Quantitative Aptitude PYQs

Shortcut Trick

Time taken by Pipe C and Pipe D to fill 3,000 L together = (6 × 4) ÷ (6 + 4) = 2.4 hr.

Since the new tank capacity (6,000 L) is exactly double the initial capacity (3,000 L), the time taken will also be doubled.

Required time = 2.4 × 2 = 4.8 hr.

∴ The correct answer is 4.8 hr.

Alternate Method

Given:

Capacity of initial tank = 3,000 L

Time for Pipe C = 6 hr | Time for Pipe D = 4 hr

Capacity of new tank = 6,000 L

Calculations:

⇒ Efficiency (Rate) of Pipe C = 3,000 ÷ 6 = 500 L/hr

⇒ Efficiency (Rate) of Pipe D = 3,000 ÷ 4 = 750 L/hr

⇒ Combined Efficiency of (C + D) = 500 + 750 = 1,250 L/hr

⇒ Time taken to fill 6,000 L tank = Total Capacity ÷ Combined Efficiency

⇒ Required Time = 6,000 ÷ 1,250

⇒ Required Time = 4.8 hr

∴ The correct answer is 4.8 hr.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to time for a constant volume of work. Formula: Work = Efficiency × Time.

Pipes and Cisterns Rule

If an inlet pipe fills in x hours and an outlet pipe empties in y hours, the net work in 1 hour = (1/x) − (1/y).

Direct Proportion in Work

If efficiency remains constant, the time required is directly proportional to the amount of work (tank capacity). If capacity doubles, time doubles.

Shortcut Trick

For a tower height h, when the angle changes from θ₁ to θ₂ and shadow increases by d:

Shortcut formula: h = d ÷ (cot θ₂ − cot θ₁)

Here, d = 20 m, θ₁ = 60°, and θ₂ = 45°

h = 20 ÷ (cot 45° − cot 60°) = 20 ÷ (1 − 1/1.7) = 20 ÷ (0.7/1.7) = 34 ÷ 0.7 ≈ 48.6 m

∴ The correct answer is 48.6 m.

Alternate Method

Let AB be the tower of height h. Let C and D be the points where the sun's altitude is 60° and 45° respectively.

Given: CD = 20 m, ∠ACB = 60°, ∠ADB = 45°, √3 = 1.7

In right-angled triangle ABC:

⇒ tan 60° = AB / BC

⇒ 1.7 = h / BC

⇒ BC = h / 1.7

In right-angled triangle ABD:

⇒ tan 45° = AB / BD

⇒ 1 = h / BD

⇒ BD = h

According to the question:

⇒ BD − BC = CD

⇒ h − h / 1.7 = 20

⇒ (1.7h − h) / 1.7 = 20

⇒ 0.7h / 1.7 = 20

⇒ 0.7h = 20 × 1.7

⇒ 0.7h = 34

⇒ h = 34 / 0.7 ≈ 48.5714

⇒ h ≈ 48.6 m

∴ The correct answer is 48.6 m.

Additional Information

Trigonometric Ratios

Commonly used ratios: tan θ = Perpendicular/Base and cot θ = Base/Perpendicular.

Angle of Elevation

As the object (Sun) moves lower in the sky (altitude decreases), the shadow of a vertical object becomes longer.

General Distance Formula

The distance between two points on the same side of a tower is given by d = h(cot θ₂ − cot θ₁), where θ₁ > θ₂.

Shortcut Trick

Use the distance relation: D = [S × (S + a) / a] × t₁ = [S × (S − b) / b] × t₂

⇒ [(S + 14) / 14] × 35 = [(S − 10) / 10] × 30

⇒ 5(S + 14) = 6(S − 10) ⇒ 5S + 70 = 6S − 60 ⇒ S = 130 km/h

⇒ D = [130 × (130 + 14) / 14] × (35 / 60) = 780 km

∴ The correct answer is 780 km.

