Search Previous Year Questions

Shortcut Trick

Assume a base average of 7.

Calculate deviations from 7: (6−7), (8−7), (7−7), (9−7), (5−7) ⇒ −1, +1, 0, +2, −2.

Net deviation = −1 + 1 + 0 + 2 − 2 = 0.

Required Mean = Base Average + (Net Deviation ÷ Count) = 7 + (0 ÷ 5) = 7.

∴ The correct answer is 7.

Alternate Method

Given:

Scores of five friends: 6, 8, 7, 9, and 5.

Total number of friends (n) = 5.

Formula Used:

Arithmetic Mean = (Sum of all observations) ÷ (Total number of observations)

Calculations:

⇒ Sum of scores = 6 + 8 + 7 + 9 + 5

⇒ Sum of scores = 35

⇒ Arithmetic Mean = 35 ÷ 5

⇒ Arithmetic Mean = 7

∴ The correct answer is 7.

Additional Information

Weighted Arithmetic Mean

When observations have different levels of importance, the mean is calculated as Σ(w_i × x_i) ÷ Σw_i, where w is the weight.

Properties of Mean

The sum of deviations of all observations from their arithmetic mean is always zero.

Median vs Mean

While the mean is the average, the median is the middle value of a sorted data set. For the set (5, 6, 7, 8, 9), the median is also 7.

Shortcut Trick

If the sum of three numbers a + b + c = 0, then a³ + b³ + c³ = 3abc.

Check the sum of given numbers: 22 + (−15) + (−7) = 22 − 22 = 0.

Apply the trick: 3 × 22 × (−15) × (−7).

Calculation: 66 × 105 = 6930.

∴ The correct answer is 6930.

Alternate Method

Given Expression: 22³ + (−15)³ + (−7)³

Let a = 22, b = −15, and c = −7.

Formula Used:

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Calculations:

⇒ First, find the sum (a + b + c):

⇒ a + b + c = 22 + (−15) + (−7)

⇒ a + b + c = 22 − 15 − 7 = 0

⇒ Since a + b + c = 0, the identity simplifies to: a³ + b³ + c³ = 3abc

⇒ Substituting the values: 3 × 22 × (−15) × (−7)

⇒ = 3 × 22 × 105

⇒ = 66 × 105

⇒ = 66 × (100 + 5) = 6600 + 330 = 6930

∴ The correct answer is 6930.

Additional Information

The Cubic Identity

The expression a³ + b³ + c³ − 3abc is always zero if a + b + c = 0 or if a = b = c.

Square of Trinomial

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca). This is used to find individual squares or sum of products.

Difference of Cubes

x³ − y³ = (x − y)(x² + xy + y²), useful for factoring two-term cubic expressions.

Shortcut Trick

Let x, y, and z be the number of ₹10, ₹5, and ₹2 coins respectively.

Interchange ₹10 & ₹2: Increase = (10 − 2) × (z − x) = 304 − 120 ⇒ 8(z − x) = 184 ⇒ z − x = 23

Interchange ₹5 & ₹2: Increase = (5 − 2) × (z − y) = 165 − 120 ⇒ 3(z − y) = 45 ⇒ z − y = 15

Substitute x = z − 23 and y = z − 15 in original eq: 10(z − 23) + 5(z − 15) + 2z = 120

17z − 305 = 120 ⇒ 17z = 425 ⇒ z = 25

∴ The correct answer is 25 coins.

Alternate Method

Given:

Total amount = ₹120

Denominations = ₹10, ₹5, ₹2

New amount (₹10 ↔ ₹2) = ₹304

New amount (₹5 ↔ ₹2) = ₹165

Formula Used:

Total Value = (Number of coins × Value of each coin)

Let the number of coins be x, y, and z for ₹10, ₹5, and ₹2 respectively.

⇒ 10x + 5y + 2z = 120 −−− (1)

When ₹10 and ₹2 are interchanged:

⇒ 2x + 5y + 10z = 304 −−− (2)

When ₹5 and ₹2 are interchanged:

⇒ 10x + 2y + 5z = 165 −−− (3)

Subtract (1) from (2):

⇒ (2x + 5y + 10z) − (10x + 5y + 2z) = 304 − 120

⇒ 8z − 8x = 184

⇒ z − x = 23 ⇒ x = z − 23 −−− (4)

Subtract (1) from (3):

⇒ (10x + 2y + 5z) − (10x + 5y + 2z) = 165 − 120

⇒ 3z − 3y = 45

⇒ z − y = 15 ⇒ y = z − 15 −−− (5)

Substitute x and y in (1):

⇒ 10(z − 23) + 5(z − 15) + 2z = 120

⇒ 10z − 230 + 5z − 75 + 2z = 120

⇒ 17z − 305 = 120

⇒ 17z = 425

⇒ z = 25

∴ The correct answer is 25 coins.

Additional Information

Linear Equations in Three Variables

Equations of the form ax + by + cz = d require at least three independent constraints to find unique values for all variables.

