Search Previous Year Questions

Shortcut Trick

Profit % = 25% = 1/4

Ratio of Price per article (CP : SP) = 4 : 5

Since Total Cost Price = Total Selling Price, the quantity ratio is inversely proportional to the price ratio.

Quantity Ratio (CP items : SP items) = 5 : 4

According to the question, 5 units = 25 articles ⇒ 1 unit = 5 articles

Therefore, x = 4 units = 4 × 5 = 20

∴ The correct answer is 20.

Alternate Method

Given:

Cost Price (CP) of 25 articles = Selling Price (SP) of x articles.

Profit percentage = 25%.

Formula Used:

Profit % = [(Articles Bought − Articles Sold) ÷ Articles Sold] × 100

Calculations:

⇒ Let the Cost Price of 1 article be ₹1.

⇒ Total Cost Price of 25 articles = 25 × 1 = ₹25.

⇒ Given, Selling Price of x articles = ₹25.

⇒ Cost Price of x articles = x × 1 = ₹x.

⇒ Profit = SP − CP = 25 − x.

⇒ Profit % = [(25 − x) ÷ x] × 100

⇒ 25 = [(25 − x) ÷ x] × 100

⇒ 25 ÷ 100 = (25 − x) ÷ x

⇒ 1 ÷ 4 = (25 − x) ÷ x

⇒ x = 4 × (25 − x)

⇒ x = 100 − 4x

⇒ 5x = 100

⇒ x = 20

∴ The correct answer is 20.

Additional Information

Profit/Loss on Articles

When the Cost Price of m items equals the Selling Price of n items, the Profit/Loss % = [(m − n) ÷ n] × 100. If positive, it is a profit; if negative, it is a loss.

Inverse Variation

For a constant total amount, the quantity of articles is inversely proportional to the price per article. Ratio of Price (a:b) ⇒ Ratio of Quantity (b:a).

Markup and Discount

The relationship between Marked Price, Cost Price, and Selling Price is defined by: SP = CP × (1 + Profit%) and SP = MP × (1 − Discount%).

Shortcut Trick

Total Distance = 380 + 340 = 720 m

Speed (m/s) = 720 ÷ 16 = 45 m/s

Speed (km/h) = 45 × (18/5) = 162 km/h

∴ The correct answer is 162 km/h.

Alternate Method

Given:

Length of the moving train (L₁) = 380 m

Length of the stationary train (L₂) = 340 m

Time taken (t) = 16 sec

Formula Used:

Speed = Total Distance ÷ Total Time

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

Calculations:

⇒ Total Distance = L₁ + L₂

⇒ Total Distance = 380 + 340 = 720 m

⇒ Speed = 720 ÷ 16

⇒ Speed = 45 m/s

⇒ Speed in km/h = 45 × (18/5)

⇒ Speed in km/h = 9 × 18

⇒ Speed in km/h = 162 km/h

∴ The correct answer is 162 km/h.

Additional Information

Speed Conversion

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

Train Crossing a Pole/Point Object

When a train crosses a pole, tree, or person, the distance covered is equal to the length of the train itself.

Relative Speed

If two objects move in the opposite direction, relative speed = S₁ + S₂. If they move in the same direction, relative speed = |S₁ − S₂|.

Shortcut Trick

The given ratio is 3 : 2 : 13.

Since the numbers in the ratio are pairwise co-prime (GCD of any two is 1).

LCM of ratio parts = 3 × 2 × 13 = 78 units.

Given LCM = 6318.

1 unit (HCF) = 6318 ÷ 78 = 81.

∴ The correct answer is 81.

Alternate Method

Given: Ratio of three numbers = 3 : 2 : 13 and their LCM = 6318.

Formula Used: LCM of (ax, bx, cx) = x × LCM(a, b, c), where x is the HCF of the numbers.

⇒ Let the three numbers be 3x, 2x, and 13x, where x is the HCF.

⇒ LCM of (3x, 2x, 13x) = x × LCM(3, 2, 13).

⇒ Since 3, 2, and 13 are prime numbers (or pairwise co-prime), their LCM is their product.

