Geometry - Mensuration - Previous Year CAT/MBA Questions

Answer: Option B

Text Explanation :

Answer: Option D

Video Explanation

Text Explanation :

Area of the equilateral triangle = √3/4 × (12)²

Let the length of each side of the regular hexagon be ‘a’.
Area of a regular hexagon with side ‘a’ = 6 × √3/4 × (a)²

Now, 6 × √3/4 × (a)² = √3/4 × (12)²

⇒ a² = 24

⇒ a = 2√6

Hence, option (d).

Answer: Option C

Video Explanation

Text Explanation :

Area of the sheet left unpainted is two-thirds of painted area i.e., area of the circles will be three-fifth of the total area.

​​​​​​​

∴ Area of the circle = 3/5 × 135 = 54 sq. in.

Let the radius of the circle be r and l be the length of the rectangle. Width of the rectangle will be 2r.

⇒ πr² = 81

⇒ r = $\frac{9}{\sqrt{\pi }}$

Area of rectangle = 135 = 2r × l

⇒ l = $\frac{135}{2r}$ = $\frac{135}{2\times \frac{9}{\sqrt{\pi }}}$ = $\frac{15\sqrt{\pi }}{2}$

∴ Perimeter of the rectangle = 2(l + r) = $2\left(\frac{15\sqrt{\pi }}{2}+\frac{18}{\sqrt{\pi }}\right)$ = $3\sqrt{\pi }\left(5 +\frac{12}{\pi }\right)$

Hence, option (c).

Answer: Option D

Video Explanation

Text Explanation :

When a cone of height ‘H’ is cut at a distance of ‘h’ from the top

⇒ $\frac{Volume of smaller cone}{Volume of original cone}$ = ${\left(\frac{h}{H}\right)}^3$

​​​​​​​

∴ $\frac{Volume of smaller cone}{Volume of original cone}$ = ${\left(\frac{9}{27}\right)}^3$ = $\frac{1}{27}$

⇒ If volume of the smaller cone = x, the volume of original cone is 27x, and volume of the frustum = 26x

⇒ 26x – x = 225

⇒ x = 9

∴ Volume of the original cone = 27x = 27 × 9 = 243.

Hence, option (d).

Answer: Option C

Video Explanation

Text Explanation :

Ratio of the three diagonals is 3 : 2√3 : √15

Let the legths of the three diagonals be 3k, (2√3)k and (√15)k.

And, the brick have length, breadth, height as x, y and z respectively. 
 
∴ x² + y² = (3k)² = 9k² ...(1)

y² + z² = [(2√3)k]² = 12k² ...(2)

z² + x² = [(√15)k]² = 15k² ...(3)

Adding (1), (2) and (3), we get;
 
x² + y² + z² = 18k² ...(4)

Using (4) along with any of (1), (2) and (3), we get;

x = k√6 , y = k√3 and z = 3k, 

Required ratio = (k√3)/3k = 1/√3.

Hence, option (c).

Answer: Option A

Video Explanation

Text Explanation :

 
In the above image, the side length of the square is equal to the side length of the equilateral triangle.
 
So, AB = BC = OC = 20 cm.
 
∆AOM is a right angled triangle with ∠AMO = 90°, AO = 20 cm. OM is the height of the pyramid.
 
AC is the diagonal of the square base, so AM = AC/2 = (20√2)/2 = 10√2.
 
So using pythagoras theorem; OM² = AO² − AM² = 20² − (10√2)² = 400 − 200 = 200.
 
∴ OM =  10√2.
 
Hence, option (b)

Answer: Option D

Video Explanation

Text Explanation :

As the total number of cylinders to be kept at a minimum, the volume of each cylinder must be maximum possible.

HCF(405, 783, 351) = 27  ⇒ Volume of each cylinder = 27 cc.
 
∴ πr²h = 27  (where r and h represent the radius and height of the cylinder)
 
Given that r = 3, so πh = 3.
 
Number of cylinders of;
 
Iron = 405/27 = 15.
 
Aluminium = 783/27 = 29.
 
Copper = 351/27 = 13.

