Quantitative Aptitude PYQs

Shortcut Trick

Net Profit% = Markup% + Discount% + (Markup × Discount) ÷ 100

Here, Markup = +40% and Discount = −10%.

Net Profit% = 40 − 10 + (40 × −10) ÷ 100

Net Profit% = 30 − 4 = 26%.

∴ The correct answer is 26%.

Alternate Method

Given: Markup Percentage = 40%, Discount Percentage = 10%.
Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = MP × (100 − Discount%) ÷ 100 | Profit% = (SP − CP) ÷ CP × 100.

⇒ Let Cost Price (CP) be 100.

⇒ Marked Price (MP) = 100 + 40% of 100 = 140.

⇒ Selling Price (SP) = 140 − 10% of 140 = 140 − 14 = 126.

⇒ Profit = SP − CP = 126 − 100 = 26.

⇒ Profit Percentage = (26 ÷ 100) × 100 = 26%.

∴ The correct answer is 26%.

Additional Information

Marked Price & Discount

The Marked Price (MP) is the price listed on the label, and Discount is always calculated on the MP to arrive at the Selling Price (SP).

Successive Percentage Change

Profit can be seen as a successive change of markup (increase) and discount (decrease): a + b + (ab ÷ 100).

Shortcut Trick

For any trigonometric identity involving secθ and tanθ, we know that sec²θ − tan²θ = 1.

This implies (secθ + tanθ)(secθ − tanθ) = 1.

If secθ + tanθ = x, then secθ − tanθ = 1/x.

Given secθ + tanθ = 4 + √15, then secθ − tanθ = 1/(4 + √15).

By rationalization, 1/(4 + √15) = (4 − √15) / (4² − (√15)²) = (4 − √15) / (16 − 15) = 4 − √15.

∴ The correct answer is 4 − √15.

Alternate Method

Given: secθ + tanθ = 4 + √15

Formula Used: sec²θ − tan²θ = 1

⇒ Let secθ + tanθ = A = 4 + √15

⇒ secθ − tanθ = 1/A

⇒ secθ − tanθ = 1 / (4 + √15)

⇒ Multiply numerator and denominator by the conjugate (4 − √15):

⇒ secθ − tanθ = (4 − √15) / [(4 + √15)(4 − √15)]

⇒ secθ − tanθ = (4 − √15) / (4² − (√15)²)

⇒ secθ − tanθ = (4 − √15) / (16 − 15)

⇒ secθ − tanθ = 4 − √15

∴ The correct answer is 4 − √15.

Additional Information

Trigonometric Identities

Standard identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Property

If cosecθ + cotθ = x, then cosecθ − cotθ = 1/x, which is useful for solving similar Cosecant-Cotangent problems.

Rationalization Technique

When the denominator is of the form (a + √b), multiplying by (a − √b) removes the radical by using the difference of squares property (a² − b).

Shortcut Trick

Apply Unitary Method directly.

Words per line = 63 ÷ 4

Total words for 52 lines = (63 ÷ 4) × 52 = 63 × 13

Calculation: 60 × 13 + 3 × 13 = 780 + 39 = 819

∴ The correct answer is 819 words.

Alternate Method

Given:

Number of lines (L₁) = 4 lines

Number of words (W₁) = 63 words

Target lines (L₂) = 52 lines

Formula Used:

Direct Proportion: W₁/L₁ = W₂/L₂

Calculations:

⇒ Let the required number of words be x.

⇒ 4 : 63 = 52 : x

⇒ 4 × x = 63 × 52

⇒ x = (63 × 52) ÷ 4

⇒ x = 63 × 13

⇒ x = 819

∴ The correct answer is 819 words.

Additional Information

Direct Proportion

When one quantity increases, the other increases at the same rate. Formula: x₁/y₁ = x₂/y₂.

Unitary Method

Finding the value of a single unit first (Value ÷ Quantity) and then multiplying by the desired number of units.

Chain Rule

Used for comparison of multiple variables: (M₁×D₁×H₁)/W₁ = (M₂×D₂×H₂)/W₂.

Shortcut Trick

Sum of ratios = 4 + 5 = 9 units.

Total age given = 36 years.

∴ 9 units = 36 years → 1 unit = 4 years.

Age of older friend (5 units) = 5 × 4 = 20 years.

∴ The correct answer is 20 years.

Alternate Method

Given:

Ratio of ages = 4 : 5

Total Sum of ages = 36 years

Let the ages be 4x and 5x.

