Quantitative Aptitude PYQs

Given:

Hall dimensions = l (length), b (breadth), h (height)

Maximum distance inside (space diagonal) = 14 m

Maximum distance on floor (floor diagonal) = 6√5 m

Formula used:

Space diagonal = √(l² + b² + h²)

Floor diagonal = √(l² + b²)

Calculations:

⇒ l² + b² + h² = 14² = 196

⇒ l² + b² = (6√5)² = 36 × 5 = 180

⇒ h² = 196 − 180 = 16

⇒ h = √16 = 4

∴ h = 4 m

Given:

PQ = 14, PR = 6√5, h = 4 (from previous)​

Concept Used:

PQ is the space diagonal; its horizontal projection on the floor is PR and the vertical component is h. So the angle α between PQ and PR has adjacent = PR and hypotenuse = PQ. Hence

cos α = adjacent/hypotenuse = PR / PQ

Calculation:

⇒ cos α = (6√5) / 14

⇒ cos α = (3√5) / 7

∴ cos α = (3√5) / 7.

Alternate Method

Given: 

Hall dimensions = l (length), b (breadth), h (height)

Maximum distance inside (space diagonal) = 14 m

Maximum distance on floor (floor diagonal) = 6√5 m

Formula used:

Space diagonal = √(l2 + b2 + h2)

Floor diagonal = √(l2 + b2)

Calculations:

⇒ l2 + b2 + h2 = 142 = 196

⇒ l2 + b2 = (6√5)2 = 36 × 5 = 180

⇒ h2 = 196 − 180 = 16

⇒ h = √16 = 4

Using dot product to find cos α

PQ = (l, b, h), PR = (l, b, 0)

PQ · PR = l² + b² + 0 = 180

|PQ| = 14, |PR| = 6√5

cos α = (PQ · PR) / (|PQ| × |PR|)

cos α = 180 / (14 × 6√5)

cos α = 180 / (84√5)

cos α = 15 / (7√5)

∴ cos α = 15 / (7√5)

Given:

Side thickness = 0.5 cm, bottom thickness = 1 cm

Internal L × B × D = 14 cm × 10 cm × 8 cm

Material density = 2000 kg/m³

Formula used:

Outer dimension = internal + 2×side thickness (for L and B), depth adds bottom thickness

Volume of material = Volume_outer − Volume_inner

1 cm³ = 10⁻⁶ m³; Mass = Density × Volume

Calculations:

⇒ Outer L = 14 + 2 × 0.5 = 15 cm

⇒ Outer B = 10 + 2 × 0.5 = 11 cm

⇒ Outer D = 8 + 1 = 9 cm

⇒ Volume_outer = 15 × 11 × 9 = 1485 cm³

⇒ Volume_inner = 14 × 10 × 8 = 1120 cm³

⇒ Volume_material = 1485 − 1120 = 365 cm³

⇒ Volume (m³) = 365 × 10⁻⁶ = 3.65 × 10⁻⁴ m³

⇒ Mass = 2000 × 3.65 × 10⁻⁴ = 0.73 kg = 730 gm

∴ Weight of material = 730 gm.

Given:

Side thickness = 0.5 cm, bottom thickness = 1 cm

Internal L × B × D = 14 cm × 10 cm × 8 cm

Material density = 2000 kg/m³

Formula used:

Outer dimension = internal + 2×side thickness (for L and B), depth adds bottom thickness

Volume of material = Volume_outer − Volume_inner

1 cm³ = 10⁻⁶ m³; Mass = Density × Volume

Calculations:

⇒ Outer L = 14 + 2 × 0.5 = 15 cm

⇒ Outer B = 10 + 2 × 0.5 = 11 cm

⇒ Outer D = 8 + 1 = 9 cm

⇒ Volume_outer = 15 × 11 × 9 = 1485 cm³

⇒ Volume_inner = 14 × 10 × 8 = 1120 cm³

⇒ Volume_material = 1485 − 1120 = 365 cm³

⇒ Volume (m³) = 365 × 10⁻⁶ = 3.65 × 10⁻⁴ m³

⇒ Mass = 2000 × 3.65 × 10⁻⁴ = 0.73 kg = 730 gm

Weight of material = 730 gm.

