Quantitative Aptitude PYQs

Given:

AB = 15 cm, BC 20 cm and AC = 25 cm

BP is perpendicular on AC.

Calculation:

Since, 15, 20, and 25 Pythagorean triplet the triangle is a right-angle triangle.

We know that,

AP = AB²/AC

⇒ AP = 15²/25 = 9cm

PC = BC²/AC

⇒ PC = 20²/25 = 16cm

BP = (AB × BC)/AC

⇒ BP = (15 × 20)/25 = 12cm

Area of ΔAPB = (AP × BP)/2

⇒ 9 × 12/2 = 54cm²

Area of ΔBPC = (BP × PC)/2

⇒ 12 × 16/2 = 96cm²

Difference between the area of both triangles,

⇒ 96 - 54 = 42cm²

∴ The correct answer is 42cm².

Formula Used:

GM of n numbers a₁, a₂....a_n = (a₁a₂a₃...a_n)^1/n

Calculation:

Let, n = 1

Positive numbers become a₁, a₂, a₃ which are in GP

According to the question, P is th GM of a₁, a₂, a₃

Therefore,

P = (a1a2a3)^1/3                 -----(1)

Again, 

Q is the GM of a₄, a₅, a_6, a₇, a₈, a₉

Q = (a4 a5 a6 a7 a8 a9)^1/6      ----(2)

GM of 3n i.e. 3 × 3 = 9 number is

(a₁ a₂ a₃ a4 a5 a6 a7 a8 a9)1/9

This can be written as 

(P³ × Q⁶)^1/9

(P^1/3 × Q2/3)

Given:

The cost price of y articles is equal to the selling price of z articles.

y ∶ z = 5 ∶ 4

Concept Used:

Profit Percentage = (Profit / Cost Price) × 100

Calculation:

Let the cost price of each article be C and the selling price of each article be S.

Given that the cost price of y articles is equal to the selling price of z articles:

⇒ yC = zS

Given y ∶ z = 5 ∶ 4, we can write:

⇒ 5C = 4S

⇒ S = (5/4)C

Profit per article = Selling Price - Cost Price

⇒ Profit = S - C

⇒ Profit = (5/4)C - C

⇒ Profit = (5C - 4C)/4

⇒ Profit = C/4

Profit Percentage = (Profit / Cost Price) × 100

⇒ Profit Percentage = (C/4C) × 100

⇒ Profit Percentage = (1/4) × 100

⇒ Profit Percentage = 25%

Hence, the correct answer is option 2.

Given:

A sum of money triples itself in 8 years.

Let the principal be P and the rate of interest be R.

It becomes n times in 20 years.

Concept Used:

Simple Interest (SI) is given by the formula:

SI = P × R × T / 100

Total Amount (A) = Principal + Simple Interest

Calculation:

When the sum triples in 8 years:

⇒ A = 3P, T = 8 years

Using the formula for Simple Interest:

SI = A - P

⇒ 3P - P = P × R × 8 / 100

⇒ 2P = 8P × R / 100

⇒ R = 25%

Now, to find n when T = 20 years:

Using the formula for Simple Interest:

SI = P × R × 20 / 100

⇒ SI = P × 25 × 20 / 100

⇒ SI = 5P

Total Amount (A) = Principal + Simple Interest

⇒ A = P + 5P = 6P

Thus, the sum becomes 6 times in 20 years.

Hence, the value of n is 6.

Given:

The work done by x men in (x + 1) days = The work done by (x + 5) men in (x - 2) days

Calculation:

Let W denote the work done by one man in one day.

The total work done by x men in (x + 1) days is W × X × (x + 1).

The total work done by (x + 5) men in (x - 2) days is: W × (x + 5) × (x - 2).

According to the problem, these two amounts of work are equal:

W × x × (x + 1) = W × (x + 5) × (x - 2)

⇒ x × (x + 1) = (x + 5) × (x - 2)

⇒ x² + x = x² - 2x + 5x - 10

⇒ x = 3x - 10

⇒ 2x = 10

⇒ x = 5

∴ The correct answer is 5.

Given:

If (a + b) ∶ (b + c) ∶ (c + a) = 5 ∶ 7 ∶ 6

Concept Used:

We use the concept of ratio and proportion.