Coin Problems Logic

The total value is the sum of (Quantity × Face Value). Interchanging quantities of denominations p and q changes the total by (p − q) × (Difference in counts).

Substitution Method

In multi-variable systems, expressing all variables in terms of one (e.g., everything in terms of z) simplifies the solution process significantly.

Shortcut Trick

If the ratio of three numbers is a : b : c and their LCM is given, then HCF = LCM ÷ LCM(a, b, c).

LCM of ratio parts (4, 3, 13) = 4 × 3 × 13 = 156 (since they are co-prime).

HCF = 4524 ÷ 156 = 29.

∴ The correct answer is 29.

Alternate Method

Given:

Ratio of three numbers = 4 : 3 : 13

LCM of these numbers = 4524

Let the HCF of the three numbers be x.

The three numbers can be represented as 4x, 3x, and 13x.

The LCM of (4x, 3x, 13x) is calculated as:

⇒ LCM = x × LCM(4, 3, 13)

Since 4, 3, and 13 have no common factors other than 1 (they are pairwise co-prime):

⇒ LCM(4, 3, 13) = 4 × 3 × 13 = 156

Now, equating to the given LCM value:

⇒ 156x = 4524

⇒ x = 4524 ÷ 156

⇒ x = 29

The HCF of the three numbers is x = 29.

∴ The correct answer is 29.

Additional Information

Relationship for Two Numbers

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 Fractions

HCF of (a/b, c/d, e/f) = HCF(a, c, e) ÷ LCM(b, d, f).

Co-prime Numbers

Two or more numbers are called co-prime if their HCF is 1. The LCM of a set of co-prime numbers is simply their product.

LCM of Fractions

LCM of (a/b, c/d, e/f) = LCM(a, c, e) ÷ HCF(b, d, f).

Shortcut Trick

Net multiplier = (1 − 0.20) × (1 + 0.05) = 0.8 × 1.05 = 0.84

Final Price = Marked Price × 0.84

Final Price = 2000 × 0.84 = 1680

∴ The correct answer is ₹1,680.

Alternate Method

Given:

Marked Price (MP) = ₹2,000

Discount = 20%

GST = 5%

Calculations:

⇒ Discount = 20% of 2000 = (20/100) × 2000 = 400

⇒ Price after Discount = 2000 − 400 = 1600

⇒ GST = 5% of 1600 = (5/100) × 1600 = 80

⇒ Final Price = 1600 + 80 = 1680

∴ The correct answer is ₹1,680.

Additional Information

Net Percentage Change

When a value is changed successively by x% and y%, the net effective change is (x + y + (xy/100))%. Note that a discount is a negative change.

Marked Price (MP)

MP is the price listed on the product. Discounts are always calculated on this value: Selling Price (SP) = MP × (100 − D%)/100.

GST Application

Taxes like GST are calculated on the actual selling price (the price after applying all discounts), not on the original marked price.

Shortcut Trick

Third Proportional c of two numbers a and b is calculated as c = b² ÷ a.

Given a = 0.25 and b = 0.15.

Third Proportional = (0.15 × 0.15) ÷ 0.25

= 0.0225 ÷ 0.25 = 0.09

∴ The correct answer is 0.09.

Alternate Method

Given:

First number (a) = 0.25

Second number (b) = 0.15

Formula Used:

If a, b, and x are in continued proportion, then a : b :: b : x.

This implies a/b = b/x, where x is the third proportional.

Calculations:

⇒ 0.25 ÷ 0.15 = 0.15 ÷ x

⇒ x = (0.15 × 0.15) ÷ 0.25

⇒ x = 0.0225 ÷ 0.25

⇒ To simplify, multiply both numerator and denominator by 10000:

⇒ x = 225 ÷ 2500

⇒ x = 9 ÷ 100

⇒ x = 0.09

∴ The correct answer is 0.09.

Additional Information

Mean Proportional

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

Fourth Proportional

The fourth proportional d of three numbers a, b, and c is calculated as d = (b × c) ÷ a where a : b = c : d.

Continued Proportion

A sequence of numbers is in continued proportion if the ratio of any two consecutive terms is constant, i.e., a/b = b/c = c/d.

Shortcut Trick

Total units of work = LCM(9, 18, 18) = 18 units.

Efficiencies: A = 2, B = 1, and C = 1.

In the first x days, all three work: Efficiency = 2 + 1 + 1 = 4.

In the next x days, only B and C work: Efficiency = 1 + 1 = 2.

Total work done = (4 × x) + (2 × x) = 6x = 18.

∴ The correct answer is 3.

Alternate Method

Given:

Time taken by A = 9 days

Time taken by B = 18 days

Time taken by C = 18 days

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Total Work = LCM of 9, 18, and 18 = 18 units.

⇒ Efficiency of (A + B + C) = 2 + 1 + 1 = 4 units/day.

⇒ Efficiency of (B + C) = 1 + 1 = 2 units/day.

⇒ Work done by all three in x days = 4 × x = 4x units.

⇒ Work done by B and C in the next x days = 2 × x = 2x units.

⇒ Total work = 4x + 2x = 6x units.

