Quantitative Aptitude PYQs

Shortcut Trick

Apply Product Law: a = 2 + 3 = 5

Apply Quotient Law: 2 − b = 1 ⇒ b = 1

Required sum a + b = 5 + 1 = 6

∴ The correct answer is 6.

Alternate Method

Given:

Equation 1: x² × x³ = x^a

Equation 2: x² ÷ x^b = x¹

Method Used:

Using Laws of Exponents for the same base x:

1. Product Rule: x^m × xⁿ = x^m+n
2. Quotient Rule: x^m ÷ xⁿ = x^m−n

Calculations:

From Equation 1:

⇒ x²⁺³ = x^a

⇒ x⁵ = x^a

⇒ a = 5

From Equation 2:

⇒ x^2−b = x¹

⇒ 2 − b = 1

⇒ b = 2 − 1 = 1

Final calculation for a + b:

⇒ a + b = 5 + 1 = 6

∴ The correct answer is 6.

Additional Information

Power of a Power Rule

When a base with an exponent is raised to another power, multiply the exponents: (x^m)ⁿ = x^{m × n}.

Zero Exponent Property

Any non-zero real number raised to the power of zero is always 1: x⁰ = 1.

Negative Exponent Property

A negative exponent indicates the reciprocal of the base: x⁻ⁿ = 1 ÷ xⁿ.

Shortcut Trick

Use the standard algebraic relation: (Difference)² = (Sum)² − 4 × (Product)

⇒ (Difference)² = 35² − 4 × 286

⇒ (Difference)² = 1225 − 1144 = 81

⇒ Difference = √81 = 9

∴ The correct answer is 9.

Alternate Method

Given: Sum of two numbers (x + y) = 35 and their Product (xy) = 286.

Formula Used: (x − y)² = (x + y)² − 4xy

⇒ Let the two positive numbers be x and y.

⇒ x + y = 35

⇒ xy = 286

⇒ (x − y)² = (35)² − 4(286)

⇒ (x − y)² = 1225 − 1144

⇒ (x − y)² = 81

⇒ x − y = √81 = 9

∴ The correct answer is 9.

Additional Information

Quadratic Form Relation

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

Basic Algebraic Identities

(a + b)² = a² + b² + 2ab and (a − b)² = a² + b² − 2ab. Subtracting these gives (a + b)² − (a − b)² = 4ab.

Factorization Method

For xy = 286, find factors such that they add to 35. Factors of 286 are 1, 2, 11, 13, 22, 26, 143, 286. Here 22 + 13 = 35.

Shortcut Trick

Use the effective percentage change formula: Net % = x + y + (x × y)/100.

Combine Mark-up (+40%) and first Discount (−18%): 40 − 18 − (40 × 18)/100 = 14.8%.

Combine result (+14.8%) and second Discount (−5%): 14.8 − 5 − (14.8 × 5)/100 = 9.8 − 0.74 = 9.06%.

Since the net result is positive, it represents a Profit Percentage.

∴ The correct answer is 9.06%.

Alternate Method

Given:

Mark-up Percentage = 40%

Successive Discounts = 18% and 5%

Formula Used:

Marked Price (MP) = CP × (1 + Mark-up/100)

Selling Price (SP) = MP × (1 − d₁/100) × (1 − d₂/100)

Profit % = [(SP − CP) / CP] × 100

⇒ Let Cost Price (CP) = 100

⇒ Marked Price (MP) = 100 + 40% of 100 = 140

⇒ Price after 18% discount = 140 × (100 − 18)/100 = 140 × 0.82

⇒ Price after 18% discount = 114.8

⇒ Price after 5% discount (SP) = 114.8 × (100 − 5)/100 = 114.8 × 0.95

⇒ Selling Price (SP) = 109.06

⇒ Profit = SP − CP = 109.06 − 100 = 9.06

⇒ Profit % = (9.06 / 100) × 100 = 9.06%

∴ The correct answer is 9.06%.

Additional Information

Successive Discount Formula

Total discount for two rates d₁ and d₂ is given by: d_net = d₁ + d₂ − (d₁ × d₂)/100.

Profit and Discount Relation

The relationship between CP and MP is: MP/CP = (100 + Profit%)/(100 − Discount%).

Markup and Profit

Mark-up is the value added to CP to reach MP. Profit is the final gain after all discounts on MP.

Shortcut Trick

Calculate the marks corresponding to 16% of the total.

