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Identify the pair where both numbers share a common factor using divisibility rules.

For the pair (171, 273):

Sum of digits of 171 = 1 + 7 + 1 = 9 (Divisible by 3).

Sum of digits of 273 = 2 + 7 + 3 = 12 (Divisible by 3).

Since both numbers are divisible by 3, their HCF is at least 3, making them not coprime.

∴ The correct answer is (171, 273).

Alternate Method

Given: Pairs of numbers to check for coprimality.

Definition: Two numbers are coprime if their Highest Common Factor (HCF) is 1.

Testing the pairs:

⇒ (173, 351): 173 is a prime number. 351 is not divisible by 173. HCF = 1 (Coprime).

⇒ (177, 827): 177 = 3 × 59. 827 is a prime number not divisible by 3 or 59. HCF = 1 (Coprime).

⇒ (171, 273):

⇒ Prime factors of 171 = 3² × 19

⇒ Prime factors of 273 = 3 × 7 × 13

⇒ HCF (171, 273) = 3

⇒ Since HCF ≠ 1, the numbers are not coprime.

⇒ (127, 257): Both 127 and 257 are prime numbers. HCF = 1 (Coprime).

∴ The correct answer is (171, 273).

Additional Information

Coprime Numbers

Two numbers a and b are called coprime or relatively prime if the only positive integer that divides both of them is 1, i.e., HCF(a, b) = 1.

Divisibility Rule of 3

A number is divisible by 3 if the sum of its digits is a multiple of 3. This is the fastest way to check for a common factor of 3 in large numbers.

Properties of Coprime Pairs

Any two consecutive integers are always coprime. Any two prime numbers are always coprime. The Least Common Multiple (LCM) of two coprime numbers is always their product (a × b).

Shortcut Trick

For a sphere inscribed inside a cube, the diameter (d) of the sphere is equal to the edge (a) of the cube (d = a).

Radius of sphere (r) = a / 2

Ratio = (Surface Area of Sphere) : (Surface Area of Cube)

Ratio = 4 × π × (a / 2)² : 6a² = 4 × π × (a² / 4) : 6a² = πa² : 6a²

Ratio = π : 6

∴ The correct answer is pi : 6.

Alternate Method

Given: Edge of the cube (a) = 10 cm.

Formula Used:

1. Surface Area of Sphere = 4 × π × r²
2. Surface Area of Cube = 6 × a²

⇒ Diameter of inscribed sphere = Edge of cube = 10 cm

⇒ Radius (r) of sphere = 10 ÷ 2 = 5 cm

⇒ Surface Area of Sphere = 4 × π × (5)² = 4 × π × 25 = 100π cm²

⇒ Surface Area of Cube = 6 × (10)² = 6 × 100 = 600 cm²

⇒ Required Ratio = (100π) : 600

⇒ Ratio = π : 6

∴ The correct answer is pi : 6.

Additional Information

Volume Ratio for Inscribed Sphere

The ratio of the volume of an inscribed sphere to the volume of the cube is π : 6.

Circumscribed Sphere

If a sphere is circumscribed around a cube, its diameter equals the body diagonal of the cube (a√3).

Cube Diagonals

Face diagonal of a cube = a√2 | Body diagonal of a cube = a√3.

Shortcut Trick

In averages, when a value is removed, the formula is:

Removed Number = Old Average + [Remaining Count × (Old Average − New Average)]

⇒ Removed Number = 18.5 + [4 × (18.5 − 16)]

⇒ Removed Number = 18.5 + [4 × 2.5] = 18.5 + 10 = 28.5

∴ The correct answer is 28.5.

Alternate Method

Given:

Initial count of numbers (n₁) = 5

Initial average (A₁) = 18.5

Remaining count of numbers (n₂) = 4

New average (A₂) = 16

Formula Used:

Sum of values = Average × Total numbers

Calculations:

⇒ Total sum of 5 numbers = 5 × 18.5 = 92.5

⇒ Total sum of remaining 4 numbers = 4 × 16 = 64

⇒ Removed number = (Sum of 5 numbers) − (Sum of 4 numbers)

⇒ Removed number = 92.5 − 64

⇒ Removed number = 28.5

∴ The correct answer is 28.5.

Additional Information

Net Deviation Rule

The sum of deviations from the average is always zero. If a value is removed, the total deviation caused by its removal must be balanced by the remaining items.

Effect of Adding/Removing

If the removed number is greater than the average, the new average decreases. If the removed number is less than the average, the new average increases.

Weighted Average Formula

Total Average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂). This is used when two groups are combined.

Shortcut Trick

Since the Simple Interest is the same for both cases, equate the product of (P × R × T).

5000 × 6 × T = 7500 × 4 × 4

⇒ 30000 × T = 120000

⇒ T = 120000 ÷ 30000

⇒ T = 4

∴ The correct answer is 4 years.

Alternate Method

Given:

First case: Principal (P₁) = ₹5,000, Rate (R₁) = 6%, Time (T₁) = T years.

Second case: Principal (P₂) = ₹7,500, Rate (R₂) = 4%, Time (T₂) = 4 years.

