Quantitative Aptitude PYQs

Shortcut Trick

Curved Surface Area of a cylinder = 2πrh

⇒ 2 × (22/7) × r × 16 = 2200

⇒ r = (2200 × 7) ÷ (44 × 16)

⇒ r = 175 ÷ 8 = 21.875 cm

⇒ r ≈ 21.88 cm (rounded to two decimal places)

∴ The correct answer is 21.88 cm.

Alternate Method

Given:

Outer curved surface area (CSA) = 2200 cm²

Length (height) of the rod (h) = 16 cm

Value of π = 22/7

r = ?

h = 16 cm

CSA = 2200 cm²

Formula Used:

Curved Surface Area of a Cylinder = 2πrh

Calculations:

⇒ 2 × (22/7) × r × 16 = 2200

⇒ 44/7 × r × 16 = 2200

⇒ r = (2200 × 7) ÷ (44 × 16)

⇒ r = (50 × 7) ÷ 16

⇒ r = 350 ÷ 16

⇒ r = 21.875 cm

Rounding to two places of decimal:

⇒ r ≈ 21.88 cm

∴ The correct answer is 21.88 cm.

Additional Information

Total Surface Area of a Cylinder

The total surface area encompasses both the curved surface and the two circular flat ends (top and base). It is given by the formula: TSA = 2πr(r + h).

Volume of a Cylinder

The total capacity enclosed within the cylinder geometry is calculated by multiplying the base area by its height: Volume = πr²h.

Volume of a Hollow Cylinder

When dealing with a hollow cylindrical pipe or tube having an outer radius R and an inner radius r, the volume of material used is determined by: Volume = πh(R² − r²).

Shortcut Trick

Let W be the time to walk one way and D be the time to drive one way.

Given: W + D = 5 hours 30 minutes.

Since replacing one walk with a drive saves 3 hours, conversely, replacing one drive with a walk will add 3 hours.

⇒ Time to walk both ways (2W) = (W + D) + 3 hours.

⇒ 2W = 5 hours 30 minutes + 3 hours = 8 hours 30 minutes.

∴ The correct answer is 8 hours 30 minutes.

Alternate Method

Given:

Time taken to walk one way and drive back (W + D) = 5 hours 30 minutes = 5.5 hours.

Time gained by driving both ways instead of mixed travel = 3 hours.

Start
Viewpoint

Walk (W)

Drive (D)

W + D = 5 hr 30 min

Calculations:

⇒ Time to drive both ways (2D) = 5.5 − 3 = 2.5 hours.

⇒ Time to drive one way (D) = 2.5 ÷ 2 = 1.25 hours = 1 hour 15 minutes.

⇒ Time to walk one way (W) = 5.5 − 1.25 = 4.25 hours = 4 hours 15 minutes.

⇒ Time to walk both ways (2W) = 2 × 4.25 = 8.5 hours.

⇒ 8.5 hours = 8 hours 30 minutes.

∴ The correct answer is 8 hours 30 minutes.

Additional Information

Average Speed (Constant Distance)

When a person travels a distance at speed S₁ and returns at speed S₂, the average speed for the whole journey is (2 × S₁ × S₂) ÷ (S₁ + S₂).

Relation between Speed and Time

When the distance is constant, speed is inversely proportional to the time taken. If the ratio of speeds is a : b, the respective ratio of time taken will be b : a.

Relative Speed Concept

When two moving bodies travel in the same direction, their relative speed is calculated as the difference of their speeds. When moving in opposite directions, it is the sum of their speeds.

Shortcut Trick
Given, 50 × SP = 19 × CP
⇒ SP / CP = 19 / 50
Here, CP = 50 units and SP = 19 units.
⇒ Loss = 50 − 19 = 31 units
⇒ Loss % = (31 / 50) × 100 = 62%
∴ The correct answer is 62% loss.

Alternate Method
Given:
Selling price of 50 books = Cost price of 19 books
Formula Used:
Loss = CP − SP
Loss % = (Loss ÷ CP) × 100

SP = 19 units
Loss = 31 units

Total CP = 50 units
Loss % = 31/50 × 100 = 62%

Calculations:
Let the Cost Price (CP) of 1 book be ₹1.
⇒ CP of 50 books = ₹50
⇒ SP of 50 books = CP of 19 books = ₹19
Since CP > SP, a loss occurs.
⇒ Loss = CP − SP
⇒ Loss = 50 − 19 = ₹31
⇒ Loss % = (Loss ÷ CP) × 100
⇒ Loss % = (31 ÷ 50) × 100
⇒ Loss % = 31 × 2 = 62%
∴ The correct answer is 62% loss.