Calculation:

Let (a + b) = 5k, (b + c) = 7k, and (c + a) = 6k

We know that:

⇒ (a + b) + (b + c) + (c + a) = 5k + 7k + 6k

⇒ 2(a + b + c) = 18k

⇒ a + b + c = 9k … (1)

From equation (1):

⇒ a = 9k - (b + c)

⇒ a = 9k - 7k = 2k … (2)

⇒ b = 9k - (c + a)

⇒ b = 9k - 6k = 3k … (3)

⇒ c = 9k - (a + b)

⇒ c = 9k - 5k = 4k … (4)

Now, we need to find (a - b + c) ∶ (a + b - c)

Using equations (2), (3), and (4):

⇒ (a - b + c) = 2k - 3k + 4k = 3k

⇒ (a + b - c) = 2k + 3k - 4k = 1k

⇒ (a - b + c) ∶ (a + b - c) = 3k ∶ 1k

⇒ 3 ∶ 1

Hence, the value of (a - b + c) ∶ (a + b - c) is 3 ∶ 1.

Given:

Principal = ₹ 1000

Rate for CI = 10%

Rate for SI = 11%

Time = 3 years

Formula Used:

Compound Interest (CI) Formula:

CI = P(1 + R/100)T - P

Simple Interest (SI) Formula:

SI = P × R × T / 100

Calculation:

For Compound Interest:

CI = 1000(1 + 10/100)3 - 1000

⇒ CI = 1000(1 + 0.1)3 - 1000

⇒ CI = 1000(1.1)3 - 1000

⇒ CI = 1000(1.331) - 1000

⇒ CI = 1331 - 1000

⇒ CI = ₹ 331

For Simple Interest:

SI = 1000 × 11 × 3 / 100

⇒ SI = 1000 × 33 / 100

⇒ SI = 330

⇒ SI = ₹ 330

Difference between CI and SI:

Difference = CI - SI

⇒ Difference = 331 - 330

⇒ Difference = ₹ 1

Hence, the correct answer is option 1.

Concept Used:

The sum of the length of the two sides of a triangle is greater than the length of the third side

 

and the difference between the two sides of a triangle is less than the length of the third side.

Calculation:

In ΔABD,

⇒ AD + AB > x

⇒ 10 + 6 > x

⇒ 16 > x

Also,

⇒ AD - AB < x

⇒ 10 - 6 < x

⇒ 4 < x

⇒ 4 < x < 16 .....

(i) In ΔDBC,

⇒ BC + DC > x

⇒ 18 + 6 > x

⇒ 24 > x

Also,

⇒ BC - DC > x

⇒ 18 - 6 > x

⇒ 12 > x

⇒ 12 < x < 24 ......

(ii) Combining both (i) and (ii),

12 < x < 16

∴ The correct answer is 12 < x < 16.

Given:

In a quarter circle of radius R, a circle of radius r is inscribed.

Concept Used:

The inscribed circle will be tangent to both the horizontal and vertical sides of the quarter circle, as well as to the arc of the quarter circle.

The center of the quarter circle is at the origin (0,0), and its radius is R.

Let's place the center of the inscribed circle at coordinates (r, r). This circle is tangent to both the x-axis and the y-axis at distances r from each axis. The distance from the center of the inscribed circle to the center of the quarter circle (0,0) is:

\(\sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}\)

Since this distance is equal to R - r (the radius R of the quarter circle minus the radius r of the inscribed circle), we have:

\(r\sqrt{2} = R - r\)

Solving for r:

\(r\sqrt{2} + r = R\)

Factor out r on the left-hand side:

\(r(\sqrt{2} + 1) = R\)

\(\frac{R}{r} = \sqrt{2} + 1\)

∴  The ratio of R to r is: \(\sqrt{2} + 1 : 1\).

Given:

In the quadrilateral ABCD:

AB = BC

CD = DA

AC (diagonal) = 6 cm

BD (diagonal) = 12 cm

Calculation:

Triangles ABC and CDA are isosceles.

Area of quadrilateral ABCD = (1/2) × d₁ × d₂

Where d₁ and d₂ are the lengths of the diagonals.

Substituting the given values:

Area = (1/2) × 6 cm × 12 cm

Area = (1/2) × 72 cm²

Area = 36 cm²

∴ The area of the quadrilateral ABCD is 36 cm².

Given:

tan(3A) = cot(A - 22°), where 3A is an acute angle,

Calculation:

cot(x) = tan(90° - x).

