Quantitative Aptitude PYQs

Shortcut Trick

The hypotenuse of the right triangle is the diameter of the inscribed circle, so AC = 10 cm.

By the right triangle altitude theorem: AB² = AD × AC

⇒ 6² = AD × 10

⇒ 36 = 10 × AD

⇒ AD = 3.6 cm

∴ The correct answer is 3.6 cm.

Alternate Method

Given:

⇒ Diameter (Hypotenuse AC) = 10 cm

⇒ Leg AB = 6 cm

O A C B D

6 cm

AD = ?

AC = 10 cm

Formula Used:

⇒ Angle in a semicircle is always a right angle, so ∠ABC = 90°.

⇒ In a right-angled triangle, the altitude drawn to the hypotenuse creates similar triangles: ΔADB ∼ ΔABC.

Calculations:

⇒ From the similarity of ΔADB and ΔABC: AB / AC = AD / AB

⇒ By cross-multiplying: AB² = AD × AC

⇒ Substituting the given values: 6² = AD × 10

⇒ 36 = 10 × AD

⇒ AD = 36 ÷ 10

⇒ AD = 3.6 cm

∴ The correct answer is 3.6 cm.

Additional Information

Right Triangle Similarity Theorem

When an altitude is drawn to the hypotenuse of a right triangle, it divides the triangle into two smaller triangles that are similar to the original triangle and to each other.

Geometric Mean Theorem

The length of the altitude drawn to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse: BD² = AD × CD.

Angle in a Semicircle

Any angle subtended by a diameter at the circumference of a circle is always exactly a right angle (90°).

Shortcut Trick When two circles intersect, they overlap each other, meaning the distance between their centers is less than the sum of their radii. Any internal (transverse) common tangent must pass between their centers, which is impossible without intersecting the overlapping area. Thus, no internal common tangents can exist for intersecting circles. Only external (direct) tangents can be drawn to touch the outer boundaries of both circles simultaneously. Consequently, exactly 2 direct common tangents can be drawn. ∴ The correct answer is Only two common tangents exist.

Alternate Method Given: Two circles intersect at two distinct points. Concept Used: The possible number of common tangents depends entirely on the distance between the circle centers (d) and their radii (r1 and r2).

C₁ C₂ Direct Common Tangent 1 Direct Common Tangent 2

Calculations: ⇒ For intersecting circles, the distance between their centers follows the condition: |r1 − r2| < d < r1 + r2. ⇒ Because d < r1 + r2, the Transverse (internal) Common Tangents are restricted as they would cut through the overlapping region of the circles instead of behaving as tangents. ⇒ Only Direct (external) Common Tangents can be drawn, touching both circles on their outer boundaries without crossing the central intersecting zone. ⇒ Therefore, exactly 2 Direct Common Tangents are geometrically possible. ∴ The correct answer is Only two common tangents exist.

Additional Information Number of Common Tangents Rule If d > r1 + r2 → 4 tangents. If d = r1 + r2 → 3 tangents.If d = |r1 − r2| → 1 tangent. If d < |r1 − r2| → 0 tangents.

Length of Direct Common Tangent (DCT) The formula to find the length of a direct tangent when circles are separated by distance d is: DCT = √[d2 − (r1 − r2)2]

Length of Transverse Common Tangent (TCT) The formula to find the length of an internal cross tangent when circles are non-intersecting is: TCT = √[d2 − (r1 + r2)2]

Shortcut Trick

When two circles with radii r₁ and r₂ touch externally, the distance between their centers is exactly d = r₁ + r₂.

By applying this to the standard Direct Common Tangent formula, it reduces to a direct standard result.

The length of the direct common tangent is the geometric mean of their diameters.

Direct Formula: T = √(4 × r₁ × r₂)

This simplifies directly to T = 2√(r₁r₂).

∴ The correct answer is 2√(r₁r₂).

Alternate Method

r₁ r₂ T (Tangent)

C₁ C₂

r₁ − r₂ T r₁ + r₂

Given:

Radius of the larger circle = r₁

Radius of the smaller circle = r₂

Since the circles touch each other externally, the distance between their centers (d) = r₁ + r₂.

Length of the direct common tangent = T

Formula Used:

In the right-angled triangle formed by drawing a horizontal line from C₂ to the radius of C₁:

⇒ Base² + Perpendicular² = Hypotenuse²

Calculations:

⇒ T² + (r₁ − r₂)² = (r₁ + r₂)²

⇒ T² = (r₁ + r₂)² − (r₁ − r₂)²

Applying the standard algebraic identity (a + b)² − (a − b)² = 4ab:

⇒ T² = 4 × r₁ × r₂

⇒ T = √(4r₁r₂)

⇒ T = 2√(r₁r₂)

∴ The correct answer is 2√(r₁r₂).