Volume of water = Volumeinner = 14 × 10 × 8 = 1120 cm3 = 1120 × 10^-6 m³

Weight of water = 1000 × 1120 × 10-6 kg = 1.12 kg

Total weight of the box with water = 0.73 kg + 1.12 kg = 1.850 kg

∴ The correct answer is option 1.

Given:

The triangle ABC is right-angled at A.

AB = 8 cm, BC = 10 cm,

D is perpendicular to BC.

Calculation:

AC² = BC² - AB²

AC = √(100 - 64)

AC = √36 = 6 cm  

AD = (AB × AC)/(BC)

AD = (6 × 8)/10 = 4.8 cm

∴ The correct answer is option 1.

Given:

 

The triangle ABC is right-angled at A.

AB = 8 cm, BC = 10 cm,

D is perpendicular to BC.

Calculation:

AC2 = BC2 - AB2

AC = √(100 - 64)

AC = √36 = 6 cm  

ΔADB ~ ΔADB

So, \(\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{6^2}{8^2}\)

⇒ \(\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{36}{64}\) 

⇒ \(\frac{Ar\ \Delta ADC}{Ar\ \Delta ADB} = \frac{9}{16}\) 

∴ The correct answer is option 2.

Given:

Annual rainfall = 40 cm = 0.40 m

Area = 1 hectare = 10,000 m²

Formula used:

Volume = Area × Rainfall depth

Calculations:

⇒ Volume = 10,000 × 0.40

⇒ Volume = 4,000 m³

∴ Volume of rainfall = 4,000 m³ per hectare (annually).

Given:

Weight of water = 1 metric tonne per m³

Formula used:

Volume = Area × Rainfall depth

Weight = Volume × Density (tonnes/m3)

Calculations:

⇒ Volume = 10,000 × 0.40

⇒ Volume = 4,000 m3

⇒ Weight = 4,000 × 1

⇒ Weight = 4,000 metric tonnes

∴ Weight of water = 4,000 metric tonnes per hectare (annually).

Calculation:

In Δ AOB,

sin 60° = p/h = OB / AB

⇒ √3/2 = r / l

⇒ r : l = √3 : 2

Hence, the correct answer is option (3).

Given:

Angle at the vertex of the conical body = 120°

Sum of slant height, height, and radius = 9 + 3√3 cm

Formula used:

Volume of the cone = (1/3)πr²h

Relationship between slant height, radius, and height = l² = r² + h²

Calculations:

Let the height of the cone be h, radius be r, and slant height be l.

We know that angle at the vertex is 120° and tan(60°) = r/h, so r = √3 × h.

We are given that r + h + l = 9 + 3√3, substituting r = √3 × h and l = √(r² + h²) = 2h:

(√3h) + h + 2h = 9 + 3√3

⇒ (√3 + 3)h = 9 + 3√3

⇒ h = (9 + 3√3) / (√3 + 3)

Rationalize the denominator:

h = (9 + 3√3)(√3 - 3) / (3 - 9)

⇒ h = -18 / -6

⇒ h = 3 cm

Now, r = √3 × h = √3 × 3 = 3√3 cm.

Now, volume of the cone = (1/3)πr²h = (1/3)π(3√3)² × 3

⇒ Volume = (1/3)π × 27 × 3 = 27π cubic cm.

∴ The volume of the cone is 27π cubic cm.