⇒ LCM(3, 2, 13) = 3 × 2 × 13 = 78.

⇒ Given that the LCM is 6318, we have: 78x = 6318.

⇒ x = 6318 ÷ 78.

⇒ x = 81.

∴ The correct answer is 81.

Additional Information

Relationship between HCF and LCM

For any two numbers a and b, the product of the numbers is equal to the product of their HCF and LCM: a × b = HCF × LCM.

HCF of Co-prime Numbers

The Highest Common Factor (HCF) of any two or more co-prime numbers is always 1.

LCM of Fractions

To find the LCM of fractions, use the formula: LCM = (LCM of Numerators) ÷ (HCF of Denominators).

Shortcut Trick

Let Income = 100, then Expenditure = 70 and Savings = 30.

Change in Income = 25% of 100 = 25.

Change in Expenditure = 10% of 70 = 7.

Change in Savings = 25 − 7 = 18.

Percentage Increase in Savings = (18 ÷ 30) × 100 = 60%.

∴ The correct answer is 60%.

Alternate Method

Given:

Initial Expenditure = 70% of Income

Increase in Income = 25%

Increase in Expenditure = 10%

Let initial Income be ₹100.

⇒ Initial Expenditure = 70% of 100 = ₹70

⇒ Initial Savings = 100 − 70 = ₹30

⇒ New Income = 100 + 25% of 100 = ₹125

⇒ New Expenditure = 70 + 10% of 70 = 70 + 7 = ₹77

⇒ New Savings = 125 − 77 = ₹48

⇒ Increase in Savings = 48 − 30 = ₹18

⇒ Percentage Increase = (18 ÷ 30) × 100 = 60%

∴ The correct answer is 60%.

Additional Information

Income, Expenditure and Savings Relation

The fundamental relation is Income = Expenditure + Savings. Any change in one or two components directly affects the third.

Percentage Change Formula

Percentage Change = [(Final Value − Initial Value) ÷ Initial Value] × 100. It is used to calculate the relative growth or decline.

Net Deviation Method

For weighted averages or income problems, Net Change in Income = (Weight of Exp × % Change in Exp) + (Weight of Sav × % Change in Sav).

Shortcut Trick

Sum of ages 5 years ago = 50 − (5 × 2) = 40 years.

Ratio of Mother : Daughter (5 years ago) = 7 : 1.

Total ratio parts = 7 + 1 = 8 units.

8 units = 40 years ⇒ 1 unit = 5 years.

Mother's age 5 years ago = 7 units = 7 × 5 = 35 years.

Mother's present age = 35 + 5 = 40 years.

∴ The correct answer is 40 years.

Alternate Method

Given:

Sum of present ages = 50 years.

Condition: 5 years ago, Mother = 7 × Daughter.

Calculations:

Let the present age of the daughter be x years.

Then, the present age of the mother = (50 − x) years.

5 years ago:

⇒ Daughter's age = (x − 5)

⇒ Mother's age = (50 − x − 5) = (45 − x)

According to the problem:

⇒ (45 − x) = 7 × (x − 5)

⇒ 45 − x = 7x − 35

⇒ 7x + x = 45 + 35

⇒ 8x = 80

⇒ x = 10

Mother's present age = 50 − 10 = 40 years.

∴ The correct answer is 40 years.

Additional Information

Sum of Ages Change

If there are n people, the sum of their ages changes by n × t years over a period of t years.

Age Difference Property

The difference between the ages of two individuals remains constant regardless of the number of years passed.

Ratio Method for Ages

When ratios at two different times are given, use the constant age difference to balance the ratio units for faster calculation.

Shortcut Trick

Convert perfect squares to integers: √2601 = 51 and (√1089 + √324) = 33 + 18 = 51.

The equation becomes: √x × 51 = 51 × 35.

Divide both sides by 51: √x = 35.

Square both sides: x = 35² = 1225.

∴ The correct answer is 1225.

