Quantitative Aptitude PYQs

Shortcut Trick

Slope (m) is the ratio of the vertical change to the horizontal change between two points.

m = (Change in y) ÷ (Change in x)

m = (7 − 3) ÷ (4 − 2)

m = 4 ÷ 2 = 2

∴ The correct answer is 2.

Alternate Method

Given:

Points on the line: (x₁, y₁) = (2, 3) and (x₂, y₂) = (4, 7)

Formula Used:

Slope of a line, m = (y₂ − y₁) / (x₂ − x₁)

X Y 0

(2, 3) (4, 7) Run = 2 Rise = 4 Slope (m) = Rise ÷ Run = 4 ÷ 2 = 2

Calculations:

⇒ Substitute the given coordinates into the slope formula.

⇒ m = (7 − 3) / (4 − 2)

⇒ m = 4 / 2

⇒ m = 2

∴ The correct answer is 2.

Additional Information

Equation of a Straight Line

The point-slope form of the line equation is given by (y − y₁) = m(x − x₁), where m is the slope.

Parallel and Perpendicular Lines

Parallel lines always have equal slopes (m₁ = m₂). Perpendicular lines have negative reciprocal slopes (m₁ × m₂ = −1).

Distance Between Two Points

The straight-line distance d between (x₁, y₁) and (x₂, y₂) is calculated using d = √((x₂ − x₁)² + (y₂ − y₁)²).

Shortcut Trick

For sectors with the same radius, their areas are directly proportional to their central angles.

Angle of the first sector = 180° = π radians.

Angle of the second sector = π/2 radians.

Ratio of areas = π ÷ (π/2) = 2 : 1.

∴ The correct answer is 2:1.

Alternate Method

Given:

Radius of both sectors (r) = 12 cm

Central angle of first sector (θ₁) = 180°

Central angle of second sector (θ₂) = π/2 radians

180° r = 12 cm Sector 1

:

π/2 rad r = 12 cm Sector 2

Formula Used:

Area of a sector (in degrees) = (θ ÷ 360°) × π × r²

Area of a sector (in radians) = (1/2) × r² × θ

Calculations:

⇒ Area of first sector = (180° ÷ 360°) × π × (12)²

⇒ Area of first sector = (1/2) × π × 144 = 72π cm²

⇒ Area of second sector = (1/2) × (12)² × (π/2)

⇒ Area of second sector = (1/2) × 144 × (π/2) = 36π cm²

⇒ Ratio = Area of first sector : Area of second sector

⇒ Ratio = 72π : 36π = 2 : 1

∴ The correct answer is 2:1.

Additional Information

Area of a Sector

When the angle θ is in degrees, Area = (θ ÷ 360°) × π × r². When θ is in radians, Area = (1/2) × r² × θ.

Arc Length

The length of an arc is given by l = (θ ÷ 360°) × 2πr in degrees, and l = r × θ in radians.

Radians to Degrees Conversion

To convert an angle from radians to degrees, multiply it by (180° ÷ π). Therefore, π radians = 180°.

Shortcut Trick

The sum of all interior angles of any polygon depends directly on its number of sides (n).

Sum = (n − 2) × 180°

For a 10-sided polygon, substitute n = 10.

Sum = (10 − 2) × 180° = 8 × 180° = 1440°

∴ The correct answer is 1440°.

Alternate Method

Decagon (10 Sides) Number of triangles = n − 2 Triangles = 10 − 2 = 8 Each triangle sums to 180° Total = 8 × 180° = 1440°

Given:

The polygon is a decagon, which has 10 sides.

Formula Used:

Sum of interior angles of an n-sided polygon = (n − 2) × 180°

Calculations:

⇒ Number of sides (n) = 10

⇒ Sum of interior angles = (10 − 2) × 180°

⇒ Sum = 8 × 180°

⇒ Sum = 1440°

∴ The correct answer is 1440°.

Additional Information

Sum of Exterior Angles

The sum of all exterior angles for any convex polygon is always 360°, regardless of the number of sides.

Interior Angle of a Regular Polygon

Each interior angle of a regular n-sided polygon can be found using the formula: [(n − 2) × 180°] ÷ n.

Number of Diagonals

The total number of diagonals that can be drawn in an n-sided polygon is given by the formula: n(n − 3) ÷ 2.

Shortcut Trick

Midpoint of the line connecting (2, 8) and (6, 4) is ((2 + 6) ÷ 2, (8 + 4) ÷ 2) = (4, 6).