\(\tan(3A) = \tan\left(90^∘ - (A - 22^∘)\right)\)

\(\tan(3A) = \tan\left(90^∘ - A + 22^∘\right)\)

\(\tan(3A) = \tan(112^∘ - A)\)

Since the tangent function is periodic with a period of 180°, we have:

\(3A = 112^∘ - A + n \cdot 180^∘\)

where n is an integer. Since 3A is an acute angle, (i.e. (0^∘  < A < 90^∘).

3A = 112^∘ - A

4A = 112^∘

A = 28^∘

∴ The correct answer is 28∘.

Given:

\(\frac{\sin θ-\cos θ+1}{\sin θ+\cos θ-1}\) = p secθ + q tan θ 

0 < θ < \(\frac{π}{2}\)

Calculation:

Since, 0 < θ < \(\frac{π}{2}\)

We can put θ = π /4 = 45^∘ 

sin 45^∘ = 1/√2

cos 45^∘ = 1/√2

sec 45^∘ = √2 

tan 45^∘ =  1

Putting all these values in the given equation,

⇒ \(\frac{1}{\frac{1}{√2} + \frac{1}{√2} - 1}\) = p(√2) + q

⇒ \(\frac{1}{√2 - 1}\) = √2p + q

⇒ \(\frac{1\times (√2 + 1)}{(√2 - 1)(√2 + 1)}\) = √2p + q

⇒ \(\frac{√2 + 1}{(1)}\) = √2p + q

Comparing both sides,

p = 1, q = 1

so, p + q = 1 + 1 = 2

∴ The correct answer is 2.

Given:

The angles of elevation of the top of a tower from two points A and B at a distance of

x and (x + 5)

Height of tower = 6 m

Calculation:

Since the angles are complementary,

angles are θ and 90 - θ 

cotθ = 6/x

tanθ = 6/(x + 5)

Multiplying both equations,

cotθ × tanθ = 6/x × 6/(x + 5) 

⇒ 1 = 36/(x)(x + 5)

⇒ x(x + 5) = 36

⇒ x² + 5x -36 = 0

⇒ x² + 9x - 4x -36 = 0

⇒ (x + 9)(x - 4) = 0

x = -9 , x = 4

∴ The correct answer is 4m.

Calculation:

Statement 1:

sin A + sin B + sin C = \(\frac{3 \sqrt{3}}{2}\)

sin A + sin B + sin C = \(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\)

It means A = B = C = 60^∘ 

The triangle can be equilateral.

Statement 2:

cos A + cos B + cos C = \(\frac{3}{2}\)

cos A + cos B + cos C = \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

It means A = B = C = 60^∘ 

The triangle can be equilateral.

∴ The correct answer is Both 1 and 2.

Given:

Train A speed = 80 km/h

Train B speed = 100 km/h

Train A departure time = 7:00 a.m.

Train B departure time = 7:50 a.m.

Concept Used:

Relative speed concept and distance calculation.

Formula Used:

Distance = Speed × Time

Calculation:

Time difference between Train A and Train B = 50 minutes = 50/60 hours = 5/6 hours

Distance covered by Train A in 5/6 hours = 80 km/h × 5/6 hours

⇒ Distance covered by Train A = 80 × 5/6 km

⇒ Distance covered by Train A = 400/6 km

Relative speed of Train B with respect to Train A = 100 km/h - 80 km/h

⇒ Relative speed = 20 km/h

Time taken for Train B to catch up Train A = Distance / Relative speed

⇒ Time taken = (400/6) km / 20 km/h = 10/3 hours

Distance traveled by Train B in (10/3) hours = 100 km/h × (10/3) hours

⇒ Distance traveled by Train B = 100 × (10/3) = 1000/3 km

Thus, the distance from Delhi where the two trains meet is:

⇒ Distance = 1000/3 km

∴ The two trains will be together after 1000/3 km from Delhi.

Given:

Initial Length (L) = L

Initial Breadth (B) = B

Initial Height (H) = H

Length increased by 10%

Breadth increased by 20%

Height increased by 50%

Concept:

The volume of a cuboid is given by the product of its length, breadth, and height.