⇒ 6x = 18

⇒ x = 18 ÷ 6 = 3

∴ The correct answer is 3.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a task. Efficiency = Work ÷ Time.

Combined Efficiency

When multiple people work together, their individual efficiencies are added to find the total work done per day.

LCM Method for Total Work

Assuming the Total Work as the LCM of the given time periods allows for calculation with whole numbers instead of fractions.

Shortcut Trick

To find the least beneficial scheme, identify the one with the lowest effective discount percentage.

(A) Successive 34% and 37% → 34 + 37 − (34 × 37 / 100) = 71 − 12.58 = 58.42%

(B) Successive 36%, 15%, 14% → 100 × 0.64 × 0.85 × 0.86 = 46.78% price → 53.22% discount

(C) Average discount for 9 toys → (6 × 33% + 3 × 9%) / 9 = (198% + 27%) / 9 = 225% / 9 = 25%

(D) 3 toys free on 9 toys bought → (Free / Total) × 100 = (3 / 9) × 100 = 33.33% (or 3/12 = 25%)

Comparing all, Scheme C provides the minimum discount of 25%.

∴ The correct answer is Scheme C.

Alternate Method

Assume the Marked Price (MP) of 1 toy is ₹100. A customer wants to buy 9 toys, so the total MP = ₹900.

Formula Used: Effective Discount % = (Total Discount / Total MP) × 100

⇒ Scheme A: Net SP = 900 × (1 − 0.34) × (1 − 0.37) = 900 × 0.66 × 0.63 = ₹374.22

⇒ Scheme B: Net SP = 900 × 0.64 × 0.85 × 0.86 = ₹421.06

⇒ Scheme C: Discount = (6 × 33) + (3 × 9) = 198 + 27 = ₹225 | Net SP = 900 − 225 = ₹675

⇒ Scheme D: Buying 9 toys means getting 3 free out of the 9. Customer pays for 6 toys. Net SP = 6 × 100 = ₹600

⇒ Comparing the final Selling Prices, Scheme C results in the highest payment (₹675), making it the least beneficial scheme for the customer.

∴ The correct answer is Scheme C.

Additional Information

Successive Discount Formula

For two successive discounts of a% and b%, the effective discount is (a + b − ab/100)%.

Buy X Get Y Free

The effective discount percentage in a "Buy X Get Y Free" offer is calculated as [Y / (X + Y)] × 100.

Effective Discount for Bulk Purchase

Net Discount % = (Total Discount Value / Total Marked Price) × 100. This is useful for tiered discount schemes.

Shortcut Trick

Present ratio (Amit : Sumit) = 2 : 1 | Ratio 5 years ago = 5 : 2.

Equate the age difference (2−1=1 and 5−2=3) by multiplying the present ratio by 3.

New Present ratio = 6 : 3 | Ratio 5 years ago = 5 : 2.

Difference in ratio parts (6−5 or 3−2) = 1 unit.

1 unit = 5 years. Therefore, Sumit present age = 3 units = 3 × 5 = 15 years.

∴ The correct answer is 15 years.

Alternate Method

Given:

Present age of Amit is twice that of Sumit.

Five years ago, the ratio of their ages was 5 : 2.

Formula Used:

Age ratio equation: (Age₁ − t) / (Age₂ − t) = Ratio_past

Let the Present age of Sumit be x years.

Then, the Present age of Amit = 2x years.

⇒ Age of Sumit five years ago = x − 5

⇒ Age of Amit five years ago = 2x − 5

⇒ (2x − 5) / (x − 5) = 5 / 2

⇒ 2 × (2x − 5) = 5 × (x − 5)

⇒ 4x − 10 = 5x − 25

⇒ 5x − 4x = 25 − 10

⇒ x = 15

∴ The correct answer is 15 years.

Additional Information

Constant Age Difference

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

Ratio Balancing

When the time gap is the same for both people, make the difference between the ratio terms equal to solve the problem faster without variables.

Concept of 'n' years ago/hence

If present age is x, age 'n' years ago is (x − n) and age 'n' years hence is (x + n).

Shortcut Trick

Solve the innermost bracket first: (26 − 13 + 5) = 18.

Multiply the result by the value in the braces: −9 × 18 = −162.

Apply the final addition: 30 + (−162) = −132.

∴ The correct answer is -132.

Alternate Method

Given: 30 + [{−9 × (26 − 13 + 5)}]

Formula Used: BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction).

⇒ 30 + [{−9 × (13 + 5)}]

⇒ 30 + [{−9 × 18}]

⇒ 30 + [−162]

⇒ 30 − 162

⇒ −132

∴ The correct answer is -132.

Additional Information

BODMAS Rule

It provides the correct sequence for solving mathematical operations: Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Types of Brackets

Solve in the following priority order: Parentheses ( ) → Curly Braces { } → Square Brackets [ ].

Sign Rules for Multiplication

Multiplying two different signs results in a negative value: (+) × (−) = (−). Multiplying two same signs results in a positive value: (−) × (−) = (+).