Slope of the given line segment = (4 − 8) ÷ (6 − 2) = −4 ÷ 4 = −1.

Slope of its perpendicular bisector = −1 ÷ (−1) = 1.

Equation of the bisector passing through (4, 6) with slope 1: y − 6 = 1 × (x − 4).

Simplifying the equation gives: y = x + 2.

∴ The correct answer is y = x + 2.

Alternate Method

Given:

Two points: A(2, 8) and B(6, 4)

Formula Used:

Midpoint formula: M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)

Slope formula: m = (y₂ − y₁) ÷ (x₂ − x₁)

Perpendicular slope condition: m₁ × m₂ = −1

Equation of a straight line: y − y₁ = m × (x − x₁)

A (2, 8) B (6, 4) M (4, 6)

y = x + 2

Calculation:

⇒ Let the line segment be AB. The perpendicular bisector passes exactly through the midpoint M of AB.

⇒ Midpoint M = ((2 + 6) ÷ 2, (8 + 4) ÷ 2)

⇒ M = (8 ÷ 2, 12 ÷ 2) = (4, 6)

⇒ Slope of line AB (m₁) = (4 − 8) ÷ (6 − 2)

⇒ m₁ = −4 ÷ 4 = −1

⇒ Let the slope of the perpendicular bisector be m₂.

⇒ m₁ × m₂ = −1

⇒ −1 × m₂ = −1 ⇒ m₂ = 1

⇒ Now, substitute the point M(4, 6) and slope m₂ = 1 into the straight line equation.

⇒ y − 6 = 1 × (x − 4)

⇒ y − 6 = x − 4

⇒ y = x − 4 + 6

⇒ y = x + 2

∴ The correct answer is y = x + 2.

Additional Information

Distance Formula

The linear distance between two points (x₁, y₁) and (x₂, y₂) on a Cartesian plane is given by D = √((x₂ − x₁)² + (y₂ − y₁)²).

Section Formula

The coordinates of a point dividing a line segment internally in the ratio m:n are calculated as ((m×x₂ + n×x₁) ÷ (m+n), (m×y₂ + n×y₁) ÷ (m+n)).

Parallel and Perpendicular Lines

Two lines are parallel if their slopes are strictly equal (m₁ = m₂). Two lines are exactly perpendicular if the product of their respective slopes is negative one (m₁ × m₂ = −1).

Shortcut Trick

The ratio of the areas of similar triangles is the square of the ratio of their perimeters.

Ratio of areas = 3² : 5² = 9 : 25.

Total area in ratio units = 9 + 25 = 34 units.

Given total area = 34 cm², which means 34 units = 34 cm² ⇒ 1 unit = 1 cm².

Area of the larger triangle = 25 units × 1 = 25 cm².

∴ The correct answer is 25 cm².

Alternate Method

Given:

Ratio of perimeters of two similar triangles = 3 : 5

Sum of their areas = 34 cm²

Formula Used:

For two similar triangles, Ratio of Areas = (Ratio of Perimeters)²

Perimeter = 3x Area = 9a

Perimeter = 5x Area = 25a

Similar Triangles Area Relationship Total Area = 9a + 25a = 34 cm²

Calculations:

⇒ Let the areas of the smaller and larger triangles be A₁ and A₂ respectively.

⇒ A₁ : A₂ = 3² : 5²

⇒ A₁ : A₂ = 9 : 25

⇒ Let A₁ = 9a and A₂ = 25a

⇒ Total Area = 9a + 25a = 34a

⇒ 34a = 34

⇒ a = 1 cm²

⇒ Area of the larger triangle = 25a = 25 × 1 = 25 cm²

∴ The correct answer is 25 cm².

Additional Information

Areas of Similar Triangles Theorem

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, altitudes, or perimeters.

Perimeters of Similar Triangles Theorem

The ratio of the perimeters of two similar triangles is exactly equal to the ratio of their corresponding sides or corresponding medians.

Volumes of Similar Figures

If two 3D geometrical figures are similar, the ratio of their volumes is equal to the cube of the ratio of their corresponding linear dimensions (sides or radii).

Shortcut Trick

If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Here, a = 19, b = 20, and c = −39. Since 19 + 20 − 39 = 0, we can apply the trick.

19³ + 20³ − 39³ = 3 × 19 × 20 × (−39) = −44460.