Alternate Method

Given:

Increase in speed (a) = 14 km/hr, Reduction in time (t₁) = 35 min = 35/60 hr

Decrease in speed (b) = 10 km/hr, Increase in time (t₂) = 30 min = 30/60 hr

Formula Used:

Distance (D) = [S × (S + ΔS) / ΔS] × Δt

Calculations:

⇒ Let original speed be S km/hr and distance be D km.

⇒ According to the first condition: D = [S(S + 14) / 14] × (35 / 60) --- (i)

⇒ According to the second condition: D = [S(S − 10) / 10] × (30 / 60) --- (ii)

⇒ Equating (i) and (ii): [S(S + 14) / 14] × (35 / 60) = [S(S − 10) / 10] × (30 / 60)

⇒ [(S + 14) / 14] × 35 = [(S − 10) / 10] × 30

⇒ [(S + 14) / 2] × 5 = (S − 10) × 3

⇒ 5S + 70 = 6S − 60

⇒ S = 130 km/hr

⇒ Substituting S = 130 in equation (i):

⇒ D = [130 × (130 + 14) / 14] × (35 / 60)

⇒ D = [130 × 144 / 14] × (7 / 12)

⇒ D = (130 / 2) × (144 / 12) = 65 × 12 = 780 km

∴ The correct answer is 780 km.

Additional Information

Relationship between Speed, Time, and Distance

Distance is directly proportional to speed when time is constant (D = S × T). For a fixed distance, speed and time are inversely proportional.

Distance Formula with Time Difference

When a person travels at two different speeds S₁ and S₂ and the difference between the time taken is Δt, the distance D = [S₁ × S₂ / (S₁ − S₂)] × Δt.

Average Speed Formula

If a certain distance is covered at x km/hr and the same distance is covered at y km/hr, the Average Speed for the whole journey is 2xy / (x + y) km/hr.

Shortcut Trick

Let Cost Price (CP) = 100 units.

Initial Selling Price (SP) = 120 units (20% profit).

New SP = 120 − (10% of 120) = 120 − 12 = 108 units.

New Profit % = 108 − 100 = 8%.

∴ The correct answer is 8%.

Alternate Method

Given:

Initial Profit = 20%

Reduction in Selling Price = 10%

Formula used:

Profit Percentage = ((SP − CP) ÷ CP) × 100

Calculations:

⇒ Let Cost Price (CP) = ₹100

⇒ Initial Selling Price (SP₁) = CP + 20% of CP = 100 + 20 = ₹120

⇒ New Selling Price (SP₂) = SP₁ − 10% of SP₁

⇒ SP₂ = 120 − (10 ÷ 100) × 120 = 120 − 12 = ₹108

⇒ New Profit = SP₂ − CP = 108 − 100 = ₹8

⇒ New Profit Percentage = (8 ÷ 100) × 100 = 8%

∴ The correct answer is 8%.

Additional Information

Profit and Loss Basics

Profit occurs when Selling Price (SP) > Cost Price (CP). Profit % is always calculated on CP unless stated otherwise.

Successive Change Effect

When SP is changed, the new profit depends on the original CP, which remains constant unless specifically mentioned.

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). SP = MP − Discount.

Shortcut Trick

Convert Interest Rate to fraction: 12.5% = 1/8. For 3 months (1/4 year), Rate = 1/8 × 1/4 = 1/32.

Principal to Amount Ratio: (8 × 8 × 32) : (9 × 9 × 33) ⇒ 2048 : 2673.

Principal 2048 units = ₹25,600 ⇒ 1 unit = ₹12.5.

Interest units = 2673 − 2048 = 625 units.

Compound Interest = 625 × 12.5 = ₹7,812.50.

∴ The correct answer is ₹7,812.50.

Alternate Method

Given: Principal (P) = ₹25,600, Rate (R) = 12.5% per annum, Time (T) = 2 years 3 months.

Formula Used: Amount (A) = P(1 + R/100)ⁿ × (1 + (R × t)/100) for fractional years.

⇒ Convert 3 months to years: 3/12 = 1/4 year.

⇒ Annual rate in fraction: 12.5/100 = 1/8.

⇒ Rate for 3 months: (1/8) × (1/4) = 1/32.