Additional Information

General Direct Common Tangent (DCT)

DCT = √[d² − (r₁ − r₂)²], where d is the straight line distance between the centers of the two circles.

Transverse Common Tangent (TCT)

TCT = √[d² − (r₁ + r₂)²]. The transverse tangent diagonally crosses the line joining the centers.

Maximum Common Tangents

When two circles touch each other externally, exactly 3 common tangents can be drawn (2 direct common tangents and 1 transverse common tangent touching at the intersection point).

Internally Touching Circles

If two circles touch each other internally, the distance between their centers is d = |r₁ − r₂| and only 1 common tangent can be drawn.

Shortcut Trick

The angle at the center is twice the angle at the circumference → 2 × 30° = 60°.

The triangle formed by the center and the chord is an isosceles triangle with a 60° vertex, making it an equilateral triangle.

Radius of the circle = Chord length = 12 cm.

The distance from the center to the chord is simply the altitude of this equilateral triangle.

Distance = (√3 ÷ 2) × Side = (√3 ÷ 2) × 12 = 6√3 cm.

∴ The correct answer is 6√3 cm.

Alternate Method

Given:

Length of the chord (AB) = 12 cm

Angle subtended at the circumference (∠ACB) = 30°

Formula Used:

Angle at the center = 2 × Angle at the circumference

Pythagoras Theorem: Hypotenuse² = Perpendicular² + Base²

60° 30°

O A B C M

AB = 12 cm d = ?

Calculation:

⇒ Angle subtended at the center (∠AOB) = 2 × 30° = 60°

⇒ In ▵AOB, OA = OB (radii of the same circle), making it an isosceles triangle.

⇒ Base angles ∠OAB = ∠OBA = (180° − 60°) ÷ 2 = 60°

⇒ Since all internal angles are exactly 60°, ▵AOB is an equilateral triangle.

⇒ Radius (OA) = AB = 12 cm

⇒ Let OM be the perpendicular drawn from the center O to the chord AB.

⇒ The perpendicular from the center bisects the chord perfectly, so AM = AB ÷ 2 = 12 ÷ 2 = 6 cm.

⇒ In right-angled ▵OMA, applying Pythagoras Theorem:

⇒ OA² = OM² + AM²

⇒ 12² = OM² + 6²

⇒ 144 = OM² + 36

⇒ OM² = 144 − 36 = 108

⇒ OM = √108 = √(36 × 3) = 6√3 cm

∴ The correct answer is 6√3 cm.

Additional Information

Angle Subtended by an Arc

The angle subtended by an arc at the center is always double the angle subtended by it at any remaining part of the circle's circumference.

Perpendicular from Center to Chord

A perpendicular drawn from the center of a circle to any chord will perfectly bisect the chord into two equal halves.

Equilateral Triangle Altitude

In an equilateral triangle where all sides are equal, the height or altitude can be calculated directly using the standard formula h = (√3 ÷ 2) × side.

Shortcut Trick

Area of segment = Area of sector − Area of triangle

Area = [(θ ÷ 360°) × πr²] − [(1 ÷ 2) × r² × sin(θ)]

Substitute r = 8 cm and θ = 60°:

Area = [(60 ÷ 360) × π × 8²] − [(1 ÷ 2) × 8² × sin(60°)]

Area = (64π ÷ 6) − (32 × √3 ÷ 2)

Area = 32π/3 − 16√3 cm²

∴ The correct answer is 32π/3 − 16√3 cm².

Alternate Method

O A B 8 cm 8 cm 60°

Segment Area = Sector OAB − Triangle OAB

Given:

Radius of the circle, r = 8 cm

Angle subtended at the center by the chord, θ = 60°

Formula Used:

Area of Segment = Area of Sector − Area of Triangle

Area of Sector = (θ ÷ 360°) × πr²

Since the two sides bounding the angle 60° are the radii of the same circle, they are equal in length, forming an isosceles triangle. With the included angle being 60°, it becomes an equilateral triangle.

Area of Equilateral Triangle = (√3 ÷ 4) × a² (where a = r)

Calculations:

⇒ Area of Sector = (60° ÷ 360°) × π × 8²

⇒ Area of Sector = (1 ÷ 6) × 64π = 32π/3 cm²

⇒ Area of Triangle = (√3 ÷ 4) × 8²

⇒ Area of Triangle = (√3 ÷ 4) × 64 = 16√3 cm²

⇒ Area of Segment = Area of Sector − Area of Triangle

⇒ Area of Segment = 32π/3 − 16√3 cm²

∴ The correct answer is 32π/3 − 16√3 cm².

Additional Information

Area of a Sector

The region bounded by two radii and an arc is called a sector. Its area is calculated as (θ ÷ 360°) × πr².

Length of an Arc

The distance along the curved line making up the arc is given by the formula (θ ÷ 360°) × 2πr.

