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Shortcut Trick

If third proportional of a and b is c, then c = b² ÷ a.

TP₁ (36, 48) = (48 × 48) ÷ 36 = 64.

TP₂ (49, 64) = (64 × 64) ÷ 49.

Ratio = 64 : (64 × 64 ÷ 49) = 1 : (64 ÷ 49) = 49 : 64.

∴ The correct answer is 49:64.

Alternate Method

Given:

First pair: a₁ = 36, b₁ = 48

Second pair: a₂ = 49, b₂ = 64

Formula Used:

For two numbers a and b, the third proportional x is calculated as: a : b = b : x ⇒ x = b² ÷ a.

Calculations:

⇒ Let x be the third proportional of 36 and 48.

⇒ x = (48 × 48) ÷ 36

⇒ x = 2304 ÷ 36 = 64

⇒ Let y be the third proportional of 49 and 64.

⇒ y = (64 × 64) ÷ 49 = 4096 ÷ 49

⇒ Required Ratio = x : y = 64 : (4096 ÷ 49)

⇒ Ratio = (64 × 49) : 4096

⇒ Ratio = 3136 : 4096

⇒ Simplifying by dividing both terms by 64: (3136 ÷ 64) : (4096 ÷ 64)

⇒ Ratio = 49 : 64

⇒ This result corresponds to the value given in Option (3).

∴ The correct answer is 49:64.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is √(a × b). It is the value x such that a : x = x : b.

Fourth Proportional

For four numbers a, b, c, and d in proportion (a : b = c : d), d is the fourth proportional, calculated as d = (b × c) ÷ a.

Ratio and Proportion Properties

If a/b = c/d, then by Invertendo b/a = d/c, and by Alternando a/c = b/d.

Shortcut Trick

Difference in ratio units between Bala and Amit = 7 − 5 = 2 units.

Given that 2 units = ₹3,600 ⇒ 1 unit = ₹1,800.

Chetan total savings = 9 units = 9 × 1,800 = ₹16,200.

∴ The correct answer is ₹16,200.

Alternate Method

Given: Savings ratio of Amit, Bala, and Chetan = 5 : 7 : 9. Bala savings − Amit savings = ₹3,600.

⇒ Let the savings of Amit, Bala, and Chetan be 5x, 7x, and 9x respectively. ⇒ According to the question, Bala savings − Amit savings = 3600. ⇒ 7x − 5x = 3600. ⇒ 2x = 3600. ⇒ x = 3600 ÷ 2 = 1800. ⇒ Chetan total savings = 9x. ⇒ Chetan total savings = 9 × 1800. ⇒ Total = ₹16,200.

∴ The correct answer is ₹16,200.

Additional Information

Ratio and Proportion

A ratio represents how many times one number contains another. For a ratio a:b:c, actual values can be expressed as ax, bx, and cx.

Unitary Method

This method involves finding the value of one unit by dividing a known quantity by its ratio parts, then multiplying by the desired ratio parts.

Compound Ratio

When two or more ratios are multiplied term by term (a:b and c:d), the resulting ratio (ac:bd) is called the compound ratio.

Shortcut Trick

Assume all 8 notebooks are of the first type.

The extra cost for the 5 Type 2 notebooks is 5 × ₹20 = ₹100.

Subtract this excess from the total: ₹420 − ₹100 = ₹320.

Cost of one Type 1 notebook = 320 ÷ 8 = ₹40.

∴ The correct answer is ₹40.

Alternate Method

Given:

Quantity of Type 1 = 3 | Quantity of Type 2 = 5

Total Cost of 3 Type 1 and 5 Type 2 = ₹420

Price Difference: Type 2 = Type 1 + ₹20

Formula Used:

Total Price = (n₁ × p₁) + (n₂ × p₂)

Let the cost of one Type 1 notebook be x.

⇒ Cost of one Type 2 notebook = x + 20

⇒ 3 × x + 5 × (x + 20) = 420

⇒ 3x + 5x + 100 = 420

⇒ 8x = 420 − 100

⇒ 8x = 320

⇒ x = 320 ÷ 8

⇒ x = 40

∴ The correct answer is ₹40.

Additional Information

Linear Equation in One Variable

An equation which can be written in the form ax + b = 0, where a and b are real numbers and a ≠ 0.

Substitution Method

A strategy for solving systems of equations by expressing one variable in terms of the other to reduce the number of variables.

Price-Quantity Relation

Total Expenditure is always the product of the Price per unit and the Quantity purchased (E = P × Q).

Shortcut Trick

For two successive discounts of x% and y%, the net percentage reduction is given by the formula: (x + y − (x × y) / 100)%.

⇒ Net Discount = (10 + 20 − (10 × 20) / 100)%

⇒ Net Discount = (30 − 2)%

⇒ Net Discount = 28%

∴ The correct answer is 28%.

