Quantitative Aptitude PYQs

Given:

External radius (R_ext) = 10.5 cm

Internal radius (R_int) = 7.7 cm

Density of the metal (ρ) = 10.3 g/cm³

π = 22/7

Formula Used:

Volume of a sphere = \(\frac{4}{3} \pi r^3 \)

Volume of the hollow sphere = Volume of external sphere - Volume of internal sphere

Mass = Density × Volume

Calculation:

Volume of external sphere (V_ext)

⇒ V_ext = \(\frac{4}{3} \pi (10.5)^3\)

⇒ V_ext = \(\frac{4}{3} \times \frac{22}{7} \times 1157.625 \)

⇒ V_ext = 4852.5 cm³

Volume of internal sphere (V_int)

⇒ V_int = \(\frac{4}{3} \pi (7.7)^3\)

⇒ V_int = \(\frac{4}{3} \times \frac{22}{7} \times 456.533 \)

⇒ V_int = 1980 cm³

Volume of the hollow sphere (V)

⇒ V = V_ext - V_int

⇒ V = 4852.5 - 1980

⇒ V = 2872.5 cm³

Mass (m)

⇒ m = ρ × V

⇒ m = 10.3 × 2872.5

⇒ m = 29587.75 g

Convert g to kg

⇒ m = 29587.75 / 1000

⇒ m ≈ 29.6 kg

Rounded off to the nearest tens, mass = 30 kg

The mass of the sphere is 30 kg.

Given:

Numbers to find HCF of: 620 and 248.

Formula Used:

Euclidean Algorithm: HCF(a, b) = HCF(b, a mod b)

Calculation:

Step 1: Apply Euclidean Algorithm:

⇒ 620 mod 248 = 124

Step 2: Apply Euclidean Algorithm again:

⇒ 248 mod 124 = 0

The HCF of 620 and 248 is 124.

Given:

Area of the base of the cone = 25π cm²

Height of the cone = 12 cm

Formula Used:

Curved Surface Area (CSA) of a cone = π × r × l

where r = radius of the base, l = slant height

Slant height, l = √(r² + h²)

Calculation:

Area of the base = π × r² = 25π

⇒ r² = 25

⇒ r = √25

⇒ r = 5 cm

Slant height, l = √(r² + h²)

⇒ l = √(5² + 12²)

⇒ l = √(25 + 144)

⇒ l = √169

⇒ l = 13 cm

Curved Surface Area (CSA) = π × r × l

⇒ CSA = π × 5 × 13

⇒ CSA = 65π cm²

The curved surface area of the cone is 65π cm².

Given:

First number: 3² × 4² × 9

Second number: 3³ × 4⁴ × 9¹⁰

Formula used:

HCF is calculated by taking the smallest powers of common prime factors.

Note:

9 = 3².

Calculation:

Rewrite the numbers in terms of prime factors:

First number = 3² × 4² × (3²) = 3⁴ × 2⁴

Second number = 3³ × 4⁴ × (3²⁰) = 3²³ × 2⁸

HCF = Take the smallest powers of 3 and 2:

For 3: min(4, 23) = 4

For 2: min(4, 8) = 4

HCF = 3⁴ × 2⁴

HCF = 81 × 16 = 1296

∴ The HCF is 1296.

Given:

Marked price = ₹500

First seller's discounts = 10% and 20%

Second seller's discount = 30%

Formula Used:

Selling Price = Marked Price - Discount

Calculation:

First Seller's Calculation:

First discount: 10% of ₹500

⇒ 10% × ₹500 = ₹50

Price after first discount: ₹500 - ₹50

⇒ ₹450

Second discount: 20% of ₹450

⇒ 20% × ₹450 = ₹90

Selling price after second discount: ₹450 - ₹90

⇒ ₹360

Second Seller's Calculation:

30% discount on ₹500

⇒ 30% × ₹500 = ₹150

Selling price after 30% discount: ₹500 - ₹150

⇒ ₹350

Difference between their selling prices:

⇒ ₹360 - ₹350

⇒ ₹10

The difference between their selling prices is ₹10.

Given:

Amount in 2 years (A1) = ₹9,680

Amount in 4 years (A2) = ₹11,712.80

Formula used:

A = P(1 + \(\frac{r}{100}\))^t

Where, r = rate

Calculations:

A2 = A1(1 + \(\frac{r}{100}\))²

11,712.80 = 9,680(1 + \(\frac{r}{100}\))²

⇒ \(\frac{11712.80}{9680} = (1+\frac{r}{100})^2 \)

⇒ \(\frac{11712.80}{9680} = 1.21 \)

⇒ (1 + \(\frac{r}{100}\))² = 1.21

⇒ 1 + \(\frac{r}{100}\) = \(\sqrt{1.21} \)

⇒ 1 + \(\frac{r}{100}\) = 1.1

⇒ \(\frac{r}{100}\) = 0.1

⇒ r = 10%

∴ The correct answer is option (2).