Area of Triangle in a Circle

For a triangle formed by two circle radii and a connecting chord, the general area can be found using (1 ÷ 2) × r² × sin(θ).

Perimeter of a Sector

The perimeter of a circular sector is the sum of its arc length and the two boundary radii: Length of Arc + 2r.

Shortcut Trick

For a homogeneous ratio expression, directly substitute the ratio values.

Put p = 4 and q = 1 in the expression (5p + 3q) : (5p − 3q).

⇒ (5 × 4 + 3 × 1) : (5 × 4 − 3 × 1)

⇒ (20 + 3) : (20 − 3)

⇒ 23 : 17

∴ The correct answer is 23 : 17.

Alternate Method

Given:

p : q = 4 : 1

p = 4k

q = k

5(4k) + 3(k) : 5(4k) − 3(k) 23k : 17k

Calculations:

⇒ Let p = 4k and q = k, where k is a common multiple.

⇒ We need to find the value of the expression: (5p + 3q) : (5p − 3q)

⇒ Substitute the values of p and q into the expression:

⇒ [5(4k) + 3(k)] : [5(4k) − 3(k)]

⇒ (20k + 3k) : (20k − 3k)

⇒ 23k : 17k

⇒ Canceling the common term k from both sides, we get 23 : 17.

∴ The correct answer is 23 : 17.

Additional Information

Homogeneous Expression

An expression where the sum of the powers of variables in each term is identical. In ratio forms like (ax + by)/(cx + dy), substituting equivalent proportional values directly always gives the correct simplified ratio.

Componendo and Dividendo

A fundamental proportion rule stating that if a : b = c : d, then (a + b) : (a − b) = (c + d) : (c − d). This is extremely useful for quickly simplifying complex rational equations.

Proportion Rule

When two ratios are equivalent, they form a proportion. For any relation a : b = c : d, the product of the extremes equals the product of the means (a × d = b × c).

Shortcut Trick

Simplify the surds first: √18 = 3√2 and √8 = 2√2.

Substitute into the expression: (3√2 + √2)² − (2√2 + √2)²

This simplifies to: (4√2)² − (3√2)²

Calculate the squares directly: 32 − 18 = 14.

∴ The correct answer is 14.

Alternate Method

Given:

(√18 + √2)² − (√8 + √2)²

Formula Used:

(a + b)² = a² + b² + 2ab

Calculations:

⇒ Expand the first term: (√18 + √2)² = 18 + 2 + 2√(18 × 2)

⇒ 20 + 2√36 = 20 + 2(6) = 32

⇒ Expand the second term: (√8 + √2)² = 8 + 2 + 2√(8 × 2)

⇒ 10 + 2√16 = 10 + 2(4) = 18

⇒ Subtract the two results: 32 − 18

⇒ 14

∴ The correct answer is 14.

Additional Information

Properties of Surds

√(a × b) = √a × √b, which helps in simplifying large square roots, e.g., √18 = √(9 × 2) = 3√2.

Algebraic Identities

Using standard formulas like (a + b)² = a² + b² + 2ab and a² − b² = (a − b)(a + b) simplifies complex algebraic and numerical calculations significantly.

Simplification Rules

According to BODMAS, always solve the expressions inside the brackets first (when possible) before evaluating powers and applying subtraction.

Shortcut Trick

First bracket: LCM of 6 and 4 is 12 → (10 + 9)/12 = 19/12

Second bracket: LCM of 8 and 3 is 24 → (21 − 8)/24 = 13/24

Divide the fractions: (19/12) × (24/13) = 19 × 2 / 13 = 38/13

∴ The correct answer is 38/13.

Alternate Method

Given:

Expression: (5/6 + 3/4) ÷ (7/8 − 1/3)

Concept Used:

Apply the BODMAS rule to solve the values inside the brackets first, then perform the division of fractions.

Calculations:

⇒ Step 1: Solve the first bracket (5/6 + 3/4)

⇒ LCM of denominators 6 and 4 is 12

⇒ (5 × 2)/(6 × 2) + (3 × 3)/(4 × 3)

⇒ 10/12 + 9/12 = 19/12

⇒ Step 2: Solve the second bracket (7/8 − 1/3)

⇒ LCM of denominators 8 and 3 is 24

⇒ (7 × 3)/(8 × 3) − (1 × 8)/(3 × 8)

⇒ 21/24 − 8/24 = 13/24

⇒ Step 3: Divide the results

⇒ (19/12) ÷ (13/24)

⇒ (19/12) × (24/13)

⇒ (19 × 24) / (12 × 13)

⇒ (19 × 2) / 13

⇒ 38/13

∴ The correct answer is 38/13.

Additional Information

BODMAS Rule

It dictates the order of operations: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction. Expressions inside brackets are always evaluated first.

Dividing Fractions

To divide by a fraction, you simply multiply by its reciprocal: (a/b) ÷ (c/d) = (a/b) × (d/c).