16% of Maximum Marks = 448 + 64 = 512 marks.

1% of Maximum Marks = 512 ÷ 16 = 32 marks.

Friend's Percentage = 672 ÷ 32 = 21%.

∴ The correct answer is 21%.

Alternate Method

Given:

Marks scored by Girish = 448

Shortfall to reach 16% = 64

Marks scored by Friend = 672

Formula Used:

Percentage = (Obtained Marks ÷ Maximum Marks) × 100

Calculations:

⇒ 16% of Maximum Marks = 448 + 64

⇒ 16% of Maximum Marks = 512

⇒ Maximum Marks = (512 ÷ 16) × 100

⇒ Maximum Marks = 32 × 100 = 3200

⇒ Friend's Percentage = (672 ÷ 3200) × 100

⇒ Friend's Percentage = 672 ÷ 32

⇒ Friend's Percentage = 21%

∴ The correct answer is 21%.

Additional Information

Maximum Marks and Base Value

The Maximum Marks acts as the base (100%) for all percentage calculations in an examination. Total Marks = (Score ÷ Percentage) × 100.

Percentage Shortfall

Shortfall is the difference between a target score and the actual score. Target Marks = Actual Marks + Shortfall.

Passing Marks Criteria

Many exam problems define Passing Marks as a specific percentage of the total. If a student fails by x marks, then Passing Marks = Student's Marks + x.

Shortcut Trick

Initial Speed of the bus = 90 ÷ 3 = 30 kmph

Increased Speed = 30 + 5 = 35 kmph

Required Time = 70 ÷ 35 = 2 hours

∴ The correct answer is 2 hr.

Alternate Method

Given:

Initial Distance (D₁) = 90 km

Initial Time (T₁) = 3 hr

Speed Increase = 5 kmph

New Distance (D₂) = 70 km

Formula Used:

Speed = Distance ÷ Time

Time = Distance ÷ Speed

Calculations:

⇒ Initial Speed = D₁ ÷ T₁

⇒ Initial Speed = 90 ÷ 3 = 30 kmph

⇒ New Speed = 30 + 5 = 35 kmph

⇒ New Time = D₂ ÷ New Speed

⇒ New Time = 70 ÷ 35

⇒ New Time = 2 hr

∴ The correct answer is 2 hr.

Additional Information

Unit Conversion

To convert speed from kmph to m/s, multiply by 5/18. To convert m/s to kmph, multiply by 18/5.

Average Speed

Average Speed = Total Distance ÷ Total Time. If distances are equal, Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Relative Speed

Objects in opposite direction: Relative Speed = S₁ + S₂. Objects in same direction: Relative Speed = |S₁ − S₂|.

Shortcut Trick

Rate 12.5% = 1/8 in fraction.

Ratio of Principal (P) to Amount (A) for 1 year = 8 : 9.

For 2 years, Ratio P : A = 8² : 9² = 64 : 81.

Compound Interest (CI) = 81 − 64 = 17 units.

64 units = ₹8,000 ⇒ 1 unit = 8,000 ÷ 64 = 125.

CI = 17 × 125 = ₹2,125.

∴ The correct answer is ₹2,125.

Alternate Method

Given:

Principal (P) = ₹8,000

Rate (R) = 12.5% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P × (1 + R/100)ⁿ

Compound Interest (CI) = Amount − Principal

Calculations:

⇒ A = 8,000 × (1 + 12.5/100)²

⇒ A = 8,000 × (1 + 1/8)²

⇒ A = 8,000 × (9/8) × (9/8)

⇒ A = 125 × 81

⇒ A = ₹10,125

⇒ CI = 10,125 − 8,000

⇒ CI = ₹2,125

∴ The correct answer is ₹2,125.

Additional Information

Effective Interest Rate

For two years at rate R%, the effective compound interest rate is R + R + (R × R)/100.

Difference between CI and SI

For 2 years, CI − SI = P × (R/100)².

Compounding Periods

If compounded half-yearly, the rate becomes R/2 and the time period becomes 2n.

Shortcut Trick

Use the direct formula for percentage reduction to restore the original value: [r / (100 + r)] × 100.

Here, r = 25.

Reduction = [25 / (100 + 25)] × 100

Reduction = (25 / 125) × 100 = (1 / 5) × 100 = 20%.

∴ The correct answer is 20%.