Additional Information
Profit and Loss Formulas
Profit = SP − CP, and Profit % = (Profit ÷ CP) × 100
Dishonest Dealer
If a dealer sells at CP but uses false weight, Profit % = (Error ÷ False Weight) × 100
Marked Price & Discount
Discount = MP − SP, and Discount % = (Discount ÷ MP) × 100

Shortcut Trick

Let the assumed average be 50 kg.

Deviations of the given weights from 50 kg are: −10, −1, +6, +24, and −14.

Net deviation = (−10) + (−1) + 6 + 24 + (−14) = +5 kg.

Average deviation = Net deviation ÷ Total members = +5 ÷ 5 = +1 kg.

Actual average = Assumed average + Average deviation = 50 + 1 = 51 kg.

∴ The correct answer is 51 kg.

Alternate Method

Given:

Weights of the family members = 36 kg, 40 kg, 49 kg, 56 kg, and 74 kg

Total number of members = 5

Formula Used:

Average = (Sum of all observations) ÷ (Total number of observations)

Average as Center of Mass3040506070803640495674Avg = 51

Calculation:

⇒ Sum of all weights = 36 + 40 + 49 + 56 + 74

⇒ Sum of all weights = 255 kg

⇒ Average weight = 255 ÷ 5

⇒ Average weight = 51 kg

∴ The correct answer is 51 kg.

Additional Information

Average of Consecutive Numbers

For numbers arranged in an arithmetic progression with a common difference, the average is simply the middle term or equivalently evaluated as (First term + Last term) ÷ 2.

Combined Average

If two distinct groups have sizes n₁ and n₂ with respective averages A₁ and A₂, their combined collective average equals (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Effect of Arithmetic Operations on Average

If a uniform constant value is added, subtracted, multiplied, or divided across every individual observation, the overall average of the set directly undergoes the exact same mathematical operation.

Shortcut Trick

The unit digit of 861616843 is 3 (an odd number), which immediately eliminates even divisors like 36 and numbers ending in zero like 50.

Between 43 and 47, we can systematically split the given number into simple chunks: 86,00,00,000 + 16,16,800 + 43.

Each chunk is visibly divisible by 43 (86 ÷ 43 = 2 | 16168 ÷ 43 = 376 | 43 ÷ 43 = 1).

Since every single part perfectly divides by 43, the entire number is directly divisible by 43.

∴ The correct answer is 43.

Alternate Method

Given:

Number = 861616843

Options provided = 43, 36, 50, 47

Method Used:

Basic divisibility rules and division by separating parts.

Calculations:

⇒ The given number 861616843 ends with the odd digit 3.

⇒ Any number ending with an odd digit cannot be completely divided by an even number.

⇒ Thus, the options 36 (even number) and 50 (ends with 0) are eliminated directly.

⇒ Now, check for divisibility by 43 by breaking the huge number into smaller, easily identifiable multiples.

⇒ 861616843 = 86,00,00,000 + 16,16,800 + 43

⇒ Part 1: 86,00,00,000 ÷ 43 = 2,00,00,000

⇒ Part 2: 16,16,800 ÷ 43 = (16168 × 100) ÷ 43 = 376 × 100 = 37600

⇒ Part 3: 43 ÷ 43 = 1

⇒ Since all individual parts yield an exact integer quotient without any remainder, the complete number is absolutely divisible by 43.

∴ The correct answer is 43.

Additional Information

Divisibility by 2

A number is perfectly divisible by 2 only if its last digit is an even number (0, 2, 4, 6, or 8).

Divisibility by 10

A number is completely divisible by 10 (and its multiples like 50) strictly if its unit digit is 0.

Prime Number Divisibility Method

For larger primes like 43 or 47, breaking the large dividend into chunks that are easily recognizable multiples of the divisor is the quickest method to verify divisibility during an exam.

