Quantitative Aptitude PYQs

Given:

p sin² α + q cos² α = m

q sin² β + p cos² β = n

p ≠ m, n and q ≠ m, n

Formula used:

sin² θ + cos² θ = 1

tan² θ = sin² θ ÷ cos² θ

Calculation:

p sin² α + q cos² α = m

⇒ p sin² α + q(1 − sin² α) = m

⇒ (p − q) sin² α = m − q

⇒ sin² α = (m − q) ÷ (p − q)

⇒ cos² α = (p − m) ÷ (p − q)

⇒ tan² α = (m − q) ÷ (p − m)

q sin² β + p cos² β = n

⇒ q sin² β + p(1 − sin² β) = n

⇒ (q − p) sin² β = n − p

⇒ sin² β = (n − p) ÷ (q − p)

⇒ cos² β = (q − n) ÷ (q − p)

⇒ tan² β = (n − p) ÷ (q − n)

⇒ (tan α ÷ tan β)² = [(m − q) ÷ (p − m)] × [(q − n) ÷ (n − p)]

⇒ (tan α ÷ tan β)² = (m − q)(n − p) ÷ (m − p)(n − q)

∴ The correct answer is option (4).

Given:

p sin² α + q cos² α = m

q sin² β + p cos² β = n

α and β are complementary ⇒ α + β = 90°

Find correct relation among m and n

Formula used:

If α + β = 90°

sin β = cos α

cos β = sin α

Calculation:

Given:

m = p sin² α + q cos² α

n = q sin² β + p cos² β

Using complementary property:

⇒ sin² β = cos² α

⇒ cos² β = sin² α

Substitute in n:

⇒ n = q cos² α + p sin² α

⇒ n = p sin² α + q cos² α

⇒ n = m

⇒ m − n = 0

∴ The correct answer is m − n = 0.

Given:

cosec θ − sin θ = p

sec θ − cos θ = q

Find p sin θ + q cos θ = ?

Formula used:

cosec θ = 1/sin θ

sec θ = 1/cos θ

Calculation:

p sin θ + q cos θ

= (cosec θ − sin θ) sin θ + (sec θ − cos θ) cos θ

= cosec θ × sin θ − sin² θ + sec θ × cos θ − cos² θ

⇒ 1 − sin² θ + 1 − cos² θ

⇒ 2 − (sin² θ + cos² θ)

⇒ 2 − 1

⇒ 1

∴ The required value is 1.

Given:

p = cosec θ − sin θ

q = sec θ − cos θ

Find p²q²(p² + q² + 3) = ?

Formula used:

sin² θ + cos² θ = 1

Calculation:

Assume θ = 45°

sin θ = cos θ = 1/√2

cosec θ = sec θ = √2

⇒ p = √2 − 1/√2 = 1/√2

⇒ q = √2 − 1/√2 = 1/√2

⇒ p² = 1/2 , q² = 1/2

⇒ p²q² = 1/4

⇒ p² + q² + 3 = 1/2 + 1/2 + 3 = 4

⇒ p²q²(p² + q² + 3) = 1/4 × 4

⇒ 1

∴ The required value is 1.

Given:

\(\frac{\sin \alpha}{\sin \beta}=\frac{4\sqrt2}{3}\)

\(\frac{\cos \alpha}{\cos \beta}=\frac{2\sqrt3}{9}\)

Find tan² α = ?

Formula used:

sin²θ + cos²θ = 1

tan θ = sin θ / cos θ

Calculation:

Let sin α = 4√2 × k , sin β = 3k

Let cos α = 2√3 × m , cos β = 9m

⇒ For α :

(4√2k)² + (2√3m)² = 1

⇒ 32k² + 12m² = 1 ……(1)

⇒ For β :

(3k)² + (9m)² = 1

⇒ 9k² + 81m² = 1 ……(2)

⇒ Solving (1) and (2)

⇒ m² = 1/108 , k² = 1/36

⇒ sin α = 4√2 × 1/6 = 2√2/3

⇒ cos α = 2√3 × 1/√108 = 1/3

⇒ tan α = (2√2/3) ÷ (1/3)

⇒ tan α = 2√2

⇒ tan² α = (2√2)²

⇒ tan² α = 8

∴ The required value is 8.