Given:

Odd numbers from 1 to 43.

The sum of n of odd number = n2

Where, n = Number of terms

Number of terms = (First terms + Last terms)/2

Average = The sum of all number/(Total number)

Calculation:

The odd numbers between  1 to 43.

1 + 3 + 5 + ..........+ 43

Number of terms = (First terms + Last terms)/2

⇒ n = (1 + 43)/2

⇒ n = 22

Given:

A can mow the field in 12 hours, so A's work rate = 1/12 of the field per hour.

B can mow the field in 10 hours, so B's work rate = 1/10 of the field per hour.

They work alternately, starting with A at 5:00 a.m.

Formula used:

Total work completed in 2 hours = Work done by A + Work done by B.

Remaining work = Total work - Work done.

Calculations:

Work done by A in 1 hour = 1/12.

Work done by B in 1 hour = 1/10.

Total work completed in 2 hours = (1/12) + (1/10) = (5 + 6)/60 = 11/60.

Total time taken:

In 2 hours, they complete 11/60 of the work.

Work remaining after 5 full cycles (10 hours):

Total work done in 10 hours = (11/60) × 5 = 55/60.

Remaining work = 1 - 55/60 = 5/60 = 1/12.

At 5:00 a.m., A starts again for the last 1/12 of the work.

Time taken by A to finish 1/12 = 1 hour.

Final time:

10 hours of work + 1 hour (A finishes the remaining work) = 11 hours.

Start time = 5:00 a.m.

End time = 5:00 a.m. + 11 hours = 4:00 p.m.

∴ The mowing will be completed at 4:00 p.m.

Given:

A salesman earns a 2% return on (1/4) of his capital, 4% on (1/2) of his capital, and 10% on the remaining capital.

Formula used:

Average rate of return = (Total return) / (Total capital)

Calculation:

Let the total capital be C.

Return on (1/4) of C = 2% of (1/4)C = 2/100 × (1/4)C = (1/200)C

Return on (1/2) of C = 4% of (1/2)C = 4/100 × (1/2)C = (2/100)C = (1/50)C

Remaining capital = C - (1/4)C - (1/2)C = (1/4)C

Return on remaining (1/4) of C = 10% of (1/4)C = 10/100 × (1/4)C = (1/10)C = (1/40)C

Total return = (1/200)C + (1/50)C + (1/40)C

To find a common denominator, we use 200:

(1/200)C = (1/200)C

(1/50)C = (4/200)C

(1/40)C = (5/200)C

Total return = (1/200)C + (4/200)C + (5/200)C

⇒ Total return = (10/200)C

⇒ Total return = (1/20)C

Average rate of return = (Total return) / (Total capital)

⇒ Average rate of return = (1/20)C / C

⇒ Average rate of return = 1/20

⇒ Average rate of return = 0.05 or 5%

∴ The correct answer is option (4).

Alternate MethodGiven:

A salesman earns 2% on 1/4 of his capital.

He earns 4% on 1/2 of his capital.

He earns 10% on the remaining 1/4 of his capital.

Concept Used:

The average rate of return is calculated as the weighted average of the returns from different portions of capital.

Calculation:

Determine the weights and respective returns:

Return on 1/4 of capital = 2%

Return on 1/2 of capital = 4%

Return on 1/4 of capital = 10%

Calculate the average rate of return:

Average return = (Weight 1 × Rate 1) + (Weight 2 × Rate 2) + (Weight 3 × Rate 3)

Weights: 1/4, 1/2, and 1/4

Rates: 2%, 4%, and 10%

Average return = (1/4 × 2) + (1/2 × 4) + (1/4 × 10)

Simplify the expression:

Average return = (2/4) + (4/2) + (10/4)

Average return = 0.5 + 2 + 2.5

Average return = 5%

The average rate of return earned on his capital is 5%.

Given:

Sides of the triangle are a = 3 cm, b = 4 cm, and c = 5 cm

Formula used:

Using Heron's formula:

Area = √(s(s - a)(s - b)(s - c))

Where s = (a + b + c) / 2

Calculations:

s = (3 + 4 + 5) / 2

⇒ s = 12 / 2

⇒ s = 6

Area = √(6 × (6 - 3) × (6 - 4) × (6 - 5))

⇒ Area = √(6 × 3 × 2 × 1)

⇒ Area = √(36)

⇒ Area = 6 cm²

∴ The correct answer is option (4).