Shortcut Trick

Effective Rate % = Rate × Time

Effective Rate % = 16% × 5.5 = 88%

Simple Interest = Effective Rate % of Principal

⇒ SI = 88% of 400 = (88 ÷ 100) × 400 = 352

∴ The correct answer is ₹352.

Alternate Method

Given:

Principal (P) = ₹400

Rate of Interest (R) = 16% per annum

Time (T) = 5.5 years

Formula Used:

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

Simple Interest Calculation

P = ₹400

T = 5.5 yrs
R = 16% p.a.

SI = ?

SI = (P × R × T) ÷ 100

Calculation:

⇒ SI = (400 × 16 × 5.5) ÷ 100

⇒ SI = 4 × 16 × 5.5

⇒ SI = 64 × 5.5

⇒ SI = 352

∴ The correct answer is ₹352.

Additional Information

Amount in Simple Interest

The total amount (A) after a certain time is the sum of the Principal (P) and the Simple Interest (SI).

Formula: A = P + SI = P × [1 + (R × T) ÷ 100]

Compound Interest (CI)

Unlike Simple Interest, Compound Interest is calculated on the initial principal and the accumulated interest of previous periods.

Formula: A = P × [1 + (R ÷ 100)]^T, and CI = A − P

Difference Between CI and SI

For 2 years, the difference between Compound Interest and Simple Interest for the same principal and rate is given by:

Difference = P × (R ÷ 100)²

Shortcut Trick

Total work = LCM(11, 15) = 165 units.

Lokesh's efficiency = 15 units/day, Manoj's efficiency = 11 units/day.

Work done together in 5 days = (15 + 11) × 5 = 130 units.

Remaining work = 165 − 130 = 35 units.

Time for Lokesh to finish remaining work = 35 ÷ 15 = 7/3 = 2¹/₃ days.

∴ The correct answer is 2¹/₃ days.

Alternate Method

Given:

Lokesh can complete the work in 11 days.

Manoj can complete the work in 15 days.

They work together for 5 days.

Method Used:

We use the LCM method to find the total work and individual efficiencies.

Calculations:

⇒ Combined efficiency of Lokesh and Manoj = 15 + 11 = 26 units/day

⇒ Work completed by both in 5 days = 26 × 5 = 130 units

⇒ Remaining work = Total work − Work completed

⇒ Remaining work = 165 − 130 = 35 units

⇒ Time taken by Lokesh to complete the remaining work = Remaining work ÷ Lokesh's efficiency

⇒ Time = 35 ÷ 15

⇒ Time = 7/3 = 2¹/₃ days

∴ The correct answer is 2¹/₃ days.

Additional Information

Total Work Concept

Total Work is assumed as the LCM of the days taken by individuals. Efficiency is the work done in one unit of time. (Total Work = Efficiency × Time).

Pipes & Cisterns

This is similar to time and work. An inlet pipe has positive efficiency while an outlet (leak) has negative efficiency.

Alternate Day Work

When individuals work on alternate days, calculate the work done in a complete cycle of 2 days and then determine the remaining work iteratively.

Work and Wages

Wages are always distributed in the ratio of the total work done by each person. If they work for the same duration, wages are distributed in the ratio of their efficiencies.

Shortcut Trick

Total work = LCM(11, 26, 12) = 1716 units.

Efficiency of Pipe A = 1716 ÷ 11 = 156 units/hour.

Efficiency of Pipe B = 1716 ÷ 26 = 66 units/hour.

Efficiency of Pipe C = 1716 ÷ 12 = 143 units/hour.

Total efficiency = 156 + 66 + 143 = 365 units/hour.

Time taken = Total work ÷ Total efficiency = 1716 ÷ 365 = 4 ²⁵⁶/₃₆₅ hours.

∴ The correct answer is 4 ²⁵⁶/₃₆₅ hours.

Alternate Method

Given:

Time taken by Pipe A = 11 hours

Time taken by Pipe B = 26 hours

Time taken by Pipe C = 12 hours

Formula Used:

Time = Total Work ÷ Total Efficiency

Calculations:

⇒ Total efficiency of A, B, and C = 156 + 66 + 143 = 365 units/hour

⇒ Time taken by all three pipes together = Total Work ÷ Total Efficiency

⇒ Time taken = 1716 ÷ 365

⇒ Time taken = 4 ²⁵⁶/₃₆₅ hours

∴ The correct answer is 4 ²⁵⁶/₃₆₅ hours.

