CMAT Geometry Questions

AC is ladder, C is base of ladder, AB is wall

Given AC = 30 m and BC = $$\frac{1}{3}AC$$ = $$\frac{1}{3}\left(30\right)$$ = 10 m

Applying pythagoras theorem,

$$AB^2+\ BC^2=\ AC^2$$

$$AB^2$$ = $$30^2-10^2$$ = 800 

AB = $$20\sqrt{\ 2}$$

Answer is option B.

Volume of copper wire melted will be equal to volume of sphere.

Volume of copper wire = Area $$\times\ $$ length = $$\pi\ \left(\frac{2}{10}\right)^2\times\ 24300\ cm^3$$

Let the radius of sphere be r

$$\frac{4}{3}\pi\ r^{3\ \ }=\ \pi\ \left(\frac{2}{10}\right)^2\times\ 24300\ cm^3$$

$$r^3$$ = 729

r = 9 cm

Diameter of sphere = 2(9) = 18 cm

Answer is option B.

we have :

Triangles PSR and PSQ are right triangles 
so using pythaghoras theorem
we get
441 = $$PS^2+9x^2$$
81 =$$PS^2+x^2$$
subtracting we get
360=$$8x^2$$
$$x^2\ =45$$
x=$$3\sqrt{\ 5}$$
QR = 4x = $$12\sqrt{\ 5}$$

Curved Surface area of roller = $$2\pi\ rh$$ 
Cost = Area covered* cost per meter square 
so we get total cost = $$2\times\ \frac{22}{7}\times\ \frac{7}{20}\times\ 4\times\ 4\times\ 2000$$
=Rs 70,400

We are given :

Now In triangle ABC 
using the mid point theorem 
we get DE || BC and DE = $$\frac{1}{2}\left(BC\right)$$
Now ADE is similar to ABC 
So Area of ADE : Area of ABC = (DE/BC)^2 = 1:4
Similarly Area of BDF :Area of ABC = 1:4
and Area of CEF : Area of ABC = 1:4
Now we can say if area of ABC = 4A
then area of ADE = area of BFD = area of CEF = a
Therefore area of DEF = 4a-3a =a
so area of DEF : area of ABC = 1:4

In a triangle, the sum of two sides is always greater than the third side.

In the last two options, sum of first two sides is less than third side, and in first two options sum is equal to third side, hence these three are not possible.

Thus, possible lengths of triangle = [5,7,9]

=> Ans - (B)

Height of cylinder = 8 units and radius = 8 units

=> Farthest distance = hypotenuse = $$\sqrt{(8)^2+(16)^2}$$

= $$8\sqrt{5}$$ units

=> Ans - (A)

In a triangle, the angular bisector divides the side which it intersects in the ratio of other two sides.

So, In the triangle ABC, the angular bisector AD divides the side BC in the ration of AB and AC

$$\Rightarrow$$ BD : DC = AB : AC

Given BC = 10 cm and DC = 6 cm 

$$\Rightarrow$$ BD = BC - DC = 10 - 6 = 4 cm

Then $$AB =\frac{BD}{DC}\times AC = \frac{4}{6}\times 4.2 = 2\times 1.4 = 2.8$$ cm 

Let us consider the fucntion f(x) = $$a\cos x + b\sin x + c$$ 

The range of f(x) if given as $$c - \sqrt{a^2 + b^2} \leq f(x) \leq c + \sqrt{a^2 + b^2}$$

Then the maximum value of  3 $$cosx+4 sinx+8$$  is = 8 + $$\sqrt{3^2 + 4^2}$$ 

= 8 + 5 = 13

Given the length, breadth and height of a rectangular cuboid are in the ratio 1: 2 : 3.

Let the length, breadth and height be k, 2k, and 3k units where 'k' is a constant.

Volume of the cuboid(V') = $$l\times b\times h$$ units = $$k\times 2k\times 3k$$  = $$6k^3$$ units

Given the length, breadth and height are increased by 100% each .

Then the new length, breadth and height will be doubled.

So new length, breadth and height of the rectangular cuboid are 2k, 4k, and 6k units.

New Volume of the cuboid (V') = $$2k\times 4k\times 6k = 48k^3$$ units.