∴ Total number of cylinders = 15 + 29 + 13 = 57.
 
Total surface area of all the cylinders = 57 × [2πr² + 2πrh] 
 
Put r = 3 and πh = 3 to get;
 
Required surface area = 1026(1 + π) sq. cm.
 
Hence, option (d).

Answer: 198

Video Explanation

Text Explanation :

∆ABE ~ ∆ACD

⇒ $\frac{\mathrm{AB}}{\mathrm{AC}}$ = $\frac{\mathrm{BE}}{\mathrm{CD}}$

AB = 3 ft, AC = 12 ft and CD = 4 ft ⇒ BE = 1 ft

The required volumn of the cone = $\frac{1}{3}\mathrm{\pi }\times 4^2\times 12$ - $\frac{1}{3}\mathrm{\pi }\times 1^2\times 3$ = $\frac{1}{3}\mathrm{\pi }\times (192-3)$ = $\frac{1}{3}\times \frac{22}{7}\times 189$ = 198 cubic ft.

Hence, 198.

Answer: Option A

Video Explanation

Text Explanation :

The area of the semicircle = 72π sq. cm. If ‘r’ is the radius of the semi-circular portion removed,

We get $\pi \times \frac{r^2}{2}=72$

Therefore r = 12 cm.
Therefore the diameter of the semi-circle = AB = 24 cm.

Area of rectangle ABCD = 768 sq. cm.

Therefore the length of side AC $=\frac{768}{24}=32 cm.$

The perimeter of the leftover portion

= Perimeter of the semicircle + l(AD) + l(DC) + l(CB)
= 12π + 32 + 24 + 32 = 88 + 12π

Hence, option (a).

Answer: Option B

Text Explanation :

The volumes of the 5 smaller cubes and the original big one are in the ratio 1 : 1 : 8 : 27 : 27 : 64.

Therefore, the sides are in the ratio 1 : 1 : 2 : 3 : 3 : 4 while the areas are in the ratio 1 : 1 : 4 : 9 : 9 : 16.

The sum of the areas of the 5 smaller cubes is 24 parts while that of the big cube is 16 parts.

The sum is 50% greater.

Hence, option (b).

Answer: 6

Text Explanation :

The height of the cylinder (h) = 3

The volume = 9π
⇒ πr²h = 9π
⇒ r = √3

The radius of the ball (R) = 2

The height of O, the centre of the ball, above the line representing the top of the cylinder is say a.

Hence, 2² = a² + (√3)²

∴ a = 1

∴ The height of the topmost point of the ball from the base of the cylinder is = height of cylinder + height of center of ball from the top of cylinder + radius of the ball = 3 + 1 + 2 = 6.

Hence, 6.

Answer: Option C

Video Explanation

Text Explanation :

Let us draw the base of the vertical pillar as shown below

In the above figure EF = AB = 10 cm

Also, as AD = BC, by rule of symmetry
DE = FC

Now DC = DE + EF + FC

Let DE = FC = x

20 = x + 10 + x
⇒ 2x + 10 = 20
⇒ x = 5

Now in ∆AED, AE = 12 cm and DE = 5 cm

∴ AD = $\sqrt{5^2+{12}^2}=13$

Also given AD = BC

∴ BC = 13 units

Now, total surface area of vertical pilar with base ABCD = Area of Rectangle with side AB and side (Height) 20 cm + Area of Rectangle side AD and side (Height) 20 cm + Area of Rectangle with side BC and side (Height) 20 cm + Area of Rectangle with side DC and side (Height) 20 cm + 2 × Area of Trapezium ABCD

⇒ 10 × 20 + 13 × 20 + 13 × 20 + 20 × 20 + 2 × $\frac{1}{2}$ × (20 + 10) × 12
⇒ 200 + 260 + 260 + 400 + 360
⇒ 1480 sq.cm.

Hence, option (c).

Answer: 200

Video Explanation

Text Explanation :

Let ‘x’ and ‘y’ be the dimensions of the rectangle

Let us suppose 2x + y = 400 … (1)

Area = xy.