Additional Information

Inlet vs Outlet Pipes

An inlet pipe fills the tank and its work is considered positive. An outlet pipe empties the tank and its work is considered negative.

Combined Efficiency Method

When multiple pipes operate together, their net rate of filling or emptying is the algebraic sum of their individual efficiencies: Net Efficiency = Σ(Inlet Efficiencies) − Σ(Outlet Efficiencies).

Work Capacity & Volume

If the flow rate (e.g., liters per minute) of a pipe is given, the total capacity of the tank can be found by multiplying the flow rate by the total time taken by that pipe to independently fill the tank.

Shortcut Trick

Total Distance = 252 + 891 + 183 = 1326 km

Total Time = (252 ÷ 63) + (891 ÷ 99) + (183 ÷ 61) = 4 + 9 + 3 = 16 hours

Average Speed = Total Distance ÷ Total Time = 1326 ÷ 16 = 82.875 km/hr

Rounding off 82.875 to two decimal places gives 82.88 km/hr.

∴ The correct answer is 82.88 km/hr.

Alternate Method

Given:

Part 1: Distance = 252 km, Speed = 63 km/hr

Part 2: Distance = 891 km, Speed = 99 km/hr

Part 3: Distance = 183 km, Speed = 61 km/hr

Formula Used:

Average Speed = Total Distance ÷ Total Time

Time = Distance ÷ Speed

252 km
891 km
183 km

63 km/hr
99 km/hr
61 km/hr

T = 4 hr
T = 9 hr
T = 3 hr

Total D = 1326 km | Total T = 16 hr

Calculations:

⇒ Total Distance = 252 + 891 + 183 = 1326 km

⇒ Total Time = (252 ÷ 63) + (891 ÷ 99) + (183 ÷ 61)

⇒ Total Time = 4 + 9 + 3 = 16 hours

⇒ Average Speed = 1326 ÷ 16

⇒ Average Speed = 82.875 km/hr

⇒ Rounding off to two decimal places, we get 82.88 km/hr.

∴ The correct answer is 82.88 km/hr.

Additional Information

Average Speed for Equal Distances

If a person travels two equal distances at speeds S₁ and S₂, then Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Relative Speed

For objects moving in the same direction, Relative Speed = S₁ − S₂. For objects moving in opposite directions, Relative Speed = S₁ + S₂.

Unit Conversion

To convert speed from km/hr to m/s, multiply the value by 5/18. To convert m/s to km/hr, multiply by 18/5.

Shortcut Trick
Total CP = Cost + Decoration = 199 + 9 = ₹208
Profit = Selling Price − Total CP = 231 − 208 = ₹23
Profit % = (Profit ÷ Total CP) × 100
Profit % = (23 ÷ 208) × 100 = 575/52%
∴ The correct answer is 575/52% profit.

Alternate Method
Given:
Cost price of the toy = ₹199
Cost of decoration = ₹9
Selling price (SP) = ₹231

Total CP = 208
199 + 9 (decor)

SP = 231

• 23 Profit

Formula Used:
Total CP = Original Cost Price + Overhead Charges
Profit = SP − Total CP
Profit % = (Profit ÷ Total CP) × 100

Calculations:
⇒ Total CP = 199 + 9 = ₹208
⇒ Since Selling Price > Total Cost Price, there is a profit.
⇒ Profit = 231 − 208 = ₹23
⇒ Profit % = (23 ÷ 208) × 100
⇒ Profit % = (23 ÷ 52) × 25
⇒ Profit % = 575/52%
∴ The correct answer is 575/52% profit.

Additional Information
Overhead Expenses
Any extra money spent on transportation, repairs, or decoration is always added to the base cost price to determine the actual effective cost price (Total CP) before calculating overall profit or loss.
Loss Percentage
When Total CP > SP, there is a loss instead of a profit. Loss = Total CP − SP, and the formula is: Loss % = (Loss ÷ Total CP) × 100.
Marked Price and Discount
A discount is universally calculated on the Marked Price (MP) of an item. The selling price after a discount can be found as: SP = MP × (100 − Discount%) ÷ 100.

Shortcut Trick

Savings percentage = 100% − 75% = 25%.

Given: 25% units = ₹4,827.