Lowest Common Multiple (LCM)

When adding or subtracting fractions with different denominators, you must find a common denominator by taking the LCM of the original denominators.

Shortcut Trick

Profit ratio = (Capital_A × Time_A) : (Capital_B × Time_B) : (Capital_C × Time_C)

Profit ratio = (24 × 12) : (36 × 8) : (40 × 6)

Profit ratio = 288 : 288 : 240 = 6 : 6 : 5

B's share = (6 ÷ 17) × 50000

B's share = 300000 ÷ 17 ≈ 17647.06

∴ The correct answer is ₹17,647.06.

Alternate Method

Given:

A's investment = ₹24,000 for 12 months

B's investment = ₹36,000 for 8 months

C's investment = ₹40,000 for 6 months

Total profit = ₹50,000

Formula Used:

Ratio of Profit = (Investment × Time Period)

A B C

24k × 12m = 288k Ratio: 6

36k × 8m = 288k Ratio: 6

40k × 6m = 240k Ratio: 5

Total Profit Ratio = 6 : 6 : 5 (Total 17 parts)

Calculations:

⇒ Profit ratio of A : B : C = (24000 × 12) : (36000 × 8) : (40000 × 6)

⇒ Profit ratio = 288000 : 288000 : 240000

⇒ Dividing by 1000: Profit ratio = 288 : 288 : 240

⇒ Dividing by 48: Profit ratio = 6 : 6 : 5

⇒ Total ratio parts = 6 + 6 + 5 = 17 parts

⇒ B's share of profit = (6 ÷ 17) × 50000

⇒ B's share = 300000 ÷ 17

⇒ B's share = 17647.0588...

⇒ B's share ≈ ₹17,647.06

∴ The correct answer is ₹17,647.06.

Additional Information

Compound Partnership

When investments are made for different time periods, profit is shared based on the product of capital and time. Profit Ratio = (C₁ × T₁) : (C₂ × T₂).

Simple Partnership

When all partners invest for the exact same time period, the profit is shared directly in the ratio of their invested capital amounts.

Working vs Sleeping Partner

A working partner manages the business and receives a fixed salary or commission from the total profit before the remaining profit is divided among all partners (including the working partner) based on investments.

Shortcut Trick

Total age increase for the 9 committee members = 9 × 3 = 27 years.

Sum of the ages of the two replaced men = 42 + 44 = 86 years.

Total age of the two incoming women = Age of replaced men + Total age increase.

Total age of the two women = 86 + 27 = 113 years.

Average age of the two women = 113 ÷ 2 = 56.5 years.

∴ The correct answer is 56.5 years.

Alternate Method

Given:

Number of members in the committee = 9

Increase in the average age = 3 years

Ages of the replaced men = 42 years and 44 years

Formula Used:

Total Sum = Average × Number of observations

New Sum = Old Sum − Sum of outgoing + Sum of incoming

Outgoing: 42 + 44 = 86

Increase: 9 × 3 = 27

Incoming 2 Women: 86 + 27 = 113

Average = 113 ÷ 2 = 56.5 years

Calculations:

⇒ Let the initial average age of the committee be A years.

⇒ Initial total age of the 9 members = 9A

⇒ After replacing the two men with two women, the new average becomes (A + 3) years.

⇒ New total age = 9 × (A + 3) = 9A + 27

⇒ Also, New total age = Initial total age − (42 + 44) + Sum of ages of the two women

⇒ 9A + 27 = 9A − 86 + Sum of ages of the two women

⇒ Sum of ages of the two women = 27 + 86

⇒ Sum of ages of the two women = 113 years

⇒ Average age of the two women = 113 ÷ 2 = 56.5 years

∴ The correct answer is 56.5 years.

Additional Information

Replacement Method in Averages

When individuals are replaced and the overall average changes, the value of the incoming individuals can be found using: New Value = Replaced Value + (Increase in Average × Total Members).

Total Sum Logic

The sum of elements in a group is strictly linked to their average by the equation: Sum = Average × Count. Any deviation in the sum proportionally affects the average.

Combined Group Average

When merging two groups of sizes n₁ and n₂ having individual averages A₁ and A₂, the resulting average is represented as (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Shortcut Trick

Let the average of the 6 integers be A ⇒ Sum = 6A.

Average of 5 integers = A + 2 ⇒ Sum = 5(A + 2) = 5A + 10.

Difference in sums is given as 4: (5A + 10) − 6A = 4.

⇒ 10 − A = 4 ⇒ A = 6.

∴ The correct answer is 6.

Alternate Method

Given:

⇒ Number of integers in the first set = 6

⇒ Number of integers in the second set = 5

⇒ The average of the 6 integers is 2 less than the average of the 5 integers.

⇒ The sum of the 5 integers is 4 more than the sum of the 6 integers.

