Quantitative Aptitude PYQs

Given:

(tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y)

Formula Used:

\(\frac{\tan x+\tan y}{1-\tan x\tan y} =\frac{\cot x +\cot y}{\cot x \cot y -1}\)

Calculation:

(tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y)

∵ \(\frac{\tan x+\tan y}{1-\tan x\tan y} =\frac{\cot x +\cot y}{\cot x \cot y -1}\)

∴ tan x + tan y = \(\frac{(\cot x +\cot y)(1 - \tan x \tan y)}{\cot x \cot y -1} \)

= \(\frac{(\cot x +\cot y)(1 - \tan x \tan y)}{\cot x \cot y -1} \) × (1 – cot x cot y)  + (cot x + cot y)(1 – tan x tan y)

= \(-\frac{(\cot x +\cot y)(1 - \tan x \tan y)}{1-\cot x \cot y } \)× (1 – cot x cot y)  + (cot x + cot y)(1 – tan x tan y)

= - (cot x + cot y)(1 – tan x tan y) + (cot x + cot y)(1 – tan x tan y)

= 0

∴ The value of (tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y) is 0.

Given:

\(\sqrt {\frac{{\sec x - \tan x}}{{\sec x + \tan x}}} \)

Formula Used:

\(\rm {\sec ^2}{\rm{x;}} - {\rm{;}}ta{n^2}{\rm{x;}} = 1\)

Calculation:

\(\sqrt {\frac{{\sec x - \tan x}}{{\sec x + \tan x}}} \)

Multiply by \((\sec x + \tan x)\) on numerator and denominator;

⇒ \(\sqrt {\frac{{{{\sec }^2}{\rm{x;}} - {\rm{;}}ta{n^2}{\rm{x;}}}}{{{{(\sec x + \tan x)}^2}}}} \)

We know that: \({\sec ^2}{\rm{x;}} - {\rm{;}}ta{n^2}{\rm{x;}}\)= 1

⇒ \(\sqrt {\frac{1}{{{{(\sec x + \tan x)}^2}}}} \)

⇒ \(\frac{1}{{\sec x + \tan x}}\)

Given:

\(\cot θ = \frac{{63}}{{16}}\)

Formula Used:

sin θ = Perpendicular/Hypotenus

cos θ = Base/ Hypotenuse

cot θ = Perpenicular/Base

Pythagorus theorem

(Hypotenus)² = (Perpendicular)² + (Base)²

Calculation:

\(\cot θ = \frac{{63}}{{16}}\) = B/P

Here, hypotenuse = \(\sqrt {{{63}^2} + {{16}^2}} = \sqrt {4225} \) = 65

So,

sinθ = P/H = 16/65

cosθ = B/H = 63/65

∴ (sin θ + cos θ) = (16/65) + (63/65) = 79/65

Concept Used:

(a + b)2 - 2ab = a2 + b2 

sin²θ + cos²θ = 1

Calculation:

We have

\(\frac{{1 - 2{{\sin }^2}θ {{\cos }^2}θ }}{{{{\sin }^4}θ + {{\cos }^4}θ }} + 4\)

Since, (a + b)2 - 2ab = a2 + b2 

⇒ \(\frac{1-2sin^2\theta \ cos^2\theta}{(sin^2\theta+cos^2\theta)^2-2sin^2\theta \ cos^2\theta}+ 4\)

We know that, sin2θ + cos2θ = 1

⇒ \(\frac{1-2sin^2\theta \ cos^2\theta}{1-2sin^2\theta \ cos^2\theta}+ 4\)

⇒ 1 + 4 = 5

∴ The value of \(\frac{{1 - 2{{\sin }^2}θ {{\cos }^2}θ }}{{{{\sin }^4}θ + {{\cos }^4}θ }} + 4\) is 5.

Given:

A rectangle is 48 cm long and 14 cm wide

Formula Used:

sec θ = Hypotenus/Base

cosec θ = Hypotenus/Perpendicula

Pythagorus theorem

(Hypotenus)² = (Perpendicular)² +(Base)²

Calculation:

Triangle ABC is right angle triangle:

Hypotenuse = AC = x = \(\sqrt {{{48}^2} + {{14}^2}} = \sqrt {2304 + 196} = \sqrt {2500}\) = 50

Here,

Sec θ = AC/AB = 50/48

Cosec θ = AC/BC = 50/14

Hence,

(sec θ + cosec θ) = (50/48) + (50/14) = 1550/336 = 775/168 

∴ The value sec θ + cosec θ of is 775/168

Shortcut Trick

Lets take x = y = z = 2 unit

One side = \(\frac{y}{z} + \frac{z}{x} = \frac{2}{2}+\frac{2}{2} = 2\)

Similarly, the other two sides will be 2 & 2.