Expenses = 75% units = 3 × 25% units.

Monthly Expenses = 3 × 4,827 = ₹14,481.

∴ The correct answer is ₹14,481.

Alternate Method

Given:

Spending = 75% of Monthly Income

Savings = ₹4,827

Formula Used:

Income = Expenditure + Savings

Calculations:

⇒ Let Total Income be x.

⇒ Expenditure = 75% of x = 0.75x

⇒ Savings = x − 0.75x = 0.25x

⇒ 0.25x = 4,827

⇒ x = 4,827 ÷ 0.25 = 4,827 × 4 = 19,308

⇒ Monthly Expenses = 75% of 19,308

⇒ Expenses = 0.75 × 19,308 = ₹14,481

∴ The correct answer is ₹14,481.

Additional Information

Percentage Change

Percentage increase or decrease is calculated as (Change ÷ Original Value) × 100.

Income-Expenditure Relation

If income increases by a% and expenditure increases by b%, the percentage change in savings depends on the initial ratio of expenditure to savings.

Base Change Rule

To find the total value when a part is known: Total = (Part Value ÷ Part Percentage) × 100.

Shortcut Trick

Simplify ratio: 4/8 : 11/16 : 18/21 → 1/2 : 11/16 : 6/7.

Multiply by LCM(2, 16, 7) = 112 → 56 : 77 : 96.

Difference between greatest (96) and smallest (56) = 40 units.

Given 40 units = 40, so 1 unit = 1. Largest number = 96 units = 96.

∴ The correct answer is 96.

Alternate Method

Given: Ratio of three numbers = 4/8 : 11/16 : 18/21

Step 1: Simplify fractions → 1/2 : 11/16 : 6/7

Step 2: To remove denominators, multiply by LCM of 2, 16, and 7, which is 112.

⇒ Number 1 = 1/2 × 112 = 56

⇒ Number 2 = 11/16 × 112 = 11 × 7 = 77

⇒ Number 3 = 6/7 × 112 = 6 × 16 = 96

Step 3: Identify greatest and smallest.

⇒ Greatest = 96x, Smallest = 56x

⇒ Difference = 96x − 56x = 40x

Step 4: Solve for x using given difference.

⇒ 40x = 40 ⇒ x = 1

⇒ Largest number = 96 × 1 = 96

∴ The correct answer is 96.

Additional Information

Ratio Simplification

To simplify a ratio involving fractions, multiply each term by the Least Common Multiple (LCM) of all the denominators to convert them into integers.

Proportional Parts

If a total value V is divided in ratio a:b:c, the largest part is calculated as [Max(a,b,c) / (a+b+c)] × V.

Difference Rule

When the difference between two parts is given, 1 unit value = (Actual Difference) ÷ (Difference in Ratio parts).

Shortcut Trick

For fourth proportional: P : 25 :: 8 : 4

Product of extremes = Product of means

P × 4 = 25 × 8

P = 25 × 2 = 50

∴ The correct answer is 50.

Alternate Method

Given:

First term = P, Second term = 25, Third term = 8, Fourth term = 4

Formula Used:

If a, b, c, d are in proportion, then a/b = c/d

Calculations:

⇒ P / 25 = 8 / 4

⇒ P / 25 = 2

⇒ P = 2 × 25

⇒ P = 50

∴ The correct answer is 50.

Additional Information

Fourth Proportional

If four quantities a, b, c, and d are in proportion, then d is called the fourth proportional. Formula: d = (b × c) / a.

Third Proportional

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

Mean Proportional

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

Shortcut Trick

The Mode is defined as the observation that occurs most frequently in the given data set.

By scanning the data, we observe the frequency of each number:

Value 48 occurs 5 times (highest frequency).

Value 42 occurs 3 times, while others occur 2 times or once.

∴ The correct answer is 48.

Alternate Method

Given Data: 41, 48, 54, 50, 54, 48, 42, 43, 48, 42, 53, 45, 44, 50, 49, 45, 43, 48, 42, 48

Formula Used: Mode = Value with the Maximum Frequency

To find the Mode, we construct a frequency distribution table:

⇒ Frequency of 41 = 1

⇒ Frequency of 42 = 3

⇒ Frequency of 43 = 2

⇒ Frequency of 44 = 1

⇒ Frequency of 45 = 2

⇒ Frequency of 48 = 5

⇒ Frequency of 49 = 1

⇒ Frequency of 50 = 2

⇒ Frequency of 53 = 1

⇒ Frequency of 54 = 2

⇒ Since 48 has the maximum frequency of 5.

