.

∴ Area of the original square = 8 × 50 = 400 cm^{2}.

**Alternately,**

If the side of the isosceles triangle is 10 cm then we get the side of the square as 20 cm on unfolding it.

Hence, the area of square = 400 cm^{2}.

Hence, option (a).

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**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)

Answer: Option A

**Explanation** :

Volume of liquid = volume of cylinder – volume of rectangular solid

Volume of cylinder = 𝜋*r*^{2}*h* = (7^{2})(15)= 735𝜋

Since each corner of the solid with a square base is tangential to the cylinder wall, the diagonal of the square base is equal to the diameter of the base of the cylinder.

∴ Diagonal of square base = 2(7) = 14

∴ Side of square = 14/√2 = 7√2

Volume of rectangular solid = (7√2) × (7√2) × 12 = 1176

∴ Volume of liquid = 735𝜋 – 1176

= 147(5𝜋 – 8)

Hence, option 1.

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**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

Answer: Option D

**Explanation** :

DC is diameter of the cylinder.

m∠ADC = 90°

∴ AC is the diameter of the sphere.

Thus, AC = 16 and AD = height of the cylinder = 14

DC2 = 162 – 142 = 60

∴ Radius2 = 60/4 = 15

∴ Volume of cylinder = πr2h = π × 15 × 14 = 660

Hence, option 4.

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**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.

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**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

Answer: Option D

**Explanation** :

Flat surface area of the cylinder = 2 × π × 7^{2} = 98 π cm^{2}

Volume of the cylinder = π × 7^{2} ×10

= 490 π cm^{3}

Volume of cone A = (3/7) × 490 π = 210 π cm^{3} = (1/3) × flat surface area of cone

A × 10

Flat surface area of cone A = 63 π cm^{2}

Volume of cone B = (4/7) × 490 π = 280 π cm^{3} = (1/3) × flat surface area of cone

B × 10

Flat surface area of cone B = 84 π cm^{2}

Total flat surface area of cones

= (63 + 84) π = 147 π cm^{2}

Percentage change in the flat surface area

= (147 – 98)/98 × 100 = 50%

Hence, option 4.****

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**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

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

= 1254 cm^{2}

Cost = 1254 ×100 = Rs. 125,400

Hence, option 2.

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**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

Answer: Option E

**Explanation** :

Let V be the volume of water-melon, S be the total surface area and t be the thickness of the skin, then volume useable for Juice is, V – St

Hence, if total surface area is minimum, then useable volume of the water-melon will be highest.

Now, for equal volume, sphere has the least surface area.

Hence, option 5.

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**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

Answer: Option E

**Explanation** :

Let *r* be the radius of the smaller sphere.

Now, the volume of the big sphere and the 1000 small spheres is same.

Hence, we have,

Hence, *r*^{3} = 1

Hence, *r* = 1

Now, total surface area of big sphere = 4 × π × 10^{2} = 400π

Total surface area of 1000 new sphere = 1000 × 4π × 1^{2} = 4000π

Hence, total surface area increases by, (4000π – 400π) / 400π = 9 times.

Hence, option 5.

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**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

Answer: Option E

**Explanation** :

Volume of the equipment is;

1/3 × π × *r*^{2} × *h* = 1/3 × 22/7 × 7 × (0.25)^{2} = 11/6 cm^{3}

Now, 11/24 cm^{3} can write 330 words.

Hence, 1 cm^{3} can write,

330 × 24/11 = 720

We know that 1 cm^{3} = 1 ml

3/5^{th} of a litre is 600 ml which equals 600 cm^{3}

Hence, 600 cm^{3} will write, 720 × 600 = 432000 words

Hence, option 5.

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