⇒ 4x + 5x = 36

⇒ 9x = 36

⇒ x = 36 ÷ 9

⇒ x = 4

Age of the older friend = 5x

⇒ 5 × 4 = 20 years

∴ The correct answer is 20 years.

Additional Information

Ratio and Proportion Basics

If a total quantity S is divided in the ratio a : b, the parts are [a/(a+b)] × S and [b/(a+b)] × S.

Age Problems Properties

The difference between ages remains constant throughout time, whereas the ratio of ages changes every year.

Direct Proportion

When the sum is constant, an increase in one ratio part leads to a proportional decrease in the other part to maintain the total.

Shortcut Trick

The volume of a cylinder depends on the square of the radius and the height (V ∝ r²h).

Effect of halving the radius: (1/2)² = 1/4

Effect of doubling the height: 2/1 = 2

Combined effect on volume: 1/4 × 2 = 1/2

∴ The correct answer is Halved.

Alternate Method

Given:

Initial radius = r | Initial height = h

New radius (r₁) = r/2 | New height (h₁) = 2h

Formula:

Volume of a cylinder (V) = πr²h

Calculations:

⇒ Initial Volume (V) = πr²h

⇒ New Volume (V₁) = π × (r/2)² × (2h)

⇒ V₁ = π × (r²/4) × 2h

⇒ V₁ = π × r² × (2/4)h

⇒ V₁ = (1/2) × πr²h

⇒ V₁ = V / 2

∴ The correct answer is Halved.

Additional Information

Curved Surface Area (CSA) of Cylinder

The area of the side surface is given by the formula 2πrh.

Total Surface Area (TSA) of Cylinder

The sum of the areas of the curved surface and the two circular bases: 2πrh + 2πr².

Volume of a Cone

A cone with the same radius and height as a cylinder has 1/3^rd of its volume: (1/3)πr²h.

Volume of a Sphere

The volume of a sphere with radius r is given by (4/3)πr³.

Shortcut Trick

Given Sum (x + y) = 12 and Product (xy) = 32.

Direct formula for x² + y² = (Sum)² − 2(Product)

⇒ 12² − 2 × 32

⇒ 144 − 64 = 80

∴ The correct answer is 80.

Alternate Method

Given:

x + y = 12

xy = 32

Formula Used:

(x + y)² = x² + y² + 2xy

Calculations:

⇒ (x + y)² = 12²

⇒ x² + y² + 2xy = 144

⇒ x² + y² + 2(32) = 144

⇒ x² + y² + 64 = 144

⇒ x² + y² = 144 − 64

⇒ x² + y² = 80

∴ The correct answer is 80.

Additional Information

Algebraic Square Identities

The difference of squares is given by (x − y)² = x² + y² − 2xy.

Relationship between Identities

(x + y)² − (x − y)² = 4xy. This is useful when the difference of the numbers is required.

Sum and Difference of Cubes

x³ + y³ = (x + y)(x² − xy + y²) and x³ − y³ = (x − y)(x² + xy + y²).

Shortcut Trick

Express the surds in their simplest form by factoring out perfect squares:

√72 = √(36 × 2) = 6√2

√12 = √(4 × 3) = 2√3

Compare 6√2 + 2√3 with a√2 + b√3 ⇒ a = 6, b = 2.

Calculate a − b = 6 − 2 = 4.

∴ The correct answer is 4.

Alternate Method

Given:

√72 + √12 = a√2 + b√3

Calculations:

⇒ Prime factorise the numbers inside the square roots to extract perfect squares.

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

⇒ √(6² × 2) + √(2² × 3) = a√2 + b√3

⇒ 6√2 + 2√3 = a√2 + b√3

⇒ By comparing the corresponding coefficients of the like surds on both sides, we get:

⇒ a = 6 and b = 2

⇒ Now substitute the values into the required expression:

⇒ a − b = 6 − 2

⇒ a − b = 4

∴ The correct answer is 4.

Additional Information

Properties of Surds

The multiplication rule of surds √(ab) = √a × √b allows extracting perfect squares from under the radical sign to simplify expressions.

Like and Unlike Surds

Surds having the same irrational factor are called like surds (e.g., 2√3 and 5√3). They can be added or subtracted directly by simply combining their coefficients.

Equating Irrational Expressions

If p√x + q√y = a√x + b√y, where √x and √y are distinctly different prime surds, then we can equate their coefficients directly, meaning p = a and q = b.