⇒ Amount (A) = 25600 × (1 + 1/8)² × (1 + 1/32)

⇒ A = 25600 × (9/8) × (9/8) × (33/32)

⇒ A = 25600 × (81/64) × (33/32)

⇒ A = (25600 ÷ 2048) × (81 × 33)

⇒ A = 12.5 × 2673 = ₹33,412.50

⇒ Compound Interest (CI) = Amount − Principal

⇒ CI = ₹33,412.50 − ₹25,600 = ₹7,812.50

∴ The correct answer is ₹7,812.50.

Additional Information

Fractional Time Period in CI

When time is in the form T years and m months, calculate for T years normally and apply rate (m/12) × R for the remaining fractional part.

Difference between CI and SI

For a 2-year period, the difference between compound and simple interest is D = P(R/100)².

Effective Rate of Interest

The effective rate for two successive years with rates r₁ and r₂ is r₁ + r₂ + (r₁ × r₂)/100.

Shortcut Trick

Speed in km/h = Distance ÷ Time = 360 ÷ 8 = 45 km/h.

To convert km/h to m/s, multiply by 5/18.

Speed = 45 × (5/18) = 2.5 × 5 = 12.5 m/s.

∴ The correct answer is 12.5 m/s.

Alternate Method

Given:

Total Distance = 360 km

Total Time = 8 hr

Formula Used:

Speed = Distance ÷ Time

Conversion factor: 1 km/h = 5/18 m/s

⇒ Speed in km/h = 360 ÷ 8

⇒ Speed = 45 km/h

⇒ Speed in m/s = 45 × (5 ÷ 18)

⇒ Speed = (5 × 5) ÷ 2

⇒ Speed = 25 ÷ 2 = 12.5 m/s

∴ The correct answer is 12.5 m/s.

Additional Information

Unit Conversion Rules

To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5.

Speed-Distance-Time Formula

Distance = Speed × Time. Ensure all units are in the same system (SI or Metric) before calculating.

Average Speed

Average Speed = (Total Distance) ÷ (Total Time). For two equal distances at speeds x and y, Average Speed = 2xy ÷ (x + y).

Shortcut Trick

Observe the common difference between the divisors and their respective remainders:

3 − 2 = 1

4 − 3 = 1

5 − 4 = 1

Smallest Number = LCM(3, 4, 5) − Common Difference

Required Number = 60 − 1 = 59.

∴ The correct answer is 59.

Alternate Method

Given:

Divisors: 3, 4, 5

Remainders: 2, 3, 4

Formula Used:

When the difference between the divisor (d) and remainder (r) is constant (k = d − r), the smallest number is given by: LCM(d₁, d₂, d₃, ...) − k.

Calculations:

⇒ Difference 1 = 3 − 2 = 1

⇒ Difference 2 = 4 − 3 = 1

⇒ Difference 3 = 5 − 4 = 1

⇒ Constant difference (k) = 1

⇒ LCM of 3, 4, 5:

⇒ 3 = 3¹

⇒ 4 = 2²

⇒ 5 = 5¹

⇒ LCM = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60

⇒ Smallest number = 60 − 1 = 59

∴ The correct answer is 59.

Additional Information

LCM and Remainder Concept

If a number leaves the same remainder 'r' in each case, the number is LCM(divisors) + r.

Successive Division

In successive division, the quotient of the first division becomes the dividend for the second division.

Chinese Remainder Theorem

A general method to find a number satisfying multiple linear congruences when divisors are coprime.

Shortcut Trick

Buy 2 Get 1 Free means the customer gets 3 items but pays for only 2 items.

Total Selling Price (SP) of 3 items = 2 × 2,700 = ₹ 5,400.

Since the trader earns 20% profit, Total Cost Price (CP) = 5,400 ÷ 1.2 = ₹ 4,500.

Cost Price of one pair of trousers = 4,500 ÷ 3 = ₹ 1,500.

∴ The correct answer is ₹ 1,500.