For xy to be maximum, 2xy should also be maximum.
Now the product 2xy will be maximum when 2x = y. [AM ≥ GM]

So y + y = 400 [From (1)]
⇒ 2y = 400 or y = 200

Substituting value of y in (1) we get,
2x = 200 or x = 100

In a rectangle, length is greater than breadth, so we take y as the length.

Hence area of the park is maximum when length is 200 ft.

Hence, 200.

Answer: Option A

Video Explanation

Text Explanation :

As shown in the figure, ABC is the cross section of a cone of height 10 cm and radius of base 4 cm.

PQRS is the cross section of a cylinder which needs to be fitted inside the cone such that one of the flat faces of the cylinder (represented by RS) coincides with the base of the cone (represented by BC).

∵ ∆ABD and ∆PBR are similar triangles.

∴ $\frac{10}{4}=\frac{h}{4-r}$

∴ h = 10 - 2.5 × r

Total surface area of the cylinder = S = 2πr × (h + r) = 2πr × (10 - 2.5 × r + r) cm²

⇒ S = πr(20 - 3r) cm²   ...(1)

Now consider two terms 3r and (20 - 3r)

Now, $\frac{3\mathrm{r}+(20-3\mathrm{r})}{2}$ ≥ $\sqrt{3r\times (20-3r)}$ [AM ≥ GM]

⇒ 10 ≥ $\sqrt{3r\times (20-3r)}$

⇒ 3r(20 - 3r) ≤ 100

∴ r(20 - 3r) ≤ 100/3

From (1), we have

⇒ S = πr(20 - 3r) ≤ π × 100/3

Hence, option (a).

Answer: Option B

Text Explanation :

Let the inner radius be r meter. Capacity of tank = (1 m³ = 1 kilolitre)

From statement A, since r ≥ 4m

∴ Capacity of tank > 256 m³

Since the capacity needed is more than 256 m³ statement A is insufficient.

From statement B,

Volume of the material of tank = mass/density = 30000kg/(3 gm/cc) = 10,000,000 cm³ = 10 m³

Hence the inner volume of tank = Outer volume – Volume of material of tank

Therefore, we can say that the tank capacity is adequate.

Hence, option (b).

Answer: Option E

Video Explanation

Text Explanation :

Let the original length, breadth and height of the room be 3x, 2x and x respectively.

∴ The new length, breadth and height are 6x, x and x/2 respectively.

Area of four walls = (2 × length × height) + (2 × breadth × height)

Original area of four walls = 6x² + 4x² = 10x²

New area of four walls = 6x² + x² = 7x²

∴ Area of wall decreases by [(10x²  − 7x²)/10x²] × 100 = 30%

Hence, option (e).

Answer: Option B

Video Explanation

Text Explanation :

Let PQRS be the square sheet and let the hole have centre O.

As P lies on the circumference of the circle and as m ∠APC = 90°, AC is a diameter.

∵ BP is a diameter, m ∠PAB = m ∠BCP = 90°.

∵ BP = AC, ABCP is a square.

∴ m ∠POC = 90° and OP = OC = 1 unit

The area of part of the circle falling outside the square sheet

= 2 × (Area of sector OPC – Area of ∆ OPC)

$$\begin{gathered} =2\times \left[\left(\frac{\pi \times 1^2}{4}\right)-\left(\frac{1}{2}\times 1^2\right)\right] \\ =\frac{\pi -2}{2} \mathrm{sq}.\mathrm{units} \end{gathered}$$

Area of part of hole on sheet = Area of hole − Area of part of the circle falling outside the square sheet

$=\pi -\left(\frac{\pi -2}{2}\right)=\frac{\pi +2}{2} \mathrm{sq}.\mathrm{units}$

Part of square remaining after punching = Area of square − Area of part of hole on sheet

$=4-\left(\frac{\pi +2}{2}\right)=\frac{6-\pi }{2} \mathrm{sq}.\mathrm{units}$

∴ Proportion of sheet area that remains after punching $=\frac{\left({\displaystyle \frac{6-\pi }{2}}\right)}{4}=\frac{6-\pi }{8}$

Hence, option (b).

