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Shortcut Trick

Ratio of Initial Salary to New Salary = 100 : 110

Initial Savings = 35% of 100 = 35 units

New Savings = 40% of 110 = 44 units (Since new Expenditure is 60%)

Difference in Savings = 44 − 35 = 9 units → ₹3,150

1 unit = 3150 ÷ 9 = 350

Present Salary = 110 × 350 = ₹38,500

∴ The correct answer is ₹38,500.

Alternate Method

Given:

Initial Expenditure = 65% of Salary

Salary increase = 10%

Savings increase = ₹3,150

New Expenditure = 60% of New Salary

Formula Used:

Salary = Expenditure + Savings

Calculations:

Let the Initial Salary be 100x.

⇒ Initial Savings = 100x − 65% of 100x = 35x

⇒ New Salary = 100x + 10% of 100x = 110x

⇒ New Expenditure = 60% of 110x = 0.60 × 110x = 66x

⇒ New Savings = 110x − 66x = 44x

⇒ Difference in Savings = 44x − 35x = 9x

⇒ 9x = 3,150

⇒ x = 3,150 ÷ 9 = 350

⇒ Present Salary = 110x = 110 × 350

⇒ Present Salary = ₹38,500

∴ The correct answer is ₹38,500.

Additional Information

Basic Salary Relation

Income is always the sum of Expenditure and Savings. Income = Expenditure + Savings.

Percentage Change Formula

Percentage Change = (Final Value − Initial Value) ÷ Initial Value × 100.

Effective Percentage increase

When a base increases by a% and the component share changes from b% to c%, the net change depends on the product of the new base and the new percentage.

Shortcut Trick

The given expression is in the standard form of the algebraic identity: (a³ + b³ + c³ − 3abc) ÷ (a² + b² + c² − ab − bc − ca).

This identity simplifies directly to (a + b + c).

Here, a = 2.3, b = 1.5, and c = 1.2.

Required Value = 2.3 + 1.5 + 1.2 = 5.0.

∴ The correct answer is 5.

Alternate Method

Given:

a = 2.3

b = 1.5

c = 1.2

Formula Used:

The sum of cubes identity states: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Calculations:

⇒ Expression = ((2.3)³ + (1.5)³ + (1.2)³ − 3 × 2.3 × 1.5 × 1.2) ÷ ((2.3)² + (1.5)² + (1.2)² − 2.3 × 1.5 − 1.5 × 1.2 − 1.2 × 2.3)

⇒ Let a = 2.3, b = 1.5, c = 1.2

⇒ Expression = (a³ + b³ + c³ − 3abc) ÷ (a² + b² + c² − ab − bc − ca)

⇒ Expression = [(a + b + c)(a² + b² + c² − ab − bc − ca)] ÷ (a² + b² + c² − ab − bc − ca)

⇒ Expression = a + b + c

⇒ Expression = 2.3 + 1.5 + 1.2

⇒ Expression = 5

∴ The correct answer is 5.

Additional Information

Special Case of the Identity

If a + b + c = 0, then the identity simplifies to a³ + b³ + c³ = 3abc.

Alternative Form of the Quadratic Part

The term (a² + b² + c² − ab − bc − ca) can also be written as 1/2[(a − b)² + (b − c)² + (c − a)²].

Basic Algebraic Identities

Remember (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), which is often used to find missing components of the cubic identity.

Shortcut Trick

For an odd number of consecutive integers, the average is always the middle term.

Middle term (3^rd term) = 10.

The sequence is: 8, 9, 10, 11, 12.

The smallest integer is the first term of this sequence.

∴ The correct answer is 8.

Alternate Method

Given:

Number of consecutive integers (n) = 5

Average of the integers = 10

Formula Used:

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

Calculations:

Let the five consecutive integers be x, (x + 1), (x + 2), (x + 3), and (x + 4).

⇒ Sum = x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 5x + 10

⇒ Average = (5x + 10) ÷ 5 = 10

⇒ x + 2 = 10

⇒ x = 10 − 2

⇒ x = 8

∴ The correct answer is 8.