Alternate Method

Given:

Marked Price (MP) per pair = ₹ 2,700

Offer: Buy 2, Get 1 Free (Total items received = 3)

Profit percentage = 20%

Calculations:

⇒ Total Selling Price (SP) for 3 items = 2 × 2,700 = ₹ 5,400

⇒ Let the Cost Price (CP) of one pair of trousers be ₹ x

⇒ Total Cost Price for 3 pairs = 3x

⇒ Profit = 20%, so Total SP = 120% of Total CP

⇒ 5,400 = 1.20 × (3x)

⇒ 5,400 = 3.6x

⇒ x = 5,400 ÷ 3.6

⇒ x = 1,500

∴ The correct answer is ₹ 1,500.

Additional Information

Discount % in 'Buy X Get Y Free' schemes

Discount % = [Free Items ÷ Total Items] × 100. In this case: [1 ÷ (2 + 1)] × 100 = 33.33%.

Relation between CP and MP

CP ÷ MP = (100 − Discount %) ÷ (100 + Profit %). This ratio helps find CP directly when MP, discount, and profit are known.

Profit and Loss Basics

Profit % = [(SP − CP) ÷ CP] × 100. Always ensure CP and SP are calculated for the same number of units.

Shortcut Trick

Net change in savings = (Increase in Income) − (Increase in Expenditure)

Increase in Income = 10% of 45,000 = ₹4,500

Increase in Expenditure = 20% of 36,000 = ₹7,200

Change in savings = 4,500 − 7,200 = −₹2,700 (Decrease)

New monthly savings = (45,000 − 36,000) − 2,700 = 9,000 − 2,700 = ₹6,300

∴ The correct answer is ₹6,300.

Alternate Method

Given: Original monthly income = ₹45,000, Original monthly expenditure = ₹36,000.

Formula Used: Savings = Income − Expenditure

⇒ New Income = 45,000 × (110 ÷ 100) = ₹49,500

⇒ New Expenditure = 36,000 × (120 ÷ 100) = ₹43,200

⇒ New Savings = New Income − New Expenditure

⇒ New Savings = 49,500 − 43,200

⇒ New Savings = ₹6,300

∴ The correct answer is ₹6,300.

Additional Information

Basic Budget Equation

Income = Expenditure + Savings. This relation holds true for both original and updated values.

Percentage Change in Savings

Percentage change in savings can be calculated as [(New Savings − Old Savings) ÷ Old Savings] × 100.

Income Increase vs Expenditure Increase

If the absolute increase in expenditure exceeds the absolute increase in income, the savings will always decrease.

Shortcut Trick

Identify factors of the product 30 that add up to 13.

Factors of 30: (1, 30), (2, 15), (3, 10), (5, 6).

Sum check: 10 + 3 = 13 (Correct).

Positive Difference: 10 − 3 = 7.

∴ The correct answer is 7.

Alternate Method

Given:

Sum of two positive numbers (x + y) = 13

Product of the numbers (xy) = 30

Formula Used:

(x − y)² = (x + y)² − 4xy

Calculations:

⇒ (x − y)² = (13)² − 4 × (30)

⇒ (x − y)² = 169 − 120

⇒ (x − y)² = 49

⇒ (x − y) = √49

⇒ x − y = 7

∴ The correct answer is 7.

Additional Information

Quadratic Roots Relation

Any two numbers can be treated as roots of a quadratic equation: z² − (Sum)z + (Product) = 0.

Algebraic Identities

Remember that (x + y)² = x² + y² + 2xy and (x − y)² = x² + y² − 2xy. Their difference is always 4xy.

Root Difference Formula

For a quadratic equation az² + bz + c = 0, the difference of roots is given by √D/|a|, where D = b² − 4ac.

Shortcut Trick

Apply the Successive Percentage formula: Net Change = a + b + (a × b ÷ 100)

Here, a = −7 (decrease in cost) and b = +20 (increase in sales).

Net Change = −7 + 20 + [(−7 × 20) ÷ 100]

Net Change = 13 − 1.4 = 11.6%

Since the result is positive, it indicates an increase.

∴ The correct answer is increase of 11.6%.