Given:

\(\frac{\sin \alpha}{\sin \beta}=\frac{4\sqrt2}{3}\)

\(\frac{\cos \alpha}{\cos \beta}=\frac{2\sqrt3}{9}\)

Find tan² β = ?

Formula used:

sin²θ + cos²θ = 1

tan θ = sin θ / cos θ

Calculation:

From previous calculation:

k² = 1/36 , m² = 1/108

⇒ sin β = 3k = 3 × 1/6 = 1/2

⇒ cos β = 9m = 9 × 1/√108

⇒ √108 = 6√3

⇒ cos β = 9/(6√3) = 3/(2√3) = √3/2

⇒ tan β = (1/2) ÷ (√3/2)

⇒ tan β = 1/√3

⇒ tan² β = (1/√3)²

⇒ tan² β = 1/3

∴ The required value is 1/3.

Given:

\(\frac{1+\sin \theta}{\cos \theta}=p+\sqrt{p^2+1}\)

Find sec θ = ?

Formula used:

(1 + sin θ)/cos θ = sec θ + tan θ

sec² θ − tan² θ = 1

Calculation:

(1 + sin θ)/cos θ = sec θ + tan θ

⇒ sec θ + tan θ = p + √(p² + 1)

Using identity:

(sec θ + tan θ)(sec θ − tan θ) = 1

⇒ sec θ − tan θ = 1/(p + √(p² + 1))

Rationalising:

⇒ sec θ − tan θ = √(p² + 1) − p

Adding both equations:

⇒ 2 sec θ = (p + √(p² + 1)) + (√(p² + 1) − p)

⇒ 2 sec θ = 2√(p² + 1)

⇒ sec θ = √(p² + 1)

∴ The correct answer is √(p² + 1).

Given:

\(\frac{1+\sin \theta}{\cos \theta}=p+\sqrt{p^2+1}\)

Find tan θ = ?

Formula used:

(1 + sin θ)/cos θ = sec θ + tan θ

(sec θ + tan θ)(sec θ − tan θ) = 1

Calculation:

sec θ + tan θ = p + √(p² + 1)

⇒ sec θ − tan θ = 1/(p + √(p² + 1))

⇒ sec θ − tan θ = √(p² + 1) − p

Subtracting:

⇒ 2 tan θ = (p + √(p² + 1)) − (√(p² + 1) − p)

⇒ 2 tan θ = 2p

⇒ tan θ = p

∴ The correct answer is p.

Given:

sin θ + cos θ = p , p ≠ 1

sec θ + cosec θ = q

Find relation between p and q

Formula used:

(sin θ + cos θ)² = 1 + 2 sin θ cos θ

sec θ + cosec θ = (sin θ + cos θ)/(sin θ cos θ)

Calculation:

(sin θ + cos θ)² = p²

⇒ 1 + 2 sin θ cos θ = p²

⇒ 2 sin θ cos θ = p² − 1

⇒ sin θ cos θ = (p² − 1)/2

Now,

q = sec θ + cosec θ

⇒ q = (sin θ + cos θ)/(sin θ cos θ)

⇒ q = p ÷ [(p² − 1)/2]

⇒ q = 2p/(p² − 1)

⇒ 2p = q(p² − 1)

∴ The correct relation is 2p = q(p² − 1).

Given:

sin θ + cos θ = p

sec θ + cosec θ = q

Find tan θ + cot θ = ?

Formula used:

tan θ + cot θ = (sin² θ + cos² θ)/(sin θ × cos θ)

sin² θ + cos² θ = 1

Calculation:

sec θ + cosec θ = 1/cos θ + 1/sin θ

⇒ q = (sin θ + cos θ)/(sin θ × cos θ)

⇒ q = p/(sin θ × cos θ)

⇒ sin θ × cos θ = p/q

tan θ + cot θ = 1/(sin θ × cos θ)

⇒ tan θ + cot θ = 1 ÷ (p/q)

⇒ tan θ + cot θ = q/p

∴ The correct answer is q/p.