∴ The correct answer is 48.

Additional Information

Empirical Relationship

For a moderately skewed distribution, the relationship between measures of central tendency is: Mode = 3 Median − 2 Mean.

Types of Mode

A data set can be Unimodal (one mode), Bimodal (two modes), or Multimodal (more than two modes). If all values occur with the same frequency, there is no mode.

Mean and Median

Mean is the sum of observations divided by their count. Median is the middle value when data is arranged in ascending or descending order.

Shortcut Trick

If x + 1/x = k, then the value of x² + 1/x² = k² − 2.

Here, the given value k = 17.

Required value = 17² − 2

Required value = 289 − 2 = 287.

∴ The correct answer is 287.

Alternate Method

Given: x + 1/x = 17

Formula Used: (a + b)² = a² + b² + 2ab

On squaring both sides of the given equation:

⇒ (x + 1/x)² = (17)²

⇒ x² + (1/x)² + 2 × x × (1/x) = 289

⇒ x² + 1/x² + 2 = 289

⇒ x² + 1/x² = 289 − 2

⇒ x² + 1/x² = 287

∴ The correct answer is 287.

Additional Information

Negative Sign Identity

If x − 1/x = k, then x² + 1/x² = k² + 2.

Cubic Algebraic Identity

If x + 1/x = k, then x³ + 1/x³ = k³ − 3k.

Power of Four Identity

If x + 1/x = k, then x⁴ + 1/x⁴ = (k² − 2)² − 2.

Shortcut Trick

Assume the smaller number is x.

The other number is 2x + 5.

Sum: x + (2x + 5) = 32 → 3x = 27.

Solving gives x = 9.

∴ The correct answer is 9.

Alternate Method

Given: Sum of two numbers = 32. Relation: Larger = 2 × Smaller + 5.

Let the smaller number be x and the larger number be y.

⇒ x + y = 32 ...

(i) ⇒ y = 2x + 5 ...

(ii) Substitute y from (ii) into (i):

⇒ x + (2x + 5) = 32

⇒ 3x + 5 = 32

⇒ 3x = 32 − 5

⇒ 3x = 27

⇒ x = 27 ÷ 3 = 9

∴ The correct answer is 9.

Additional Information

Linear Equations in One Variable

A linear equation involves a single variable with a power of 1, typically solved by isolating the variable on one side.

Substitution Method

This involves expressing one variable in terms of another and substituting it into a second equation to find the value.

Sum and Difference Problems

If the sum (S) and difference (D) are known, larger number = (S+D)/2 and smaller number = (S−D)/2.

Shortcut Trick

The difference in ages remains constant (2 years).

Ratio after 18 years = 21 : 22. Difference in ratio parts = 22 − 21 = 1 unit.

1 unit = 2 years → Gopal = 21 × 2 = 42 years, Manoj = 22 × 2 = 44 years (after 18 years).

Sum 10 years ago = (42 + 44) − 28 (for 18 years back to now + 10 years back) = 86 − 56 = 30 years.

∴ The correct answer is 30 years.

Alternate Method

Let the present ages of Gopal and Manoj be G and M respectively.

Given: M − G = 2 years (Since Manoj is older in the ratio 21:22).

After 18 years: (G + 18) / (M + 18) = 21 / 22

⇒ 22(G + 18) = 21(M + 18)

⇒ 22G + 396 = 21M + 378

⇒ 21M − 22G = 18

Substitute M = G + 2:

⇒ 21(G + 2) − 22G = 18 ⇒ 21G + 42 − 22G = 18

⇒ −G = 18 − 42 ⇒ G = 24 years, M = 26 years.

Sum of ages 10 years ago = (24 − 10) + (26 − 10) = 14 + 16 = 30 years.

∴ The correct answer is 30 years.

Additional Information

Constant Age Difference

The numerical difference between the ages of two individuals remains the same at any point in time (past, present, or future).

Ratio Balancing

If the age difference is 'd' and the ratio is a:b, then (b − a) units = d years. This helps in finding ages without complex algebra.