Alternate Method

Given: Decrease in cost = 7% | Increase in sales = 20%

Formula Used: Revenue = Price × Quantity

⇒ Let the initial cost of one article be 100 units.

⇒ Let the initial number of articles sold be 100 units.

⇒ Initial Revenue = 100 × 100 = 10,000

⇒ New cost = 100 − 7% of 100 = 93

⇒ New number of articles sold = 100 + 20% of 100 = 120

⇒ New Revenue = 93 × 120 = 11,160

⇒ Increase in Revenue = 11,160 − 10,000 = 1,160

⇒ Percentage Increase = (1,160 ÷ 10,000) × 100 = 11.6%

∴ The correct answer is increase of 11.6%.

Additional Information

Net Percentage Change Formula

For any product P = A × B, if A changes by x% and B changes by y%, then P changes by (x + y + xy/100)%.

Revenue Components

Total Revenue is directly proportional to both Price per unit and Total Sales volume.

Fraction Method

Changes can be expressed as multipliers: 7% decrease = 0.93 multiplier; 20% increase = 1.20 multiplier. Net change = 0.93 × 1.20 = 1.116, which is an 11.6% increase.

Shortcut Trick

A number is divisible by 12 if and only if it is divisible by its coprime factors, which are 3 and 4.

The question states that the number 7549A is already divisible by both 4 and 3.

Since it satisfies both conditions, the number must be perfectly divisible by 12 (LCM of 3 and 4).

For any number that is perfectly divisible, the remainder is always 0.

∴ The correct answer is 0.

Alternate Method

Given: A five-digit number 7549A is divisible by 4 and 3.

⇒ Step 1: Check Divisibility by 4. The last two digits 9A must be divisible by 4.

⇒ Multiples of 4 in the 90s are 92 and 96. Thus, A can be 2 or 6.

⇒ Step 2: Check Divisibility by 3. The sum of digits must be divisible by 3.

⇒ Sum = 7 + 5 + 4 + 9 + A = 25 + A.

⇒ If A = 2, Sum = 25 + 2 = 27 (Divisible by 3).

⇒ If A = 6, Sum = 25 + 6 = 31 (Not divisible by 3).

⇒ Step 3: Determine the Remainder. The number is 75492.

⇒ Since 75492 is divisible by both 3 and 4, it is divisible by 12.

⇒ 75492 ÷ 12 = 6291 with no remainder.

∴ The correct answer is 0.

Additional Information

Divisibility Rule of 12

A number is divisible by 12 if it is divisible by both 3 and 4. This is because 12 is a composite number and 3 and 4 are its coprime factors.

Coprime Numbers

Two numbers are coprime if their Highest Common Factor (HCF) is 1. If a number is divisible by two coprime numbers x and y, it is also divisible by their product xy.

Divisibility Rule of 3 and 9

For both 3 and 9, the rule depends on the sum of the digits. If the sum is divisible by 3 or 9, the entire number is divisible by 3 or 9 respectively.

Shortcut Trick

Total reduction in combined age 5 years ago = 5 years × 2 people = 10 years.

Sum of ages 5 years ago = 34 − 10 = 24 years.

The ratio of Anika's age to Riya's age was 2 : 1.

Anika's age 5 years ago = 24 × (2 / 3) = 16 years.

Anika's present age = 16 + 5 = 21 years.

∴ The correct answer is 21 years.

Alternate Method

Given: Combined age = 34 years; Age relation 5 years ago = 2 : 1.

Let Riya's age 5 years ago be x and Anika's age be 2x.

⇒ Present age of Anika = 2x + 5

⇒ Present age of Riya = x + 5

⇒ (2x + 5) + (x + 5) = 34

⇒ 3x + 10 = 34

⇒ 3x = 34 − 10

⇒ 3x = 24

⇒ x = 8

⇒ Anika's current age = 2(8) + 5 = 16 + 5 = 21

∴ The correct answer is 21 years.

Additional Information

Constant Age Difference

The difference between the ages of two people always remains constant, regardless of the number of years passed or gone by.

