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The greatest number that divides the given numbers exactly is their Highest Common Factor (HCF).

Find the differences between the consecutive numbers:

105 − 75 = 30

150 − 105 = 45

The HCF must divide these differences. The greatest common divisor of 30 and 45 is 15.

Since 15 also divides 75, 105, and 150 exactly, the greatest divisor is 15.

∴ The correct answer is 15.

Alternate Method

Given:

The numbers are 75, 105, and 150.

Formula/Concept Used:

To find the greatest number that divides a set of numbers exactly, we calculate their Highest Common Factor (HCF) using the Prime Factorization Method.

Shortcut Trick

Final Selling Price = Marked Price × (1 − d₁/100) × (1 − d₂/100)

⇒ SP = 3000 × (1 − 15/100) × (1 − 25/100)

⇒ SP = 3000 × 0.85 × 0.75

⇒ SP = 3000 × 0.6375 = ₹1,912.50

∴ The correct answer is ₹1,912.50.

Alternate Method

Given:

Marked Price (MP) = ₹3,000

First Discount (d₁) = 15%

Second Discount (d₂) = 25%

Calculations:

⇒ Price after first discount = 3000 − (15% of 3000)

⇒ Price after first discount = 3000 − (15/100 × 3000) = 3000 − 450 = ₹2,550

⇒ Price after second discount = 2550 − (25% of 2550)

⇒ Price after second discount = 2550 − (25/100 × 2550) = 2550 − 637.50

⇒ Final Selling Price = ₹1,912.50

∴ The correct answer is ₹1,912.50.

Additional Information

Single Equivalent Discount Formula

For two successive discounts of x% and y%, the single equivalent discount is (x + y − xy/100)%.

Order of Successive Discounts

The final selling price remains the same regardless of the order in which the successive discounts are applied (e.g., 15% then 25% is the same as 25% then 15%).

Marked Price vs Selling Price

Selling Price = Marked Price − Discount. Discounts are always calculated on the Marked Price (or the remaining balance in successive cases).

Shortcut Trick

To find the larger value (Elder Sibling) when the Sum and Difference are given:

Elder Sibling Age = (Sum + Difference) ÷ 2

⇒ Age = (28 + 8) ÷ 2

⇒ Age = 36 ÷ 2 = 18 years

∴ The correct answer is 18 years.

Alternate Method

Given:

Sum of ages = 28 years

Difference of ages = 8 years

Method: Linear Equations

⇒ Let the age of the elder sibling be x years and the younger sibling be y years.

⇒ x + y = 28 --- (Equation 1)

⇒ x − y = 8 --- (Equation 2)

⇒ Adding Equation 1 and Equation 2:

⇒ (x + y) + (x − y) = 28 + 8

⇒ 2x = 36

⇒ x = 36 ÷ 2

⇒ x = 18 years

∴ The correct answer is 18 years.

Additional Information

Sum and Difference Formula

For any two numbers x and y, if Sum = S and Difference = D, then the larger number x = (S + D) ÷ 2 and the smaller number y = (S − D) ÷ 2.

Age Ratio Problems

If the ratio of ages of two people is a : b, their ages can be taken as ax and bx. The difference between their ages remains constant throughout their lives.

Average Age

The average age of n persons is the sum of their ages divided by n. If each person's age increases by 'k' years, the average also increases by 'k' years.

Shortcut Trick

Ratio of Amount to Principal = 1728 : 1000.

Since interest is compounded semi-annually for 1.5 years, there are 3 half-yearly periods.

Take the cube root for one period: √³1728 : √³1000 = 12 : 10.

Rate per half-year = (12 − 10)/10 × 100 = 20%.

Annual Interest Rate = 20% × 2 = 40%.

∴ The correct answer is 40%.

Alternate Method

Given: Principal (P) = ₹1,000, Amount (A) = ₹1,728, Time = 1 year 6 months.

Formula Used: A = P(1 + r/200)²ⁿ (for semi-annual compounding, where r is annual rate and n is years).

⇒ Total periods (n) = 1.5 years × 2 = 3 half-years.