Formula Used:

⇒ Sum of observations = Average × Number of observations

Sum of 5 integers = 5 × (A + 2)

A+2

A+2

A+2

A+2

A+2

Sum of 6 integers + 4 = 6A + 4

A

A

A

A

A

A

+4

Equal Sum

5(A + 2) = 6A + 4

Calculations:

⇒ Let the average of the 6 integers be x.

⇒ Therefore, the average of the 5 integers = x + 2.

⇒ Sum of the 6 integers = 6 × x = 6x.

⇒ Sum of the 5 integers = 5 × (x + 2) = 5x + 10.

⇒ According to the problem statement: Sum of 5 integers = Sum of 6 integers + 4.

⇒ 5x + 10 = 6x + 4

⇒ 6x − 5x = 10 − 4

⇒ x = 6

∴ The correct answer is 6.

Additional Information

Average Formula

The average (or arithmetic mean) is defined as the sum of all observations divided by the total number of observations. Conversely, Sum = Average × Total Count.

Average of Consecutive Integers

For any sequence of consecutively spaced numbers, the average is always equal to the exact middle value of the data set. Additionally, the average equals (First Term + Last Term) ÷ 2.

Properties of Averages

If every observation in a dataset is increased, decreased, multiplied, or divided by a constant value 'k', the new resulting average also undergoes the exact same mathematical operation by 'k'.

Shortcut Trick

Let the total number be 12 units (the LCM of denominators 4 and 3).

75% (or 3/4) of 12 units = 9 units.

2/3 of 12 units = 8 units.

Difference = 9 units − 8 units = 1 unit.

Since 1 unit represents 15, the total number = 12 × 15 = 180.

∴ The correct answer is 180.

Alternate Method

Given:

75% of a number is 15 more than 2/3 of that number.

Method Used:

Convert percentage to fraction: 75% = 75/100 = 3/4

Total Number = 12 Units

12 Units (LCM of 4 and 3)

75% (or 3/4) = 9 Units

9 Units

2/3 = 8 Units

8 Units

Diff = 15

Calculations:

Let the required number be x.

⇒ 75% of x = 3x/4

⇒ Two-thirds of x = 2x/3

According to the problem statement:

⇒ 3x/4 − 2x/3 = 15

Taking the LCM of the denominators 4 and 3 (LCM = 12):

⇒ (9x − 8x) ÷ 12 = 15

⇒ x / 12 = 15

⇒ x = 15 × 12

⇒ x = 180

∴ The correct answer is 180.

Additional Information

Fractions and Percentages Equivalents

Percentages can be converted into fractions by dividing by 100. Important standard equivalents to remember: 75% = 3/4, 66.66% = 2/3, 25% = 1/4, and 50% = 1/2.

LCM Method for Fractions

When solving problems with multiple fractional parts, assuming the total initial value as the Least Common Multiple (LCM) of all denominators allows for faster, integer-based calculations.

Percentage Change

If a value changes from an initial base A to a new value B, the overall percentage change is given by the formula: ((B − A) ÷ A) × 100%.

Shortcut Trick

Interest rate = 10% = 1/10 ⇒ The amount to be paid back becomes 11/10 times the principal each year.

Since the rate multiplier has 11 in the numerator, the annual instalment value must be completely divisible by 11.

Checking the given options, only ₹6,655 is exactly divisible by 11 (6655 ÷ 11 = 605).

Alternatively, the remaining principal = 19650 − 3100 = ₹16,550.

Each Instalment = 16550 × (1331 / 3310) = 5 × 1331 = ₹6,655.

∴ The correct answer is ₹6,655.

Alternate Method

PV = ₹16,550

Year 1: ₹X

Year 2: ₹X

Year 3: ₹X

PV&sub1; = X / 1.1 PV&sub2; = X / 1.21 PV&sub3; = X / 1.331

Given:

⇒ Total price of TV = ₹19,650

⇒ Down payment = ₹3,100

⇒ Rate of interest (R) = 10% per annum

⇒ Number of equal instalments (n) = 3

Formula Used:

⇒ Present Value (P) = X / (1 + R/100) + X / (1 + R/100)² + X / (1 + R/100)³

Calculations:

⇒ Principal to be paid in instalments (P) = 19,650 − 3,100 = ₹16,550

⇒ Let the equal annual instalment be ₹X.

⇒ 16550 = X / (1 + 10/100) + X / (1 + 10/100)² + X / (1 + 10/100)³

⇒ 16550 = X / 1.1 + X / 1.21 + X / 1.331

⇒ 16550 = X × [10/11 + 100/121 + 1000/1331]

⇒ 16550 = X × [(1210 + 1100 + 1000) / 1331]

⇒ 16550 = X × (3310 / 1331)

⇒ X = (16550 × 1331) / 3310

⇒ X = 5 × 1331 = 6655

∴ The correct answer is ₹6,655.