Additional Information

Average of Arithmetic Progression

For any sequence in Arithmetic Progression, Average = (First term + Last term) ÷ 2.

Sum of Consecutive Integers

The sum of n consecutive integers starting from x is given by n × x + [n × (n − 1) ÷ 2].

Middle Term Property

If n is odd, the average is exactly the {(n + 1) ÷ 2}^th term of the consecutive sequence.

Shortcut Trick

Product of the ratio parts = 17 × 13 × 8 = 1768.

Since the ratio parts 17, 13, and 8 are pairwise co-prime, their LCM is their product multiplied by the HCF.

HCF = Given LCM ÷ Product of ratio parts.

HCF = 7072 ÷ 1768 = 4.

∴ The correct answer is 4.

Alternate Method

Given: Ratio of numbers = 17 : 13 : 8, LCM = 7072.

Formula: If numbers are in ratio a : b : c and their HCF is x, then the numbers are ax, bx, and cx. Their LCM is x × LCM(a, b, c).

⇒ Let the three numbers be 17x, 13x, and 8x, where x is the HCF.

⇒ The LCM of 17x, 13x, and 8x is calculated by taking the common factor x and the LCM of (17, 13, 8).

⇒ LCM = x × 17 × 13 × 8

⇒ 7072 = 1768 × x

⇒ x = 7072 ÷ 1768

⇒ x = 4

∴ The correct answer is 4.

Additional Information

Product Property of HCF and LCM

For two numbers a and b: Product of numbers = HCF × LCM. This property does not directly apply to three or more numbers.

Co-prime Numbers

Numbers that have only 1 as their common factor are called co-prime. The LCM of a set of pairwise co-prime numbers is always their product.

HCF and LCM of Fractions

HCF = (HCF of Numerators) ÷ (LCM of Denominators) | LCM = (LCM of Numerators) ÷ (HCF of Denominators).

Shortcut Trick

Third proportional to 4 and 6 = (6 × 6) ÷ 4 = 9 units.

Base of the rectangle = 9 units.

Height of the rectangle = 4 units.

Area of Rectangle = Base × Height = 9 × 4 = 36 unit².

∴ The correct answer is 36 unit².

Alternate Method

Given:

First number (a) = 4

Second number (b) = 6

Formula Used:

1. If a, b, and c are in continuous proportion, then a : b :: b : c, where c is the third proportional (c = b² ÷ a).
2. Area of a rectangle = Base × Height.

Calculations:

⇒ Let the third proportional be x.

⇒ 4 : 6 :: 6 : x

⇒ x = (6 × 6) ÷ 4

⇒ x = 36 ÷ 4 = 9 units.

⇒ Base of the rectangle = 9 units.

⇒ Height of the rectangle = 4 units (given as the first number).

⇒ Area = 9 × 4 = 36 unit².

∴ The correct answer is 36 unit².

Additional Information

Mean Proportional

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

Fourth Proportional

For three numbers a, b, and c, the fourth proportional d is calculated such that a : b = c : d, which means d = (b × c) ÷ a.

Perimeter of Rectangle

The perimeter of a rectangle is the total distance around its boundary, calculated as 2 × (Base + Height).

Shortcut Trick

For 2 years, the effective Compound Interest rate = R + R + (R×R)/100.

Effective Rate = 2 + 2 + (2×2)/100 = 4.04%.

Compound Interest = 4.04% of ₹25,000.

CI = 25,000 × (4.04/100) = 250 × 4.04 = ₹1,010.

∴ The correct answer is ₹1,010.

Alternate Method

Given:

Principal (P) = ₹25,000

Rate (R) = 2% per annum

Time (T) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)^T

Compound Interest (CI) = Amount − Principal

Calculations:

⇒ A = 25,000 × (1 + 2/100)²

⇒ A = 25,000 × (102/100) × (102/100)

⇒ A = 25,000 × 1.02 × 1.02

⇒ A = 25,000 × 1.0404

⇒ A = 26,010

⇒ CI = 26,010 − 25,000

⇒ CI = 1,010

∴ The correct answer is ₹1,010.

