Quantitative Aptitude PYQs

Given:

p ∝ q/r2

Formula used:

% Change = [(Final value - Initial Value)/ Initial Value] × 100

Calculation:

According to the question

p ∝ q/r² 

⇒ p = kq/r²; k = constant

Now,

q increases by 20% & r decrease by 20%

p' = \(\frac{k(1+\frac{20}{100})q}{(1-\frac{20}{100}r)^2}=\frac{1.2kq}{0.64r^2}\)

⇒ p' = 1.875kq/r²

Hence, 

% Change = (Final value - Initial Value)/ Initial Value

% Change = \(\frac{\frac{1.875kq}{r^2}-\frac{kq}{r^2}}{\frac{kq}{r^2}}× 100\)

%Change = (0.875) × 100

∴ Change in p = 87.5%

Given:

Total days of work = 30 days

Earning on present-day = Rs. 500

Losing on absent day = Rs. 100

Total amount received = Rs. 11400

Calculation:

Let he was present for p days

Then, according to the question:

500p - 100(30 - p) = 11400

⇒ 500p - 3000 + 100p = 11400

⇒ 600p = 14400

⇒ p = 24

∴ He worked for 24 days

Given:

Cost price(CP) of a chair and a table = Rs. 750

Gain at chair = 5%

Gain at table = 20%

Total gain = 16%

Formula used:

Gain = SP - CP

Gain% = (Gain/CP) × 100

SP = Selling price

Calculation:

Let CP of chair be c and that of table be t

According to the question:

Total SP = 1.16 × 750 = 870

⇒ Total SP = 870

Again according to the question

c + t = 750

Multiplying it by 6

⇒ 6c + 6t = 4500     ----(i)

Again according to the question

1.05c + 1.2t = 870

Multiplying it by 5

⇒ 5.25c + 6t = 4350     ----(ii)

By subtracting (ii) from (i), we get

⇒ 0.75c = 150

⇒ c = 200

⇒ t = 550

∴ Original cost of table = Rs. 550

Given:

A person rode one-third of a journey at 60 km/hr

One-third at 50 km/hr and the rest at 40 km/hr

Had the person ridden half of the journey at 60 km/hr and the rest at 40 km/hr,

He would have taken 4 minutes longer to complete the journey

Formula used:

Speed = distance/time

Calculation:

Let total distance be 6x

Time is taken in 1^st case

t₁ = 2x/60 + 2x/50 + 2x/40

⇒ t₁ = 37x/300

Time taken in 2^nd case

t₂= 3x/60 + 3x/40

⇒ t₂ = 5x/40

According to the question:

t₂  = t₁  + 4/60    (∵ 1 hr = 60 min)

⇒ 5x/40 = 37x/300 + 1/15

⇒ 5x/8 = 37x/60 + 1/3

⇒ 75x = 74x + 40

⇒ x = 40

∴ Total distance = 6x = 240 km

Given:

The person saves Rs. 1000 more each next year, while his first-year savings is Rs. 2000

Formula used:

Sn = (n/2)[2a + (n - 1)d]

S_n = sum of n terms of an Arithmetic Progression

n = number of terms

a = first term

d = common difference

Calculation:

a = 2000

d = 1000

S_n = 170000

Sn = (n/2)[2a + (n - 1)d]

⇒ 170000 = (n/2)[2 × 2000 + (n - 1)1000]

⇒ 340 = n(4 + n - 1)

⇒ n² + 3n - 340 = 0

⇒ (n - 17)(n + 20) = 0

⇒ n = 17, - 20; negative rejected

∴ He will save Rs. 170000 in 17 years

Shortcut Trick

For trigonometric expression "acos3θ + bsec3θ" 

Case I: If a ≤ b then the minimum value is a + b 

Case II: If a > b then the minimum value is 2√ab

Here in this question a = 1 and b = 1 

So, According to case I,

The minimum value is 1 + 1 = 2

 

Alternate Method 

Given: 

cos3θ + sec3θ  where 0° ≤ 0 < 90°

Concept Used:

To find the minimum or maximum value we have to find dy/dθ = 0 

Calculation:

Let y = cos3θ + sec3θ

⇒ dy/dθ = 3cos2θ (-sinθ) + 3sec2θ . secθ. tanθ 

⇒ dy/dθ = -3sinθ cos2θ + 3 sec³θ sinθ/cosθ 

⇒ dy/dθ = sinθ(-3cos3θ + 3 sec3θ)/cosθ 

⇒ dy/dθ = tanθ.(-3cos3θ + 3 sec3θ)

Now, Put dy/dθ = 0 

⇒ tanθ.(-3cos3θ + 3 sec3θ) = 0

⇒ tanθ = 0 or (-3cos3θ + 3 sec3θ) = 0 

⇒ θ = 0° or

⇒ cosθ = secθ ⇒ θ = 0° 

Now, The minimum value at θ = 0° 

y = cos³0° + sec³0° 

⇒ y = 1 + 1 = 2

∴ The minimum value of the trigonometric expression is 2.