Alternate Method

Given:

⇒ Increase in price = 25%

Formula Used:

⇒ Percentage Reduction = (Difference / Increased Price) × 100

Calculations:

⇒ Let the original price of the article be ₹100.

⇒ Increased price = 100 + 25% of 100 = 100 + 25 = ₹125.

⇒ To return to the original price (₹100), the reduction required = 125 − 100 = ₹25.

⇒ Required percentage reduction = (25 / 125) × 100

⇒ Required percentage reduction = (1 / 5) × 100

⇒ Required percentage reduction = 20%

∴ The correct answer is 20%.

Additional Information

Base Change Rule

If quantity A is r% more than B, then B is [r / (100 + r)] × 100% less than A. If A is r% less than B, then B is [r / (100 − r)] × 100% more than A.

Fractional Equivalents

25% is equivalent to 1/4. When a value increases by 1/n, it must be decreased by 1/(n+1) to return to its original value. Here, n=4, so reduction = 1/(4+1) = 1/5 = 20%.

Net Percentage Change

For two successive percentage changes a% and b%, the effective change is [a + b + (ab / 100)]%. To result in zero net change, the formula must equate to 0.

Shortcut Trick

Combine the decimal parts first: 5.25 + 2.75 = 8.

Simplify the product of fractions: (2/5) × (10/20) = (2/5) × (1/2) = 1/5 = 0.2.

Convert the remaining fraction to decimal: 3/5 = 0.6.

Perform the final subtraction: 8 − 0.6 − 0.2 = 7.2.

∴ The correct answer is 7.2.

Alternate Method

Given: 5.25 + 2.75 − (3/5) − (2/5) × (10/20)

Method: Solve using the BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction).

⇒ 5.25 + 2.75 − 0.6 − (2/5 × 1/2)

⇒ 8.00 − 0.6 − (1/5)

⇒ 8 − 0.6 − 0.2

⇒ 8 − 0.8

⇒ 7.2

∴ The correct answer is 7.2.

Additional Information

BODMAS Rule

It defines the order of operations: Brackets, Of (orders), Division, Multiplication, Addition, and Subtraction.

Fraction to Decimal Conversion

To convert a fraction to a decimal, divide the numerator by the denominator (e.g., 3 ÷ 5 = 0.6).

Simplification Shortcut

Always look for common factors in fractions before multiplying to reduce the complexity of the numbers.

Shortcut Trick

Convert the given trigonometric expression into tangent and secant forms: 15 + 12 tan² A − 12 sec² A.

Factor out the common term 12: 15 + 12(tan² A − sec² A).

Apply the Pythagorean identity: Since sec² A − tan² A = 1, then tan² A − sec² A = −1.

The expression simplifies to: 15 + 12(−1) = 15 − 12 = 3.

∴ The correct answer is 3.

Alternate Method

Given: 15 + 12/cot² A − 12/cos² A

Formula Used: 1/cot A = tan A and 1/cos A = sec A; sec² A − tan² A = 1

⇒ Expression = 15 + 12 × (1/cot² A) − 12 × (1/cos² A)

⇒ Expression = 15 + 12 tan² A − 12 sec² A

⇒ Expression = 15 + 12(tan² A − sec² A)

⇒ From the identity sec² A − tan² A = 1, we know tan² A − sec² A = −1

⇒ Expression = 15 + 12(−1)

⇒ Expression = 15 − 12 = 3

∴ The correct answer is 3.

Additional Information

Pythagorean Identities

There are three fundamental identities: sin² A + cos² A = 1, 1 + tan² A = sec² A, and 1 + cot² A = cosec² A.

Reciprocal Relations

Trigonometric ratios are related as follows: tan A = 1/cot A, sec A = 1/cos A, and cosec A = 1/sin A.

Simplification Rule

When simplifying trigonometric expressions, always look to convert terms into sine/cosine or use established identities to reduce variables.

Shortcut Trick

Find the LCM of divisors 6, 14, and 7, which is 42.

The largest 4-digit multiple of 42 is 9996 (calculated as 9999 − 3).

Adding the remainder: 9996 + 5 = 10001 (This is a 5-digit number).

Subtract the LCM to find the previous valid number: 10001 − 42 = 9959.

∴ The correct answer is 9959.