Additional Information

Compound Interest Instalments

When a loan P is paid back in n equal annual instalments X at a compound interest rate R%, the formula used is: P = X / (1 + R/100) + X / (1 + R/100)² + ... + X / (1 + R/100)ⁿ.

Simple Interest Instalments

If a debt A is discharged in n equal annual instalments X at a simple interest rate R%, the formula is: A = nX + [RX × n(n − 1)] / 200.

Present Value Concept

The sum of the present values of all future instalment payments is always exactly equal to the total principal loan amount taken initially.

Shortcut Trick

Let the Cost Price (CP) of the mobile phone be 100 units.

Since the initial profit is 20%, the original selling price P = 120 units.

The Marked Price (MP) for the festive offer is 1.5P = 1.5 × 120 = 180 units.

After a 10% discount on MP, the final Selling Price = 180 − (10% of 180) = 162 units.

Net Profit % = [(162 − 100) ÷ 100] × 100 = 62%.

∴ The correct answer is 62%.

Alternate Method

Given:

Initial Profit = 20%

Initial Selling Price = ₹P

Marked Price (MP) = 1.5P

Discount = 10%

Formula Used:

Profit = Final Selling Price − Cost Price

Profit % = (Profit ÷ Cost Price) × 100

  CP = 100 P = 120 New SP = 162 MP = 180   +20% Profit   Marked at 1.5P   -10% Discount   Final Profit = 62%

Calculations:

⇒ Let Cost Price (CP) = 100x

⇒ Original Selling Price (P) = CP + 20% of CP = 100x + 20x = 120x

⇒ Marked Price (MP) = 1.5 × P = 1.5 × 120x = 180x

⇒ Discount given = 10% of MP = (10 ÷ 100) × 180x = 18x

⇒ Final Selling Price = MP − Discount = 180x − 18x = 162x

⇒ Profit = Final Selling Price − CP = 162x − 100x = 62x

⇒ Profit % = (62x ÷ 100x) × 100 = 62%

∴ The correct answer is 62%.

Additional Information

Marked Price and Discount

Marked price is the price printed on an article. Discount is always provided on the Marked Price. Formula: SP = MP × (100 − Discount%) ÷ 100

Mark-up Percentage

It is the percentage by which the cost price is increased to set the marked price. Formula: MP = CP × (100 + Mark-up%) ÷ 100

Successive Discounts

When two discounts d₁% and d₂% are given consecutively, the equivalent single discount is [d₁ + d₂ − (d₁ × d₂) ÷ 100]%.

Shortcut Trick

If 20% profit corresponds to ₹450, the Selling Price (120% of CP) = ₹450 × 6 = ₹2700.

This SP is after a 15% discount, meaning 85% of Marked Price (MP) = ₹2700.

We need the new Selling Price at a 25% discount, which is 75% of MP.

New SP = ₹2700 × (75 ÷ 85) = ₹2700 × (15 ÷ 17)

New SP = 40500 ÷ 17 ≈ ₹2382.35

∴ The correct answer is ₹2382.35.

Alternate Method

Given:

Discount₁ = 15%

Profit = 20%

Profit amount = ₹450

Discount₂ = 25%

Formula Used:

Cost Price (CP) = Profit × (100 ÷ Profit%)

Selling Price (SP) = CP + Profit

Marked Price (MP) = SP × [100 ÷ (100 − Discount%)]

CP

SP₁ = 2700

MP

SP₂ = ?

+20% (₹450)

-15%

-25% Discount

Calculations:

⇒ Cost Price (CP) = 450 × (100 ÷ 20) = ₹2250

⇒ Selling Price (SP₁) = 2250 + 450 = ₹2700

⇒ SP₁ is after a 15% discount, meaning it is 85% of MP.

⇒ 85% of MP = ₹2700

⇒ MP = 2700 × (100 ÷ 85) = 54000 ÷ 17

⇒ New Selling Price (SP₂) at a 25% discount is 75% of MP.

⇒ SP₂ = (54000 ÷ 17) × (75 ÷ 100)

⇒ SP₂ = 40500 ÷ 17 ≈ ₹2382.35

∴ The correct answer is ₹2382.35.

Additional Information

Marked Price (MP)

The price printed on an article. Discount is always calculated on the MP unless explicitly stated otherwise.

Direct Relation (CP & MP)

The ratio between Marked Price and Cost Price can be directly found using: MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%).

Successive Discounts

If two successive discounts of a% and b% are given, the equivalent single discount is (a + b − ab ÷ 100)%.

Shortcut Trick

Total work = LCM(20, 30, 60) = 60 units.

Efficiencies: A = 60 ÷ 20 = 3, B = 60 ÷ 30 = 2, C = 60 ÷ 60 = 1.

Pattern: A works for 2 days, and A + B + C work on the 3^rd day.

Work done in 3 days = (3) + (3) + (3 + 2 + 1) = 3 + 3 + 6 = 12 units.