Shortcut Trick

Income ratio = 5 : 9 | Expenditure ratio = 2 : 4 (Simplified to 1 : 2)

Cross-multiplication method: (5 × 2) − (9 × 1) units = ₹36,000 × (2 − 1)

1 unit = ₹36,000

Rupesh's income (5 units) = 5 × ₹36,000 = ₹1,80,000

∴ The correct answer is ₹1,80,000.

Alternate Method

Given: Ratio of Income = 5 : 9, Ratio of Expenditure = 2 : 4, Savings = ₹36,000 each.

Formula: Income − Savings = Expenditure

⇒ Let Rupesh's income be 5x and Manjit's income be 9x.

⇒ (5x − 36,000) / (9x − 36,000) = 2 / 4

⇒ (5x − 36,000) / (9x − 36,000) = 1 / 2

⇒ 2(5x − 36,000) = 1(9x − 36,000)

⇒ 10x − 72,000 = 9x − 36,000

⇒ 10x − 9x = 72,000 − 36,000

⇒ x = 36,000

⇒ Rupesh's monthly income = 5 × 36,000 = ₹1,80,000

∴ The correct answer is ₹1,80,000.

Additional Information

Income-Expenditure Relationship

The fundamental relation is Income = Expenditure + Savings. When savings are equal, the difference between the ratio parts of income and expenditure can often be balanced.

Ratio Balancing Method

If the difference between income ratio parts (9−5=4) and expenditure ratio parts (4−2=2) is not same, multiply ratios by constants to make the saving units (Income − Expenditure) equal for both individuals.

Shortcut Trick

When distance is constant, speed is inversely proportional to time.

Speed ratio (Usual : New) = 4 : 3

Time ratio (Usual : New) = 3 : 4

Difference in time units = 4 − 3 = 1 unit → 5 minutes.

Usual time = 3 units = 3 × 5 = 15 minutes.

∴ The correct answer is 15 min.

Alternate Method

Given:

New speed = 3/4 × Usual speed

Time delay = 5 minutes

Formula: Distance = Speed × Time

⇒ Let usual speed be S and usual time be T.

⇒ Distance (D) = S × T

⇒ New speed = 3/4 S

⇒ New time = T + 5

⇒ Since Distance is same: S × T = (3/4 S) × (T + 5)

⇒ T = 3/4 (T + 5)

⇒ 4T = 3T + 15

⇒ T = 15 minutes

∴ The correct answer is 15 min.

Additional Information

Inverse Proportion Rule

If speed becomes a/b of its original value, the time taken becomes b/a of the original time.

Late/Early Concept

Usual Time = (Late time × a) ÷ (b − a) where new speed is a/b of usual speed.

Average Speed

For same distance, Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Shortcut Trick

Total base SR = SB + BR = 8 + 14 = 22 cm (since BR = radius RQ = 14 cm).

Height h of the trapezium = QR × sin 60° = 14 × (√3/2) = 7√3 cm.

Area of Trapezium PQRS = 1/2 × (14 + 22) × 7√3 ≈ 218.23 cm².

Area of Sector QRB = (60/360) × (22/7) × 14² ≈ 102.67 cm².

Required Area PQABSP = 218.23 − 102.67 = 115.56 ≈ 116 cm².

∴ The correct answer is 116 cm².

Alternate Method

Given: PQ = 14 cm, QR = 14 cm, SB = 8 cm, ∠QRS = 60°.

Formula Used: Area of Trapezium = 1/2 × (Sum of parallel sides) × Height; Area of Sector = (θ/360) × πr².

⇒ Radius RB = RQ = 14 cm (since B is on the arc with centre R).

⇒ Side SR = SB + BR = 8 + 14 = 22 cm.

⇒ Height of trapezium = QR × sin 60° = 14 × (√3/2) = 7 × 1.732 = 12.124 cm.

⇒ Area of Trapezium PQRS = 1/2 × (14 + 22) × 12.124 = 18 × 12.124 = 218.232 cm².

⇒ Area of Sector QRB = (60/360) × (22/7) × 14 × 14 = (1/6) × 22 × 28 = 308/3 = 102.667 cm².

⇒ Area of region PQABSP = Area(PQRS) − Area(Sector QRB) = 218.232 − 102.667 = 115.565 cm².

⇒ Rounding to the nearest integer, we get 116 cm².