Now, add 118 to this result: −44460 + 118 = −44342.

∴ The correct answer is −44342.

Alternate Method

Given:

19³ + 20³ − 39³ + 118

Formula Used:

If a + b + c = 0, then a³ + b³ + c³ = 3abc

Calculations:

⇒ Let a = 19, b = 20, and c = −39

⇒ Sum of terms: a + b + c = 19 + 20 − 39 = 0

⇒ Therefore, 19³ + 20³ + (−39)³ = 3 × 19 × 20 × (−39)

⇒ 3 × 19 = 57

⇒ 57 × 20 = 1140

⇒ 1140 × (−39) = −44460

⇒ Substituting this value back into the original expression:

⇒ −44460 + 118 = −44342

∴ The correct answer is −44342.

Additional Information

Algebraic Identity for Cubes

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). If a + b + c = 0, the right-hand side becomes 0, leading to the direct shortcut: a³ + b³ + c³ = 3abc.

Sum of Cubes Formula

a³ + b³ = (a + b)(a² − ab + b²)

Difference of Cubes Formula

a³ − b³ = (a − b)(a² + ab + b²)

BODMAS Rule

Always follow the correct order of operations in algebraic expressions: Brackets, Orders (powers/roots), Division × Multiplication, Addition − Subtraction.

Shortcut Trick

Using the trigonometric identity: cosec²A = 1 + cot²A

Substitute cot A = x + 1/x directly into the identity.

cosec²A = 1 + (x + 1/x)²

cosec²A = 1 + x² + 1/x² + 2 = x² + 1/x² + 3

∴ The correct answer is x² + 1/x² + 3.

Alternate Method

Given:

cot A = x + 1/x

Formula Used:

cosec²θ − cot²θ = 1

(a + b)² = a² + b² + 2ab

Calculations:

⇒ cosec²A = 1 + cot²A

⇒ Substitute the given value of cot A:

⇒ cosec²A = 1 + (x + 1/x)²

⇒ Expanding the term using the algebraic identity:

⇒ (x + 1/x)² = x² + (1/x)² + 2 × x × (1/x)

⇒ (x + 1/x)² = x² + 1/x² + 2

⇒ Now substitute this back into the expression for cosec²A:

⇒ cosec²A = 1 + x² + 1/x² + 2

⇒ cosec²A = x² + 1/x² + 3

∴ The correct answer is x² + 1/x² + 3.

Additional Information

Fundamental Trigonometric Identities

1. sin²θ + cos²θ = 1

2. sec²θ − tan²θ = 1

3. cosec²θ − cot²θ = 1

Important Algebraic Identities

1. (a + b)² = a² + b² + 2ab

2. (a − b)² = a² + b² − 2ab

Reciprocal Trigonometric Relations

sin θ = 1 / cosec θ | cos θ = 1 / sec θ | tan θ = 1 / cot θ

Shortcut Trick

A tangent line forms a 90° angle with the radius at the point of contact, creating a right-angled triangle.

The sides of this triangle are the radius (8 cm), the tangent length, and the distance to the center (17 cm).

These values form a standard Pythagorean triplet: (8, 15, 17).

Since the radius is 8 cm and the hypotenuse is 17 cm, the tangent must be the remaining leg, which is 15 cm.

∴ The correct answer is 15 cm.

Alternate Method

O P A

8 cm 17 cm PA = ?

ΔOAP forms a right-angled triangle

Given:

Radius of the circle, OA = 8 cm

Distance from the external point P to the center O, OP = 17 cm

Formula Used:

By Pythagoras Theorem: Hypotenuse² = Base² + Perpendicular²

Calculations:

⇒ In right-angled ΔOAP, OP is the hypotenuse because the angle at the point of tangency (angle OAP) is 90°.

⇒ OP² = OA² + PA²

⇒ 17² = 8² + PA²

⇒ 289 = 64 + PA²

⇒ PA² = 289 − 64

⇒ PA² = 225

⇒ PA = √225 = 15 cm

∴ The correct answer is 15 cm.

Additional Information

Tangent-Radius Theorem

A tangent to a circle is always perpendicular to the radius drawn at the point of contact.

Tangents from an External Point

The lengths of the two distinct tangents drawn from an external point to a circle are always equal.

Secant-Tangent Theorem

If a tangent segment PA and a secant segment PBC are drawn to a circle from an exterior point P, then PA² = PB × PC.