Sum of Ages Shift

If there are n people, the sum of their ages t years ago was (Current Sum − n × t) and t years from now will be (Current Sum + n × t).

Ratio of Ages

When ages are given in a ratio a : b, they can be represented as ax and bx to simplify algebraic equations.

Shortcut Trick

शुद्ध सदस्यता आय = 7643 − 1105 (प्रारंभिक बकाया) + 916 (अंतिम बकाया) = ₹7,454

कुल आय = 7454 + 2198 (पुरानी बिक्री) = ₹9,652

कुल व्यय = 5287 + 3120 = ₹8,407

अधिशेष = कुल आय − कुल व्यय = 9652 − 8407 = ₹1,245

∴ The correct answer is ₹1,245 अधिशेष.

Alternate Method

दिया गया है:

प्राप्त सदस्यता शुल्क = ₹7,643

वर्ष के प्रारंभ में बकाया राशि = ₹1,105

वर्ष के अंत में बकाया राशि = ₹916

पुराने खेल उपस्कर की बिक्री = ₹2,198

खेल सामग्री पर व्यय = ₹5,287

अनुरक्षण पर व्यय = ₹3,120

सूत्र:

वर्ष की वास्तविक आय = (प्राप्त सदस्यता शुल्क − प्रारंभिक बकाया + अंतिम बकाया) + अन्य आय

गणना:

⇒ वास्तविक सदस्यता आय = 7643 − 1105 + 916

⇒ वास्तविक सदस्यता आय = ₹7,454

⇒ कुल वार्षिक आय = 7454 + 2198 = ₹9,652

⇒ कुल वार्षिक व्यय = 5287 + 3120 = ₹8,407

⇒ अधिशेष = कुल आय − कुल व्यय

⇒ अधिशेष = 9652 − 8407 = ₹1,245

चूँकि आय व्यय से अधिक है, इसलिए यह एक अधिशेष है।

∴ The correct answer is ₹1,245 अधिशेष.

Additional Information

अधिशेष और घाटा (Surplus and Deficit)

जब किसी गैर-लाभकारी संगठन की कुल आय उसके कुल व्यय से अधिक होती है, तो उसे अधिशेष कहा जाता है। इसके विपरीत स्थिति को घाटा कहा जाता है।

उपार्जन आधार (Accrual Basis)

आय और व्यय खाते में केवल वर्तमान लेखा वर्ष से संबंधित मदों को शामिल किया जाता है। पिछले या अगले वर्ष के बकाये को समायोजित किया जाता है।

बकाया सदस्यता (Outstanding Subscription)

वर्ष के अंत में बकाया सदस्यता चालू वर्ष की आय में जोड़ी जाती है, जबकि वर्ष के प्रारंभ में बकाया (जो पिछले वर्ष का है) घटाई जाती है।

Shortcut Trick

Initial Ratio (Milk : Water) = 5 : 3

Final Ratio (Milk : Water) = 5 : 5

Since only water is added, the milk units remain constant (5 units).

Increase in water = 5 − 3 = 2 units

2 units = 48 liters ⇒ 1 unit = 24 liters

Initial Milk = 5 units = 5 × 24 = 120 liters

∴ The correct answer is 120 liters.

Alternate Method

Given: Initial ratio 5:3, Final ratio 5:5, Water added = 48 liters.

Formula Used: Ratio Proportionality | Quantity = Ratio Part × Scaling Factor (x)

⇒ Let the initial quantity of milk and water be 5x and 3x respectively.

⇒ Total quantity of water after adding 48 liters = 3x + 48

⇒ New ratio of milk to water = 5x / (3x + 48)

⇒ Given new ratio = 5 / 5 = 1 / 1

⇒ 5x / (3x + 48) = 1 / 1

⇒ 5x = 3x + 48

⇒ 2x = 48

⇒ x = 24

⇒ Initial quantity of milk = 5 × 24 = 120 liters.

∴ The correct answer is 120 liters.

Additional Information

Constant Component Principle

In mixture problems, if only one component is added or removed, the quantity of the other component remains unchanged in absolute terms. Adjust ratios to reflect this.

