Quantitative Aptitude PYQs

Let’s say

C = Circumference of a circle

R = Radius of circle

P = Perimeter of square

L = Length of n side of square

Given that,

12 m long wire is converted into circle & square

Then,

C + P = 12       ----(1)

C = 2πR       ----(2)

From figure

L = 2R       ----(3)

P = 4 × D = 8 × R       ----(4)

4 equations & 4 unknowns

2πR + 8R = 12

⇒ πR + 4R = 6

R (π + 4) = 6

\(\Rightarrow R = \frac{6}{{\pi + 4}}\)

Angle are is ratio 1 : 1 : 4

Then let’s say angles are x, x, & 4x

Sum of angles of triangles is 180°

6x° = 180°

x = 30°

so, angles are 30°, 30° & 120°

We know that

\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = P\)

\(\Rightarrow a = P\sin A = P\sin 120^\circ = \frac{P{\sqrt 3 }}{2}\)

\(b = P\sin B = P\sin 30^\circ = \frac{P}{2}\)

\(C = \frac{P}{2}\)

Now perimeters of the triangle is k times its largest side

⇒ a + b + c = k × a

\(\frac{{\sqrt 3 }}{2} + 1 = k \times \frac{{\sqrt 3 }}{2}\)

\(\Rightarrow k = \frac{{\sqrt 3 + 2}}{{\sqrt 3 }};or;1 + \frac{2}{{\sqrt 3 }}\)

Let’s say side of right-angled triangle are a, b & 10

From Pythagoras

a² + b² = 100       ----(1)

\(Area = \frac{1}{2} \times a \times b = 24\)

⇒ a × b = 48       ----(2)

(a + b)² = a² + b² + 2ab

(a + b)² = 100 + 96

= 196

⇒ a + b = 14

∴ a × b = 48

a (14 - a) = 48

14a - a² = 48

⇒ a² - 14a + 48 = 0

⇒ a² - 8a - 6a + 48 = 0

a (a - 8) - 6(a - 8) = 0

a = 6 or a = 8

if a = 6 ⇒ b = 14 - a = 8

if a = 8 ⇒ b = 14 - 8 = 6

so, (a, b) is (6, 8) or (8, 6)

We can take any once of the above possibilities

⇒ a = 8 & b = 6 

Now shorter side is halved & longest side is doubled

Then \(a' = 2a;& ;b' = \frac{b}{2 }\) 

⇒ a' = 16 & b' = 3

So, hypotenuse \(= \sqrt {{{16}^2} + {3^2}} = \sqrt {265 } \).

From the property of circle angle submitted by diameters at the circumference of circle is 90°

∠ACD = ∠ABD = 90°,

From geometry

CD = BD

By Pythagoras theorem

CD² = BD² = AD² - AC²

= 16² - 12²

= 112

⇒ CD = BD = 4√7

Now let’s say AE = x and BE = EC = y

⇒ ED² = (16 - x)²

ED² = DB² - BE²

\({\left( {16 - x} \right)^2} = {\left( {4\sqrt 7 } \right)^2} - {y^2}\)

(16 - x)² = 112 - y²       ----(1)

AE² = AB² - EB²

X² = 12² - y²

x² + y² = 144       ----(2)

From equation (1) & (2)

144 - x² = 112 - (16 - x)²

144 - x² = 112 - 250 - x² + 32x

32x = 288

x = 9

\(y = \sqrt {144 - {9^2}}\)

\(\Rightarrow y = 3\sqrt 7\)

So, length of chord DC = 2 × y = 2 × 3 √7

= 6√7

1. An isosceles trapezium is always cyclic

So, AM = DN & AB = CD

So, by hypo-side test

ΔAMB is congruent to ΔDNC

⇒ ∠B = ∠C       ----(1)

By interior angler property of parallel lines

∠A + ∠B = 180°

⇒ ∠A + ∠C = 180°

So, by converse of cyclic quadrilateral theorem.

As isosceles trapezium is always cyclic

⇒ Statement 1 is true.

I. Any cyclic parallelogram is a rectangle

So, ABCD is a cyclic parallelogram

So, we have to proof that one angle of parallelogram is 90°

Since ABCD is a parallelogram

∠A = ∠C

In a cyclic parallelogram ABCD

∠A + ∠C = 180°

∠A + ∠A = 180°

⇒ ∠A = 90°

Hence, any cyclic parallelogram is a rectangle.

So, both statements are true.

Calcualation:

Here, AD and BE are the same ladders.

So, AD = BE

In Δ ACD,

AD² = AC² + CD² = (y + 0.8)² + 2.5²

In Δ BCE,

BE² = BC²+ CE² = y² + (2.5 + 1.4)² = y² + 3.9²

Since, AD = BE

So, AD² = BE²

(y + 0.8)2 + 2.52 =  y2 + 3.92

⇒ y² + (0.8)² + 2 × y × 0.8  + 6.25 = y² + 15.21

⇒ 0.64 + 1.6y + 6.25 = 15.21

⇒ 1.6y = 15.21 - 0.64 - 6.25 = 8.32

⇒ y = 8.32 / 1.6 = 5.2

So,

AD² =  (y + 0.8)2 + 2.52 = (5.2 + 0.8)² + 2.5² = 36 + 6.25 = 42.25

So, Length of the ladder = AD = √42.25 = 6.5 m

∴ The correct answer is option (2).

