Quantitative Aptitude PYQs

Concept:

For the interval θ ∈ [0, π/2]:

If θ₁ > θ_2, Then

sin θ₁ > sin θ₂    

cos θ1 < cos θ2   

Complementary angle formula:

sin(π/2 - θ) = cos θ 

cos(π/2 - θ) = sin θ 

Calculation:

1. sin 1° < cos 57°

⇒ sin 1° = sin (90 - 89) = cos 89°

⇒ cos 89° < cos 57°

This statement is true.

2. cos 60° > sin 57°

⇒ sin 57° = sin (90 - 57) = cos 33°

⇒ cos 60° > cos 33°

This statement is false.

∴ From above, option 1 is correct.

Given:

p = sec θ - tan θ and

q = cosec θ + cot θ

Formulae Used:

• sec θ = 1/cos θ
• tan θ = sin θ/cos θ
• cot θ = cos θ/sin θ
• sin²θ + cos²θ = 1

 

Calculation:

Let the required value be x.

⇒ x = p + q(p - 1)

⇒ x = p + qp - q     ---(1)

⇒ p = sec θ - tan θ

⇒ p = \(\frac{1-sin\ θ}{cos\ θ}\)       ---(2)

⇒ q = cosec θ + cot θ

⇒ q = \([\frac{1+cos\ θ}{sin\ θ}]\)    ----(3)

Put the value of p, q in equation (1)

⇒ x = \(\frac{1-sin\ θ}{cos\ θ}\) + [\(\frac{1-sin\ θ}{cos\ θ}\)] × [\(\frac{1+cos\ θ}{sin\ θ}\)] - \([\frac{1+cos\ θ}{sin\ θ}]\)

⇒ x = \(\frac{1-sin\ θ}{cos\ θ}\) - \([\frac{1+cos\ θ}{sin\ θ}]\) + [\(\frac{1-sin\ θ}{cos\ θ}\)] × [\(\frac{1+cos\ θ}{sin\ θ}\)] 

⇒ x = \(\frac{sin\ θ-sin^2\ θ-cos\ θ-cos^2\ θ}{sin\ θ \ cos\ θ}\) + \(\frac{1-sin^\ θ+cos\ θ-sin\ θ \ cos\ θ}{sin\ θ \ cos\ θ}\)

⇒ x = \(\frac{sin\ θ-1-cos\ θ+1-sin^\ θ+cos\ θ-sin\ θ \ cos\ θ}{sin\ θ \ cos\ θ}\)

⇒ x = \(\frac{-sin\ θ \ cos\ θ}{sin\ θ \ cos\ θ}\)

∴ x = -1 

Shortcut Trick Put θ = 45° 

⇒ p = sec 45° - tan 45° = √2 - 1

​⇒ q = cosec 45° + cot 45° = √2 + 1

⇒ p + q(p - 1) = p - q + pq

⇒ (√2 - 1) - (√2 + 1) + (√2 - 1)(√2 + 1)

​⇒ -2 + 1

​⇒ -1 

∴ The correct answer is option (1).

Identity used:

cosec²θ – cot²θ = 1

Calculation:

⇒ cosec θ - cot θ = m      ---- (1)

⇒ (cosec θ - cot θ)(cosec θ + cot θ)/(cosec θ + cot θ) = m

⇒ (cosec²θ - cot²θ)/(cosec θ + cot θ) = m

⇒ 1/(cosec θ + cot θ) = m

⇒ cosec θ + cot θ = 1/m      ---- (2)

By adding (1) and (2), we'll get

2cosecθ = m + 1/m

∴ cosec θ = m/2 + 1/2m

Calculation:

⇒ tan A = BC/AC = a/b

⇒ tan B = AC/BC = b/a

⇒ tan A + tan B = a/b + b/a = (a² + b²)/ab

Using pythagoras theorem,

⇒ c² = a² + b²

∴ tan A + tan B = c²/ab

Formula Used:

sin²A + cos²A = 1

cos (A - B) = cos A cos B + sin A sin B

Calculation:

We have,

cos α + cos β = 2

Squaring,

⇒ cos²α + 2cosα .cosβ + cos²β = 4 ......(1)

⇒ sin α + sin β = 0

Squaring,

⇒ sin²α + 2sinα.sinβ + sin²β = 0......(2)

Adding (1) & (2),

2cos α . cos β + 2 sin α . sin β 

2(cos α . cos β + sin α . sin β)

By using the above formula 

⇒ 2cos(α - β) = 2

⇒ cos(α - β) = 1

⇒ α - β = cos^-1(1) = 0

⇒ α = β

⇒ ? = cos 2α - cos 2β

∴ ? = 0

Shortcut Trick 

Put θ = 60^°

2 + (1/2) = 5/2 (LHS = RHS)

Sin²θ = (√3/2)² = 3/4

∴ The correct answer is 3/4.