Alternate Method

Given:

Divisors: 6, 14, 7

Remainder: 5

Target: Largest 4-digit number

Formula Used:

Required Number = [LCM(Divisors) × k] + Remainder

Calculations:

⇒ LCM(6, 14, 7) = 42

⇒ Largest 4-digit number = 9999

⇒ 9999 ÷ 42 = 238 (Quotient), 3 (Remainder)

⇒ Largest 4-digit multiple of 42 = 9999 − 3 = 9996

⇒ Check 4-digit constraint: 9996 + 5 = 10001 (Not a 4-digit number)

⇒ Previous multiple of 42 = 9996 − 42 = 9954

⇒ Required number = 9954 + 5 = 9959

∴ The correct answer is 9959.

Additional Information

LCM and Remainder Concept

When a number N is divided by a, b, c and leaves the same remainder 'r' in each case, then N = LCM(a, b, c) × k + r, where k is an integer.

Divisibility by 14

A number is divisible by 14 if it is simultaneously divisible by its co-prime factors 2 and 7.

Successive Division vs Single Remainder

If remainders are different (r₁, r₂, r₃), and the difference (divisor − remainder) is constant 'd', then the number N = LCM(divisors) × k − d.

Shortcut Trick

Divide the parts into ratios based on percentage: 50% : 20% : 30% → 5 : 2 : 3 (Total = 10 units).

Apply net percentage change formula: Net % = [5 × 21 + 2 × 36 + 3 × (−29)] / 10.

Net % = (105 + 72 − 87) / 10 = 90 / 10 = 9%.

Since the overall profit is x%, the value of x is 9.

∴ The correct answer is 9.

Alternate Method

Let the total quantity of goods be 100 units and the cost price per unit be ₹1.

Given:

First part: 50% of 100 = 50 units sold at 21% profit.

Second part: 40% of remaining (100 − 50 = 50) = 20 units sold at 36% profit.

Third part: Remaining (50 − 20) = 30 units sold at 29% loss.

Calculations:

⇒ Profit on 1^st part = 50 × 21/100 = ₹10.5

⇒ Profit on 2^nd part = 20 × 36/100 = ₹7.2

⇒ Loss on 3^rd part = 30 × 29/100 = ₹8.7

⇒ Total Net Profit = 10.5 + 7.2 − 8.7 = ₹9

⇒ Overall Profit % = (Total Profit / Total CP) × 100

⇒ x = (9 / 100) × 100 = 9

∴ The correct answer is 9.

Additional Information

Weighted Average Profit %

When goods are sold in parts with different profit/loss rates, the overall profit % = (w₁p₁ + w₂p₂ + ... + w_np_n) / (w₁ + w₂ + ... + w_n), where w represents weight/quantity and p represents profit/loss rate.

Successive Percentage Change

For two successive percentage changes a% and b%, the effective change is [a + b + (ab/100)]%. This is useful when the price is changed multiple times on the same base.

Basic Profit and Loss

Profit % = (Profit / CP) × 100 and Loss % = (Loss / CP) × 100. Always ensure the denominator is the Cost Price (CP) unless specified otherwise.

Shortcut Trick

Sum of Length and Breadth = Half of Perimeter = 56 ÷ 2 = 28 cm.

Let Breadth be x, then Length = 2x + 4.

⇒ (2x + 4) + x = 28

⇒ 3x = 24 ⇒ x = 8 cm.

∴ The correct answer is 8 cm.

Alternate Method

Given:

⇒ Perimeter of Rectangle = 56 cm

⇒ Length (L) = 4 + 2 × Breadth (B)

Formula Used:

⇒ Perimeter = 2 × (Length + Breadth)

Calculations:

⇒ 2 × (L + B) = 56

⇒ L + B = 56 ÷ 2

⇒ L + B = 28

Substitute L = 2B + 4 in the equation:

⇒ (2B + 4) + B = 28

⇒ 3B + 4 = 28

⇒ 3B = 28 − 4

⇒ 3B = 24

⇒ B = 24 ÷ 3

⇒ B = 8 cm

∴ The correct answer is 8 cm.

Additional Information

Area of Rectangle

The total space occupied by the rectangle is given by the product of its dimensions: Area = Length × Breadth.

Diagonal of Rectangle

The distance between opposite corners can be found using the Pythagorean theorem: Diagonal = √(Length² + Breadth²).

Perimeter of a Square

A square is a special rectangle where Length = Breadth = s. Its perimeter is 4 × s.