Answer: Option D

Video Explanation

Text Explanation :

We have calculated this value while solving the previous question.

The area of part of the circle falling outside the square sheet

= 2 × (Area of sector OPC – Area of ∆OPC)

$=\frac{\pi -2}{2} \mathrm{sq}.\mathrm{units}$

Hence, option (d).

Answer: Option D

Video Explanation

Text Explanation :

​​​​​​​

Let r be the radius of the circular tracks.

Length and breadth of the rectangular track are 4r and 2r respectively.

Length (perimeter) of the rectangular track = 12r

Length of the two circular tracks (figure of eight) = 4πr

If A and B have to reach their starting points at the same time,

 $\frac{12r}{a}=\frac{4\pi r}{b}$
(where a and b are the speeds of A and B respectively)

∴ $\frac{b}{a}=\frac{4\pi }{12}$

⇒ $\frac{(b-a)}{a}=\frac{4\pi -12}{12}$

∴ (b - a) × $\frac{100}{a}$ = 0.047 × 100 = 4.7%

Hence, option (d).

Answer: Option C

Video Explanation

Text Explanation :

This problem can be solved by trying different ways of placing the tiles on the floor. The maximum number of tiles that can be accommodated is 6, as shown in the figure.

Hence, option (c).

Answer: Option B

Video Explanation

Text Explanation :

Let each square tile have side = 1 unit

Let the length of the rectangular floor be m units and the breadth be n units.

Number of red tiles = (m – 2)(n – 2)

Number of white tiles = mn − (m – 2)(n – 2)

Now, (m – 2)(n – 2) = mn − (m – 2)(n – 2)
⇒ 2(mn – 2m – 2n + 4) − mn = 0
⇒ mn – 4m – 4n + 8 = 0
⇒ n(m – 4) = 4m − 8 
⇒ n = 4(m – 2)/(m – 4)

 Now, consider the options.

Option (a): If m = 10, n = 32/6, which is not possible as n is an integer

Option (b): If m = 12, n = 40/8 = 5, which is possible

Option (c): If m = 14, n = 48/10, which is not possible as n is an integer

Option (d): If m = 16, n = 56/12, which is not possible as n is an integer

Hence, option (b).

Answer: Option B

Video Explanation

Text Explanation :

Let the length and breadth of the rectangle be x and y respectively.

x > y

By conditions, $\frac{x}{y}=\frac{y}{{\displaystyle \frac{x}{2}}}$

∴  $\frac{x^2}{2}=y^2$

∴  $\frac{x^2}{2}$ = 4 (∵ y = 2)

∴ x² = 8

∴ x = 2√2

∴ Area of the smaller rectangle = 2√2 sq.units.

Hence, option (b).

Answer: Option A

Video Explanation

Text Explanation :

Let the side of the cube be a units.

DF, AG and CE are body diagonals each of length a√3 units.

∴ Circumradius of the equilateral triangle = $\frac{a\sqrt{3}}{\sqrt{3}}$ = a units

∴ Circumradius = side of cube

Hence, option (a).

Answer: Option D

Text Explanation :

Ratio of areas of two spheres = s₁ : s₂ = 4 : 1

∴ Ratio of their radii = r₁ : r₂ = 2 : 1

∴ Ratio of their volumes = v₁ : v₂ = 8 : 1

Volume of A is 12.5% (1/8th) of the volume of B.

But, volume of A is k% less than B.

∴ k = (100 − 12.5) = 87.5%

Hence, option (d).

Answer: Option C

Text Explanation :

The first layer has 1 ball.

Second layer has 1 + 2 = 3 balls

Third layer has 1 + 2 + 3 = 6 balls

∴ The nth layer of the stack would have $\left[\frac{n(n+1)}{2}\right]$ balls

∴ Total balls in all the layers = $\sum _{}^{}\left[\frac{n(n+1)}{2}\right]$ = 8436

∴ $\frac{1}{2}\times \left[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]$ = 8436

Only n = 36 satisfies the above equation.

Hence, option (c).