Shortcut Trick

When the angle between two tangents from an external point is 90°, the tangents and the radii to the points of contact form a square.

The side of this square is exactly equal to the radius of the circle (r = 5 cm).

The distance from the external point to the center is simply the diagonal of this square.

Distance OP = r × √2 = 5 × √2 = 5√2 cm.

∴ The correct answer is 5√2 cm.

Alternate Method

P O A B

5 cm 5 cm 90°

PO = 5√2 cm

Given:

⇒ Radius of the circle (OA = OB) = 5 cm.

⇒ Angle between tangents PA and PB = ∠APB = 90°.

Formula Used:

⇒ Diagonal of a square = Side × √2

Calculations:

⇒ A tangent to a circle is always perpendicular to the radius drawn through the point of contact.

⇒ ∴ ∠OAP = 90° and ∠OBP = 90°.

⇒ In quadrilateral OAPB, the sum of all interior angles is 360°.

⇒ ∠AOB = 360° − (90° + 90° + 90°) = 90°.

⇒ Since all four angles of the quadrilateral are 90° and the adjacent sides are equal (OA = OB = 5 cm), OAPB is a square.

⇒ The distance from point P to the center O is represented by the diagonal OP of the square OAPB.

⇒ Length of diagonal OP = Side × √2 = 5 × √2 = 5√2 cm.

∴ The correct answer is 5√2 cm.

Additional Information

Tangents from an External Point

When two tangents are drawn to a circle from a single external point, their lengths are always strictly equal (PA = PB).

Tangent-Radius Theorem

The tangent line is always perpendicular to the radius of the circle drawn to the exact point of tangency.

Properties of a Square

A quadrilateral where all four angles are 90° and at least two adjacent sides are equal is a square. The diagonal bisects the internal angles and has a length of Side × √2.

Shortcut Trick

Step 1: Ignore the decimals and multiply the integers directly: 45 × 6 = 270.

Step 2: Count the total number of decimal places in the given numbers: 2 (in 0.45) + 1 (in 0.6) = 3 decimal places.

Step 3: Place the decimal point 3 positions from the right in the product: 0.270 = 0.27.

∴ The correct answer is 0.27.

Alternate Method

Given:

The expression to simplify is 0.45 × 0.6

Method Used:

Convert the decimal numbers into fractions, multiply the numerators and denominators, and then convert the result back to a decimal.

Calculations:

⇒ 0.45 = 45 / 100

⇒ 0.6 = 6 / 10

⇒ 0.45 × 0.6 = (45 / 100) × (6 / 10)

⇒ 0.45 × 0.6 = (45 × 6) / (100 × 10)

⇒ 0.45 × 0.6 = 270 / 1000

⇒ 0.45 × 0.6 = 27 / 100

⇒ 0.45 × 0.6 = 0.27

∴ The correct answer is 0.27.

Additional Information

Multiplication of Decimals

When multiplying decimal numbers, multiply them exactly as if they are whole numbers. The number of decimal places in the final product will always be equal to the total sum of the decimal places present in the original factors.

Converting Decimals to Fractions

To convert a decimal to a fraction, write the decimal number without the point as the numerator. For the denominator, write 1 followed by as many zeros as there are decimal places in the original number (e.g., 0.85 = 85/100).

BODMAS Rule

When simplifying mathematical expressions containing multiple operations, always follow the BODMAS sequence: Bracket, Order (Powers/Roots), Division, Multiplication, Addition, and Subtraction.

Shortcut Trick

Expand the given expression using the formula (a + b)² = a² + b² + 2ab

⇒ (√3 + 2)² = (√3)² + (2)² + 2 × √3 × 2

⇒ 3 + 4 + 4√3 = ? + 4√3

⇒ 7 + 4√3 = ? + 4√3

⇒ ? = 7

∴ The correct answer is 7.

Alternate Method

Given:

(√3 + 2)² = ? + 4√3

Formula Used:

(a + b)² = a² + b² + 2ab

Calculations:

Let a = √3 and b = 2

⇒ LHS = (√3 + 2)²

⇒ LHS = (√3)² + (2)² + 2 × √3 × 2

⇒ LHS = 3 + 4 + 4√3

⇒ LHS = 7 + 4√3

Now, substituting this value back into the original equation:

⇒ 7 + 4√3 = ? + 4√3

Subtracting 4√3 from both sides:

⇒ ? = 7

∴ The correct answer is 7.