Alternate Method 

Formula used:

• cos θ = 1/sec θ
• sin2 θ + cos2 θ = 1

 

Calculation:

We have,

⇒ sec θ + cos θ = 5/2

⇒ 1/cos θ  + cos θ = 5/2

Above equation satisfy for θ = 60°, 

Hence,

sin²θ = sin²60° = (√3/2)² 

sin2θ = 3/4

Alternate Method

We have,

⇒ sec θ + cos θ = 5/2

⇒ 1/cos θ  + cos θ = 5/2

⇒ (1 + cos² θ)/cos θ = 5/2

⇒ 2 + 2cos² θ = 5cosθ

⇒ 2cos²θ - 5cosθ + 2 = 0

⇒ 2 cos2θ - 4cos θ - cos θ  + 2 = 0

⇒ 2cos θ(cos θ - 2) -1(cos θ - 2) = 0

⇒ (2cos θ - 1)(cos θ - 2) = 0

⇒ cosθ = 1/2

⇒ cos² θ = 1/4

We know that,

⇒ sin² θ + cos² θ = 1

∴  sin² θ = 1 - 1/4 = 3/4

Shortcut Trick 

Put θ = 45

⇒ (1 + 1 - √2)(1 + 1 + √2)

⇒ (2² - √2²) = (4 - 2) = 2

∴ (1 + cot θ - cosec θ)(1 + tan θ + sec θ) = 2

Alternate Method 

Formula used:

• sin θ/cos θ = tan θ
• cos θ/sin θ = cot θ
• sin2θ + cos2θ = 1
• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• (a - b)² = a² - 2ab + b²

 

Calculation:

Let the required value be x.

⇒ x = (1 + cot θ - cosec θ)(1 + tan θ + sec θ)

⇒ x = (1 + cosθ/sinθ - 1/sinθ)(1 + sinθ/cosθ + 1/cosθ)

⇒ x = (sinθ + cosθ - 1)(cosθ + sinθ + 1)/(sinθ.cosθ)

⇒ x = [(sin θ + cosθ)² - 1]/(sinθ.cosθ)

⇒ x = [sin²θ + cos²θ + 2sinθ.cosθ - 1]/(sinθ.cosθ)

∴ (1 + cot θ - cosec θ)(1 + tan θ + sec θ) = 2

Formula used:

• (a - b)² = a² - 2ab + b²
• (a + b)2 = a2 + 2ab + b2
• sec2θ - tan2θ = 1

 

Calculation:

Given that,

6 + 8 tan θ = sec θ 

Multiply with 6 on both sides,

36 + 48 tan θ = 6sec θ       ----(1)

Also, we have

8 - 6 tan θ = k sec θ

Multiply with 8 on both sides,

64 - 48 tan θ = 8k sec θ      -----(2)

Adding equation (1) & (2), we get

36 + 48 tan θ + 64 - 48 tan θ  = 6sec θ + 8k sec θ

⇒ 100 = sec θ(6 + 8k) 

⇒ sec θ = 100/(6 + 8k)      -----(3)

Since, 6 + 8 tan θ = sec θ 

⇒ 6 + 8 tan θ = 100/(6 + 8k)  

⇒ 8 tan θ = 100/(6 + 8k)  - 6

⇒ 8 tan θ = (100 - 36 - 48k)/(6 + 8k)

⇒ 8 tan θ = 8(8 - 6k)/(6 + 8k)

⇒ tan θ = (8 - 6k)/(6 + 8k)       ----(4)

We know that, 

sec²θ - tan²θ = 1

Hence, from equations  (3) and (4)

[100/(6 + 8k)]² -  [(8 - 6k)/(6 + 8k)]² = 1 

⇒ [100²  - (8 - 6k)²]/(6 + 8k)² = 1

⇒ 10000 - 64 - 36k² + 96k = 36 + 64k² + 96k

⇒ 10000 - 100 = 100k²

⇒ 9900 = 100k²

⇒ k² = 99

∴ The value of k² is 99.