Shortcut Trick

Simplify using BODMAS in segments:

Segment 1: (14/24 ÷ 14/9) = (14/24 × 9/14) = 3/8

Segment 2: (3/7 × 14/9 + 5/8) = (2/3 + 5/8) = 31/24

Segment 3: (9/10) ÷ (31/24 of 24/10) = (9/10) ÷ (31/10) = 9/31

Calculation: (3/8 ÷ 31/24) + 9/31 = (3/8 × 24/31) + 9/31 = 9/31 + 9/31 = 18/31

∴ The correct answer is 18/31.

Alternate Method

Given: { (14/24) ÷ (14/9) } ÷ (3/7 × 14/9 + 5/8) + 9/10 ÷ 31/24 of 24/10

Simplify using the BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction):

⇒ Solve the first set of brackets: (14/24) × (9/14) = 9/24 = 3/8

⇒ Solve the inner multiplication in the second bracket: (3/7 × 14/9) = (1 × 2) / (1 × 3) = 2/3

⇒ Complete the second bracket by addition: 2/3 + 5/8 = (16 + 15) / 24 = 31/24

⇒ Solve the 'of' operator: 31/24 of 24/10 = 31/24 × 24/10 = 31/10

⇒ Rewrite the expression: (3/8 ÷ 31/24) + (9/10 ÷ 31/10)

⇒ Apply division by multiplying by the reciprocal: (3/8 × 24/31) + (9/10 × 10/31)

⇒ 9/31 + 9/31 = 18/31

∴ The correct answer is 18/31.

Additional Information

BODMAS Rule

It defines the order of operations as Brackets, Of, Division, Multiplication, Addition, and Subtraction.

'Of' Operator Precedence

The 'of' operator is treated as a multiplication but must be performed before any division or standard multiplication in the expression.

Division of Fractions

To divide one fraction by another, multiply the first fraction by the reciprocal of the second (a/b ÷ c/d = a/b × d/c).

Shortcut Trick

Scheme A: Net Discount = 5 + 30 − (5 × 30)/100 = 33.5%

Scheme B: Net Discount ≈ 18 + 15 + 5 = 38% (approx, but effectively around 33.8%)

Scheme C: For exactly 5 toys, only the 11% slab applies ⇒ 11%

Scheme D: Buy 5 Get 2 Free ⇒ Discount = (2 / 7) × 100 ≈ 28.57%

Comparing all, Scheme C (11%) provides the minimum discount, making it the least beneficial.

∴ The correct answer is Scheme C.

Alternate Method

Let the Marked Price (MP) of each toy be ₹100. For 5 toys, Total MP = ₹500.

Scheme A (Successive 5% and 30%):

⇒ Price after 5% = 500 − 25 = ₹475

⇒ Price after 30% = 475 − 142.5 = ₹332.5

⇒ Total Discount = 500 − 332.5 = ₹167.5 (33.5%)

Scheme B (Successive 18%, 15%, 5%):

⇒ Net Discount % = 100 − [100 × (0.82 × 0.85 × 0.95)]

⇒ Net Discount % = 100 − 66.215 = 33.785%

Scheme C (11% on first 5 toys):

⇒ Since the customer buys exactly 5 toys, the discount is strictly 11%.

⇒ Total Discount = 11% of 500 = ₹55

Scheme D (2 toys free on buying 5):

⇒ Customer pays for 5 toys and receives 7 toys.

⇒ Effective Discount % = (Free Items ÷ Total Items) × 100

⇒ Discount % = (2 ÷ 7) × 100 ≈ 28.57%

Comparing the effective discount percentages, Scheme C offers the lowest value (11%), which means the customer pays the highest price under this scheme.

∴ The correct answer is Scheme C.

Additional Information

Successive Discount Formula

For two successive discounts d₁% and d₂%, the single equivalent discount is (d₁ + d₂ − d₁d₂/100)%.

Buy X Get Y Free

The effective discount percentage in such schemes is calculated as [Y ÷ (X + Y)] × 100.

Least Beneficial Scheme

For a customer, the least beneficial scheme is the one that offers the lowest net discount or results in the highest Selling Price (SP).

Shortcut Trick

Given Remainder = 15.

Divisor = 4 × 15 = 60.

Quotient = 60 ÷ 5 = 12.

Dividend = (60 × 12) + 15 = 720 + 15 = 735.

∴ The correct answer is 735.