∴ The correct answer is 116 cm².

Additional Information

Area of a Trapezium

The area of a trapezium is calculated as 1/2 × (a + b) × h, where 'a' and 'b' are the lengths of the parallel sides and 'h' is the perpendicular distance between them.

Area of a Circular Sector

The area of a sector of a circle with radius 'r' and central angle 'θ' is given by the formula (θ/360) × πr².

Trigonometric Height

In any trapezium, the height can be found using the non-parallel side (c) and the angle it makes with the base (α) as h = c × sin(α).

Shortcut Trick

Prime numbers between 20 and 40 are 23, 29, 31, and 37. Their average is 120 ÷ 4 = 30.

Multiples of 2 form an Arithmetic Progression. Average of 10, 12, ..., 30 = (First + Last) ÷ 2 = (10 + 30) ÷ 2 = 20.

Required Difference = 30 − 20 = 10.

∴ The correct answer is 10.

Alternate Method

Given:

Range 1: Prime numbers between 20 and 40.

Range 2: Multiples of 2 between 10 and 30.

Formula Used:

Average = Sum of observations ÷ Number of observations

For an Arithmetic Progression (AP), Average = (a + l) ÷ 2, where a = first term and l = last term.

Calculations:

⇒ Prime numbers between 20 and 40 = 23, 29, 31, 37

⇒ Sum = 23 + 29 + 31 + 37 = 120

⇒ Average₁ = 120 ÷ 4 = 30

⇒ Multiples of 2 between 10 and 30 = 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30

⇒ Average₂ = (10 + 30) ÷ 2 = 40 ÷ 2 = 20

⇒ Difference = Average₁ − Average₂ = 30 − 20 = 10

∴ The correct answer is 10.

Additional Information

Average of AP

The average of any Arithmetic Progression is the mean of the first and the last term: (a + l) ÷ 2.

Prime Number Property

A prime number has exactly two distinct factors: 1 and itself. Note that 1 is neither prime nor composite, and 2 is the only even prime number.

Sum of First n Even Numbers

The sum of the first n consecutive even numbers is given by the formula n × (n + 1).

Shortcut Trick

Compare the total time taken by each person to complete 100% of the work.

A: 13 × 6 = 78 days

B: (17 ÷ 34) × 100 = 50 days

C: 19 × 3 = 57 days

D: (15 ÷ 3) × 4 = 20 days

The person with the minimum number of days completes the work first.

∴ The correct answer is D.

Alternate Method

Given:

Work done by A = 1/6 in 13 days

Work done by B = 34% in 17 days

Work done by C = 1/3 in 19 days

Work done by D = 3/4 in 15 days

Formula Used:

Total time = (Time taken) ÷ (Part of work completed)

Calculations:

⇒ Total days for A = 78 days

⇒ Total days for B = 50 days

⇒ Total days for C = 57 days

⇒ Total days for D = 20 days

⇒ Comparing all: 20 < 50 < 57 < 78

∴ The correct answer is D.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a work. The person who takes the least time is the most efficient.

Work from Part to Whole

If a person completes 1/x part of work in 't' days, the total time taken for the full work is (t × x) days.

Percentage to Fraction Conversion

For calculations involving percentages like 34%, always convert to fractions (34/100) or use the unitary method: (17 ÷ 34) × 100.

Shortcut Trick

Total Capacity (LCM of 12 and 48) = 48 units

Efficiency of P + Q = 48 ÷ 12 = 4 units/hour

Efficiency of P = 48 ÷ 48 = 1 unit/hour

Efficiency of Q = 4 − 1 = 3 units/hour

Time for Q to fill one-fourth of the cistern = (48 ÷ 4) ÷ 3 = 4 hours

∴ The correct answer is 4 hours.

Alternate Method

Given:

Time taken by P and Q together = 12 hours

Time taken by P alone = 48 hours

Formula Used:

Total Work = LCM of individual times

Efficiency = Total Work ÷ Time

Calculations:

⇒ Efficiency of Q = Efficiency of (P + Q) − Efficiency of P

⇒ Efficiency of Q = 4 − 1 = 3 units/hour

⇒ One-fourth of the cistern capacity = 48 ÷ 4 = 12 units

⇒ Time taken by Q alone to fill 12 units = 12 ÷ 3 = 4 hours

∴ The correct answer is 4 hours.