Given:

14 sin2θ + 10 cos2θ = 11

where 0° < θ < 90°,

Formula used:

sin2θ + cos2θ = 1

Calculation:

14 sin2θ + 10 cos2θ = 11

⇒ 4 sin2θ + 10 sin2θ + 10 cos2θ = 11

⇒ 4 sin2θ + 10 (sin2θ + cos2θ) = 11

⇒ 4 sin2θ + 10 = 11

⇒ sin2θ = 1/4

⇒ sinθ = 1/2, - 1/2

Since 0° < θ < 90°, hence negative value rejected

⇒ perpendicular/hypotenuse = 1/2

⇒ base² = 2² - 1² = 3

⇒ base = √3

⇒ tanθ + cotθ = (1/√3) + (√3)

∴ tanθ + cotθ = 4/√3

Given:

\(\frac{\sin^3 θ + \cos^3 θ }{\sin θ + \cos θ} + \frac{\sin^3 θ - \cos^3 θ}{\sin θ - \cos θ}\)

Formula Used:

a³ + b³ = (a + b) (a² + b² - ab)

a3 - b3 = (a - b) (a2 + b2 + ab)

sin²θ + cos²θ = 1

Calculation:

We have 

\(\frac{\sin^3 θ + \cos^3 θ }{\sin θ + \cos θ} + \frac{\sin^3 θ - \cos^3 θ}{\sin θ - \cos θ}\)

\(⇒ \frac{(\sin θ + \cos θ)(\sin^2 θ + \cos^2 θ - sinθ. cosθ) }{(\sin θ + \cos θ)} + \frac{(\sin θ - \cos θ)(\sin^2 θ + \cos^2 θ + sinθ. cosθ)}{(\sin θ - \cos θ)}\)

⇒ (sin2θ + cos2θ) - sinθ.cosθ + (sin2θ + cos2θ) + sinθ.cosθ 

⇒ 1 + 1 = 2

∴ The value of \(\frac{\sin^3 θ + \cos^3 θ }{\sin θ + \cos θ} + \frac{\sin^3 θ - \cos^3 θ}{\sin θ - \cos θ}\)is 2.

Shortcut Trick

Let the height of the flagstaff be h.

The ladder of 10 m reaches (h − 10) m height from the ground.

In a 30°−60°−90° triangle, the side opposite to 60° is √3 times the base x, so h = x√3.

Using Pythagoras theorem for the ladder: x² + (h − 10)² = 10².

Substitute x = h/√3 → h²/3 + h² − 20h + 100 = 100 → 4h²/3 = 20h → 4h = 60 → h = 15 m.

∴ The correct answer is 15 m.

Alternate Method

Given:

Length of ladder (BD) = 10 m

Distance from top AB = 10 m

Angle of elevation of top (A) from D = 60°

Formula Used:

In ΔACD, tan 60° = AC / CD

In ΔBCD, BC² + CD² = BD²

Calculations:

Let AC = h and CD = x

⇒ tan 60° = h / x

⇒ √3 = h / x ⇒ x = h / √3

In right ΔBCD, BC = AC − AB = h − 10

⇒ (h − 10)² + (h / √3)² = 10²

⇒ h² − 20h + 100 + h² / 3 = 100

⇒ 4h² / 3 − 20h = 0

⇒ 4h² = 60h

⇒ h = 15 m

∴ The correct answer is 15 m.

Additional Information

Trigonometric Ratios

For a right-angled triangle, tan θ = Perpendicular / Base and sin θ = Perpendicular / Hypotenuse.

Pythagoras Theorem

In any right-angled triangle, Base² + Height² = Hypotenuse².

Standard Angles

The value of tan 60° = √3, tan 30° = 1/√3, and tan 45° = 1.