Hence, the triangle will be an equilateral triangle, 

Since, are of equilateral triangle = \(\frac{\sqrt 3}{4}\)(side2) 

⇒ Area of given triangle = \(\frac{\sqrt 3}{4}\)(22) = √3

Putting values of x, y and z in options only option 3 is satisfying.

∴ \(\sqrt {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \) will be the area of required tirangle. 

Alternate Method

Formula Used:

S = (a + b + c)/2

Area of triangle = √[s(s – a)(s – b)(s – c)]

Calculation:

Sides: \(\frac{y}{z} + \frac{z}{x},\frac{z}{x} + \frac{x}{y},\frac{x}{y} + \frac{y}{z}\)

Semi perimeter = s = \(2\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \right)/2 = \left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \right)\)

Area of triangle = √[s(s – a)(s – b)(s – c)]

So,

Area = \({\sqrt {\left[ {\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \right)\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x} - \frac{y}{z} - \frac{z}{x}} \right)\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x} - \frac{z}{x} - \frac{x}{y}} \right)\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x} - \frac{x}{y} - \frac{y}{z}} \right)} \right]} }\)

Area = \(\sqrt {\left[ {\left( {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \right)\left( {\frac{x}{y}} \right)\left( {\frac{y}{z}} \right)\left( {\frac{z}{x}} \right)} \right]} \)

∴ Area is \(\sqrt {\frac{x}{y} + \frac{y}{z} + \frac{z}{x}} \)

Given:

Two angles are 30° and 45°

Sides = \(\left( {\sqrt 3 + 1} \right)cm\)

Concept Used:

The sum of angle of triangle = 180° 

Formula Used:

Area of triangle = \(\frac{{{a^2}}}{2}; \times \frac{{\left[ {SinB;SinC} \right]}}{{SinA}}\)

So, 3^rd angle = 180° – 30° – 45° = 105°

a = \(\left( {\sqrt 3 + 1} \right)cm\)      [Given]

Area of triangle =  \(\frac{{{a^2}}}{2}; \times \frac{{\left[ {SinB;SinC} \right]}}{{SinA}}\)

= \(\frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{2} \times \frac{{\left[ {Sin30^\circ ;Sin45^\circ } \right]}}{{Sin105^\circ }}\)

Here, Sin105° = Sin(60° + 45°)

= Sin60°Cos45° + Sin45°Cos60°

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

⇒ A =  \(\frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{2}; \times ;\frac{{\frac{1}{2}; \times \frac{1}{{\sqrt 2 }}}}{{\frac{{\sqrt 3 + 1}}{{2\sqrt {2;} }}}}\) 

= \(\frac{1}{2}\left( {\sqrt 3 + 1} \right);c{m^2}\)

∴ The Area of triangle is \(\frac{1}{2}\left( {\sqrt 3 + 1} \right);c{m^2}\).

If we suppose the complete area of the parallelogram is 8 units.

Then, Diagonal BD will divide it in 2 equal parts:

So, Area ABD = Area BCD = 4 units

Further CO will divide triangle BCD in 2 equal parts:

So, Area DOC = 4/2 = 2 units

Now,

Given EF || AD so, AD = EF

EO will divide DOC in 2 equal parts;

∴ Area of triangle DOE = 2/2 = 1 unit

When DOE is cut out from the plate ABCD;

∴ Ratio of area of remaining portion of the plate to the whole = (8 – 1) : 8 = 7 : 8

Given:

Given:

PA = 20 cm, QA = 16 cm and PQ = d cm

Concept Used:

Using Pythagorus Theorem 

(Hypotenus)² = (Base)² + (Perpendicular)²

Calculation:

In the figure:

P and Q are centres of 2 circles.

