MAH MBA CET Geometry Questions

Volume of cone = $$\ \frac{\ 1}{3}\pi\ r^2h$$ 
Let initial height be 'h' and radius be 'r' of the cone.
So, Initial volume =  $$\ \frac{\ 1}{3}\pi\ r^2h$$
New height H = h(1+200%) = h(1+$$\ \frac{\ 200}{100}$$) = 3h
New radius R = r(1-50%) = r(1-0.5) = $$\ \frac{\ r}{2}$$
New volume = $$\ \ \frac{\ 1}{3}\pi\ \times\ \left(\frac{r}{2}\right)^2\times\ 3h$$ = $$\ \frac{\ \pi\ r^2h}{4}$$
Change in volume = $$\frac{\left(\frac{\ \pi\ r^2h}{4}-\frac{\ \pi\ r^2h}{3}\right)}{\ \frac{\ \pi\ r^2h}{3}}\times\ 100$$ = -25%
So, volume decreased by 25%.
Option B is the answer.

We can use the Angle Bisector Theorem:

In triangle ABC, if AD bisects ∠A, then $$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$

We know: BC=10 and DC=6

So, BD = BC − DC = 10 − 6 = 4

Applying the theorem

$$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$ 

Given
AC=4.2, so:$$AB=\dfrac{2}{3}\times4.2=2.8\ cm$$

Volume remains conserved during melting.
Volume before melting = Volume after melting
Before melting, the wire was in cylindrical shape and after melting, it became sphere.
So, Vol. before = $$\pi r^2l$$ = Vol. after = $$\ \frac{\ 4}{3}\pi\ R^3$$, where R is the radius of sphere formed.
$$\left(2\times\ 10^{-3}\right)^2\times\ 243\ =\ \frac{\ 4}{3}R^3$$ 
$$R^3=729\times\ 10^{-6}$$
R=9cm.
So, diameter of the sphere = 18cm.
Option C is the answer.

Let the radius of sphere A be 'r' and sphere B be 'R'
Surface area of A = $$4\pi r^2$$
Surface area of B = $$4\pi R^2$$ = (1+300%) of A = 4 * $$4\pi r^2$$
R=2r
Volume of A = $$\frac{4}{3}\pi r^3$$ 
Volume of B = $$\frac{4}{3}\pi R^3\ =\ \ \frac{\ 4}{3}\pi\ \left(2r\right)^3\ =\ \ \frac{\ 32}{3}\pi\ r^3$$
Required percent = $$\ \frac{\ \left(\frac{32}{3}\pi\ r^3\ -\ \frac{4}{3}\pi\ r^3\right)}{\ \frac{32}{3}\pi\ r^3}\times\ 100\ =87.5\ \%\ $$
Option C is the answer.

Let the sides of the isosceles triangle be (a, a, b), where a and b are integers.

Case 1: The sum of the two equal sides is 12

a + a = 12 ⇒ a = 6

a + a = 12 ⇒ a = 6

Triangle inequality: 6 + 6 > b ⇒ b < 12

Also b > 0 and integer.

So, b = 1, 2, 3, …, 11

So, Total = 11 triangles

Case 2: One equal side + base = 12 

a + b = 12 ⇒ b = 12 − a

Triangle inequality: a + a > b ⇒ 2a > 12 − a ⇒ 3a > 12 ⇒ a > 4

Also: b > 0 ⇒ 12 − a > 0 ⇒ a < 12

So, a = 5,6,7,8,9,10,11

Hence, Total = 7 triangles

But when a=6, the triangle is (6,6,6), which was already counted in Case 1. So subtract 1 duplicate.

So final count: 11 + 7 - 1 = 17

Let us assume the original side to be 10 units.
The length of side is mistakenly taken as 10*1.05 = 10.5 units.
The initial area was 10*10 = 100 sq units and the incorrect area is 10.5*10.5 = 110.25 units.
The change in area is 10.25%.

The LCM of 180 and 105 = 15.
The number of tiles required for each row = 180/15 = 12 and,
the number of tiles required for each column = 105/15 = 7.
So, total number of tiles required = 12*7 = 84.

The well is in the form of a cylinder with radius = 3.5 m and depth = 22.5 m. 
Then, the earth dug out is spread around to form an embarkment. Volume of earth dug out = Volume of earth on the embarkment.
Volume of earth on the embarkment = Area of embarkment * height.
So, $$\pi\ \times\ 3.5^2\times\ 22.5\ =\ Area\times\ height$$.
Area of embarkment = Area of larger circle - Area of inner circle = $$\pi\ \times\ 14^2-\pi\ \times\ 3.5^2\ =\pi\ \left(14^2-3.5^2\right)$$.
Putting this value of area in the above equation gives us the value of height to be 1.5 m.