Radius of both circles are ‘r’

Length of common chord BD is \(10\sqrt 3 \)

AC = AB = BC = CD = DA = r

So, ABCD a is a rhombus

So, ΔABC & ΔACD are equilateral triangle

Then area of equilateral triangle \(= \frac{{\sqrt 3 }}{4}{\left( {side} \right)^3}\)

\(= \frac{{\sqrt 3 }}{4} \times {r^2}\)

So, sum of area of both triangle \(ABC;& ;ADC = 2 \times \frac{{\sqrt 3 }}{4} \times {r^2}\)

\(= \frac{{\sqrt 3 }}{2}{r^2}\)       ----(1)

Now area of rhombus \(= \frac{1}{2} \times diagonal;AC \times diagonal;BD\)

\(= \frac{1}{2} \times r \times 10\sqrt 3 \)

\(= 5\sqrt 3 r\)       ----(2)

Now equate 1^st & 2^nd equation

\(\frac{{\sqrt 3 }}{2}{r^2} = 5\sqrt 3 ;r\)

⇒ r = 10 cm

Hence, diameter = 2r = 20 cm

1. Number of circles drawn from three non-collinear point is only 1

⇒ statement 1 is wrong

2. Angle formed by minor segment of a circle is obtuse & for major segment it’s an acute angle

So, statement 2 is wrong

So, neither 1 nor 2

Sum of two sides of any triangle is greater than third side

1. AC < BC + AB true

2. BC < AB + AC true

3. AB < AC + BC true

So, statement 1, 2 & 3 all are correct

Let A, BC be the triangle & D, E & F are midpoints

Now sum of two sides of a triangle is greater than twice the radius

So, AB + AC > 2 AD

Same for

BC + AC > 2CF

BC + AB > 2BE

Odd above 3 equations

2(A.B + BC + AC) > 2(AD + BE + CF)

⇒ AB + BC + AC > AD + BE + CF

So, perimeters of triangle is greater than sum of its medians

⇒ Statement 1 is correct

D is a midpoint on BC means AD is a median

We know that,

Sum of two sides a triangle is greater the twice of the median

⇒ AB + AC > 2AD

Since BC is positive number

⇒ AB + AC + BC > 2AD

So, statement 2 is correct

⇒ Both 1 & 2 are correct

a) Surface area of sphere = 4 πr²

\(= 4\pi {\left( {\frac{{\sqrt 3 a}}{2}} \right)^2}\)

= 3πa²

Curved surface area of cone = πrl

\(= \pi \left( {\frac{a}{2}} \right)\sqrt {{{\left( {\frac{a}{2}} \right)}^2} + {a^2}} \)

\(= \frac{{\pi {a^2}\sqrt 5 }}{{2\sqrt 2 }}\)

So, surface area of sphere is not √5 times curved surface area cone

b) Curved surface area of cylinder = 2πrh

\(= 2\pi \left( {\frac{a}{2}} \right) \times a\)

= πa²

Surface area cube = 6a²

So, they are not equal

So, neither a nor b

Volume of the cube to that of the cylinder

\(\frac{{{a^3}}}{{\frac{{\pi {a^3}}}{4}}} = \frac{4}{\pi } = \frac{{14}}{{11}}\)

Cube is inscribed in a sphere if length of cube is ‘a’ then length of diagonal of cube will be \(\sqrt 3 a\)

& From geometry length of diagonal of the cube = Diameters of sphere then the radius of sphere \(= \frac{{\sqrt 3 a}}{2}\)

Volume of cube = a³           ----(1)

Volume of sphere \(= \frac{4}{3}\pi {\left( {\frac{{\sqrt 3 a}}{2}} \right)^3}\)

\(= \frac{4}{3} \times \frac{{3\sqrt 3 ;{a^3}}}{8}\)

\(= \frac{{\sqrt 3 ;{a^3}}}{2}\) ----(2)

Now, cylinder is inside a cube

Height of cylinder = length cube = a

Radius of cylinder \(= \frac{a}{2}\)

\(\Rightarrow Volume;of;cylinder = \pi {\left( {\frac{a}{2}} \right)^2} \times a\)

\(= \frac{{\pi ;{a^3}}}{4}\)

Now cone is inside cylinders with common base & same height

⇒ Volume of cone \(= \frac{1}{3}\pi {\left( {\frac{a}{2}} \right)^2} \times a\)

\(= \frac{{\pi ;{a^3}}}{{12}}\)

So, ratio volume of sphere to that of the cone is

\(\frac{{\frac{{\sqrt 3 }}{2}{a^3}\pi }}{{\frac{{\pi ;{a^3}}}{{12}}}} = \frac{{6\sqrt 3 }}{1} \Rightarrow 6\sqrt 3 ;:;1\)