Number of 3-day cycles needed = 60 ÷ 12 = 5 cycles.

Total time = 5 × 3 = 15 days.

∴ The correct answer is 15 days.

Alternate Method

Given:

Time taken by A = 20 days

Time taken by B = 30 days

Time taken by C = 60 days

Calculations:

Let the Total Work be the LCM of the given days.

⇒ Work done by A on the 1^st day = 3 units

⇒ Work done by A on the 2^nd day = 3 units

⇒ Work done by A, B, and C together on the 3^rd day = 3 + 2 + 1 = 6 units

⇒ Total work done in one 3-day cycle = 3 + 3 + 6 = 12 units

⇒ Number of cycles required to complete 60 units of work = 60 ÷ 12 = 5 cycles

⇒ Total days required = 5 cycles × 3 days/cycle = 15 days

∴ The correct answer is 15 days.

Additional Information

Total Work Rule

Total Work is often considered as the Least Common Multiple (LCM) of the days taken by different individuals to complete the given task.

Work Efficiency

Efficiency = Total Work ÷ Total Days. It represents the units of work done by a person in a single day.

Alternate Day Work

When people work in a specific periodic pattern, sum their combined work into a cycle and find the total time by finding the number of full cycles needed.

Combined Efficiency

When individuals work together on any given day, their combined efficiency is simply the sum of their individual daily efficiencies.

Shortcut Trick

Mean Cost Price = Selling Price × [100 ÷ (100 + Profit%)]

Mean Cost Price = 132 × (100 ÷ 120) = ₹110/kg

Using Alligation Cross-Multiplication directly:

Quantity Ratio = (120 − 110) : (110 − 80)

Quantity Ratio = 10 : 30 = 1 : 3

∴ The correct answer is 1 : 3.

Alternate Method

Given:

Cost price of Grade 1 rice (CP₁) = ₹80/kg

Cost price of Grade 2 rice (CP₂) = ₹120/kg

Selling price of the mixture (SP) = ₹132/kg

Overall profit margin = 20%

Formula Used:

Cost Price (CP) = (SP × 100) ÷ (100 + Profit%)

Alligation Ratio = (CP₂ − Mean CP) : (Mean CP − CP₁)

Calculations:

⇒ Mean CP of mixture = (132 × 100) ÷ (100 + 20)

⇒ Mean CP = 13200 ÷ 120 = ₹110/kg

80 CP₁

120 CP₂

110 Mean CP

10

30

Ratio = 10 : 30

⇒ Quantity of Grade 1 : Quantity of Grade 2 = 10 : 30

⇒ Simplest Ratio = 1 : 3

∴ The correct answer is 1 : 3.

Additional Information

Rule of Alligation

It allows finding the specific ratio in which two or more ingredients at given prices must be mixed to produce a mixture of a desired average price. Ratio = (Dearer Price − Mean Price) : (Mean Price − Cheaper Price).

Mean Price Concept

The cost price of a unit quantity of the mixture is called its mean price. When using the alligation method, always ensure that all values (CP₁, CP₂, and Mean Price) are purely Cost Prices. Never use Selling Price directly in the alligation cross.

Repeated Replacement

If a vessel contains 'x' units of a pure liquid and 'y' units are taken out and replaced by water 'n' times, the quantity of pure liquid remaining = x × (1 − y ÷ x)ⁿ.

Shortcut Trick

Speed of the bus = 320 ÷ 4 = 80 km/h.

Given the speed ratio of Car : Bus = 3 : 4.

Car's speed = (3 ÷ 4) × 80 = 60 km/h.

∴ The correct answer is 60 km/h.

Alternate Method

Given:

Distance covered by the bus = 320 km

Time taken by the bus = 4 hours

Ratio of speeds (Car : Bus) = 3 : 4

Formula Used:

Speed = Distance ÷ Time

Speed Ratio → Car : Bus = 3 : 4

BUS 320 km ÷ 4 h = 80 km/h 4 units = 80

CAR Speed = ? 3 units = 60

Car Speed = 60 km/h

Calculations:

⇒ Speed of the bus = 320 ÷ 4 = 80 km/h

⇒ Let the speed of the car and the bus be 3x and 4x respectively.

⇒ Speed of the bus = 4x

⇒ 4x = 80

⇒ x = 20

⇒ Speed of the car = 3x = 3 × 20 = 60 km/h

∴ The correct answer is 60 km/h.

Additional Information

Relative Speed

When two bodies are moving in the same direction, relative speed = a − b. When moving in opposite directions, relative speed = a + b.

Average Speed Formula

When equal distances are covered at speeds a and b, Average Speed = (2 × a × b) ÷ (a + b). For varying distances, Average Speed = Total Distance ÷ Total Time.