Given:

Profit on article X = 15%

Selling price of article X = ₹34,500

Loss on article Y = 10%

No overall gain or loss (Net profit = ₹0)

Formula used:

Cost price (CP) = \(\frac{\text{Selling Price (SP)}}{1+\frac{\text{Profit%}}{100}}\)

Cost price (CP) = \(\frac{\text{Selling Price (SP)}}{1-\frac{\text{Loss%}}{100}}\)

Calculations:

For article X:

CP of article X = \(\frac{34500}{1+\frac{15}{100}}\)

⇒ CP of article X = \(\frac{34500}{1.15}\)

⇒ CP of article X = ₹30,000

For article Y:

Total cost price of X and Y = Total selling price of X and Y

⇒ CP of X + CP of Y = SP of X + SP of Y

⇒ ₹30,000 + CP of Y = ₹34,500 + SP of Y

⇒ CP of Y = SP of Y - ₹4,500

CP of article Y = \(\frac{\text{SP of Y}}{1-\frac{10}{100}}\)

⇒ SP of Y - ₹4,500 = \(\frac{\text{SP of Y}}{0.9}\)

⇒ 0.9(SP of Y - ₹4,500) = SP of Y

⇒ 0.9SP of Y - ₹4,050 = SP of Y

⇒ SP of Y - 0.9SP of Y = ₹4,050

⇒ 0.1SP of Y = ₹4,050

⇒ SP of Y = ₹4,050 × 10

⇒ SP of Y = ₹40,500

∴ The correct answer is option (2).

Given:

Quantity = 100 quintals

Total = 10 metric tonnes

Conversion:

1 quintal = 100 kg

1 metric tonne = 1000 kg

So,

100 quintals = 100 × 100 = 10,000 kg

10 metric tonnes = 10 × 1000 = 10,000 kg

Calculations:

⇒ (10000 / 10000) × 100 = 100%

∴ The correct answer is: 100%

Given:

Radius of the circle (r) = 2 cm

The circle is inscribed in an equilateral triangle.

Formula used:

Area of an equilateral triangle = \(\dfrac{\sqrt{3}}{4}a^2\)

Where, a = side of the equilateral triangle

Relation between radius (r) and side (a) of an equilateral triangle:

\(\text{r} = \dfrac{\sqrt{3}}{6}a\)

Calculation:

\(\text{r} = \dfrac{\sqrt{3}}{6}a\)

⇒ 2 = \(\dfrac{\sqrt{3}}{6}a\)

⇒ \(\text{a} = \dfrac{2 \times 6}{\sqrt{3}} = \dfrac{12}{\sqrt{3}} = 4\sqrt{3}\)

Area of the triangle:

\(\text{Area} = \dfrac{\sqrt{3}}{4}a^2\)

⇒ \(\text{Area} = \dfrac{\sqrt{3}}{4}(4\sqrt{3})^2\)

⇒ \(\text{Area} = \dfrac{\sqrt{3}}{4} \times 16 \times 3\)

⇒ \(\text{Area} = 12\sqrt{3}\)

∴ The correct answer is option (1).

Given:

The sides of a triangle are k, 1.5k, and 2.25k.

Formula used:

The sum of the squares of medians of a triangle is given by:

\(\text{Sum of squares of medians} = \frac{3}{4} (a^2 + b^2 + c^2)\)

Where, a, b, and c are the sides of the triangle.

Calculation:

Let a = k, b = 1.5k = 3k/2, c = 2.25k = 9k/4.

\(a^2 + b^2 + c^2 = k^2 + \frac{9}{4}k^2 + \frac{81}{16}k^2\)

\(a^2 + b^2 + c^2 = \frac{16}{16}k^2 + \frac{36}{16}k^2 + \frac{81}{16}k^2 = \frac{133}{16}k^2\)

Sum of squares of medians = \(\frac{3}{4} \times \frac{133}{16}k^2\)

Sum of squares of medians = \(\frac{3 \times 133}{4 \times 16}k^2 = \frac{399}{64}k^2\)

Therefore, the sum of the squares of the medians is: \(\frac{399}{64}k^2\)

Given:

2s = a + b + c ⇒ s = (a + b + c)/2

s(s − a)(s − b)(s − c) × \(\left[\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}\right]\)