Calculation:

⇒ Let length of pole be AB metre.

⇒ ∠ ACD and ∠ BCD = ∠m

BD = DA, as D is the midpoint of AB

Δ ADC and Δ BDC are congruent triangles.

⇒ ∠ ADC = ∠ BDC

⇒ ∠ ADC + ∠ BDC = 180 ......(Linear pair angles)

⇒ ∠ ADC = ∠ BDC = 90

In Δ BDC,

⇒∠ BCD = m = 180 - 90 - 30 = 60

⇒ sin 60 = BD/x

⇒ BD = √3x/2

⇒ AB = 2BD = √3x

∴ Length of pole is √3x metre.

Calculation:

The values of tangent can be found using the given table.

Let AB is a height of pole and ECBD is a rectangular plot.

⇒ cot 60 = BD/AB

⇒ 1/√3 = BD/AB

⇒ cot 45 = BC/AB

⇒ BC = AB

As, ECBD is a rectangle and BE is a diagonal.

BD = EC and DE = BC

Using pythagoras theorem,

⇒ BE² = BD² + DE²

⇒ BE² = BD² + BC²

Then,

⇒ cot θ = BE/AB

⇒ cot θ = √(BD² + BC²)/AB

⇒ cot θ = √(AB²/3 + AB²)/AB

∴ cot θ = 2/√3

Formulae Used:

log₁₀(m)ⁿ = n × log10 (m)

log10(ab) = log10(a) + log10(b)

log10(a/b) = log10(a) – log10(b)

Calculation:

log10 31.25 = log₁₀ (3125/100)

⇒ log₁₀ (3125) – log₁₀ (100)

⇒ log10 (5)⁵ – log10 (10)²

⇒ 5[log10 (5)] – 2[log10 (10)]

⇒ 5[log10 (10/2)] – 2

⇒ 5[log10 (10) – log10 (2)] – 2

⇒ 5 – 5[log10 (2)] – 2

⇒ 3 – 5[log10 (2)]

∴ log10 31.25 = 3 – 5 log10 2

Shortcut Trick 

√(15 - 4√14) = √(8 + 7 - 4√14)

⇒ √{(2√2)² + (√7)² - 2 × 2√2 × √7}

⇒ √(2√2 - √7)² = (2√2 - √7)

∴ The square root of 915 – 4√14) is (2√2 – √7)

Alternate Method 

Let 15 – 4√14 = a

⇒ a = 8 + 7 – (2 × 2 × √14)

⇒ a = 8 + 7 – [2 × √(2² × 14)]

⇒ a = (√8)² + (√7)² – (2 × √56)

⇒ a = (√8)2 + (√7)2 – (2 × √8 × √7)

⇒ a = (√8)2 – (2 × √7 × √8) + (√7)2 

⇒ a = (√8 – √7)²

Taking square root on both the sides, we get:

√a = √8 – √7 = 2√2 – √7 

∴ The square root of 915 – 4√14) is (2√2 – √7)

Concept:

Natural numbers: Positive integers which start from 1 and ends at

infinity, such as: 1,2,3,4,5,6,7,8,9,10....∞ 

Calculation:

Let the two alternate numbers be x and (x + 2)

Reciprocals  of two number is

\(\frac{1}{x}+\frac{1}{x+2}=\frac{7}{24}\)

⇒ \(\frac{x+2+x}{x(x+2)}=\frac{7}{24}\)

⇒ 24 × (2x + 2) = 7 × (x2 + 2x)

⇒ 48x + 48 = 7x2 + 14x

⇒ 48x – 14x = 7x2 – 48

⇒ 34x = 7x2 – 48

⇒ 7x2 – 34x – 48 = 0

⇒ (x – 6) [x + (8/7)] = 0

⇒ x = 6, -(8/7)

We are given natural numbers only. So, no negative number will be allowed for x.