Unit Conversion

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

Shortcut Trick

From the given conditions, R = 2.5P and R = 3Q. The ratio of investments P : Q : R is 6 : 5 : 15.

Let the investments in the three schemes be 600, 500, and 1500 units respectively.

Total interest in 1 year = 7% of 600 + 9% of 500 + 11% of 1500 = 42 + 45 + 165 = 252 units.

We are given total interest = ₹2520, so 252 units = 2520 ⇒ 1 unit = 10.

Amount invested in Scheme P = 600 units = 600 × 10 = ₹6000.

∴ The correct answer is ₹6000.

Alternate Method

Given:

Rate of interest for P = 7% p.a., Q = 9% p.a., R = 11% p.a.

Total Simple Interest (SI) earned in 1 year = ₹2520

Investment in R = 250% of P ⇒ R = (5 ÷ 2) × P ⇒ P ÷ R = 2 ÷ 5 = 6 ÷ 15

Investment in R = 300% of Q ⇒ R = 3 × Q ⇒ Q ÷ R = 1 ÷ 3 = 5 ÷ 15

Formula Used:

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

Scheme P P = 6x R = 7% SI = 0.42x

Scheme Q P = 5x R = 9% SI = 0.45x

Scheme R P = 15x R = 11% SI = 1.65x

Total Interest = 2.52x = ₹2520

Calculations:

⇒ Let the principal amounts be P = 6x, Q = 5x, and R = 15x.

⇒ SI from Scheme P = (6x × 7 × 1) ÷ 100 = 0.42x

⇒ SI from Scheme Q = (5x × 9 × 1) ÷ 100 = 0.45x

⇒ SI from Scheme R = (15x × 11 × 1) ÷ 100 = 1.65x

⇒ Total Simple Interest = 0.42x + 0.45x + 1.65x = 2.52x

⇒ We are given Total SI = ₹2520.

⇒ 2.52x = 2520

⇒ x = 2520 ÷ 2.52 = 1000

⇒ Amount invested in Scheme P = 6x = 6 × 1000 = ₹6000

∴ The correct answer is ₹6000.

Additional Information

Simple Interest (SI)

Simple interest is calculated only on the principal amount for the entire duration. Formula: SI = (P × R × T) ÷ 100.

Compound Interest (CI)

Compound interest is calculated on the principal plus any accumulated interest. Amount = P × (1 + R ÷ 100)^T.

Equal Simple Interest Rule

If a sum is divided into parts P₁, P₂, P₃ such that their simple interests are equal, their ratio is 1 ÷ (R₁T₁) : 1 ÷ (R₂T₂) : 1 ÷ (R₃T₃).

Shortcut Trick

Horizontal Distance = Height of tower × cot(bottom depression angle)

⇒ Distance = 50 × cot(45°) = 50 × 1 = 50 m

Height of building = Height of tower × [1 − cot(bottom angle) × tan(top angle)]

⇒ Height = 50 × [1 − cot(45°) × tan(30°)] = 50 × [1 − 1 × (1 ÷ √3)]

⇒ Height ≈ 50 × [1 − 0.577] = 50 × 0.423 = 21.15 m ≈ 21 m

∴ The correct answer is Height = 21 m, Distance = 50 m.

Alternate Method

30° 45° 30° 45°

50 m x = 50 m h ≈ 21 m

A B C D E F

Given:

Height of tower AB = 50 m

Angle of depression to the top of the building ∠FAC = 30°

Angle of depression to the bottom of the building ∠FAD = 45°

Calculations:

From the property of alternate interior angles, ∠ACE = ∠FAC = 30° and ∠ADB = ∠FAD = 45°.

In right-angled ΔABD:

⇒ tan(45°) = AB ÷ BD

⇒ 1 = 50 ÷ BD

⇒ BD = 50 m

So, the horizontal distance separating the tower from the building is 50 m.

Since CE is parallel to the ground BD, CE = BD = 50 m.

In right-angled ΔAEC:

⇒ tan(30°) = AE ÷ CE

⇒ 1 ÷ √3 = AE ÷ 50

⇒ AE = 50 ÷ √3 ≈ 50 ÷ 1.732 ≈ 28.87 m

The height of the building (CD) = EB = AB − AE

⇒ CD = 50 − 28.87 = 21.13 m

⇒ Approximated height of the building ≈ 21 m

∴ The correct answer is Height = 21 m, Distance = 50 m.

Additional Information

Angles of Elevation & Depression

An angle of elevation is measured upwards from the horizontal observer's line, whereas an angle of depression is measured downwards. They are numerically equivalent via the alternate interior angles property.

Common Trigonometric Values

tan(30°) = 1 ÷ √3 | tan(45°) = 1 | tan(60°) = √3. Standard approximate value used for √3 is 1.732.

Multi-Angle Height Formula

If top and bottom of an object are observed at respective depression angles α and β from height H, object height h = H × [1 − cot(β) × tan(α)].