⇒ 1728 = 1000 × (1 + r/200)³

⇒ 1728/1000 = (1 + r/200)³

⇒ 1.728 = (1 + r/200)³

⇒ √³(1.728) = 1 + r/200

⇒ 1.2 = 1 + r/200

⇒ r/200 = 1.2 − 1 = 0.2

⇒ r = 0.2 × 200 = 40%.

∴ The correct answer is 40%.

Additional Information

Compound Interest Compounding Frequency

For semi-annual compounding, the rate is halved (r/2) and the time periods are doubled (2n). For quarterly, the rate is divided by 4 and periods multiplied by 4.

Effective Interest Rate

The actual interest earned over a year when compounding occurs more than once. Formula: E = (1 + r/k)^k − 1, where k is compounding frequency per year.

Rule of 72

A quick way to estimate the time required to double an investment: Years to double ≈ 72 ÷ Annual Interest Rate.

Shortcut Trick

Let Cost Price (CP) = 100.

Marked Price (MP) is 150% above CP ⇒ 100 + 150 = 250.

To gain 110%, Selling Price (SP) must be 100 + 110 = 210.

Discount = MP − SP = 250 − 210 = 40.

Discount % = (40 ÷ 250) × 100 = 16%.

∴ The correct answer is 16%.

Alternate Method

Given:

Markup = 150% above Cost Price

Required Profit = 110%

Calculations:

⇒ Assume CP = 100 units

⇒ Marked Price (MP) = 100 + (150% of 100) = 100 + 150 = 250 units

⇒ Selling Price (SP) = 100 + (110% of 100) = 100 + 110 = 210 units

⇒ Discount Amount = MP − SP = 250 − 210 = 40 units

⇒ Discount % = (Discount Amount ÷ MP) × 100

⇒ Discount % = (40 ÷ 250) × 100

⇒ Discount % = 4 ÷ 25 × 100 = 16%

∴ The correct answer is 16%.

Additional Information

Relationship between CP, MP, Profit, and Discount

The ratio of MP to CP can be expressed as: MP/CP = (100 + Profit%) ÷ (100 − Discount%).

Markup Percentage

Markup % is the percentage by which the cost price is increased to set the marked price: ((MP − CP) ÷ CP) × 100.

Effective Discount

For two successive discounts d₁ and d₂, the single equivalent discount is (d₁ + d₂ − (d₁ × d₂) ÷ 100)%.

Shortcut Trick

Identify factors of the product 175 that add up to the sum 40.

Factors of 175: 1 × 175, 5 × 35, 7 × 25.

Check sum: 5 + 35 = 40 (Correct).

Calculate positive difference: 35 − 5 = 30.

∴ The correct answer is 30.

Alternate Method

Given: Sum (x + y) = 40 and Product (xy) = 175.

Formula: (x − y)² = (x + y)² − 4xy

⇒ (x − y)² = 40² − 4 × 175

⇒ (x − y)² = 1600 − 700

⇒ (x − y)² = 900

⇒ x − y = √900 = 30

∴ The correct answer is 30.

Additional Information

Algebraic Identity

The relationship between sum, product, and difference is given by the identity: (a+b)² − (a−b)² = 4ab.

Quadratic Equations

If the sum (S) and product (P) are known, the numbers are roots of the equation: t² − St + P = 0.

Difference of Roots

For a quadratic equation ax² + bx + c = 0, the positive difference of roots is given by √D / |a|, where D = b² − 4ac.

Given:

3^x − 3^x−1 = 18

Formula used:

a^m−n = a^m / aⁿ

Calculation:

3^x − 3^x−1 = 18

⇒ 3^x − (3^x / 3) = 18

⇒ 3^x (1 − 1/3) = 18

⇒ 3^x × (2/3) = 18

⇒ 3^x = (18 × 3) / 2

⇒ 3^x = 9 × 3

⇒ 3^x = 27

⇒ 3^x = 3³

⇒ x = 3

Substituting x in the expression (2x − 1) / (2x + 3):

⇒ (2 × 3 − 1) / (2 × 3 + 3)

⇒ (6 − 1) / (6 + 3)

⇒ 5/9

∴ The correct answer is option (2).