Shortcut Trick

Identify the Highest Common Factor (HCF) first to reduce the numbers.

HCF of 780, 2080, and 2600 is 260.

Divide numbers by 260: 780/260 = 3, 2080/260 = 8, 2600/260 = 10.

LCM of 3, 8, and 10 is 120.

Final LCM = 260 × 120 = 31200.

∴ The correct answer is 31200.

Alternate Method

Given: Numbers are 780, 2080, and 2600.

Formula Used: Prime Factorization Method for finding LCM.

Prime factorization of each number:

⇒ 780 = 2 × 2 × 3 × 5 × 13 = 2² × 3¹ × 5¹ × 13¹

⇒ 2080 = 2 × 2 × 2 × 2 × 2 × 5 × 13 = 2⁵ × 5¹ × 13¹

⇒ 2600 = 2 × 2 × 2 × 5 × 5 × 13 = 2³ × 5² × 13¹

⇒ LCM = Product of highest powers of all prime factors

⇒ LCM = 2⁵ × 3¹ × 5² × 13¹

⇒ LCM = 32 × 3 × 25 × 13

⇒ LCM = 96 × 325 = 31200

∴ The correct answer is 31200.

Additional Information

LCM (Least Common Multiple)

The smallest positive integer that is divisible by each of the given numbers. Found using division method or prime factorization.

HCF and LCM Relationship

For two numbers A and B: Product of A × B = HCF(A,B) × LCM(A,B).

LCM of Fractions

LCM of fractions = LCM of Numerators ÷ HCF of Denominators.

Shortcut Trick

Total Distance = Sum of lengths = 248 + 142 = 390 m

Relative Speed = Sum of speeds (opposite direction) = 34 + 74 = 108 kmph

Convert kmph to m/s: 108 × (5/18) = 30 m/s

Time = Distance ÷ Relative Speed = 390 ÷ 30 = 13 sec

∴ The correct answer is 13 sec.

Alternate Method

Given:

Length of first train (L₁) = 248 m

Length of second train (L₂) = 142 m

Speed of first train (S₁) = 34 km/hr

Speed of second train (S₂) = 74 km/hr

Formula Used:

Relative Speed (Opposite Direction) = S₁ + S₂

Time = (L₁ + L₂) ÷ Relative Speed

Calculations:

⇒ Total distance = 248 + 142 = 390 m

⇒ Relative speed = 34 + 74 = 108 km/hr

⇒ Speed in m/s = 108 × (5/18) = 6 × 5 = 30 m/s

⇒ Time taken = 390 ÷ 30

⇒ Time taken = 13 seconds

∴ The correct answer is 13 sec.

Additional Information

Relative Speed Concepts

When objects move in opposite directions, relative speed = S₁ + S₂. When moving in the same direction, relative speed = |S₁ − S₂|.

Unit Conversion

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

Train Crossing a Point Object

If a train crosses a pole or a stationary man, the distance traveled is equal to the length of the train itself.

Shortcut Trick

Speed ratio (New : Normal) = 3 : 4

Since distance is constant, Time ratio (New : Normal) = 4 : 3

Difference in time ratio parts = 4 − 3 = 1 unit

Given, 1 unit = 19 minutes

Usual time = 3 units = 3 × 19 minutes = 57 minutes

∴ The correct answer is 57 min.

Alternate Method

Given:

New Speed = 3/4 × Normal Speed

Late time = 19 minutes

Formula Used:

Distance = Speed × Time

Calculations:

⇒ Let normal speed be S and usual time be T.

⇒ Normal Distance = S × T

⇒ New Speed = 3S/4

⇒ New Time = T + 19

⇒ Since distance is same: S × T = (3S/4) × (T + 19)

⇒ T = (3/4) × (T + 19)

⇒ 4T = 3T + 57

⇒ T = 57 minutes

∴ The correct answer is 57 min.

Additional Information

Inverse Proportion Rule

When distance is constant, speed is inversely proportional to time (S ∝ 1/T). If speed becomes a/b of its original, time becomes b/a of the original.

General Formula for Late/Early

Usual Time = (Late time × Numerator) ÷ (Difference between Denominator and Numerator) when speed is a/b of usual speed.

Average Speed

If a person covers equal distances at speeds x and y, then Average Speed = 2xy/(x + y).