Increase in the volume of the cuboid = V' - V = $$48k^3 - 6k^3 = 42k^3 = 7 [6k^3] = 7V$$

Therefore the volume has increased 7 times the initial volume.

let the length of the halls be 'a' and 'a+1' m respectively and height of the halls be 'h' m.

Area of 4 walls of first square hall (A1) = $$(4\times a\times h) m^2$$

Given, cost of painting is Rs. 5 per sq.m

$$\Rightarrow$$  cost of painting 4 walls of the first square hall = A1 * 5 = $$4\times a\times h\times 5 = Rs. 20*a*h$$.

Given that this cost of painting = Rs. 6000

$$\Rightarrow$$ 20*a*h = 6000...........(1)

Similarly,

Area of 4 walls of second square hall (A2) = $$(4\times a+1\times h) m^2$$

$$\Rightarrow$$  cost of painting 4 walls of the second square hall = A2 * 5 = $$4\times a+1\times h\times 5 = Rs. 20*(a+1)*h$$.

Given that this cost of painting = Rs. 6100

$$\Rightarrow$$ 20*(a+1)*h = 6100............(2)

Dividing (2) with (1), we get

$$\frac{a+1}{a} = \frac{6100}{6000}$$

$$\Rightarrow 1 + \frac{1}{a} = \frac{61}{60}$$

$$\Rightarrow a = 60$$

Substituting this value in equation (1), we get

20*60*h = 6000

$$\Rightarrow$$ h = 5 m.

So, the height of the two walls is 5 m.

Volume left out in the rectangular box = Total volume of the rectangular box - Volume of the cylinder placed within it.

Total volume of the rectangular box of width(a), length(b), and height(c) is given as (V) = $$a\times b\times c$$ cubic units.

= $$6\times 8\times 10$$

= 480 cubic units.

Volume of the solid cylinder of height 'c' and of diameter 'a' is given by (V') = $$\frac{\pi}{4}*a^2*c$$ cubic units.

= $$\frac{\pi}{4} * 6^2 * 10$$

= 90$$\pi$$ cubic units.

Therefore left out volume = 480 - 90$$\pi$$ cubic units.

Whenever some cubes are melted to form a single larger cube, the total volume always remains constant.

So, Sum of the volumes of three small cubes = Volume of the larger cube.

Let the side of the lager cube so formed be 'a' cm.

Then $$1^3 + 6^3 + 8^3 = a^3$$

$$\Rightarrow a^3 = 729$$

$$\Rightarrow a = 9$$ cm.

Surface area of cube so formed = 6*$$a^2$$ = $$6\times 9^2 = 6\times 81 = 486  cm^2$$

Curved surface area of cylindrical pillar is given as A = 2$$\pi$$rh, where 'r' is base radius and 'h' is height of the cylinder.

So curved surface area of each pillar A = $$2\times \pi\times 0.35\times 5 = 11 m^2$$

Given there are 30 cylindrical pillars.

The curved surface area of half the number of pillars = 15 * A = 165 $$m^2$$.

Given rate of painting is Rs. 10 per $$m^2$$

The cost of painting curved surface of half the number of pillars = Curved surface area * rate of painting

= $$165\times 10$$

= Rs. 1650. 

Volume(V)  of the sphere with radius R=8 cm is given by V= $$\frac{4}{3} \pi  R^{3}$$.

                                                                                                       = $$\frac{4}{3} \pi(8^{3})$$= $$\frac{4}{3} \pi (512) cm^{3}$$.                                                                                                

and Volume (v) of each iron ball with radius r= 1cm is given by v= $$\frac{4}{3} \pi r^{3}$$.

                                                                                                              = $$\frac{4}{3} \pi (1^{3})$$= $$\frac{4}{3} \pi cm^{3}$$.

Let say 'n' iron balls each of volume 'v' are required to form a sphere of volume 'V'.

=> Total volume of 'n' iron balls = Volume of the sphere

=> $$n\times v$$ = V

=> n= $$V\div v$$ = $$\frac{4}{3}  \pi (512)\div \frac{4}{3}  \pi$$ = 512. 

Hence 512 iron balls are required in total to form the sphere.

In $$\triangle$$AOB, AO = OB (Equal radii)

∴ $$\triangle$$AOB is isosceles.