Additional Information

Pipes and Cisterns Efficiency Rule

If an inlet pipe fills a tank in X hours and another inlet pipe fills it in Y hours, their combined 1-hour work is (1/X + 1/Y). For outlet pipes, the efficiency is taken as negative.

Fraction of Work Formula

Time required to complete a fraction F of work = (F × Total Work) ÷ Efficiency of the individual or group performing the work.

Shortcut Trick

Net effective rate for 2 years at 15% = 15 + 15 + (15 × 15)/100 = 32.25%.

Total Amount = Principal + 32.25% of Principal.

Amount = 2000 + (32.25/100) × 2000 = 2000 + 645 = ₹2,645.

∴ The correct answer is ₹2,645.

Alternate Method

Given:

Principal (P) = ₹2,000

Rate of Interest (R) = 15% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

⇒ A = 2000 × (1 + 15/100)²

⇒ A = 2000 × (1 + 3/20)²

⇒ A = 2000 × (23/20)²

⇒ A = 2000 × (529/400)

⇒ A = 5 × 529

⇒ A = ₹2,645

∴ The correct answer is ₹2,645.

Additional Information

Successive Percentage Change

For two successive changes of x% and y%, the net effective change is given by the formula: x + y + (xy/100)%.

Difference between CI and SI

For a period of 2 years, the difference between Compound Interest and Simple Interest is: CI − SI = P(R/100)².

Compound Interest Calculation

Compound Interest is calculated using the formula: CI = A − P = P[(1 + R/100)ⁿ − 1].

Shortcut Trick

Given Discount% = 9%, which means Selling Price (SP) is 91% of Marked Price (MP).

Using unitary method: 91 units = 53,053.

1 unit = 53,053 ÷ 91 = 583.

Marked Price (100 units) = 583 × 100 = 58,300.

∴ The correct answer is ₹58,300.

Alternate Method

Given: Selling Price (SP) = ₹53,053, Discount (D%) = 9%.

Formula: MP = [SP × 100] ÷ [100 − Discount%]

⇒ MP = [53,053 × 100] ÷ [100 − 9]

⇒ MP = [53,053 × 100] ÷ 91

⇒ MP = 583 × 100

⇒ MP = 58,300

∴ The correct answer is ₹58,300.

Additional Information

Marked Price & Discount

The Marked Price (MP) is the price printed on an article. Discount is always calculated on the MP. Selling Price = MP − Discount.

Discount Percentage

Discount% = (Discount ÷ MP) × 100. It represents the reduction offered on the tag price.

Shortcut Trick

For a tower of height H, if the angles of elevation from two points at distances d₁ and d₂ from its base are complementary, the height is given directly by H = √(d₁ × d₂).

Given the distances from the base are d₁ = b² m and d₂ = a² m.

Applying the formula, height of the tower H = √(b² × a²) = ab.

∴ The correct answer is ab.

Alternate Method

X
Y
Top (T)
Base (B)
b² m
a² m
h

θ

90° - θ

Given:

Distance of point X from the base of the tower, XB = b² m

Distance of point Y from the base of the tower, YB = a² m

Let the height of the vertical tower TB be h.

The angles of elevation at X and Y are complementary. Let the angle at X be θ and the angle at Y be 90° − θ.

Formula Used:

In a right-angled triangle, tan(θ) = Perpendicular ÷ Base

Calculations:

In right-angled ΔTBX:

⇒ tan(θ) = TB ÷ XB

⇒ tan(θ) = h ÷ b²   ... (Equation 1)

In right-angled ΔTBY:

⇒ tan(90° − θ) = TB ÷ YB

⇒ cot(θ) = h ÷ a²   ... (Equation 2)

Multiplying Equation 1 and Equation 2:

⇒ tan(θ) × cot(θ) = (h ÷ b²) × (h ÷ a²)

⇒ 1 = h² ÷ (a² × b²)

⇒ h² = a²b²

⇒ h = √(a²b²)

⇒ h = ab

∴ The correct answer is ab.

Additional Information

Complementary Angles Property in Towers

When the angles of elevation of a tower from two points at distances x and y from its base are complementary, the height is always H = √(x × y).

Angles of Elevation and Depression

The angle of elevation is formed when looking up from the horizontal line of sight, while the angle of depression is formed when looking down. Both are alternate interior angles and are equal in magnitude.

Trigonometric Co-functions

For any angle θ, the tangent and cotangent functions are complementary values: tan(90° − θ) = cot(θ) and cot(90° − θ) = tan(θ).

