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Shortcut Trick

Speed of 2^nd train = 300 ÷ 3 = 100 km/h.

Ratio of speeds (Train 1 : Train 2) = 5 : 8.

Since 8 units = 100 km/h ⇒ 1 unit = 12.5 km/h.

Speed of 1^st train = 5 units = 5 × 12.5 = 62.5 km/h.

∴ The correct answer is 62.5 km/h.

Alternate Method

Given: Ratio of speeds = 5 : 8 | Distance (Train 2) = 300 km | Time (Train 2) = 3 hours.

Formula Used: Speed = Distance ÷ Time

⇒ Speed of 2^nd train = Distance ÷ Time

⇒ Speed of 2^nd train = 300 ÷ 3 = 100 km/h.

⇒ Let the speed of the 1^st train be 5x and the 2^nd train be 8x.

⇒ 8x = 100

⇒ x = 100 ÷ 8 = 12.5

⇒ Speed of 1^st train = 5x = 5 × 12.5 = 62.5 km/h.

∴ The correct answer is 62.5 km/h.

Additional Information

Speed-Distance Relationship

When time is constant, Speed is directly proportional to Distance (S₁/S₂ = D₁/D₂).

Unit Conversion

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

Relative Speed

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

Shortcut Trick

Identify the mode by finding the observation with the highest frequency.

Number 2 appears 4 times, 3 appears 3 times, 4 appears 2 times, and 5 appears 1 time.

The value 2 occurs the maximum number of times.

∴ The correct answer is 2 hr.

Alternate Method

Given: Data points for 10 students = 2, 3, 4, 2, 5, 3, 2, 4, 2, 3.

Formula: Mode is the value that appears most frequently in a data set.

To find the mode, we construct a frequency table:

⇒ Data value 2: Frequency = 4

⇒ Data value 3: Frequency = 3

⇒ Data value 4: Frequency = 2

⇒ Data value 5: Frequency = 1

⇒ Comparing frequencies: 4 > 3 > 2 > 1

⇒ Since the value 2 hr occurs with the maximum frequency of 4.

∴ The correct answer is 2 hr.

Additional Information

Arithmetic Mean

The average of the data set calculated as (Sum of all observations) ÷ (Total number of observations).

Median

The middle value of a data set when arranged in ascending or descending order. For even n, it is the mean of (n/2)^th and (n/2 + 1)^th terms.

Empirical Relationship

For moderately skewed distributions: Mode = 3 × Median − 2 × Mean.

Bimodal and Multimodal

A data set with two modes is called bimodal, and one with more than two modes is called multimodal.

Shortcut Trick

If two numbers x and y are given, their Third proportional is calculated as y² ÷ x.

Here, let x = a² − b² and y = a + b.

Third proportional = (a + b)² ÷ (a² − b²)

Applying identity a² − b² = (a − b)(a + b), we get (a + b)² ÷ ((a − b)(a + b)).

Simplified result = (a + b) / (a − b).

∴ The correct answer is ((a+b))/((a-b)).

Alternate Method

Given:

First term = a² − b²

Second term = a + b

Condition: a ≠ b

We know that if x, y, z are in Continued Proportion, then:

x ÷ y = y ÷ z

⇒ z = y² ÷ x

Substitute the given expressions:

⇒ z = (a + b)² ÷ (a² − b²)

Using the algebraic identity (a² − b²) = (a + b)(a − b):

⇒ z = ((a + b) × (a + b)) ÷ ((a + b) × (a − b))

Canceling the common term (a + b) from the numerator and denominator:

⇒ z = (a + b) ÷ (a − b)

∴ The correct answer is ((a+b))/((a-b)).

Additional Information

Mean Proportional

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

Fourth Proportional

If four quantities a, b, c, d are in proportion (a:b :: c:d), then the fourth proportional d is (b × c) ÷ a.

Continued Proportion

Three quantities a, b, c are in continued proportion if a/b = b/c, which implies b² = ac.

Shortcut Trick

To find the Ratio of Boys to Teachers, equate the common entity (Girls) in both ratios.

Boys : Girls = 5 : 7 and Girls : Teachers = 14 : 1

Multiply the first Ratio by 2 to make the Girls count equal to 14: (5 × 2) : (7 × 2) = 10 : 14.

Now, the combined Ratio of Boys : Girls : Teachers = 10 : 14 : 1.

Extracting the required terms, Boys : Teachers = 10 : 1.