Given:

Speed of Train X = 100 km/hr

Speed of Train Y = 60 km/hr

Relative speed of Train X with respect to Train Y = (100 - 60) km/hr = 40 km/hr

Time taken to cross = 9 seconds

Formula used:

Length of Train X = Relative Speed × Time

Note:

To convert speed from km/hr to m/s, multiply by (5/18)

Calculations:

Relative Speed = 40 km/hr = 40 × (5/18) m/s = 100/9 m/s

⇒ Length of Train X = (Relative Speed) × Time

⇒ Length of Train X = (100/9) × 9

⇒ Length of Train X = 100 m

∴ The correct answer is option (2).

Given:

Speed of X = 4.8 km/hr

Speed of Y = 6 km/hr

Time difference = 10 minutes = \(\frac{10}{60}\) hours = \(\frac{1}{6}\) hours

Formula used:

Distance = Speed × Time

Relative Speed (when Y catches up with X) = Speed of Y - Speed of X

Calculation:

Relative Speed = 6 - 4.8 = 1.2 km/hr

Distance covered by X in \(\frac{1}{6}\) hours:

Distance = Speed × Time

⇒ Distance = 4.8 × \(\frac{1}{6}\) = 0.8 km

Now, time taken by Y to catch up with X:

Time = Distance / Relative Speed

⇒ Time = 0.8 / 1.2 = \(\frac{2}{3}\) hours

Distance covered by X in 2/3 hour = 2/3 × 4.8 = 3.2 km

So, distance from P when both will meet = 0.8 + 3.2 = 4 km

∴ The correct answer is option (3).

Given:

X's salary is increased by 12% and then decreased by 10%.

Y's salary is increased by 10% and then decreased by 12%.

X and Y's current salaries are equal.

Formula used:

Net salary after increase and decrease = Initial salary × (1 + increase%/100) × (1 - decrease%/100)

Ratio of initial salaries = Final salary of X / Final salary of Y

Calculations:

Net salary of X: \(\text{Initial salary of X} \times (1+\frac{12}{100}) \times (1-\frac{10}{100})\)

⇒ Net salary of X = Initial salary of X × 1.12 × 0.9

⇒ Net salary of X = Initial salary of X × 1.008

Net salary of Y: \(\text{Initial salary of Y} \times (1+\frac{10}{100}) \times (1-\frac{12}{100})\)

⇒ Net salary of Y = Initial salary of Y × 1.1 × 0.88

⇒ Net salary of Y = Initial salary of Y × 0.968

Since the final salaries are equal:

⇒ Initial salary of X × 1.008 = Initial salary of Y × 0.968

⇒ Ratio of initial salaries = \(\frac{\text{Initial salary of X}}{\text{Initial salary of Y}} = \frac{0.968}{1.008}\)

⇒ Ratio = \(\frac{968}{1008}\)

⇒ Simplify the ratio:

⇒ Ratio = \(\frac{121}{126}\)

∴ The correct answer is option (3): 121 : 126.

Given:

Village population (p) = total number of persons

x% = percentage of total persons who can read and write

y% = percentage of males who can read and write

z% = percentage of females who can read and write

x > z and y > z

Formula used:

Number of males = \(\dfrac{p(x - z)}{y - z}\)

Calculation:

Given formula is:

⇒ Number of males = \(\dfrac{p(x - z)}{y - z}\)

∴ The correct answer is option (1).

Given:

Alloy X: Cu : Zn = 5 : 4

Alloy Y: Cu : Zn = 5 : 13

Equal quantities of X and Y are mixed to form alloy Z.

Formula used:

In equal quantities mixing, the average ratio of Cu and Zn is taken.

Cu in Z = (Cu in X + Cu in Y) ÷ 2

Zn in Z = (Zn in X + Zn in Y) ÷ 2

Calculation:

Let the quantity of each alloy (X and Y) be 9 units (LCM of 5+4 and 5+13).