Given:

The radius of the base of a right circular cylinder is increased by 20%

The height is decreased by 30%

Formula used:

Volume of cylinder (V) = πr²h

Calculation:

Let the original radius be r and the original height be h

Original volume, V₁ = πr²h

New radius = r × 1.20

New height = h × 0.70

New volume, V₂ = π(r × 1.20)²(h × 0.70)

⇒ V₂ = π(1.20²r²)(0.70h)

⇒ V₂ = π(1.44r²)(0.70h)

⇒ V₂ = πr²h(1.44 × 0.70)

⇒ V₂ = πr²h(1.008)

Percentage change in volume = ((V₂ - V₁)/V₁) × 100

⇒ Percentage change in volume = ((1.008V₁ - V₁)/V₁) × 100

⇒ Percentage change in volume = (0.008V₁/V₁) × 100

⇒ Percentage change in volume = 0.8%

∴ The correct answer is option (1).

Given:

Simple interest on ₹800 for 4 years + Simple interest on ₹600 for 2 years = ₹280

Formula used:

Simple Interest (SI) = (P × R × T) / 100

Where, P = Principal, R = Rate of interest, T = Time

Calculations:

(800 × R × 4) / 100 + (600 × R × 2) / 100 = 280

⇒ (3200R / 100) + (1200R / 100) = 280

⇒ 4400R / 100 = 280

⇒ 44R = 280

⇒ R = 280 / 44

⇒ R = 70 / 11

⇒ R = 6 (4/11)%

Answer: The rate of interest is \(6 \frac{4}{11}%\)

Given:

7/8, 6/7, 11/13, 17/20, 23/25

Formula used:

To find the middle term in descending order.

Calculation:

Convert each fraction to decimal for easy comparison:

7/8 = 0.875

6/7 = 0.857

11/13 = 0.846

17/20 = 0.85

23/25 = 0.92

Now arrange them in descending order:

0.92, 0.875, 0.857, 0.85, 0.846

The term exactly in the middle is 6/7.

∴ The correct answer is option (3).

Given:

The marked price of a chair = ₹1,200

This price was 20% above its cost price.

The chair was sold at a discount of 5% on the marked price.

Formula used:

Cost Price (C.P) = Marked Price / (1 + Markup Percentage)

Selling Price (S.P) = Marked Price × (1 - Discount Percentage)

Profit Percentage = ((S.P - C.P) / C.P) × 100

Calculation:

Marked Price = ₹1,200

Markup Percentage = 20%

Discount Percentage = 5%

⇒ Cost Price = 1200 / (1 + 0.20)

⇒ Cost Price = 1200 / 1.20

⇒ Cost Price = ₹1,000

⇒ Selling Price = 1200 × (1 - 0.05)

⇒ Selling Price = 1200 × 0.95

⇒ Selling Price = ₹1,140

⇒ Profit Percentage = ((1140 - 1000) / 1000) × 100

⇒ Profit Percentage = (140 / 1000) × 100

⇒ Profit Percentage = 14%

∴ The correct answer is option (4).

Given:

In a 150 m race:

R beats S by 9 m.

Head start given to S = 15 m.

Formula used:

Speed ratio of R and S = Distance covered by R : Distance covered by S.

For tie, adjusted distance covered by R = adjusted distance covered by S.

Calculation:

Speed ratio of R and S:

Speed of R : Speed of S = 150 : (150 - 9)

⇒ Speed of R : Speed of S = 150 : 141

If S takes a head start of 15 m, let the race length be L m. For a tie:

Distance covered by S = Distance covered by R.

⇒ L - 15 : L = 141 : 150

⇒ (L - 15) / L = 141 / 150

⇒ 150(L - 15) = 141L

⇒ 150L - 2250 = 141L

⇒ 150L - 141L = 2250

⇒ 9L = 2250

⇒ L = 2250 / 9

⇒ L = 250

∴ The race length should be 250 m for a tie.

Given:

Numbers = 48, 92, 140

Formula used:

To find the largest number that can exactly divide 48, 92, and 140, we need to find the Greatest Common Divisor (GCD) of these numbers.

Calculation:

Prime factorization of 48: 48 = 2⁴ × 3

Prime factorization of 92: 92 = 2² × 23

Prime factorization of 140: 140 = 2² × 5 × 7

The common prime factor is 2, and the lowest power of 2 is 2².