Additional Information

Algebraic Identities

Basic expansion formulas: (a − b)² = a² + b² − 2ab and a² − b² = (a + b)(a − b).

Rules of Surds

Squaring a square root gives the original base value: (√a)² = a. Also, √a × √b = √(ab).

Shortcut Trick

Square roots of numbers ending in 25 follow a simple pattern: n × (n+1) followed by 25 gives a root of (n5).

√7225 → 72 = 8 × 9 ⇒ Root is 85.

√1225 → 12 = 3 × 4 ⇒ Root is 35.

⇒ (85 − 35) × 1/5

⇒ 50 ÷ 5 = 10

∴ The correct answer is 10.

Alternate Method

Given:

(√7225 − √1225) × (1/5)

Calculations:

To calculate √7225:

We know that 80² = 6400 and 90² = 8100.

Since 7225 is between these and ends in 25, the square root must end in 5.

⇒ √7225 = 85

To calculate √1225:

We know that 30² = 900 and 40² = 1600.

Since 1225 is between these and ends in 25, the square root must end in 5.

⇒ √1225 = 35

Substituting the square root values back into the expression:

⇒ (85 − 35) × (1/5)

⇒ 50 × (1/5)

⇒ 50 ÷ 5

⇒ 10

∴ The correct answer is 10.

Additional Information

BODMAS Rule

It defines the standard order of operations: Brackets, Orders (powers and roots), Division, Multiplication, Addition, and Subtraction.

Square Root Shortcut for Ends in 25

For a perfect square ending in 25, remove the last two digits. Factor the remaining number as n × (n + 1). The square root will simply be n followed by 5.

Shortcut Trick

Given 25% of (P + Q) = 75% of (P − Q).

Taking the inverse ratio: (P + Q) / (P − Q) = 75% / 25% = 3 / 1

Applying the Componendo and Dividendo rule:

P / Q = (3 + 1) / (3 − 1)

⇒ P / Q = 4 / 2 = 2 / 1

∴ The correct answer is 2 : 1.

Alternate Method

Given:

25% of (P + Q) = 75% of (P − Q)

P (2 parts) Q (1 part) P + Q (3 parts) P − Q (1 part) Ratio P : Q = 2 : 1

Calculations:

⇒ (25 / 100) × (P + Q) = (75 / 100) × (P − Q)

⇒ 1 × (P + Q) = 3 × (P − Q)

⇒ P + Q = 3P − 3Q

Bringing all P terms to one side and Q terms to the other:

⇒ Q + 3Q = 3P − P

⇒ 4Q = 2P

⇒ P / Q = 4 / 2

⇒ P : Q = 2 : 1

∴ The correct answer is 2 : 1.

Additional Information

Componendo and Dividendo

If a / b = c / d, then (a + b) / (a − b) = (c + d) / (c − d). This rule is extremely useful for rapidly solving equations that involve sums and differences of the same two variables.

Inverse Ratio Logic

If A% of X = B% of Y, the ratio of the variables is inversely proportional to their respective percentages. Therefore, X / Y = B / A.

Percentage Simplification

Converting standard percentages to simple fractions (e.g., 25% = 1 / 4, 75% = 3 / 4) makes cross-multiplication faster and prevents errors with larger numerical values.

Shortcut Trick

Value ratio = (Coin Ratio) × (Coin Denomination)

Value ratio of ₹5, ₹2, and ₹1 coins = (2 × 5) : (3 × 2) : (4 × 1) = 10 : 6 : 4

Total value in ratio units = 10 + 6 + 4 = 20 units

20 units = ₹200 ⇒ 1 unit = 10

Number of coins = 2 × 10, 3 × 10, and 4 × 10 = 20, 30, and 40 respectively.

∴ The correct answer is 20 coins of ₹5, 30 coins of ₹2, and 40 coins of ₹1.

Alternate Method

Given:

Ratio of number of ₹5, ₹2, and ₹1 coins = 2 : 3 : 4

Total amount in the bag = ₹200

Formula Used:

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

₹5

₹2

₹1

Ratio: 2x Ratio: 3x Ratio: 4x

Value: 10x Value: 6x Value: 4x

Total Value = 10x + 6x + 4x = 20x = ₹200

Calculations:

⇒ Let the number of ₹5, ₹2, and ₹1 coins be 2x, 3x, and 4x respectively.