Formula Used:

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

Calculations:

⇒ SI₁ = (5000 × 6 × T) ÷ 100 = 300T

⇒ SI₂ = (7500 × 4 × 4) ÷ 100 = 75 × 16 = 1200

⇒ According to the question, SI₁ = SI₂

⇒ 300T = 1200

⇒ T = 1200 ÷ 300

⇒ T = 4 years

∴ The correct answer is 4 years.

Additional Information

Simple Interest Definition

Simple Interest is calculated only on the initial Principal amount of a loan or deposit. Formula: SI = (P × R × T) ÷ 100.

Total Amount

The Total Amount (A) accumulated at the end of the time period is given by the sum of the Principal and the Simple Interest: A = P + SI.

Rate of Interest conversion

Always ensure the Rate of Interest (R) and Time (T) are in the same units (usually per annum and years). If time is in months, divide by 12; if in days, divide by 365.

Shortcut Trick

Scheme A: Net discount = 43 + 19 − (43 × 19)/100 = 53.83%

Scheme B: Successive discounts of 41%, 21% and 49% results in > 70% discount.

Scheme C: For exactly 5 toys, the discount is fixed at 35%.

Scheme D: Discount % = [Free / Total] × 100 = [4 / (5 + 4)] × 100 ≈ 44.44%

Comparing all: 35% (C) is the minimum discount, making it the least beneficial.

∴ The correct answer is Scheme C.

Alternate Method

Given: A customer buys 5 toys of the same Marked Price (MP).

Formula Used:

1. Successive Discount (x%, y%) = [x + y − (xy / 100)]%
2. Buy X Get Y Free Discount % = [Y / (X + Y)] × 100

⇒ Scheme A: Successive discounts of 43% and 19%

⇒ Net Discount = 43 + 19 − (43 × 19)/100 = 62 − 8.17 = 53.83%

⇒ Scheme B: Successive 41%, 21% and 49%

⇒ First two: 41 + 21 − (41 × 21)/100 = 62 − 8.61 = 53.39%

⇒ With 49%: 53.39 + 49 − (53.39 × 49)/100 = 102.39 − 26.16 = 76.23%

⇒ Scheme C: Customer buys 5 toys. According to the scheme, the discount is 35% on the first 5 toys. Since the purchase is exactly 5 toys, the effective discount is 35%.

⇒ Scheme D: 4 toys free on buying 5 toys. Total toys = 5 + 4 = 9.

⇒ Discount % = (4 / 9) × 100 ≈ 44.44%

⇒ The least beneficial scheme is the one with the lowest discount percentage.

⇒ Minimum value among {53.83%, 76.23%, 35%, 44.44%} is 35%.

∴ The correct answer is Scheme C.

Additional Information

Successive Discounts

For three successive discounts a%, b%, and c%, the final multiplier is (1 − a/100) × (1 − b/100) × (1 − c/100). The effective discount is (1 − Multiplier) × 100.

Buy X Get Y Free

In this offer, the customer pays for X items and receives X + Y items. The discount percentage is calculated as [Free Items / Total Items] × 100.

Profit and Loss Relationship

Discount is always calculated on the Marked Price (MP). Selling Price (SP) = MP × (100 − Discount%)/100.

Shortcut Trick

Total Capacity = LCM(12, 18) = 36 units.

Efficiency of Pipe A = +3 and Pipe B = −2.

Work done by A alone in 4 hr = 3 × 4 = 12 units.

Remaining work = 36 − 12 = 24 units.

Combined efficiency (A + B) = 3 − 2 = 1 unit/hr.

∴ The correct answer is 28 hr.

Alternate Method

Given:

Time taken by Pipe A to fill = 12 hr

Time taken by Pipe B to empty = 18 hr

Initial time Pipe A runs alone = 4 hr

Calculations:

⇒ Total Capacity of cistern = LCM(12, 18) = 36 units.

⇒ Part filled by Pipe A in the first 4 hours = 3 × 4 = 12 units.

⇒ Remaining part to be filled = 36 − 12 = 24 units.

⇒ Combined efficiency when both pipes are open = 3 − 2 = 1 unit/hr.

⇒ Time taken to fill the remaining part = 24 ÷ 1 = 24 hr.

⇒ Total time taken = Initial 4 hr + Next 24 hr = 28 hr.

∴ The correct answer is 28 hr.

Additional Information

Filling vs Emptying Pipes

Filling pipes have positive efficiency (+), while emptying pipes (leaks) have negative efficiency (−).

Net Efficiency

When multiple pipes work together, the net efficiency is the algebraic sum of their individual efficiencies.

Pipes and Cisterns Formula

If pipe A fills in 'x' hours and pipe B empties in 'y' hours, net work in 1 hr = (1/x) − (1/y).

Shortcut Trick

Net Simple Interest percentage = Rate × Time = 5% × 3 = 15%.

Total Amount percentage = 100% (Principal) + 15% (Interest) = 115%.

According to the question, 115% = ₹4,600.

Therefore, 100% (Sum) = (4,600 ÷ 115) × 100 = 40 × 100 = ₹4,000.