Formula Used:

Volume of cuboid = Length × Breadth × Height

Calculation:

Initial Volume (V_initial) = L × B × H

New Length (L_new) = L × (1 + 10/100) = 1.1L

New Breadth (B_new) = B × (1 + 20/100) = 1.2B

New Height (H_new) = H × (1 + 50/100) = 1.5H

New Volume (V_new) = L_new × B_new × H_new

V_new = 1.1L × 1.2B × 1.5H

V_new = 1.98LBH

Percentage Increase in Volume = [(V_new - V_initial)/V_initial] × 100

⇒ Percentage Increase = [(1.98LBH - LBH)/LBH] × 100

⇒ Percentage Increase = [0.98LBH/LBH] × 100

⇒ Percentage Increase = 0.98 × 100

⇒ Percentage Increase = 98%

⇒ Hence, the percentage increase in volume of the cuboid is 98%.

Given:

Total amount = ₹ 9400

After deductions, the ratio of shares is 3 ∶ 4 ∶ 5

Deductions: ₹ 93 from P's share, ₹ 24 from Q's share, ₹ 55 from R's share

Concept Used:

Use the concept of ratios and linear equations to solve for individual shares.

Calculation:

Let the shares of P, Q, and R be p, q, and r, respectively.

Given, (p - 93) ∶ (q - 24) ∶ (r - 55) = 3 ∶ 4 ∶ 5

Assume k is the constant of proportionality.

⇒ p - 93 = 3k

⇒ q - 24 = 4k

⇒ r - 55 = 5k

Total amount: p + q + r = 9400

Substitute the values of p, q, and r:

⇒ (3k + 93) + (4k + 24) + (5k + 55) = 9400

⇒ 3k + 4k + 5k + 93 + 24 + 55 = 9400

⇒ 12k + 172 = 9400

⇒ 12k = 9400 - 172

⇒ 12k = 9228

⇒ k = 9228 / 12

⇒ k = 769

The share of P before deduction:

⇒ p = 3k + 93

⇒ p = 3 × 769 + 93

⇒ p = 2307 + 93

⇒ p = 2400

Since 93 is deducted from P's share, the new share of P = 2400 - 93 = 2307

∴ The share of P is ₹ 2307.

Given:

P² varies as R and Q² varies as R, (P ≠ Q).

Concept:

If a variable X varies as Y, it implies X = kY for some constant k.

Calculation:

Let P² = k₁R and Q² = k₂R, where k₁ and k₂ are constants.

1. To check if P² + Q² varies as R:

⇒ P² + Q² = k₁R + k₂R

⇒ P² + Q² = (k₁ + k₂)R

Since k₁ + k₂ is a constant, P² + Q² varies as R. 2. To check if PQ varies as R:

⇒ P = √(k₁R) and Q = √(k₂R)

⇒ PQ = √(k₁R) × √(k₂R)

⇒ PQ = √(k₁k₂)R

Since √(k₁k₂) is a constant, PQ varies as R. 3. To check if P² - Q² varies as R:

⇒ P² - Q² = k₁R - k₂R

⇒ P² - Q² = (k₁ - k₂)R

Since k₁ - k₂ is a constant, P² - Q² varies as R.

∴ All three statements are correct.

Given:

p number of men can finish a piece of work in q days.

There are 50% more men, then the work will be finished 12 days earlier.

Concept Used:

Work is inversely proportional to the number of men and the number of days.

Formula Used:

Work = Men × Days

Calculation:

Initial work:

Work = p × q

With 50% more men:

Number of men = p + 0.5p = 1.5p

New days = q - 12

Since the work done is the same:

p × q = 1.5p × (q - 12)

Cancel out p from both sides:

q = 1.5(q - 12)

Expand and solve for q:

⇒ q = 1.5q - 18

⇒ q - 1.5q = -18

⇒ -0.5q = -18

⇒ q = -18 / -0.5

⇒ q = 36

∴ The value of q is 36.

Concept Used:

Minimum value of \((x+\frac{1}{x}) = 2\)

Calculation:

\(\left(\frac{a^2+3 a+1}{a}\right)\) \(\left(\frac{b^2+3 b+1}{b}\right)\) = \((a+3+\frac{1}{a})\) \((b+3+\frac{1}{b})\)

Minimum value of  \((a+3+\frac{1}{a})\) \((b+3+\frac{1}{b})\),

⇒ (2 +3)(2 + 3) = 5 × 5 = 25

∴ The correct answer is 25.