Perimeters of triangle ABC = 9 + 40 + 41

= 90 cm

Perimeter of ΔADC = 28 + 15 + 41

= 84 cm

So, their difference is 90 – 84 = 6 cm

Given ∠ABC = 90°

By Pythagoras theorem

AC2 = AB2 + BC2

= 92 + (40)2

\(= \sqrt {81 + 61600}\)

\(= \sqrt {1681}\)

⇒ AC = 41 cm

Area of \({\rm{\Delta }}ABC = \frac{1}{2} \times base \times height\)

\(= \frac{1}{2} \times 40 \times 9\)

= 180 cm2

Now, for ΔABC, we use heron’s formula

\(s = \frac{{15 + 28 + 41}}{2} = 42\)

So, area of \({\rm{\Delta }}ACD = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \) 

\(= \sqrt {42\left( {42 - 28} \right)\left( {42 - 41} \right)\left( {42 - 15} \right)} \)

\(= \sqrt {42 \times 14 \times 1 \times 27}\)

⇒ Area of ΔACD = 126 cm2

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD

= 180 + 126

= 306 cm²

Given ∠ABC = 90°

By Pythagoras theorem

AC² = AB² + BC²

= 9² + (40)²

\(= \sqrt {81 + 1600}\)

\(= \sqrt {1681}\)

⇒ AC = 41 cm

Area of \({\rm{\Delta }}ABC = \frac{1}{2} \times base \times height\)

\(= \frac{1}{2} \times 40 \times 9\)

= 180 cm²

Now, for ΔABC, we use heron’s formula

\(s = \frac{{15 + 28 + 41}}{2} = 42\)

So, area of \({\rm{\Delta }}ACD = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \) 

\(= \sqrt {42\left( {42 - 28} \right)\left( {42 - 41} \right)\left( {42 - 15} \right)} \)

\(= \sqrt {42 \times 14 \times 1 \times 27}\)

⇒ Area of ΔACD = 126 cm²

Now centroid of a triangle cuts the median into 2 : 1

In triangle BOD

BD = BC/2 = 60/2  = 30 cm

\(tan30^\circ = \frac{{OD}}{{BD}} = \frac{{OD}}{{30}}\)

\(OD = \frac{{30}}{{\sqrt 3 }} = 10\sqrt 3\)

\({{AO}}:{{OD}} = \frac{2}{1}\)

\(\Rightarrow \frac{{AO}}{{10\sqrt 3 }} = \frac{2}{1}\)

⇒ AO = 20√3

So, centroid is 20√3 cm away from vertex OA

Angle of an equilateral triangle 60° ⇒ ∠BAC = 60°

From the property of a circle angle subtended by a chord of a circle at its center is equal to doubts the angle subtended at any point on the circle

\(\angle ;BAC = \frac{1}{2}*\angle BOC\)

∠BOC = 120°

Now, draw perpendicular bisects on line BC at point D

\(\Rightarrow \angle BOD = \angle DOC = \frac{{120^\circ }}{2} = 60^\circ \)

Δ BOD is a right-angled triangle with BO = 20√3

\(\Rightarrow \sin 60^\circ = \frac{{BD}}{{BO}} = \frac{{BD}}{{20\sqrt 2 }}\)

\(\Rightarrow BD = 20\sqrt 3 \times \frac{{\sqrt 3 }}{2} = 30;cm\)

⇒ BC = 30 × 2 = 60 cm

So, side of an equilateral triangle is 60 cm.

Concept:

If l, b & h are the length, width & height of the cuboid then

Diagonal = √(l² +  b² + h²)

Formula Used:

(a + b + c)² = a2 + b2 + c2 + 2(ab + bc + ca)

a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

Calculation:

Let, the a, b & c be the length, width & height of the cuboid

Therefore,

a + b + c = 22          

(a + b + c)² = 484    -----(1)

Also

√(a² + b² + c²) = 14² 

(a2 + b2 + c2) = 196   -----(2)

If S is the sum of the cubes of the dimension of the cuboid

⇒ s = a³ + b³ + c³

V is the volume of cuboid

⇒ V = abc

Since, 

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca

From equation (1) & (2)

⇒ 196 + 2(ab + bc + ca) = 484

⇒ (ab + bc + ca) = 144         -----(1)

Now we have an identity

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

⇒ (S – 3V) = 22 × (196 – 144) = 1144 cm³

Let’s say length g a cuboid is a, b, c

Then, a + b + c = 22 cm         ----(1)

Diagonal = 14 cm

\(\Rightarrow \sqrt {{a^2} + {b^2} + {c^2}} = 14\)

a² + b² + c² = 196

now, squaring both in equation 1

(a + b + c)² = 484

⇒ a² + b² + c² + 2(ab + bc + ca) = 484

196 + 2(ab + bc + ca) = 484

2(ab + bc + ca) = 288

Surface area of the cuboid is 2(ab + bc + ca)

⇒ The surface area of the cuboid is 288 cm²