∴ The correct answer is ₹4,000.

Alternate Method

Given:

Amount (A) = ₹4,600

Time (T) = 3 years

Rate (R) = 5% per annum

Formula Used:

Amount (A) = Principal (P) + Simple Interest (SI)

A = P + (P × R × T) ÷ 100

Calculations:

⇒ 4,600 = P + (P × 5 × 3) ÷ 100

⇒ 4,600 = P + (15P ÷ 100)

⇒ 4,600 = (100P + 15P) ÷ 100

⇒ 4,600 = 115P ÷ 100

⇒ P = (4,600 × 100) ÷ 115

⇒ P = 40 × 100

⇒ P = 4,000

∴ The correct answer is ₹4,000.

Additional Information

Simple Interest Formula

The basic formula for Simple Interest is SI = (P × R × T) ÷ 100, where P is the principal sum.

Relationship between SI and Amount

Amount is calculated as A = P + SI, which can also be written as A = P[1 + (R × T) ÷ 100].

Calculating Rate and Time

To find Rate (R): (SI × 100) ÷ (P × T). To find Time (T): (SI × 100) ÷ (P × R).

Shortcut Trick

Convert the middle term to the same base (14/6): (6/14)³ = (14/6)⁻³.

Total power of LHS with respect to base (14/6) = 8 − 3 + 10 = 15.

Convert LHS to base (6/14): (14/6)¹⁵ = (6/14)⁻¹⁵.

Equate exponents: 9m + 9 = −15 ⇒ 9m = −24 ⇒ m = −8/3.

∴ The correct answer is −8/3.

Alternate Method

Given:

Equation: (14/6)⁸ × (6/14)³ × (14/6)¹⁰ = (6/14)^9m+9

Formula Used:

1. Multiplication Law: a^x × a^y = a^x+y
2. Negative Exponent Law: (a/b)ⁿ = (b/a)⁻ⁿ
3. Equality of Bases: If a^x = a^y, then x = y

Calculations:

⇒ (14/6)⁸ × (14/6)⁻³ × (14/6)¹⁰ = (6/14)^9m+9

⇒ (14/6)^{(8 − 3 + 10)} = (6/14)^9m+9

⇒ (14/6)¹⁵ = (6/14)^9m+9

⇒ (6/14)⁻¹⁵ = (6/14)^9m+9

⇒ −15 = 9m + 9

⇒ 9m = −15 − 9

⇒ 9m = −24

⇒ m = −24/9 = −8/3

∴ The correct answer is −8/3.

Additional Information

Product of Powers Rule

When multiplying two powers with the same base, add their exponents: a^m × aⁿ = a^m+n.

Negative Exponent Rule

To change the sign of an exponent, take the reciprocal of the base: (a/b)⁻ⁿ = (b/a)ⁿ.

Property of Equality for Exponential Equations

If the bases on both sides of an equation are equal, the exponents must also be equal, provided the base is not 0, 1, or −1.

Shortcut Trick

Extra runs over average = 85 − 55 = 30

Distributed increase per match = 30 ÷ 10 = 3

New average = 55 + 3 = 58

∴ The correct answer is 58.

Alternate Method

Given: Average of 9 matches = 55, 10^th match score = 85

Formula: Total = Average × Number of items

⇒ Total runs in 9 matches = 55 × 9 = 495

⇒ Total runs in 10 matches = 495 + 85 = 580

⇒ New average = 580 ÷ 10 = 58

∴ The correct answer is 58.

Additional Information

Net Deviation Concept

The sum of deviations of all observations from the average is always zero. This is used to find the effect of adding or removing an item from the data set.

Weighted Average Formula

Combined Average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where n is the quantity and A is the average of each group.

Shortcut Trick

When the initial values are equal, the overall percentage change is simply the average of the individual percentage changes.

Mathematics change = +20% | Physics change = −10%

Overall % Change = (+20% − 10%) ÷ 2

Overall % Change = 10% ÷ 2 = 5% increase.

∴ The correct answer is 5% increase.

Alternate Method

Given: Initial marks in Mathematics = Initial marks in Physics.

⇒ Let initial marks in Mathematics = 100 and Physics = 100.

⇒ Initial Total Marks = 100 + 100 = 200.

⇒ New Mathematics marks = 100 + (20% of 100) = 100 + 20 = 120.

⇒ New Physics marks = 100 − (10% of 100) = 100 − 10 = 90.

⇒ New Total Marks = 120 + 90 = 210.

⇒ Net change in total marks = 210 − 200 = 10.

⇒ Percentage Change = (Change ÷ Initial Total) × 100

⇒ Percentage Change = (10 ÷ 200) × 100 = 5%.

∴ The correct answer is 5% increase.

Additional Information

Percentage Change Formula

The percentage change is calculated as [(Final Value − Initial Value) ÷ Initial Value] × 100.

Weighted Average Percentage

If components have different weights w₁ and w₂, the net change is (w₁×p₁ + w₂×p₂) ÷ (w₁ + w₂).

Successive Percentage Change

When a value changes by a% and then b%, the net effective change is [a + b + (a×b)/100]%.