∴ ∠ABO = ∠BAO = (180°-114°)/2 = 33°

Hence, option D.

SInce PQ is the diameter, the angle subtended by it at R is 90 deg. i.e., $$\angle$$ PRQ = 90 deg.

Let $$\angle$$ RPQ = $$\theta$$, then $$\angle$$RQP = 90 - $$\theta$$

As the angles subtended by a chord in same segment are equal, $$\angle$$RPQ = $$\angle$$RSQ = $$\theta$$

In triangle RSQ, $$\angle$$QRS + $$\angle$$RSQ + $$\angle$$RQS = 180 

$$\Rightarrow$$ $$\angle$$QRS + $$\theta$$ + 35 + 90 - $$\theta$$ = 180

$$\Rightarrow$$ $$\angle$$QRS = 180 - 125 = 55 deg.

Hence $$\angle$$QRS = 55 deg.

60 minutes = 360 degrees or 2$$\pi$$ radians.

$$\Rightarrow$$ 1 minute = 60  degrees.

$$\Rightarrow$$ 30 minutes = 180  degrees  or  $$\pi$$ radians.

Therefore area (A) swept by minute hand of length 42 mm in 30 minutes =  area of semi-circle of radius (r = 42 mm) .

$$\Rightarrow A = \pi r^2\div 2 = 3.14\times 42^2\div 2 = 2769.5 mm^2$$.

Area of triangular metal plate with base(b) = 88 cm and altitude(h) = 64 cm is given as A = $$\frac{1}{2} b\times h = \frac{1}{2} 88\times 64 = 2816 cm^2$$

Given this area is to be reduced to one-fourth by making a hole in the shape of circle

$$\Rightarrow$$ Reduction in the area of the triangle = Area of the circular hole

$$\Rightarrow\frac{3}{4}\times A=\pi r^2$$

$$\Rightarrow \frac{\frac{3}{4}\times 2816}{\pi} = r^2$$

$$\Rightarrow r = \sqrt672 = 4\sqrt42 $$

So, the radius of the circular hole = $$4\sqrt42$$ cm.

Let the breadth(b) of the room be 'x' metres.

then, length(l) of the room = x+2 metres.

Area(A) = $$l\times b$$ = x(x+2) $$m^2$$

Given, length is increased by 4 meters and the breadth decreased by 2 meters

Then, new length(l') of the room = x+6 metres

new breadth(b') of the room = x-2 metres

New Area(A') of the room = $$l'\times b'$$ = (x+6)(x-2) $$m^2$$

Also given that, A = A'

$$\Rightarrow x(x+2) = (x+6)(x-2)$$

$$\Rightarrow x^2+2x = x^2+4x-12$$

$$\Rightarrow 2x = 12$$

$$\Rightarrow x = 6$$

Therefore the length of the room (l) = 8 metres

and breadth of the room (b) = 6 metres

and given height of the room (h) = 3 metres

Since the room will be in the shape of a cuboid, Surface area = 2 ($$l\times b+b\times h+l\times h$$)

But the Surface area of Walls = Total Surface area - Area of Roof and Floor = 2 ($$l\times b+b\times h+l\times h) - 2 (l\times b) = 2 (8\times 3+6\times 3) = 84 m^2$$

Hence, Surface Area of walls = 84 $$m^2$$.

The Volume of the right circular cone of base radius 'r' and height 'h' is given by 'V' = $$\frac{1}{3} \pi r^2h$$

Given 'h' has been increased by 200%

$$\Rightarrow$$ New height h' = h[1 +$$\frac{200}{100}$$] = 3h

also,radius of the base is reduced by 50%

$$\Rightarrow$$ New base radius r' = r[1 - $$\frac{50}{100}$$] = $$\frac{r}{2}$$  

New Volume of the cone with new base radius r' and new height h' is given by V' = $$\frac{1}{3} \pi r'^2h'$$ = $$\frac{1}{3} \pi (\frac{r}{2})^2(3h) = \frac{3V}{4}$$.

Change in Volume = $$\frac{New Volume - Old Volume}{Old Volume}\times 100  = \frac{\frac{3V}{4} - V}{V}\times 100 = -25$$

Hence the new volume decreased by 25 %.