⇒ Value of ₹5 coins = 5 × 2x = 10x

⇒ Value of ₹2 coins = 2 × 3x = 6x

⇒ Value of ₹1 coins = 1 × 4x = 4x

⇒ Total value of all coins = 10x + 6x + 4x = 20x

⇒ According to the question, 20x = 200

⇒ x = 200 ÷ 20 = 10

⇒ Number of ₹5 coins = 2x = 2 × 10 = 20

⇒ Number of ₹2 coins = 3x = 3 × 10 = 30

⇒ Number of ₹1 coins = 4x = 4 × 10 = 40

∴ The correct answer is 20 coins of ₹5, 30 coins of ₹2, and 40 coins of ₹1.

Additional Information

Case 1: When Total Amount is Given

If the ratio of the number of coins is given alongside the total value, always multiply the coin ratio by their respective denominations to find the ratio of their values.

Case 2: When Total Number of Coins is Given

If the ratio of the values of the coins is given alongside the total number of coins, always divide the value ratio by their respective denominations to find the ratio of the number of coins.

Direct Relation

Number of coins = Total Value ÷ Value of one coin.

Shortcut Trick

We know the direct ratio relation between Cost Price (CP) and Marked Price (MP):

⇒ CP / MP = (100 − Discount%) / (100 + Profit%)

⇒ CP / 800 = (100 − 15) / (100 + 20) = 85 / 120

⇒ CP = 800 × (85 / 120) = 1700 / 3

⇒ CP ≈ 566.67

∴ The correct answer is ₹566.67.

Alternate Method

Given:

Marked Price (MP) = ₹800

Discount = 15%

Profit = 20%

Formula Used:

Selling Price (SP) = MP − (Discount% of MP)

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

CP = ?

SP = ₹680

MP = ₹800

+20% Profit

−15% Discount

CP = ₹680 × (100 / 120) = ₹566.67

Calculations:

⇒ SP = 800 − (15 / 100) × 800

⇒ SP = 800 − 120 = ₹680

⇒ CP = (SP × 100) / (100 + 20)

⇒ CP = (680 × 100) / 120

⇒ CP = 68000 / 120

⇒ CP = 1700 / 3 ≈ ₹566.67

∴ The correct answer is ₹566.67.

Additional Information

Direct CP and MP Ratio

A very useful formula linking Cost Price and Marked Price directly without calculating the Selling Price intermediate is: CP / MP = (100 − Discount%) / (100 + Profit%).

Successive Discounts

If two successive discounts of x% and y% are offered, the equivalent single discount is given by the successive percentage formula: (x + y − xy / 100)%.

MarkUp Percentage

MarkUp is the initial amount added to the Cost Price to set the Marked Price. It is always calculated on the base of Cost Price: MarkUp% = ((MP − CP) / CP) × 100.

Shortcut Trick

Profit = 10% of CP = ₹300 ⇒ CP = ₹3000

First Selling Price (SP₁) = CP + Profit = 3000 + 300 = ₹3300

Since SP₁ is after a 20% discount, SP₁ corresponds to 80% of Marked Price (MP).

New Selling Price after a 30% discount will correspond to 70% of MP.

New SP = SP₁ × (70 ÷ 80) = 3300 × (7 ÷ 8) = ₹2887.50

∴ The correct answer is ₹2887.50.

Alternate Method

Given:

Profit % = 10%

Discount % = 20%

Profit earned = ₹300

Formula Used:

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

Selling Price (SP) = CP + Profit = Marked Price (MP) × (100 − Discount%) ÷ 100

+10% (Profit)

-20% (Discount)

-30% Discount

CP ₹3000

SP (1^st) ₹3300

MP ₹4125

New SP ₹2887.50

Calculations:

⇒ 10% of CP = 300

⇒ CP = 300 × (100 ÷ 10) = ₹3000

⇒ SP₁ = CP + Profit = 3000 + 300 = ₹3300

Let Marked Price be MP. Given the initial discount is 20%:

⇒ SP₁ = MP × (100 − 20)%

⇒ 3300 = MP × 0.8

⇒ MP = 3300 ÷ 0.8 = ₹4125

If the product is sold at a 30% discount:

⇒ New SP = MP × (100 − 30)%

⇒ New SP = 4125 × 0.7 = ₹2887.50

∴ The correct answer is ₹2887.50.

Additional Information

Marked Price and Discount

Discount is always calculated on the marked price of the article. Formula: SP = MP × (100 − D%) ÷ 100.