Additional Information

Difference between CI and SI for 2 years

The difference (D) between CI and SI for two years is given by D = P(R/100)².

Compounding Half-Yearly

When interest is compounded half-yearly, the rate is halved (R/2) and the time is doubled (2n).

Effective Rate Formula

For successive percentage changes (like CI), the net change is a + b + (ab/100). For two years, it is R + R + (R²/100).

Shortcut Trick

Evaluate Discount % for all schemes:

(A) 35 + 4 − (35 × 4)/100 = 39 − 1.4 = 37.6%

(B) Successive 45%, 46%, 50% will be > 85% (highest discount).

(C) For 9 toys: (3 × 15% + 6 × 5%) ÷ 9 = (45 + 30) ÷ 9 = 8.33%

(D) 2 free on 9 bought (Total 11): (2 ÷ 11) × 100 ≈ 18.18%

∴ The correct answer is Scheme C.

Alternate Method

Assume the Marked Price (MP) of each toy = ₹100. For 9 toys, Total MP = ₹900.

Formula Used:

1. Equivalent Discount (2 successive) = x + y − (xy/100)
2. Discount % (Buy X get Y free) = [Y / (X + Y)] × 100

Calculations:

Scheme A: Successive 35% and 4%

⇒ Total Discount % = 35 + 4 − (35 × 4)/100 = 39 − 1.4 = 37.6%

Scheme B: Successive 45%, 46% and 50%

⇒ Final Multiplier = 0.55 × 0.54 × 0.50 = 0.1485

⇒ Total Discount % = (1 − 0.1485) × 100 = 85.15%

Scheme C: 15% on first 3, 5% on next 6

⇒ Discount on first 3 = 15% of 300 = ₹45

⇒ Discount on next 6 = 5% of 600 = ₹30

⇒ Total Discount % = [(45 + 30) / 900] × 100 = 75 / 9 = 8.33%

Scheme D: 2 free on buying 9

⇒ Total toys received = 9 + 2 = 11

⇒ Total Discount % = (2 / 11) × 100 ≈ 18.18%

Comparing all, 8.33% is the minimum discount, making it the least beneficial for the customer.

∴ The correct answer is Scheme C.

Additional Information

Successive Discount

When two discounts d₁ and d₂ are given, effective discount = d₁ + d₂ − (d₁ × d₂)/100. For more than two, apply the formula to the first two and then use the result with the third.

Buy X Get Y Free

In such offers, the discount is always calculated on the total number of items (Paid + Free). Discount % = [Free Items / (Paid + Free)] × 100.

Marked Price & Discount

Discount is always calculated on the Marked Price (MP). Selling Price (SP) = MP − Discount. To be least beneficial for a customer, the SP should be the highest (minimum discount).

Shortcut Trick

Assume Total Work as the LCM of 1.5, 3, and 4.5 = 9 units.

Efficiency of (A+B+C) = 9 ÷ 1.5 = 6 units/min.

Efficiency of (A+B) = 3 units/min and (B+C) = 2 units/min.

Efficiency of Tap A = 6 − 2 = 4; Efficiency of Tap C = 6 − 3 = 3.

Efficiency of (A+C) = 4 + 3 = 7 units/min.

∴ The correct answer is 9/7 min.

Alternate Method

Given:

Time taken by (A + B + C) = 1.5 min

Time taken by (A + B) = 3 min

Time taken by (B + C) = 4.5 min

Formula Used:

Efficiency = Total Work ÷ Time

Calculations:

⇒ Efficiency of Tap C = Efficiency(A+B+C) − Efficiency(A+B)

⇒ Efficiency of Tap C = 6 − 3 = 3 units/min

⇒ Efficiency of Tap A = Efficiency(A+B+C) − Efficiency(B+C)

⇒ Efficiency of Tap A = 6 − 2 = 4 units/min

⇒ Combined Efficiency of (A + C) = 4 + 3 = 7 units/min

⇒ Time taken by (A + C) = Total Work ÷ Combined Efficiency

⇒ Time taken by (A + C) = 9 ÷ 7 = 9/7 min

∴ The correct answer is 9/7 min.