Given:

Trigonometric expression is 1 + 2sin2θ cos2θ - sin4θ - cos4θ where 0° < θ < 90°

Formula Used:

2sinθ.cosθ = sin2θ 

sin²θ + cos²θ = 1

a² + b² = (a + b)² - 2ab

Calculation:

We have 1 + 2sin2θ cos2θ - sin4θ - cos4θ

⇒ 1 + 2sin2θ cos2θ - (sin4θ + cos4θ)

⇒ 1 + 2sin2θ cos2θ - [(sin2θ)² + (cos2θ)²]

⇒ 1 + 2sin2θ cos2θ - [(sin2θ + cos2θ)² - 2sin2θcos2θ]

⇒ 1 + 2sin2θ cos2θ - 1 + 2sin2θcos2θ

⇒ 4sin2θ cos2θ

⇒ (2sinθ.cosθ)²

⇒ (sin2θ)²

⇒ sin²2θ

The maximum value of sin²x = 1 at x = 90° 

∴ The maximum value of sin22θ = 1 at 2θ = 90°

i.e θ = 45° which satisfy the given condition

Given:

Height of airplane from river = 1200 m

Width of the river = 3000 m

Angles of depression on opposite points of two banks = 30º and θ respectively

Calculation:

We have height of aeroplane from river = AD = 1200 m

Width of river = BC = 3000 m

Now, In ΔABD

cot 30º = BD/1200

⇒ BD = 1200 cot 30º

⇒ BD = 1200√3 m

And, CD = BC - BD

⇒ CD = 3000 - 1200√3

⇒ CD = 600(5 - 2√3) ≈ 922 m

Now, In ΔADC

tanθ = AD/CD

⇒ tanθ = 1200/922 ≈ 1.301

We know that tan45° = 1 and tan60° = √3 = 1.732

And 1< 1.301 < 1.732

∴ 45° < θ < 60°.

Given:

\(\dfrac{\cos^2 θ - 3 \cos θ + 2}{\sin^2 θ} = 1\)

Caculation:

We have \(\dfrac{\cos^2 θ - 3 \cos θ + 2}{\sin^2 θ} = 1\)

⇒ cos²θ - 3 cosθ + 2 = sin2θ 

⇒ cos2θ - 3 cosθ + 2 = 1 - cos2θ

⇒ 2cos2θ - 3 cosθ + 1 = 0

⇒ 2cos2θ - 2 cosθ - cosθ + 1 = 0

⇒ 2 cosθ (cosθ - 1) - 1 (cosθ - 1) = 0

⇒ (cosθ - 1) (2cosθ - 1) = 0 

So, (cosθ - 1) = 0 or (2cosθ - 1) = 0 

θ = 0° or θ = 60° 

Since 0° < θ < 90° 

So, θ = 60° is the possible angle 

Now, we have to find the value of sin2θ + cosθ at θ = 60°

⇒ sin260° + cos60°

⇒ 3/4 + 1/2 = 5/4

∴ The required value is 5/4.

Formula used:

sin²x + cos²x = 1

1 + tan²x = sec²x

Calculation:

Consider 1:

sin4θ - sin2θ = cos4θ - cos2θ

⇒ sin²θ(sin²θ - 1) = cos²θ(cos²θ - 1)

⇒ sin2θ(- cos2θ) = cos2θ(- sin2θ)

Since LHS = RHS, hence true

Consider 2:

sin4θ + cos4θ = 1 + 2sin2θ cos2θ

⇒ sin4θ + cos4θ - 2sin2θ cos2θ = 1

⇒ (sin2θ - cos²θ)² = 1

Since, LHS is not equal to RHS, hence false

Consider 3:

tan4θ + tan2θ = sec4θ - sec2θ 

⇒ tan2θ(tan2θ + 1) = sec2θ(sec2θ - 1)

⇒ tan2θ sec2θ = sec2θ tan2θ

Since, LHS = RHS, hence true

∴ Only 1 and 3 are correct.

Formula used:

sin(90 - A) = cos A

tan(90 - A) = cot A

Calculation:

sin 24° sin66° - cos24° cos66° + tan24° tan 66° - cot 24° cot 66°

⇒ sin(90° - 66°) sin(90° - 24°) - cos24° cos66° + tan(90° - 66°) tan(90° - 24°) - cot 24° cot 66°

⇒ cos 66° cos 24º - cos24° cos66° + cot 24° cot 66° - cot 24° cot 66° = 0

∴ The value of sin 24° sin66° - cos24° cos66° + tan24° tan 66° - cot 24° cot 66° is 0.