Profit and Loss

Profit and loss are always calculated on the cost price unless specified otherwise. Formula: SP = CP × (100 ± P/L%) ÷ 100.

Direct Relation between CP and MP

The ratio of Cost Price to Marked Price is directly derived as CP : MP = (100 − D%) : (100 + P%).

Successive Discounts

If two successive discounts of x% and y% are given, the equivalent single discount is given by (x + y − xy ÷ 100)%.

Shortcut Trick

For a cubic equation Aa³ + Ba² + Ca + D = 0, the sum of its roots is −B/A.

Comparing the given equation a³ − 4a² + 5a − 2 = 0, we get A = 1 and B = −4.

Sum of roots (x + y + z) = −(−4) ÷ 1 = 4.

∴ The correct answer is 4.

Alternate Method

Given:

Cubic equation: a³ − 4a² + 5a − 2 = 0

Roots of the equation: x, y, and z

Method Used:

Factorisation Method

⇒ Let P(a) = a³ − 4a² + 5a − 2 = 0

⇒ By trial and error, put a = 1 → P(1) = 1³ − 4(1)² + 5(1) − 2 = 1 − 4 + 5 − 2 = 0

⇒ Since P(1) = 0, (a − 1) is a factor of the equation.

⇒ Dividing P(a) by (a − 1) gives a quotient of (a² − 3a + 2).

⇒ Factorising the quotient: a² − 3a + 2 = a² − 2a − a + 2 = a(a − 2) − 1(a − 2) = (a − 1)(a − 2)

⇒ The complete factorisation is: (a − 1)(a − 1)(a − 2) = 0

⇒ The three roots are a = 1, a = 1, and a = 2.

⇒ Let x = 1, y = 1, and z = 2.

⇒ x + y + z = 1 + 1 + 2 = 4

∴ The correct answer is 4.

Additional Information

Vieta's Formulas for Cubic Equations

For a cubic equation Aa³ + Ba² + Ca + D = 0 with roots x, y, z: Sum of roots (x + y + z) = −B/A, sum of roots taken two at a time (xy + yz + zx) = C/A, and product of roots (xyz) = −D/A.

Factor Theorem

If P(x) is a polynomial and P(c) = 0 for some value c, then (x − c) is a factor of the polynomial P(x).

Roots of Polynomials

A polynomial of degree 'n' always has exactly 'n' complex roots (including real roots), and the sum of its roots is always calculated directly as −(Coefficient of xⁿ⁻¹) ÷ (Coefficient of xⁿ).

Shortcut Trick

Time duration from 15 March to 15 September is exactly 6 months (1/2 year).

Simple Interest = Principal × Rate × Time ÷ 100

Interest = 6000 × 10 × (1/2) ÷ 100 = ₹300

Total Amount = Principal + Interest = 6000 + 300 = ₹6300

∴ The correct answer is ₹6300.

Alternate Method

Given:

Principal (P) = ₹6000

Rate of Interest (R) = 10% per annum

Time (T) = 15 March to 15 September

15 March 15 September

P = ₹6000 Amount = ?

Time = 6 months (1/2 year)

SI = ₹300

Formula Used:

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

Amount = P + SI

Calculations:

⇒ The period from 15 March to 15 September is exactly 6 months.

⇒ Time (T) in years = 6 ÷ 12 = 1/2 year

⇒ SI = (6000 × 10 × 1/2) ÷ 100

⇒ SI = (6000 × 5) ÷ 100

⇒ SI = 30000 ÷ 100 = ₹300

⇒ Total Amount = 6000 + 300 = ₹6300

∴ The correct answer is ₹6300.

Additional Information

Simple Interest vs Compound Interest

Simple interest is calculated on the original principal amount only, whereas compound interest is calculated on both the principal and the accumulated interest of previous periods.

Compound Interest Formula

Amount = P × (1 + R÷100)^T, where P is the principal, R is the rate of interest per annum, and T is the time in years.

Time Calculation by Exact Dates

When interest is calculated day-by-day, the starting day is excluded and the ending day is included. For an exact same date across months, the number of months is directly considered as the exact time period (e.g., exactly 6 months here).

Shortcut Trick

When the simple interest from different schemes is equal, the ratio of investments is inversely proportional to the product of Rate and Time.