Alternate Method

Given:

First successive discount = 10%

Second successive discount = 20%

Let the Marked Price (MP) of the item be 100.

⇒ Price after first discount (10%) = 100 − (10% of 100)

⇒ Price after first discount = 100 − 10 = 90

⇒ Price after second discount (20%) = 90 − (20% of 90)

⇒ Price after second discount = 90 − (0.20 × 90) = 90 − 18

⇒ Final Selling Price (SP) = 72

⇒ Net percentage reduction = ((MP − Final SP) / MP) × 100

⇒ Net reduction = ((100 − 72) / 100) × 100 = 28%

∴ The correct answer is 28%.

Additional Information

Three Successive Discounts

For three successive discounts of a%, b%, and c%, first find the net discount of a and b, then use that result with c to find the total net discount.

Successive Percentage Change Formula

The effective change for successive variations x% and y% is (x + y + (xy/100))%. Use positive values for increase and negative for decrease/discount.

Multiplier Method

Final Value = Initial Value × (1 − d₁/100) × (1 − d₂/100). For 10% and 20% discounts, the multiplier is 0.9 × 0.8 = 0.72, implying a 28% reduction.

Shortcut Trick

Tap 2 fills the cistern in 13 hr and Tap 3 empties it in 13 hr, so their efficiencies cancel out.

The net filling is effectively done by Tap 1 alone at its original rate.

Tap 1 fills the entire cistern in 4 hr.

Time required to fill one-fourth of the cistern = Total time ÷ 4 = 4 hr ÷ 4 = 1 hr.

∴ The correct answer is 1 hr.

Alternate Method

Given:

Time taken by Tap A to fill = 4 hr

Time taken by Tap B to fill = 13 hr

Time taken by Tap C to empty = 13 hr

Formula Used:

Efficiency = Total Work ÷ Time Taken

Calculations:

⇒ Total capacity of the cistern = LCM(4, 13, 13) = 52 units

⇒ Net efficiency when all taps are open = 13 + 4 − 4 = 13 units/hr

⇒ Target to fill one-fourth of the cistern = 52 ÷ 4 = 13 units

⇒ Time taken = Target Work ÷ Net Efficiency

⇒ Time taken = 13 ÷ 13 = 1 hr

∴ The correct answer is 1 hr.

Additional Information

Negative Efficiency

In pipe and cistern problems, the efficiency of an outlet pipe (or emptying tap) is always treated as negative because it reduces the total work done.

Net Filling Rate

When multiple pipes work together, the net rate of filling is the algebraic sum of their individual rates: Net Rate = ∑(Inlet Rates) − ∑(Outlet Rates).

Total Capacity via LCM

Assume the total capacity of the tank/cistern as the Least Common Multiple (LCM) of the individual times taken by each pipe to simplify efficiency calculations.

Given:

The exponential equation is 6^x − 6^x−1 = 30

Formula used:

a^m−n = a^m ÷ aⁿ

Calculations:

6^x − 6^x−1 = 30

⇒ 6^x − (6^x ÷ 6¹) = 30

⇒ 6^x × (1 − 1/6) = 30

⇒ 6^x × (5/6) = 30

⇒ 6^x = (30 × 6) ÷ 5

⇒ 6^x = 6 × 6

⇒ 6^x = 36

⇒ 6^x = 6²

⇒ x = 2

To find the value of (2x − 1) ÷ (2x + 3), we substitute x = 2 in the expression:

⇒ (2 × 2 − 1) ÷ (2 × 2 + 3)

⇒ (4 − 1) ÷ (4 + 3)

⇒ 3/7

∴ The correct answer is option (4).

Shortcut Trick

To check divisibility by 11, calculate the difference between the sum of digits at odd places and the sum of digits at even places.

For 978626: Sum of digits at odd places (1^st, 3^rd, 5^th) = 9 + 8 + 2 = 19.

Sum of digits at even places (2^nd, 4^th, 6^th) = 7 + 6 + 6 = 19.

Difference = 19 − 19 = 0. If the difference is 0 or a multiple of 11, the number is divisible by 11.

∴ The correct answer is 978626.

Alternate Method

Given: Options (1) 869756, (2) 978626, (3) 447255, (4) 112044.

Formula: A number is divisible by 11 if (Sum of digits at odd places) − (Sum of digits at even places) = 0, 11, 22, ...

Let us check each option systematically:

⇒ For 869756: (8 + 9 + 5) − (6 + 7 + 6) = 22 − 19 = 3 (Not divisible by 11)

⇒ For 978626: (9 + 8 + 2) − (7 + 6 + 6) = 19 − 19 = 0 (Divisible by 11)

⇒ For 447255: (4 + 7 + 5) − (4 + 2 + 5) = 16 − 11 = 5 (Not divisible by 11)

⇒ For 112044: (1 + 2 + 4) − (1 + 0 + 4) = 7 − 5 = 2 (Not divisible by 11)

∴ The correct answer is 978626.