For Alloy X:

Cu = (5/9) × 9 = 5 units

Zn = (4/9) × 9 = 4 units

For Alloy Y:

Cu = (5/18) × 9 = 2.5 units

Zn = (13/18) × 9 = 6.5 units

In Alloy Z:

Total Cu = 5 + 2.5 = 7.5 units

Total Zn = 4 + 6.5 = 10.5 units

Cu : Zn in Z = 7.5 : 10.5

Divide by 1.5:

⇒ Cu : Zn = 5 : 7

∴ The correct answer is option (2).

Given:

Score of X = Score of Y + 20

Score of X = 75% of (Score of X + Score of Y)

Formula used:

Let Score of Y = y

Score of X = y + 20

Score of X = 75% × (Score of X + Score of Y)

Calculations:

Let Score of Y = y

⇒ y + 20 = 75% × (y + 20 + y)

⇒ y + 20 = 75% × (2y + 20)

⇒ y + 20 = (75/100) × (2y + 20)

⇒ y + 20 = (3/4) × (2y + 20)

⇒ 4 × (y + 20) = 3 × (2y + 20)

⇒ 4y + 80 = 6y + 60

⇒ 80 - 60 = 6y - 4y

⇒ 20 = 2y

⇒ y = 10

Score of Y = 10

Score of X = y + 20 = 10 + 20 = 30

Ratio of Score of X to Score of Y = X : Y

⇒ 30 : 10

⇒ 3 : 1

∴ The correct answer is option (3).

Given:

Equation: 2x² − 5p x + 2p² = 0

One root exceeds the other by 4

Formula used:

Difference between roots = √(Discriminant) ÷ a

D = b² − 4ac

Calculations:

a = 2, b = −5p, c = 2p²

Difference of roots = 4

⇒ √D / a = 4

⇒ √D / 2 = 4

⇒ √D = 8

⇒ D = 64

Now compute Discriminant:

D = b² − 4ac

= (−5p)² − 4 × 2 × 2p²

= 25p² − 16p²

= 9p²

So, 9p² = 64

⇒ p² = 64/9

⇒ p = ±8/3

∴ The value of p = 8/3.

Given:

Principal (P) = ₹10,000

Rate of interest (r) = 10% per annum

Time (t) = 3 years

Installments = 3 equal annual payments

Formula used:

Installment Amount = \(\dfrac{P \times r}{1-(1+r)^{-n}}\)

Where, n = number of installments

Calculation:

Installment Amount = \(\dfrac{10000 \times 0.10}{1-(1+0.10)^{-3}}\)

⇒ Installment Amount = \(\dfrac{1000}{1-(1.10)^{-3}}\)

⇒ Installment Amount = \(\dfrac{1000}{1 - 0.7513}\)

⇒ Installment Amount = \(\dfrac{1000}{0.2487}\)

⇒ Installment Amount ≈ ₹4021

∴ The correct answer is option (3).

Given:

5x + 3 ≤ 8x - 9 and 2x + 20 > 5x + 2

Formula used:

Simplify inequalities to find the range of x.

Calculation:

5x + 3 < 8x - 9    and    2x + 20 > 5x + 2

⇒ 8x - 5x > 12    and    18 > 3x

⇒ x >4  and  6 > x

Option (1) is correct.

The correct answer is – 80% profit

Successive discounts

For discounts of 10%, 20%, and 25%, the net factor is 0.9 × 0.8 × 0.75 = 0.54.

This indicates that the final selling price is 54% of the marked price.

Application of given profit

After all discounts, the article is sold at a profit of 8% on cost price.

Assuming cost price = 100, the selling price becomes 108.

Thus, the marked price is calculated as 108 ÷ 0.54 = 200.

Profit after first discount

After only the  first discount of 10%, the selling price equals 90% of the marked price.

Selling price = 0.9 × 200 = 180.

Profit = 180 −  100 = 80, which is 80% profit.