Ratio Balancing

To balance a ratio when a value is added to only one side, multiply the entire ratio by a factor so that the constant part matches its previous unit count.

Percentage to Ratio

A mixture ratio of 5:5 implies 50% milk and 50% water. This conversion is often useful for solving complex mixture-alligation problems.

Shortcut Trick

Total time = (Distance ÷ Speed₁) + (Distance ÷ Speed₂)

Total time = (78 ÷ 13) + (78 ÷ 26)

Total time = 6 + 3 = 9 hr

∴ The correct answer is 9 hr.

Alternate Method

Given:

Distance from Town A to Town B = 78 km

Speed during onward journey (A to B) = 13 km/h

Speed during return journey (B to A) = 26 km/h

Formula: Time = Distance ÷ Speed

⇒ Time taken to travel from Town A to Town B (T₁) = 78 ÷ 13 = 6 hr

⇒ Time taken to travel from Town B to Town A (T₂) = 78 ÷ 26 = 3 hr

⇒ Total time for the round trip = T₁ + T₂

⇒ Total time = 6 + 3 = 9 hr

Therefore, the total time taken is 9 hr, which corresponds to Option (2).

∴ The correct answer is 9 hr.

Additional Information

Average Speed for Round Trip

When the distance covered is the same for both ways, the Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂). For this trip, Average Speed = (2 × 13 × 26) ÷ (13 + 26) = 17.33 km/h.

Speed-Time Relationship

For a constant distance, Speed is inversely proportional to Time (S ∝ 1/T). Since the return speed is double the onward speed, the return time is half of the onward time.

Shortcut Trick

The mean proportional (b) between two numbers a and c is calculated as b = √(a × c).

Given a = 25 and c = 100.

Second number = √(25 × 100) = √2500 = 50.

∴ The correct answer is 50.

Alternate Method

Given:

First number (a) = 25

Third number (c) = 100

Second number (b) is the mean proportional.

Formula Used:

If three numbers are in continued proportion, then a/b = b/c ⇒ b² = ac

⇒ b² = 25 × 100 ⇒ b² = 2500 ⇒ b = √2500 ⇒ b = 50

∴ The correct answer is 50.

Additional Information

Mean Proportional (Geometric Mean)

For two numbers a and b, the mean proportional x is defined such that a:x = x:b, which results in x = √(ab).

Third Proportional

In a continued proportion a:b = b:c, the number c is called the third proportional and is calculated as c = b² / a.

Fourth Proportional

For four numbers a, b, c, and d, if a:b = c:d, then d is the fourth proportional and d = (b × c) / a.

Shortcut Trick

ध्वजदंड की ऊंचाई (h) = आधार दूरी × (tan θ₁ − tan θ₂)

जहाँ θ₁ = 60° और θ₂ = 30°

h = 25 × (tan 60° − tan 30°) = 25 × (√3 − 1/√3) = 25 × (2/√3)

h = 50 / 1.7 = 29.411... ≈ 29 m

∴ The correct answer is 29 m.

Alternate Method

दिया गया है:

आधार से दूरी (AB) = 25 मीटर

ध्वजदंड के शीर्ष का उन्नयन कोण = 60°

इमारत के शीर्ष का उन्नयन कोण = 30°

उपयोग करें: √3 = 1.7

माना इमारत की ऊंचाई BC = h मीटर और ध्वजदंड की ऊंचाई CD = x मीटर है।

समकोण त्रिभुज ABC में:

⇒ tan 30° = BC / AB

⇒ 1 / √3 = h / 25

⇒ h = 25 / √3

समकोण त्रिभुज ABD में:

⇒ tan 60° = BD / AB

⇒ √3 = (h + x) / 25

⇒ h + x = 25√3

अब, ध्वजदंड की ऊंचाई (x) ज्ञात करने के लिए:

⇒ x = (h + x) − h

⇒ x = 25√3 − 25 / √3

⇒ x = 25 × (√3 − 1/√3) = 25 × (3 − 1) / √3

⇒ x = 50 / √3

⇒ x = 50 / 1.7 ≈ 29.41 m

निकटतम पूर्णांक मान 29 मीटर है।

∴ The correct answer is 29 m.