Alternate Method

Given Expression: √x of √2601 = (√1089 + √324) × 35

Formula Used: √a × √b = √(a × b) and (n5)² = [n × (n+1)]25

⇒ First, calculate the square root of the constants:

⇒ √2601 = 51

⇒ √1089 = 33

⇒ √324 = 18

⇒ Substitute these values into the equation: √x × 51 = (33 + 18) × 35

⇒ √x × 51 = 51 × 35

⇒ √x = (51 × 35) ÷ 51

⇒ √x = 35

⇒ x = 35²

⇒ x = 1225

∴ The correct answer is 1225.

Additional Information

Squaring Numbers ending in 5

For a number ending in 5 (say A5), the square is calculated by multiplying A with (A + 1) and suffixing 25. Example: 35² = (3 × 4)25 = 1225.

Square Root Estimation

To find √2601, note that 50² = 2500. Since 2601 ends in 1, the possible root is 51 or 59. Since 2601 is very close to 2500, the root is 51.

BODMAS Rule

Always follow the order: Brackets, Of, Division, Multiplication, Addition, and Subtraction for multi-step calculations.

Shortcut Trick

Total distance = Sum of lengths = 104 + 94 = 198 m.

Relative Speed (kmph) = (Distance ÷ Time) × (18 ÷ 5)

Relative Speed = (198 ÷ 16.2) × (18 ÷ 5) = 110/9 × 18/5 = 44 kmph.

In same direction, Speed Difference = 44 kmph.

Speed of second train = 61 − 44 = 17 kmph.

∴ The correct answer is 17 km/hr.

Alternate Method

Given:

Length of Train 1 (L₁) = 104 m

Length of Train 2 (L₂) = 94 m

Speed of Train 1 (S₁) = 61 kmph

Time taken (T) = 16.2 seconds

Formula Used:

Relative Speed = Total Distance ÷ Total Time

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

Calculations:

⇒ Total Distance = L₁ + L₂ = 104 + 94 = 198 m

⇒ Relative Speed (V_rel) = Distance ÷ Time

⇒ V_rel = 198 ÷ 16.2 m/s

⇒ V_rel in kmph = (198 ÷ 16.2) × (18 ÷ 5)

⇒ V_rel = (1980 ÷ 162) × (18 ÷ 5)

⇒ V_rel = (110 ÷ 9) × (18 ÷ 5)

⇒ V_rel = 22 × 2 = 44 kmph

⇒ Since trains are in the same direction, V_rel = S₁ − S₂ (assuming Train 1 is faster)

⇒ 44 = 61 − S₂

⇒ S₂ = 61 − 44 = 17 kmph

∴ The correct answer is 17 km/hr.

Additional Information

Relative Speed Rules

When two objects move in the same direction, Relative Speed = S₁ − S₂. When they move in the opposite direction, Relative Speed = S₁ + S₂.

Unit Conversion

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

Train Crossing Distance

When a train crosses another train or a bridge, the total distance covered is the sum of the lengths of both entities.

Shortcut Trick

The Area of a Quadrilateral given one diagonal (d) and two offsets (h₁ and h₂) is calculated using the formula: 1/2 × d × (h₁ + h₂).

Given: diagonal (d) = 36 cm, perpendiculars (h₁) = 20 cm and (h₂) = 10 cm.

Area = 1/2 × 36 × (20 + 10)

Area = 18 × 30 = 540 cm².

∴ The correct answer is 540 cm².

Alternate Method

Given:

Length of the diagonal (AC) = 36 cm

Perpendicular distance (h₁) from vertex B to AC = 20 cm

Perpendicular distance (h₂) from vertex D to AC = 10 cm

Formula Used:

Area of quadrilateral = Area of ΔABC + Area of ΔADC

Area of a triangle = 1/2 × base × height

Calculations:

⇒ Area of ΔABC = 1/2 × 36 × 20

⇒ Area of ΔABC = 36 × 10 = 360 cm²

⇒ Area of ΔADC = 1/2 × 36 × 10

⇒ Area of ΔADC = 18 × 10 = 180 cm²

⇒ Total Area = 360 + 180

⇒ Total Area = 540 cm²

∴ The correct answer is 540 cm².