Additional Information

Divisibility Rule of 11

A number is divisible by 11 if the alternating sum of its digits (d₁ − d₂ + d₃ − d₄ ...) is a multiple of 11 (including 0).

Divisibility Rule of 9

A number is divisible by 9 if the sum of all its digits is divisible by 9.

Divisibility Rule of 3

A number is divisible by 3 if the sum of its digits is a multiple of 3.

Divisibility Rule of 7

Double the last digit and subtract it from the rest of the number; if the result is divisible by 7, the original number is also divisible by 7.

Shortcut Trick

Both numbers must be divisible by their HCF (15), which eliminates options 2 and 4.

The Product of two numbers must equal HCF × LCM = 15 × 225 = 3375.

Testing Option 1: 45 × 75 = 3375 (Matches).

Testing Option 3: 15 × 75 = 1125 (Incorrect).

∴ The correct answer is 45 and 75.

Alternate Method

Given: HCF = 15 and LCM = 225.

Formula Used: Product of two numbers = HCF × LCM

⇒ Let the numbers be 15a and 15b, where a and b are co-prime.

⇒ 15a × 15b = 15 × 225

⇒ 225 × ab = 3375

⇒ ab = 3375 ÷ 225 = 15

⇒ Possible co-prime pairs for (a, b) are (1, 15) and (3, 5).

⇒ Case 1: (15 × 1, 15 × 15) = (15, 225)

⇒ Case 2: (15 × 3, 15 × 5) = (45, 75)

⇒ Comparing with the given options, the pair is 45 and 75.

∴ The correct answer is 45 and 75.

Additional Information

HCF and LCM Relation

For any two positive integers a and b, the relationship is always HCF(a, b) × LCM(a, b) = a × b.

Divisibility Property

The HCF of any group of numbers is always a factor of their LCM.

Co-prime Factorization

If HCF = h, the numbers can be expressed as hx and hy, where LCM = hxy and x, y are co-prime integers.

Shortcut Trick

Assume the base average is 24.

Calculate deviations: (18 − 24) = −6, (20 − 24) = −4, (25 − 24) = +1, (27 − 24) = +3.

Net deviation of 4 numbers = −6 − 4 + 1 + 3 = −6.

To maintain the average, the fifth number must balance this: 24 − (−6) = 30.

∴ The correct answer is 30.

Alternate Method

Given: Average of 5 numbers = 24, Four numbers are 18, 20, 25, and 27.

Formula: Sum = Average × Count of numbers

⇒ Total sum of 5 numbers = 24 × 5 = 120

⇒ Sum of 4 given numbers = 18 + 20 + 25 + 27 = 90

⇒ Let the fifth number be x.

⇒ 90 + x = 120

⇒ x = 120 − 90

⇒ x = 30

∴ The correct answer is 30.

Additional Information

Weighted Average

When two groups with averages A₁ and A₂ and sizes n₁ and n₂ are combined, the combined average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Arithmetic Progression Average

For numbers in an AP (equal gaps), the average is simply (First term + Last term) ÷ 2 or the middle term.

Effect of Constant Change

If each number in a set is increased, decreased, multiplied, or divided by a constant k, the new average also changes by the same constant k.

Shortcut Trick

Given: 1.2x = 0.5y ⇒ x/y = 0.5/1.2 = 5/12

Let x = 5 and y = 12

Required value = (x + y) / (x − y) = (5 + 12) / (5 − 12) = 17 / (−7)

∴ The correct answer is −17/7.

Alternate Method

Given: 1.2x = 0.5y

Formula Used: Ratio and Proportion properties.

⇒ 1.2x = 0.5y

⇒ x / y = 0.5 / 1.2

⇒ x / y = 5 / 12

⇒ Let x = 5k and y = 12k

⇒ (x + y) / (x − y) = (5k + 12k) / (5k − 12k)

⇒ (x + y) / (x − y) = 17k / (−7k)

⇒ (x + y) / (x − y) = −17 / 7

∴ The correct answer is −17/7.

Additional Information

Componendo and Dividendo

If a/b = c/d, then (a + b)/(a − b) = (c + d)/(c − d). This is useful for solving fractional equations quickly.

Ratio Substitution

In homogeneous expressions like (x+y)/(x-y), the actual values of x and y are not needed; only their ratio is sufficient to find the result.

Decimal to Fraction

Always convert decimals to simple fractions (e.g., 0.5 = 1/2, 1.2 = 6/5) to avoid calculation errors in ratios.