Additional Information

Efficiency and Time

Efficiency is inversely proportional to the time taken to complete the work (E = W/T).

Combined Work

When multiple pipes or taps work together, their individual efficiencies are added to find the net draining or filling rate.

LCM Method

Using the Least Common Multiple (LCM) of the given times as the total work simplifies complex fraction calculations into simple integer arithmetic.

Shortcut Trick

In a right-angled triangle, the hypotenuse acts as the diameter of the circumcircle.

Hypotenuse = √(13² + 84²) = √(169 + 7056) = 85 m.

Outer radius (R) = 85 ÷ 2 = 42.5 m.

Inner radius (r) = 42.5 − 2 = 40.5 m (subtracting the 2 m track width).

Perimeter of the inner boundary = 2 × π × 40.5 = 81π m.

∴ The correct answer is 81π m.

Alternate Method

Given:

Shorter sides of the right-angled triangle = 13 m and 84 m.

Width of the circular track = 2 m.

Calculations:

⇒ First, find the hypotenuse (c) using Pythagoras theorem:

⇒ c = √(13² + 84²)

⇒ c = √(169 + 7056) = √7225 = 85 m.

⇒ The outer boundary coincides with the circumcircle, so the outer radius (R) is:

⇒ R = Hypotenuse ÷ 2 = 85 ÷ 2 = 42.5 m.

⇒ The track has a uniform width of 2 m, so the inner radius (r) is:

⇒ r = R − width = 42.5 − 2 = 40.5 m.

⇒ The perimeter (circumference) of the inner boundary is given by:

⇒ Perimeter = 2 × π × r

⇒ Perimeter = 2 × π × 40.5 = 81π m.

∴ The correct answer is 81π m.

Additional Information

Circumradius of a Right Triangle

For any right-angled triangle, the circumradius (R) is always equal to half of the hypotenuse. The circumcenter is located at the midpoint of the hypotenuse.

Inradius of a Right Triangle

The radius of the inscribed circle (inradius) of a right triangle with legs a and b and hypotenuse c is given by r = (a + b − c) ÷ 2.

Area of a Circular Track

The area of a circular track with outer radius R and inner radius r is given by Area = π(R² − r²).

Shortcut Trick

For two angles of elevation θ₁ and θ₂ from two points at a distance d apart on the same side of a tree of height h:

h = d / (cot θ₁ − cot θ₂)

Here, d = 10 m, θ₁ = 30°, and θ₂ = 60°.

⇒ h = 10 / (cot 30° − cot 60°) = 10 / (√3 − 1/√3)

⇒ h = 10 / ((3 − 1) / √3) = (10 × √3) / 2 = 5√3 m.

∴ The correct answer is 5√3 m.

Alternate Method

Given: Angle of elevation at point A = 30°, angle at point B = 60°, and distance AB = 10 m.

Formula Used: tan θ = Perpendicular / Base

⇒ Let the height of the tree CD be h and distance BC be x.

⇒ In right ΔBCD: tan 60° = h / x

⇒ √3 = h / x ⇒ x = h / √3

⇒ In right ΔACD: tan 30° = h / (AB + BC)

⇒ 1 / √3 = h / (10 + x)

⇒ 1 / √3 = h / (10 + h / √3)

⇒ 10 + h / √3 = h√3

⇒ 10 = h√3 − h / √3

⇒ 10 = (3h − h) / √3

⇒ 10 = 2h / √3

⇒ 2h = 10√3 ⇒ h = 5√3 m.

∴ The correct answer is 5√3 m.

Additional Information

Angle of Elevation

The angle formed by the line of sight with the horizontal when an observer looks at an object above the horizontal level.

Trigonometric Ratios for 30° and 60°

tan 30° = 1/√3 and tan 60° = √3. Note that as the observer moves towards the object, the angle of elevation increases.

General Formula for Distance

The distance d between two points of observation is given by d = h(cot θ₁ − cot θ₂), where h is the height.