Additional Information

Area of a Rhombus

If the diagonals of a quadrilateral are perpendicular to each other (like in a rhombus), the area is 1/2 × d₁ × d₂.

Area of a Trapezium

For a quadrilateral with one pair of parallel sides, Area = 1/2 × (sum of parallel sides) × height.

Heron's Formula for Quadrilaterals

For a cyclic quadrilateral with sides a, b, c, and d, Area = √((s−a)(s−b)(s−c)(s−d)), where s is the semi-perimeter.

Shortcut Trick

Identify the number that repeats the maximum number of times (highest frequency).

Count occurrences: 40 (2), 42 (2), 43 (2), 45 (2), 46 (2), and 51 (3).

Since 51 has the highest frequency (3), it is the mode.

∴ The correct answer is 51.

Alternate Method

Given: Data set = 40, 41, 42, 51, 45, 43, 45, 43, 51, 40, 51, 54, 46, 50, 55, 47, 42, 46

Formula Used: Mode = Value with the maximum frequency.

⇒ Frequency of 40 = 2

⇒ Frequency of 42 = 2

⇒ Frequency of 43 = 2

⇒ Frequency of 45 = 2

⇒ Frequency of 46 = 2

⇒ Frequency of 51 = 3

⇒ Other values (41, 54, 50, 55, 47) appear only 1 time each.

⇒ Maximum frequency = 3, which corresponds to the value 51.

∴ The correct answer is 51.

Additional Information

Definition of Mode

The mode is the observation that occurs most frequently in a data set. A set can have no mode, one mode (unimodal), or multiple modes (bimodal/multimodal).

Empirical Relationship

The relationship between the three measures of central tendency is: Mode = 3 × Median − 2 × Mean.

Median

The median is the middle value of a data set when it is arranged in ascending or descending order.

Mean

The arithmetic mean is the sum of all observations divided by the total number of observations.

Shortcut Trick

A number is divisible by 24 if it is divisible by its coprime factors, 3 and 8.

For divisibility by 8, the last three digits 76B must be divisible by 8 ⇒ B = 0 (760) or B = 8 (768).

For divisibility by 3, the sum of digits (25 + A + B) must be divisible by 3.

Case 1: If B = 0, then 25 + A + 0 = 27 ⇒ A = 2. Here, A + B = 2.

Case 2: If B = 8, then 25 + A + 8 = 33 ⇒ A = 0. Here, A + B = 8.

Comparing both cases, the minimum value of (A + B) is 2.

∴ The correct answer is 2.

Alternate Method

Given: The 7-digit number 32A776B is divisible by 24.

Formula Used:

1. Divisibility by 8: The number formed by the last three digits must be divisible by 8.
2. Divisibility by 3: The sum of all digits must be divisible by 3.

Calculations:

⇒ For 32A776B to be divisible by 24, it must be divisible by 8 and 3.

⇒ Checking divisibility by 8 for 76B: 760 ÷ 8 = 95 (rem = 0) and 768 ÷ 8 = 96 (rem = 0). Thus, B ∈ {0, 8}.

⇒ Checking divisibility by 3: Sum = 3 + 2 + A + 7 + 7 + 6 + B = 25 + A + B.

⇒ If B = 0, Sum = 25 + A + 0 = 25 + A. For 25 + A to be divisible by 3, A can be 2, 5, or 8.

⇒ Value of (A + B) for B = 0: (2 + 0) = 2, (5 + 0) = 5, (8 + 0) = 8.

⇒ If B = 8, Sum = 25 + A + 8 = 33 + A. For 33 + A to be divisible by 3, A can be 0, 3, 6, or 9.

⇒ Value of (A + B) for B = 8: (0 + 8) = 8, (3 + 8) = 11, (6 + 8) = 14, (9 + 8) = 17.

⇒ The possible values for (A + B) are {2, 5, 8, 11, 14, 17}.

⇒ The minimum value among these is 2.

∴ The correct answer is 2.

Additional Information

Divisibility Rule of 3

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

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 000.

Composite Divisibility

To check divisibility by a composite number N, check divisibility by its coprime factors a and b where a × b = N and HCF(a, b) = 1.