PA = 20 cm, QA = 16 cm and PQ = d cm

Common chord AB = 24 cm

Now, we know that the line joining the centres of two intersecting circles is perpendicular to the common chord and it also divides the chord in 2 equal parts;

So, AR = 24/2 = 12 cm

In triangle PAR using Pythagoras theorem

AP² = AR² + PR²

⇒ PR² = AP² – AR²

⇒ PR² = 20² – 12²

⇒ PR² = 400 – 144 = 256

⇒ PR = 16 cm

Similarly, In triangle QAR,

⇒ QR² = 16² – 12²

⇒ QR² = 256 – 144 = 112

⇒ QR = 10.58 cm

Hence, required distance 

PQ = d = PR + RQ = 16 + 10.58

⇒ d = 26.58 cm

∴ 26 cm < d < 27 cm

Given:

Radius = 5 cm

Chords AB = AC = 8 cm

Concept Used:

Using pythagorus theorem

(Hypotenus)² = (Perpendicular)² + (Base)²

Calculation:

In the triangle, O is the centre of the circle.

AB = AC = 8 cm

Radius AO = BO = CO = 5 cm

We know that the perpendicular from the centre of a circle on the chord divides the chord in 2 equal parts.

Here OD is perpendicular from centre on the chord BC:

So, BD = DC = y cm (Suppose)

Now triangles ABD and BOD are right angle triangles

y² = 8² – (5 + x)²

and

y² = 5² – x² 

Equating both of them:

8² – (5 + x)² = 5² – x² 

⇒ 64 – 25 – x² – 10x = 25 – x²

⇒ 10x = 14

⇒ x = 7/5 cm

Now,

y² = 5² – (7/5)²

⇒ y² = 25 – 49/25

⇒ y² = 576/25

⇒ y = 24/5

∴ Chord BC = 2y = 48/5 = 9.6 cm

Given:

Sum of their areas is 136π cm²

Distance between their centres = 4 cm

Calculation:

Suppose radius of 2 circles are R and r cm.

Sum of their areas is 136π cm²

So,

π (R² + r²) = 136π

⇒ (R² + r²) = 136

Now we can solve it with the help of options by just putting the values of R and r.

Option 2: 10 cm, 6 cm

⇒ 10² + 6² = 100 + 36 = 136

It satisfies the given condition. Also, 10 – 6 = 4 satisfies the second condition.

∴ Option 2 will be correct.

Given:

Area of a circle and a square are same

Formula Used:

Area of circle = πr²

The perimeter of circle = 2πr

Perimeter of square = 4a 

Here a = side of square

Calculation:

Suppose radius of circle is ‘r’ and side of square is ‘a’ cm;

Area of a circle and a square are same

So,

πr² = a²

⇒ r/a = 1/\(\sqrt π \)

Now,

Perimeter of circle = 2πr

Perimeter of square= 4a

So,

Ratio of perimeters = 2πr/4a = π/2 × 1/\(\sqrt π \) = \(\frac{{\sqrt π }}{2}\)

∴ Ratio of perimeter is \(\frac{{\sqrt π }}{2}\).

Given:

Diameter = 8 cm

Formula Used:

Radius = Diameter/2

Calculation:

In the figure:

A and B are the centres of the circles.

Radius AP = AQ = 8/2 = 4 cm

Suppose radius of smaller circle = x cm

Now,

In square APOQ:

Diagonal, AO = 4√2 cm

In square BCOD:

Diagonal, BO = x√2 cm

Both the circles touch each other at point S:

So, AS = 4 cm and SB = x cm

Now, we can say:

AO = AS + SB + BO

⇒ 4√2 = 4 + x + x√2

⇒ x(1 + √2) = 4(√2 – 1)

⇒ x = 4(√2 – 1)/(√2 + 1)

⇒ x = 4(√2 – 1)²/(2 – 1)

⇒ x = 4(3 – 2√2)

∴ Diameter of smaller circle = 2x = 8(3 – 2√2) cm

Given:

Inner radius (r) = 3 feet and thickness = 1 foot

Height (h) = 7 feet

Formula Used:

Surface area of whole cylinder = 2πRh + 2πrh + 2π (R2 – r2)

Calculation:

Inner radius (r) = 3 feet and thickness = 1 foot

So, outer radius (R) = 3 + 1 = 4 feet

Surface area of whole cylinder = 2πRh + 2πrh + 2π (R² – r²)

= 2π × 4 × 7 + 2π × 3 × 7 + 2π × (4² – 3²)