Alternate Method

Given:

In a division sum:

Remainder (R) = 15

Divisor (D) = 4 × R

Divisor (D) = 5 × Quotient (Q)

Formula Used:

Dividend = (Divisor × Quotient) + Remainder

Calculations:

⇒ Divisor = 4 × 15 = 60

⇒ 60 = 5 × Quotient

⇒ Quotient = 60 ÷ 5 = 12

⇒ Dividend = (60 × 12) + 15

⇒ Dividend = 720 + 15

⇒ Dividend = 735

∴ The correct answer is 735.

Additional Information

Division Algorithm

For any two integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

Properties of Remainder

The remainder is always a non-negative integer and is strictly less than the divisor. If the remainder is 0, the dividend is perfectly divisible by the divisor.

Verification Rule

To verify a division result, subtract the remainder from the dividend; the result must be the product of the divisor and the quotient.

Shortcut Trick

Effective Compound Interest rate for 2 years = 5 + 5 + (5 × 5)/100 = 10.25%

Given CI = 10.25% of Principal = ₹102.50

Principal = (102.50 ÷ 10.25) × 100

Principal = 10 × 100 = ₹1,000

∴ The correct answer is ₹1,000.

Alternate Method

Given:

Time (n) = 2 years, Rate (R) = 5% per annum, Compound Interest (CI) = ₹102.50

Formula Used:

CI = P[(1 + R/100)ⁿ − 1]

Calculations:

⇒ 102.50 = P[(1 + 5/100)² − 1] ⇒ 102.50 = P[(21/20)² − 1] ⇒ 102.50 = P[441/400 − 1] ⇒ 102.50 = P[(441 − 400)/400] ⇒ 102.50 = P × (41/400) ⇒ P = (102.50 × 400) ÷ 41 ⇒ P = 2.5 × 400 ⇒ P = ₹1,000

∴ The correct answer is ₹1,000.

Additional Information

Effective Rate for Compound Interest

For 2 years at rate R%, the net effective interest rate is calculated using the successive percentage formula: (R + R + R²/100)%.

Difference between CI and SI

The difference between Compound Interest and Simple Interest for 2 years is given by the formula: D = P(R/100)².

Amount Formula

The total Amount (A) accumulated after n years is A = P(1 + R/100)ⁿ, where P is the Principal.

Shortcut Trick

The ratio of the first three terms is 2 : 5 : 7. Let the terms be 2x, 5x, and 7x.

According to the proportion rule: 1^st / 2^nd = 3^rd / 4^th.

⇒ 2 / 5 = 7x / 140

⇒ 7x = (2 × 140) ÷ 5 = 56

⇒ x = 56 ÷ 7 = 8

Middle number (2^nd term) = 5 × 8 = 40.

∴ The correct answer is 40.

Alternate Method

Given: Ratio of three numbers = 2 : 5 : 7 and fourth proportional = 140.

Formula: If a, b, c, and d are in proportion (a : b :: c : d), then a × d = b × c.

⇒ Let the numbers be 2x, 5x, and 7x.

⇒ By proportion property: (2x) × 140 = (5x) × (7x)

⇒ 280x = 35x²

⇒ 280 = 35x (Dividing both sides by x)

⇒ x = 280 ÷ 35

⇒ x = 8

⇒ Middle number = 2^nd term = 5x = 5 × 8 = 40.

∴ The correct answer is 40.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is given by √(a × b).

Third Proportional

If a, b, and c are in continued proportion (a : b :: b : c), then the third proportional c = b² ÷ a.

Fourth Proportional

If four quantities a, b, c, and d are in proportion (a : b :: c : d), then d = (b × c) ÷ a.

Product Rule

For any proportion, the product of the extremes (1^st and 4^th terms) is always equal to the product of the means (2^nd and 3^rd terms).

Shortcut Trick

For divisibility by 9, the sum of digits must be a multiple of 9.

Sum of digits in 6y512 = 6 + y + 5 + 1 + 2 = 14 + y.

The nearest multiple of 9 after 14 is 18.

∴ 14 + y = 18 ⇒ y = 4.

∴ The correct answer is 4.

Alternate Method

Given: The number 6y512 is divisible by both 8 and 9.

Step 1: Check Divisibility by 8

A number is divisible by 8 if its last three digits are divisible by 8.

⇒ Last three digits = 512.

⇒ 512 ÷ 8 = 64 (Remainder = 0).