⇒ GCD = 2² = 4

∴ The correct answer is option 1.

Given:

David purchases 10 bananas for ₹25

David sells 9 bananas for ₹45

Formula used:

Gain Percentage = \(\dfrac{Selling Price - Cost Price}{Cost Price} \times 100\)

Calculations:

Cost Price (CP) of 1 banana = 25 ÷ 10 = ₹2.5

Selling Price (SP) of 1 banana = 45 ÷ 9 = ₹5

Gain per banana = SP - CP = 5 - 2.5 = ₹2.5

Gain Percentage = \(\dfrac{2.5}{2.5} \times 100\)

⇒ Gain Percentage = 100%

∴ The correct answer is option (2).

Shortcut Trick

HCF of 448 and 336 → 112 × 4 = 448 and 112 × 3 = 336 ∴ HCF₁ = 112.

HCF of 48 and 36 → 12 × 4 = 48 and 12 × 3 = 36 ∴ HCF₂ = 12.

Required times = 112 ÷ 12 = 28/3.

∴ The correct answer is 28/3.

Alternate Method

Given: Numbers are (448, 336) and (48, 36).

Formula Used: HCF is the highest common factor that divides the numbers exactly.

⇒ Step 1: Find HCF of 448 and 336.

⇒ 448 = 2 × 2 × 2 × 2 × 2 × 2 × 7 = 2⁶ × 7

⇒ 336 = 2 × 2 × 2 × 2 × 3 × 7 = 2⁴ × 3 × 7

⇒ HCF_{(448, 336)} = 2⁴ × 7 = 16 × 7 = 112.

⇒ Step 2: Find HCF of 48 and 36.

⇒ 48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3

⇒ 36 = 2 × 2 × 3 × 3 = 2² × 3²

⇒ HCF_{(48, 36)} = 2² × 3 = 4 × 3 = 12.

⇒ Step 3: Calculate the ratio.

⇒ Times = 112 ÷ 12

⇒ Times = (4 × 28) ÷ (4 × 3) = 28/3.

∴ The correct answer is 28/3.

Additional Information

HCF (Highest Common Factor)

The greatest number that divides two or more numbers without leaving a remainder. It is also known as GCD (Greatest Common Divisor).

Relationship between HCF and LCM

For any two numbers 'a' and 'b', the product of the numbers is equal to the product of their HCF and LCM: a × b = HCF(a, b) × LCM(a, b).

Division Method for HCF

Divide the larger number by the smaller one, then divide the previous divisor by the remainder. Continue until the remainder is 0; the last divisor is the HCF.

Shortcut Trick

Radius of the basketball (r) = 14 ÷ 2 = 7 cm.

Surface Area of sphere = 4πr² = 4 × (22/7) × 7 × 7 = 616 cm².

Total Painting Cost = Area × Rate = 616 × 2.5 = ₹1540.

∴ The correct answer is ₹1540.

Alternate Method

Given:

Diameter of the basketball = 14 cm

Rate of painting = ₹2.5/cm²

Value of π = 22/7

Formula Used:

Surface Area of a Sphere = 4πr²

Total Cost = Surface Area × Cost per unit area

Calculations:

⇒ Radius (r) = Diameter ÷ 2

⇒ r = 14 ÷ 2 = 7 cm

⇒ Surface Area of the basketball = 4 × (22/7) × 7 × 7

⇒ Surface Area = 4 × 22 × 7

⇒ Surface Area = 88 × 7 = 616 cm²

⇒ Total Painting Cost = 616 × 2.5

⇒ Total Painting Cost = 616 × (5/2)

⇒ Total Painting Cost = 308 × 5 = ₹1540

∴ The correct answer is ₹1540.

Additional Information

Volume of a Sphere

The total space occupied by a sphere is given by the formula: V = (4/3)πr³.

Hemisphere Surface Area

For a hemisphere, Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr².

Relation between Diameter and Radius

The diameter is always twice the radius of a circle or sphere (D = 2r).

Given:

The numbers are 30, 48, 96, 150.

Formula used:

LCM (Least Common Multiple) of given numbers

Calculation:

Prime factorization of 30 = 2 × 3 × 5

Prime factorization of 48 = 2⁴ × 3

Prime factorization of 96 = 2⁵ × 3

Prime factorization of 150 = 2 × 3 × 5²

LCM = 2⁵ × 3 × 5²

⇒ LCM = 32 × 3 × 25

⇒ LCM = 2400

Sum of the digits of 2400 = 2 + 4 + 0 + 0 = 6

∴ The correct answer is option (3).