Ratio of investments = 1/R₁ : 1/R₂ : 1/R₃ (since time is 1 year for all)

Ratio = 1/4 : 1/5 : 1/10 = 5 : 4 : 2

Total parts = 5 + 4 + 2 = 11 parts = ₹2200 ⇒ 1 part = ₹200

Amount invested at 5% corresponds to 4 parts = 4 × 200 = ₹800

∴ The correct answer is ₹800.

Alternate Method

Total = ₹2200

Part 1 @ 4% for 1 yr

Part 2 @ 5% for 1 yr

Part 3 @ 10% for 1 yr

Interest is EQUAL across all three parts (I1 = I2 = I3)

Given:

Total sum invested = ₹2200

Rate of interest for scheme 1 (R₁) = 4%

Rate of interest for scheme 2 (R₂) = 5%

Rate of interest for scheme 3 (R₃) = 10%

Time period (T) = 1 year

Formula Used:

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

Calculations:

Let the amounts invested in the three schemes be P₁, P₂, and P₃.

⇒ P₁ + P₂ + P₃ = 2200

According to the question, the simple interest earned from all three schemes is equal:

⇒ (P₁ × 4 × 1) ÷ 100 = (P₂ × 5 × 1) ÷ 100 = (P₃ × 10 × 1) ÷ 100

⇒ 4P₁ = 5P₂ = 10P₃

Dividing the equation by the LCM of 4, 5, and 10 (which is 20):

⇒ 4P₁ ÷ 20 = 5P₂ ÷ 20 = 10P₃ ÷ 20

⇒ P₁ ÷ 5 = P₂ ÷ 4 = P₃ ÷ 2

⇒ Ratio of investments (P₁ : P₂ : P₃) = 5 : 4 : 2

Sum of the ratio parts = 5 + 4 + 2 = 11 parts

⇒ 11 parts = 2200

⇒ 1 part = 2200 ÷ 11 = 200

The amount invested at 5% is P₂, which corresponds to 4 parts.

⇒ Amount invested at 5% = 4 × 200 = ₹800

∴ The correct answer is ₹800.

Additional Information

Equal Interest Condition in SI

If a total sum is divided into multiple parts such that the Simple Interest earned on each part is equal, the ratio of the principal amounts is directly given by: P₁ : P₂ : P₃ = (1/R₁T₁) : (1/R₂T₂) : (1/R₃T₃).

Equal Amount Condition in SI

If a sum is divided such that the Final Amount (Principal + Interest) earned on each part is equal, the ratio of the principal investments is given by: P₁ : P₂ : P₃ = 1/(100 + R₁T₁) : 1/(100 + R₂T₂) : 1/(100 + R₃T₃).

Simple vs Compound Interest Concept

In Simple Interest, the interest is calculated only on the original principal for the entire time period. In Compound Interest, the interest is calculated on both the principal and the accumulated interest from previous periods.

Shortcut Trick

When objects are on opposite sides of a tower of height h with angles of depression α and β:

Distance between them = h × (cot α + cot β)

⇒ Distance = 120 × (cot 30° + cot 45°)

⇒ Distance = 120 × (√3 + 1)

∴ The correct answer is 120(√3 + 1) m.

Alternate Method

Given:

Height of the tower (AB) = 120 m

Angle of depression of first car (C) = 30° ⇒ Angle of elevation from C = 30°

Angle of depression of second car (D) = 45° ⇒ Angle of elevation from D = 45°

30°

45°

30°

45°

A B C D

120 m

x = 120√3 m y = 120 m

Calculations:

In right-angled triangle ABC:

⇒ tan 30° = AB ÷ BC

⇒ 1 ÷ √3 = 120 ÷ BC

⇒ BC = 120√3 m

In right-angled triangle ABD:

⇒ tan 45° = AB ÷ BD

⇒ 1 = 120 ÷ BD

⇒ BD = 120 m

Total distance between the two cars = BC + BD

⇒ Total distance = 120√3 + 120 = 120(√3 + 1) m

∴ The correct answer is 120(√3 + 1) m.

Additional Information

Heights and Distances on the Same Side

If both objects are on the same side of the tower, the distance between them is given by d = h(cot α − cot β), where α is the smaller angle of elevation.

Angle of Elevation vs Depression

The angle of depression from the top of an object is always equal to the angle of elevation from the bottom due to alternate interior angles formed by parallel horizontal lines.

Standard Trigonometric Values

tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. Cotangent is always the reciprocal of tangent.