= 56π + 42π + 14π

= 112π

It is given on the question that ‘To paint the inner surface it requires one litre of a particular colour’;

Area of inner surface = 2πrh = 42π

Paint required for 42π area = 1 litre

∴ Paint required for 112π area = 1 × 112/42 = 8/3 litre

Given:

One side of the rectangle numerically equal to the square of the side of the square

Formula Used:

Area of square = (side)²

Area of rectangle = length × breadth 

Calculation:

Suppose length and breadth of rectangle are ‘L’ and ‘B’ respectively and side of square is ‘x’ cm;

One side of the rectangle numerically equal to the square of the side of the square;

∴ L = x²

Also, area of rectangle and square are equal.

∴ LB = x²

⇒ x² × B = x²

⇒ B = 1

∴ Other side of the rectangle is of unit length.

Given:

The length and breadth of a rectangle are in the ratio 4 : 3

Formula Used:

area of triangle = ½ × base × height

Calculation:

Consider the rectangle ABCD shown below,

Let the length and breadth of the rectangle be ‘4x’ units and ‘3x’ units respectively

⇒ AB = CD = 4x

⇒ BC = AD = 3x

∵ O is the intersection point of diagonals & is the centre of the rectangle,

⇒ OX = AB/2 = 4/2 = 2x

⇒ OY = BC/2 = 3x/2

Now, area of triangle = ½ × base × height

⇒ Area of ∆COD = ½ × CD × OY = ½ × 4x × 3x/2 = 3x²

⇒ Area of ∆BOC = ½ × BC × OX = ½ × 3x × 2x = 3x²

∴ Required ratio = 3x²∶ 3x² = 1 ∶ 1

Given:

θ = 20° and Radius = 30 feet

Formula Used:

Area of circle - πr²  (∵ r = radius)

Area of sector with angle θ = (θ/360°) × π × (radius)2 = (θ/360°) × Area of circle

Calculation:

Area of circle = π × (radius)² = π × (30)²

⇒ Area of circle = 900π square feet

Area of sector with angle θ = (θ/360°) × π × (radius)²

⇒ Area of sector = (θ/360°) × Area of circle

Given , θ = 20°

⇒ Area of sector removed = (20°/360°) × 900π

⇒ Area of sector removed = 50π square feet

∴ Area of new region = 900π – 50π = 850π square feet

Given:

ABCD is a parallelogram where AC and BD are the diagonals

∠BAD = 60°, ∠ADB = 90°

Formula Used:

sin θ = Perpendicular/Hypotenuse

sin 60° = √3/2

Calculation:

The given parallelogram can be drawn as shown below,

Considering ∆ADB,

⇒ sin 60° = Perpendicular/Hypotenuse

⇒ √3/2 = BD/AB

Squaring both sides,

⇒ 3/4 = BD²/AB²

⇒ BD² = (3/4) AB²

∴ BD2 is (3/4) AB2

Concept Used:

∆ABQ and ∆PCQ are also similar triangles

Area (ABQ)/Area (PCQ) = (AB/PC)² = (BQ/CQ)² = (AQ/PO)²

Calculation:

In the figure given below,

P is the mod-point of CD, hence, PD = PC

Also, ∠PAD = ∠PQC (alternate angles)

Similarly, ∠ADP = ∠QCP (alternate angles)

⇒ ∆PDA and ∆PCQ are congruent triangles

⇒ Area (PDA) = Area (PCQ)

Hence, statement 1 is correct

Similarly,

∆ABQ and ∆PCQ are also similar triangles

⇒ Area (ABQ)/Area (PCQ) = (AB/PC)²

But, P is the mod-point of CD & AB = CD, hence, AB = 2PC

⇒ Area (ABQ)/Area (PCQ) = (2/1)²

⇒ Area (ABQ) = 4 × Area (PCQ)

Hence, statement 2 is not correct

∴ Only statement 1 is correct

Given

Radius = 1.4 m, Depth = 5 m

Formula Used:

Volume of cylinder = π × (radius)2 × depth

Calculation

A well is in the shape of a right circular cylinder,

⇒ Quantity of earth dug out = Volume of cylinder = π × (radius)² × depth

Quantity of earth dug out = 22/7 × 1.4 × 1.4 × 5 = 30.8 cubic metre

∴ The quantity of earth dug out is 30.8 cubic metre.