∴ The correct answer is 10:1.

Alternate Method

Given:

⇒ Ratio of Boys to Girls = 5 : 7

⇒ Ratio of Girls to Teachers = 14 : 1

Formula Used:

⇒ If A:B and B:C are given, then A:C = (A/B) × (B/C)

Calculations:

⇒ Let Boys be B, Girls be G, and Teachers be T.

⇒ B / G = 5 / 7

⇒ G / T = 14 / 1

⇒ To find B / T, we multiply the two ratios: (B / G) × (G / T)

⇒ B / T = (5 / 7) × (14 / 1)

⇒ B / T = (5 × 14) / (7 × 1)

⇒ B / T = 70 / 7

⇒ B / T = 10 / 1

⇒ B : T = 10 : 1

∴ The correct answer is 10:1.

Additional Information

Compound Ratio

When two or more Ratios are multiplied together, the result is called a Compound Ratio. For ratios a:b and c:d, the Compound Ratio is (ac):(bd).

Duplicate Ratio

The Duplicate Ratio of any ratio a:b is calculated by squaring both terms, resulting in a² : b².

Proportion

A Proportion is an equation which states that two Ratios are equal (a/b = c/d), where the product of extremes equals the product of means (ad = bc).

Shortcut Trick

HCF = LCM of numbers ÷ LCM of ratio parts

LCM of ratio parts 10, 13, and 11 = 10 × 13 × 11 = 1430

HCF = 2860 ÷ 1430 = 2

∴ The correct answer is 2.

Alternate Method

Given:

Ratio of three numbers = 10 : 13 : 11

LCM of the numbers = 2860

Formula Used:

If numbers are in ratio a : b : c, then their LCM = HCF × LCM(a, b, c).

Calculations:

⇒ Let the HCF of the three numbers be x.

⇒ The numbers are 10x, 13x, and 11x.

⇒ Since 10, 13, and 11 have no common factors (they are co-prime), their LCM = 10 × 13 × 11 = 1430.

⇒ LCM of the numbers = x × 1430

⇒ 2860 = 1430x

⇒ x = 2860 ÷ 1430

⇒ x = 2

∴ The correct answer is 2.

Additional Information

Relationship for Two Numbers

For any two numbers a and b, Product of numbers (a × b) = HCF(a, b) × LCM(a, b).

HCF (Highest Common Factor)

The HCF is the largest positive integer that divides each of the given numbers without leaving a remainder.

LCM (Least Common Multiple)

The LCM is the smallest positive integer that is divisible by each of the given numbers.

Co-prime Numbers

Two or more numbers are said to be co-prime if their HCF is 1. Their LCM is simply the product of the numbers.

Shortcut Trick

The age difference between two people always remains constant over time.

Ratio of X and Y = 4 : 5 ⇒ Difference = 5 − 4 = 1 unit.

Actual difference = 3 years (given for 5 years later, but it is constant).

∴ 1 unit = 3 years.

Sum of present ages = (4 + 5) units = 9 units = 9 × 3 = 27 years.

∴ The correct answer is 27 years.

Alternate Method

Given: Present age ratio of X and Y = 4 : 5. Future age difference = 3 years.

Formula Used: Age = Ratio Part × Value of 1 Unit; Age difference is invariant.

⇒ Let the present ages of X and Y be 4k and 5k respectively.

⇒ Age of X after 5 years = (4k + 5) years.

⇒ Age of Y after 5 years = (5k + 5) years.

⇒ Difference = (5k + 5) − (4k + 5) = 3

⇒ k = 3.

⇒ X's present age = 4 × 3 = 12 years.

⇒ Y's present age = 5 × 3 = 15 years.

⇒ Sum of present ages = 12 + 15 = 27 years.

∴ The correct answer is 27 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals remains the same throughout their lives, regardless of how many years pass.

Sum of Ages in Future/Past

If the current sum of ages of 'n' people is S, then after 't' years, the sum of ages will be S + (n × t).

Ratio Proportionality

In age problems, if the ratio is a : b and the difference is 'd', the units can be directly mapped: (b − a) units = d.

Shortcut Trick

Ratio of Cost Price (CP) to Selling Price (SP) based on price: 25 : (32 × 0.8) = 25 : 25.6

Ratio based on Weight fraud: 800 : 1000 = 4 : 5

Combined Ratio (CP : SP) = (25 × 4) : (25.6 × 5) = 100 : 128

Profit % = 128 − 100 = 28%

∴ The correct answer is 28% profit.