Given:

x = psinA cosB

y = psinA sinB

z = pcos A

Formula used:

sin²θ + cos²θ = 1

Calculation:

x2 + y2 + z² = (psinA cosB)² + (psinA sinB)² + (pcosA)²

⇒ x2 + y2 + z2 = p²(sin²A cos²B + sin²A sin²B + cos²A)

⇒ x2 + y2 + z2 = p2[sin2A(cos2B + sin2B) + cos2A]

⇒ x2 + y2 + z2 = p2[sin2A + cos2A]

∴ x2 + y2 + z2 = p²

Given:

x = m secA + n tanA 

y = m tanA + n secA

Formula used:

sec2x - tan2x = 1

Calculation:

x = m secA + n tanA 

⇒ x² = m² sec²A + n² tan²A + 2mn secA.tanA

y = m tanA + n secA

⇒ y² = m² tan²A + n² sec²A + 2mn tanA.secA

x2 - y² = (m2 sec2A + n2 tan2A + 2mn secA.tanA) - (m2 tan2A + n2 sec2A + 2mn secA.tanA)

⇒ x2 - y2 = m2 sec2A + n2 tan2A + 2mn secA.tanA - m2 tan2A - n2 sec2A - 2mn secA.tanA

⇒ x2 - y2 = m²(sec2A - tan2A) + n²(tan2A - sec2A)

⇒ x2 - y2 = m2(sec2A - tan2A) - n2(sec2A - tan2A )

∴ x2 - y2 = m2 - n2

Given:

tanθ = 1; 0º < θ < 90º

Formula used:

tan 45º = 1

sin 45º = 1/√2

cos 45º = 1/√2

Calculation:

We have tanθ = 1

⇒ tanθ = tan45° 

⇒ θ = 45º

Now, we have to find the value of sin2θ - 2sinθ cosθ at θ = 45º

⇒ sin245º - 2 sin45º cos45º

⇒ (1/2) - 2(1/√2)(1/√2)

∴ sin2θ - 2sinθ cosθ = - 1/2

Formula used:

sin²x + cos²x = 1

Calculation:

\(\dfrac{2 \sin^3 θ - \sin θ}{\cos θ - 2 \cos ^3 θ }\) = \(\dfrac{sin θ}{cos θ}[\dfrac{2sin^2θ - 1}{1 - 2cos^2θ}]\)

⇒ \(\dfrac{2 \sin^3 θ - \sin θ}{\cos θ - 2 \cos ^3 θ }\) = \(tanθ[\dfrac{2sin^2θ - sin^2θ - cos^2θ}{sin^2θ + cos^2θ - 2cos^2θ}]\)

⇒ \(\dfrac{2 \sin^3 θ - \sin θ}{\cos θ - 2 \cos ^3 θ }\) = \(tanθ[\dfrac{sin^2θ - cos^2θ}{sin^2θ - cos^2θ}]\)

∴ \(\dfrac{2 \sin^3 θ - \sin θ}{\cos θ - 2 \cos ^3 θ }\) = tanθ

Given:

A, B and C are interior angles of the triangle ABC

Formula used:

sin(90º - x) = cos x

tan(90º - x) = cot x

Calculation:

Since, A, B and C are interior angles of the triangle ABC

Then, A + B + C = 180º

⇒ B + C = 180º - A

By multiplying both sides by 1/2

⇒ (B + C)/2 = 90º - A/2

Taking both sides sin

⇒ sin(B + C)/2 = sin(90º - A/2)

⇒ sin(B + C)/2 = cos A/2

Similarly, it can be proved that tan(B + C)/2 = cot A/2

Now, we have to find the value of \(\tan \left( \frac{B + C}{2} \right) + \sin \left( \frac{B + C}{2} \right) - \cot \left( \frac{A}{2} \right) - \cos \left( \frac{A}{2} \right)\) 

⇒ cot A/2 + cos A/2 - cot A/2 - cos A/2 = 0

∴ \(\tan \left( \frac{B + C}{2} \right) + \sin \left( \frac{B + C}{2} \right) - \cot \left( \frac{A}{2} \right) - \cos \left( \frac{A}{2} \right)\) = 0.

Given:

In right-angled triangle, ∠B = 90º

AB + BC = 10(1 + √3) cm; AC = 20 cm

Formula Used:

In a right-angled triangle 

P² + B² = H²

Where, P = Perpendicular, B = Base, H = Hypotenuse

Calculation:

AB + BC = 10(1 + √3) cm

⇒ AB + BC = (10 + 10√3) cm

If we take AB = 10 cm, BC = 10√3 cm

AB² + BC² = (10)² + (10√3)²  

⇒ 100 + 300 = 400

⇒ 20² = (Hypo)²

So, we can take AB = 10 cm, BC = 10√3 cm

Now, tan A = BC/AB and tan C = AB/BC

tan A + tan C = BC/AB + AB/BC

⇒ tan A + tan C = \(\dfrac{BC^2 + AB^2}{AB.BC}\)

⇒ tan A + tan C = \(\dfrac{AC^2}{AB.BC}\)

⇒ tan A + tan C = \(\dfrac{20^2}{10(10√3 )}\)

∴ tan A + tan C = 4/√3