Hence, the next natural number = 6 + 2 = 8

∴ The sum of the numbers = 6 + 8 = 14

Given equation is 52n - 1

On substituting the value of n = 1,

⇒ 5⁽2×1) - 1 = 25 – 1 = 24

We have:

24 = 2 × 2 × 2 × 3

The factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Hence, total number of factors = 8

∴  52n - 1 is always divisible by a minimum of eight natural numbers

Formulae Used:

log₁₀ x^m = m × log₁₀ x 

log₁₀ (ab) = log10 a + log10 b 

log10 (a/b) = log10 a – log10 b

Calculation:

5x - 3 = 8

5x - 3 = 2³

Taking log₁₀ on both sides,

(x – 3)log₁₀ 5 = 3log₁₀ 2

⇒ x – 3 =  3log10 2/log10 5

⇒ x = (3log10 2/log10 5) + 3

⇒ x = 3[(log10 2/log10 5) + 1]

⇒ x = 3[(log10 2 + log10 5)/log10 5]

⇒ x = 3{[log10 (2 × 5)]/log10 5}

⇒ x = 3(1/log10 5)

⇒ x = 3/log10 5

⇒ x = 3/[log10 (10/2)]

⇒ x = 3/(log10 10 – log₁₀ 2)

∴ x = 3/(1 – log10 2)

Given:

3sin2 θ + 4cos2 θ

Concept used:

sin2 θ + cos2 θ = 1

Calculation:

⇒ 3sin2 θ + 3cos2 θ + cos² θ 

⇒ 3 × (sin2 θ + cos2 θ) + cos2 θ

⇒ (3 × 1) + cos2 θ = 3 + cos2 θ

The least value that cos² θ can take is 0

So, the required least value = 3 + 0 = 3

∴ The least value of 3sin2 θ + 4cos2 θ = 3

Important Points

It must be remembered here that:

Although cos θ can take any value from -1 to +1, but the value of cos² θ can never be negative.

Hence, the minimum value of the cos² θ will be zero.

Concept:

2sin θ cos θ = sin 2θ 

-1 ≤ sin θ ≤ 1

Calculation:

We have,

sin θ cos θ = k

⇒ \(\frac{1}{2}\) (2sin θ cos θ) = k

⇒ sin 2θ = 2k       ---(1)

We know that,

-1 ≤ sin 2θ ≤ 1

But, \(0 \le \theta \le \frac{\pi }{2}\)

0 ≤ sin 2θ ≤ 1

⇒ 0 ≤ 2k ≤ 1

⇒ 0 ≤ k ≤ 0.5    

∴  0 ≤ k ≤ 0.5

Additional Information

We have:

p = sin2 θ + cos4 θ, for 0 ≤ θ ≤ (π/2)

The values of sin θ and cos θ vary according to the following table:

For θ = 0° ⇒ p = 0 + 1 = 1

For θ = 30° = (π/6) radians ⇒ p = (1/2)² + (√3/2)⁴ = (1/4) + (9/16) = 13/16 = 0.8125

For θ = 45° = (π/4) radians ⇒ p = (1/√2)2 + (1/√2)4 = (1/2) + (1/4) = 3/4 = 0.75

For θ = 60° = (π/3) radians ⇒ p = (√3/2)2 + (1/2)4 = (3/4) + (1/16) = 13/16 = 0.8125

For θ = 90° = (π/2) radians ⇒ p = 1 + 0 = 1

Hence the range of values for p is:

0.75 ≤ p ≤ 1 

So, the value of p can neither be less than (3/4) nor greater than 1.

∴ Neither the statement 1 nor statement 2 is correct

Important Points

Remember about the conversion from degrees to radians:

180° = π radians

So, 1° = (π/180) radians

Formula used:

(a + b)² = a² + 2ab + b²

Calculation:

Let 

y = 2 – 2 sin x – sin2 x

⇒ y = 3 - (sin2 x + 2 sin x + 1)

We know that,

(a + b)2 = a2 + 2ab + b2

⇒ y = 3 - (sin x + 1)²    ----(1)

We know that,

-1 ≤ sin θ ≤ 1

Also, we have 0 ≤ x ≤ (π/2). Hence,

sin x ∈ [0, 1]

For maxima, sin x = 0

From equation (1), 

(y)_max = 3 - (1)² = 2

For minima, sin x = 1

⇒ (y)_min = 3 - (1 + 1)² = -1

Hence, required ratio is

y_max/y_min = 2/-1 = -2

∴ ratio of the greatest to the smallest value is -2.

The given equation is:

x2 + y2 - 2xy sin2 θ = 0

⇒ x2 + y2 = 2xy sin2 θ      ----(i)

We can use the following table for the values of sin θ for the various values of θ:

From the table, it is clear that:

If we substitute any value of θ other than 90°, we don not get any real solution for x and y.

If θ = 90°, ⇒ x2 + y2 = 2xy

For, x = y, ⇒ x2 + x2 = 2x² 

∴ The condition is satisfied for x = y, for real values of x and y.