Let us take:

a = b = c = 2 ⇒ a + b + c = 6 ⇒ 2s = 6 ⇒ s = 3

Then: s − a = s − b = s − c = 1

Now evaluate:

s(s − a)(s − b)(s − c) × \(\left[\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}\right]\)

= 3 × 1 × 1 × 1 × [1/1 + 1/1 + 1/1 − 1/3]

= 3 × (3 − 1/3) = 3 × (8/3) = 8

Check which option gives 8:

abc = 2 × 2 × 2 = 8 (Match)

2abc = 16 (Not matching)

4abc = 32 (Not matching)

ab + bc + ca = 12 (Not matching)

Therefore, the correct answer is option 1.

Given:

Principal (P) = ₹10,000

Rate of Interest (r) = 4% per annum

Time (t) = 1 year

Compounding frequency = Quarterly (every 3 months), so n = 4

Formula used:

A = P\((1+\frac{r}{n×100})^{n×t}\)

Calculation:

A = 10,000\((1+\frac{4}{4×100})^{4×1}\)

⇒ A = 10,000\((1+\frac{1}{100})^4\)

⇒ A = 10,000\((1+0.01)^4\)

⇒ A = 10,000\((1.01)^4\)

⇒ A = 10,000 × 1.04060401

⇒ A ≈ ₹10,406

∴ The correct answer is option (1).

Given:

p = 0.099999...

Calculation:

Let p = \(0.0{\overline9}\) = x. Multiply both sides by 10:

10x = 0.9999

100x = 9.99999...

100x − 10x = 9.99999... − 0.99999...

90x = 9

x = 0.1

So, x = 1. Therefore, p = 1.

p = 0.1 into the expression 70p² + 43p − 5:

70p² + 43p − 5 = 70(0.1)² + 43(0.1) − 5

= 0.70 + 4.3 − 5

= 0

∴ The value of 70p² + 43p − 5 is 0.

Given:

Find the remainder when 2¹⁰¹ is divided by 101

Formula used:

By Fermat's Little Theorem:

If p is a prime and a is not divisible by p, then:

a^p−1 ≡ 1 (mod p)

So, 2¹⁰⁰ ≡ 1 (mod 101)

Calculations:

2¹⁰¹ = 2 × 2¹⁰⁰

⇒ 2¹⁰¹ ≡ 2 × 1 ≡ 2 (mod 101)

∴ The remainder is 2.

Given:

The quadratic equation is: x² + p·x + q = 0

Roots of the equation are: p ≠ 0 and q ≠ 0 (same as the coefficients)

Formula used:

Let α and β be the roots of the equation:

α + β = −p

α · β = q

But it's given that roots are p and q themselve

p + q = −p  ...(i)

p × q = q  ...(ii)

Calculations:

From (ii): p × q = q

Divide both sides by q (since q ≠ 0):

⇒ p = 1

Substitute p = 1 into (i):

⇒ 1 + q = −1 ⇒ q = −2

Now find: p² + q²

= 1² + (−2)² = 1 + 4 = 5

∴ The correct answer is option 4.

Given:

The equations: x² + px + q = 0 and x² + qx + p = 0 (p ≠ q) have a common root.

Formula used:

If the equations have a common root, let the common root be α. Substituting α in both equations, we have:

α² + pα + q = 0

α² + qα + p = 0

By subtracting these equations, we eliminate α² and solve for p and q.

Calculation:

From the two equations:

α² + pα + q = 0

α² + qα + p = 0

Subtract the second equation from the first:

⇒ (pα + q) - (qα + p) = 0

⇒ pα - qα + q - p = 0

⇒ α(p - q) = p - q

If p ≠ q, we can cancel (p - q):

⇒ α = 1

Substitute α = 1 in the first equation:

1² + p(1) + q = 0

⇒ 1 + p + q = 0

⇒ p + q = -1

∴ The correct answer is option (1).