Alternate Method

Given: CP = ₹25/kg, MP = ₹32/kg, Discount = 20%, False weight = 800 g instead of 1000 g.

Formula Used: Profit % = [(Total SP − Total CP) ÷ Total CP] × 100

⇒ Selling Price (SP) per kg claimed = 32 − (20% of 32)

⇒ SP = 32 − 6.4 = ₹25.6

⇒ Since the shopkeeper uses 800 g instead of 1 kg, his Actual CP is for 800 g.

⇒ Actual CP = (25 ÷ 1000) × 800 = ₹20

⇒ Profit = 25.6 − 20 = ₹5.6

⇒ Overall Profit % = (5.6 ÷ 20) × 100

⇒ Profit % = 0.28 × 100 = 28%

∴ The correct answer is 28% profit.

Additional Information

Dishonest Dealer Weight Gain

When a dealer sells at CP but uses less weight, Profit % = [Error ÷ (True Value − Error)] × 100.

Marked Price and Discount

Marked Price (MP) = CP × [100 + Markup%] ÷ 100. Selling Price (SP) = MP × [100 − Discount%] ÷ 100.

Successive Percentage Change

Overall change for two successive percentages a and b is [a + b + (a × b)/100] %. Use negative values for discounts or losses.

Shortcut Trick

To find the maximum possible length of each equal piece from ropes of lengths 54 m, 72 m, and 90 m, we directly calculate their Highest Common Factor (HCF).

Lengths of the ropes: 54 m, 72 m, 90 m

HCF(54, 72, 90) = 18

Thus, each rope can be cut into smaller pieces of exactly 18 m with no leftover rope.

∴ The correct answer is 18 m.

Alternate Method

Given:

Length of the first rope = 54 m

Length of the second rope = 72 m

Length of the third rope = 90 m

Formula Used:

Maximum possible length of each piece = HCF of (54, 72, 90)

54 m 18m 18m

Shortcut Trick

Price after first 15% discount = 2000 − 300 = ₹1700.

Required second discount amount = 1700 − 1530 = ₹170.

Second discount % = (170 ÷ 1700) × 100 = 10%.

∴ The correct answer is 10%.

Alternate Method

Given:

Marked Price (MP) of the jacket = ₹2,000

Final Selling Price (SP) = ₹1,530

First Discount (D₁) = 15%

Formula Used:

Intermediate Price = MP × (100 − D₁) / 100

D₂ % = [(Intermediate Price − Final SP) / Intermediate Price] × 100

Calculations:

⇒ First discount amount = 15% of 2000 = (15/100) × 2000 = ₹300

⇒ Price after first discount = 2000 − 300 = ₹1700

⇒ Let the second discount percentage be x%

⇒ Second discount amount = 1700 − 1530 = ₹170

⇒ x = (Second discount amount / Price after first discount) × 100

⇒ x = (170 / 1700) × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Successive Discount Formula

The single equivalent discount for two successive discounts of a% and b% is given by (a + b − ab/100)%.

Selling Price and Marked Price Relation

SP = MP × (1 − d₁/100) × (1 − d₂/100), where d₁ and d₂ are the successive discount percentages.

Marked Price Calculation

Marked Price = [Cost Price × (100 + Profit%)] / (100 − Discount%).

Shortcut Trick

Simplify √20 as 2√5.

The expression becomes (√5 + 2√5)².

Add like surds: (3√5)².

Square the term: 3² × (√5)² = 9 × 5 = 45.

∴ The correct answer is 45.

Alternate Method

Given: The expression to simplify is (√5 + √20)².

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

⇒ Let a = √5 and b = √20.

⇒ (√5)² + (√20)² + 2 × √5 × √20

⇒ 5 + 20 + 2 × √(5 × 20)

⇒ 25 + 2 × √100

⇒ 25 + 2 × 10

⇒ 25 + 20

⇒ 45

∴ The correct answer is 45.

Additional Information

Algebraic Identities

The square of a sum is given by (a + b)² = a² + 2ab + b², while the square of a difference is (a − b)² = a² − 2ab + b².

Properties of Surds

For any positive real numbers x and y, the product rule states √x × √y = √(xy) and the power rule states (√x)² = x.

Simplifying Square Roots

A square root is simplified by extracting the largest perfect square factor, such as √20 = √(4 × 5) = 2√5.