Additional Information

उन्नयन कोण (Angle of Elevation)

जब कोई प्रेक्षक ऊपर की ओर स्थित किसी वस्तु को देखता है, तो क्षैतिज रेखा और दृष्टि रेखा के बीच बनने वाला कोण उन्नयन कोण कहलाता है।

त्रिकोणमितीय अनुपात (Trigonometric Ratios)

tan θ = लंब / आधार | sin θ = लंब / कर्ण | cos θ = आधार / कर्ण।

महत्वपूर्ण मान

tan 30° = 1/√3 ≈ 0.577, tan 45° = 1, tan 60° = √3 ≈ 1.732।

Given:

{(12/27) ÷ (12/2)} ÷ ((1/11) × (22/3) + (5/9)) + (2/9) ÷ (33/22) of (22/9)

Formula used:

The simplification follows the VBODMAS rule:

V: Vinculum, B: Brackets/Braces, O: Of, D: Division, M: Multiplication, A: Addition, S: Subtraction.

Calculations:

Step 1: Simplify the terms inside the first braces:

⇒ {(12/27) ÷ (12/2)} = (12/27 × 2/12)

⇒ 2/27

Step 2: Simplify the terms inside the parentheses:

⇒ ((1/11) × (22/3) + (5/9)) = (2/3 + 5/9)

Taking the LCM of 3 and 9, which is 9:

⇒ (6/9 + 5/9) = 11/9

Step 3: Simplify the of operator part (which takes priority over division):

⇒ (33/22) of (22/9) = (33/22 × 22/9)

⇒ 33/9 = 11/3

Step 4: Substitute these values back into the expression:

⇒ (2/27 ÷ 11/9) + (2/9 ÷ 11/3)

Step 5: Apply division by multiplying by the reciprocal:

⇒ (2/27 × 9/11) + (2/9 × 3/11)

⇒ 2/33 + 2/33

Step 6: Perform the final addition:

⇒ 4/33

∴ The correct answer is option (4).

Shortcut Trick

Total capacity (LCM of 10, 15, 12) = 60 units.

Efficiencies: Pipe A = 6, Pipe B = 4, Pipe C = 5.

Draining in 2 hours by A and B = (6 + 4) × 2 = 20 units.

Remaining water = 60 − 20 = 40 units.

Net draining speed (A + B − C) = 10 − 5 = 5 units/hr.

Total time = 2 hr + (40 ÷ 5) hr = 10 hours.

∴ The correct answer is 10 hr.

Alternate Method

Given: Draining time of Pipe A = 10 hr, Pipe B = 15 hr, and Filling time of Pipe C = 12 hr.

Formula Used: Total Work = LCM of time; Efficiency = Work ÷ Time.

⇒ Total draining efficiency of A and B = 6 + 4 = 10 units/hr.

⇒ Water drained by A and B in first 2 hours = 10 × 2 = 20 units.

⇒ Remaining water in the full tank = 60 − 20 = 40 units.

⇒ Now, all pipes are open, Net Efficiency = (Efficiency of A + B) − Efficiency of C

⇒ Net Efficiency = 10 − 5 = 5 units/hr (Draining).

⇒ Time taken to drain remaining water = 40 ÷ 5 = 8 hours.

⇒ Total time taken to empty the tank = 2 + 8 = 10 hours.

∴ The correct answer is 10 hr.

Additional Information

Negative Efficiency Concept

In Pipes and Cisterns, inlet pipes have positive efficiency while outlet/drain pipes have negative efficiency when calculating net flow.

Time to Empty vs Time to Fill

If the net efficiency is positive, the tank will eventually fill. If the net efficiency is negative, the tank will eventually empty.

Pipes and Cisterns Formula

If pipe X fills in 'x' hours and pipe Y empties in 'y' hours, net work in 1 hour = (1/x − 1/y).