Shortcut Trick

To identify co-prime numbers, check if the two numbers have any common factor other than 1.

(15, 25) → Both are divisible by 5.

(14, 35) → Both are divisible by 7.

(21, 49) → Both are divisible by 7.

(9, 28) → 9 = 3×3 and 28 = 2×2×7. No common prime factor exists.

∴ The correct answer is (9, 28).

Alternate Method

Given: Pairs of numbers (15, 25), (14, 35), (9, 28), and (21, 49).

Formula Used: Two numbers are co-prime if their Highest Common Factor (HCF) is 1.

⇒ Factors of 15: 1, 3, 5, 15 | Factors of 25: 1, 5, 25. HCF = 5.

⇒ Factors of 14: 1, 2, 7, 14 | Factors of 35: 1, 5, 7, 35. HCF = 7.

⇒ Factors of 21: 1, 3, 7, 21 | Factors of 49: 1, 7, 49. HCF = 7.

⇒ Factors of 9: 1, 3, 9 | Factors of 28: 1, 2, 4, 7, 14, 28. HCF = 1.

∴ The correct answer is (9, 28).

Additional Information

Definition of Co-prime Numbers

A set of numbers is co-prime if they have no common positive factor other than 1. Their HCF is always 1.

Consecutive Numbers

Any two consecutive natural numbers (e.g., 14 and 15, or 20 and 21) are always co-prime to each other.

LCM of Co-prime Numbers

The Least Common Multiple (LCM) of two co-prime numbers is always equal to their product (a × b).

Prime Numbers Property

Any two prime numbers are always co-prime to each other since their only common factor is 1.

Shortcut Trick

Find the LCM of the given divisors: LCM(12, 8, 11) = 264.

Divide the largest 4-digit number (9999) by 264 to find the remainder: 9999 ÷ 264 gives a remainder of 231.

Subtract this remainder from 9999 to get the largest multiple: 9999 − 231 = 9768.

Add the required common remainder of 3 to the multiple: 9768 + 3 = 9771.

∴ The correct answer is 9771.

Alternate Method

Given:

Divisors = 12, 8, 11

Common remainder in each case = 3

Formula Used:

Required Number = (LCM of divisors × k) + Remainder, where k is an integer.

4-Digit Numbers 5-Digit

Shortcut Trick

Apply the Direct Distance Formula for Total Travel Time: Distance = (S₁ × S₂) ÷ (S₁ + S₂) × Total Time.

Here, S₁ = 60 km/hr, S₂ = 80 km/hr, and Total Time = 3.5 hr.

⇒ Distance = (60 × 80) ÷ (60 + 80) × 3.5

⇒ Distance = (4800 ÷ 140) × 3.5 = (480 ÷ 14) × 3.5

⇒ Distance = 480 ÷ 4 = 120 km.

∴ The correct answer is 120 km.

Alternate Method

Given: Speed from Gwalior to Jhansi = 60 km/hr, speed from Jhansi to Gwalior = 80 km/hr, and Total travel time = 3.5 hr.

Formula Used: Time = Distance ÷ Speed

Let the one-way distance be x km.

⇒ Time taken to travel from Gwalior to Jhansi = x ÷ 60

⇒ Time taken to travel from Jhansi to Gwalior = x ÷ 80

⇒ Total Time = (x ÷ 60) + (x ÷ 80) = 3.5

⇒ (4x + 3x) ÷ 240 = 3.5

⇒ 7x ÷ 240 = 3.5

⇒ 7x = 3.5 × 240

⇒ 7x = 840

⇒ x = 840 ÷ 7 = 120 km.

∴ The correct answer is 120 km.

Additional Information

Average Speed Formula

When a person travels the same distance at two different speeds S₁ and S₂, the Average Speed is given by (2 × S₁ × S₂) ÷ (S₁ + S₂).

Speed and Time Relation

For a constant distance, Speed is inversely proportional to Time (S₁ ÷ S₂ = T₂ ÷ T₁).

Distance when Time Difference is Given

If the difference in time (ΔT) is given instead of total time, the formula becomes: Distance = (S₁ × S₂) ÷ |S₁ − S₂| × ΔT.