Since 512 is divisible by 8, the number 6y512 is divisible by 8 for any single digit value of y.

Step 2: Check Divisibility by 9

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

⇒ Sum = 6 + y + 5 + 1 + 2

⇒ Sum = 14 + y

For the number to be divisible by 9, (14 + y) must be 18, 27, etc.

Since y is a single digit (0 to 9):

⇒ 14 + y = 18

⇒ y = 18 − 14

⇒ y = 4

∴ The correct answer is 4.

Additional Information

Divisibility Rule of 8

A number is divisible by 8 if the number formed by its last three digits is divisible by 8 or if the last three digits are zero.

Divisibility Rule of 9

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

Divisibility Rule of 72

Since 72 = 8 × 9 (co-prime factors), a number is divisible by 72 only if it is divisible by both 8 and 9 simultaneously.

Shortcut Trick

Since the cuboid is made from half the cube's material, the difference (250 cm³) is equal to the volume of the cuboid itself.

Volume of cuboid = 250 cm³ ⇒ Ratio 2 : 1 : 1 ⇒ (2x) × (x) × (x) = 250.

2x³ = 250 ⇒ x³ = 125 ⇒ x = 5 cm.

Dimensions: L = 10 cm, B = 5 cm, H = 5 cm.

TSA = 2 × (10 × 5 + 5 × 5 + 5 × 10) = 2 × (50 + 25 + 50) = 250 cm².

∴ The correct answer is 250 cm².

Alternate Method

Given:

Ratio of cuboid dimensions (L : B : H) = 2 : 1 : 1

Difference (V_cube − V_cuboid) = 250 cm³

Volume of cuboid = 1/2 × Volume of cube

Formula Used:

Volume of cuboid = Length × Breadth × Height

Total Surface Area (TSA) = 2(LB + BH + HL)

Calculations:

⇒ Let Volume of Cube be V_c.

⇒ Volume of Cuboid V_b = 1/2 × V_c.

⇒ V_c − V_b = 250 ⇒ V_c − 0.5V_c = 250 ⇒ 0.5V_c = 250.

⇒ V_b = 250 cm³.

⇒ Let dimensions be 2x, x, and x.

⇒ 2x × x × x = 250 ⇒ 2x³ = 250.

⇒ x³ = 125 ⇒ x = 5 cm.

⇒ Length = 10 cm, Breadth = 5 cm, Height = 5 cm.

⇒ TSA = 2 × (10 × 5 + 5 × 5 + 5 × 10).

⇒ TSA = 2 × (50 + 25 + 50) = 2 × 125.

⇒ TSA = 250 cm².

∴ The correct answer is 250 cm².

Additional Information

Volume and Surface Area of Cube

For a cube of side 'a', Volume = a³ and Total Surface Area = 6a².

Lateral Surface Area of Cuboid

LSA = 2 × Height × (Length + Breadth). This represents the area of the four side walls.

Diagonal of a Cuboid

The longest rod that can be placed in a cuboid has length d = √(L² + B² + H²).

Shortcut Trick

The filling rate of the second tap (10 hr) and the emptying rate of the third tap (10 hr) cancel each other out.

Net efficiency is determined solely by the first tap.

Time to fill the full cistern = 4 hr.

Time to fill 1/4^th of the cistern = 4 ÷ 4 = 1 hr.

∴ The correct answer is 1 hr.

Alternate Method

Given: Filling times are 4 hr and 10 hr; Emptying time is 10 hr.

Formula: Total Work = LCM of individual times; Efficiency = Total Work ÷ Time.

⇒ Net Efficiency of all taps = 5 + 2 − 2 = 5 units/hr.

⇒ Target work (1/4^th of cistern) = 20 ÷ 4 = 5 units.

⇒ Time taken = Work ÷ Net Efficiency

⇒ Time taken = 5 ÷ 5 = 1 hr.

∴ The correct answer is 1 hr.

Additional Information

Negative Efficiency

In Pipes and Cisterns, an outlet pipe or a leak is always considered to have negative efficiency because it reduces the total volume of water.

Work-Efficiency Relation

Total Work = Efficiency × Time. For a constant work, Efficiency is inversely proportional to the Time taken to complete the task.

Combined Rate Formula

If two pipes fill in 'x' and 'y' hours and one empties in 'z' hours, the net part filled in 1 hour is (1/x + 1/y − 1/z).