Shortcut Trick

Solve mentally using the BODMAS rule directly:

Parentheses → (3 − 6) = −3

Multiplication → 9 × (−3) = −27

Curly braces → {30 − 27} = 3

Division → 954 ÷ 3 = 318

∴ The correct answer is 318.

Alternate Method

Given:

Expression = [954 ÷ {30 + 9 × (3 − 6)}]

Concept Used:

BODMAS Rule: Brackets, Orders, Division, Multiplication, Addition, Subtraction.

Calculations:

⇒ [954 ÷ {30 + 9 × (3 − 6)}]

First, solve the inner parenthesis (3 − 6):

⇒ [954 ÷ {30 + 9 × (−3)}]

Next, perform the multiplication inside the curly braces:

⇒ [954 ÷ {30 − 27}]

Now, perform the subtraction inside the curly braces:

⇒ [954 ÷ 3]

Finally, divide 954 by 3:

⇒ 318

∴ The correct answer is 318.

Additional Information

BODMAS Rule

BODMAS stands for Brackets, Orders, Division, Multiplication, Addition, and Subtraction. It strictly dictates the left-to-right order of operations for solving mathematical expressions.

V-BODMAS Concept

An extension of BODMAS where 'V' stands for Vinculum (or bar). If a mathematical bar is present over an expression, that part must be evaluated before moving on to any brackets.

Bracket Solving Order

Nested brackets should always be solved from the innermost to the outermost boundary in this exact sequential order: small (), curly {}, and then square [].

Shortcut Trick

Since the distance is constant, the ratio of time taken is inversely proportional to the ratio of speeds.

Ratio of speeds (Prashant : Nisha) = 117 : 48 = 39 : 16

Ratio of time (Prashant : Nisha) = 16 : 39

Given Prashant takes 16 hours, this precisely matches the 16 units of time.

Thus, the time taken by Nisha is 39 units = 39 hours.

∴ The correct answer is 39 hours.

Alternate Method

Given:

Speed of Prashant = 117 km/hr

Time taken by Prashant = 16 hours

Speed of Nisha = 48 km/hr

Formula Used:

Distance = Speed × Time

Time = Distance ÷ Speed

Delhi
Kanpur

Prashant (117 km/hr, 16 hrs)

Nisha (48 km/hr, ? hrs)

Calculations:

⇒ Total distance from Delhi to Kanpur = 117 × 16

⇒ Total distance = 1872 km

⇒ Time taken by Nisha = Total distance ÷ Speed of Nisha

⇒ Time taken by Nisha = 1872 ÷ 48

⇒ Time taken by Nisha = 39 hours

∴ The correct answer is 39 hours.

Additional Information

Relative Speed

When two bodies move in the same direction, Relative Speed = S₁ − S₂. When moving in opposite directions, Relative Speed = S₁ + S₂.

Average Speed

For equal distances covered at speeds S₁ and S₂, Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂). For varying distances, Average Speed = Total Distance ÷ Total Time.

Unit Conversion

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

Shortcut Trick

Rate = 5% = 1/20. Let Principal = (20)² = 400 units.

1^st year CI = 20 units, 2^nd year CI = 20 + 1 = 21 units.

Given 2^nd year CI = ₹2,352 ⇒ 21 units = 2352 ⇒ 1 unit = 112.

Principal = 400 × 112 = ₹44,800.

∴ The correct answer is ₹44,800.

Alternate Method

Given:

Rate of Interest (R) = 5% per annum

Compound Interest for 2^nd year (I₂) = ₹2,352

Principal (P)

1^st Year CI (I₁)
5% of P

2^nd Year CI (I₂)
₹ 2,352

I₂ = I₁ × 1.05

Formula Used:

Interest for 2^nd year = Interest for 1^st year + Interest on 1^st year's interest

Simple Interest for 1^st year = (Principal × Rate × Time) ÷ 100

Calculations:

⇒ Let I₁ be the interest for the first year.

⇒ I₂ = I₁ + (5% of I₁) = 1.05 × I₁

⇒ 1.05 × I₁ = 2352

⇒ I₁ = 2352 ÷ 1.05 = ₹2,240

⇒ Since 1^st year interest is 5% of the Principal (P):

⇒ 5% of P = 2240

⇒ P × (5 ÷ 100) = 2240

⇒ P = 2240 × 20 = ₹44,800

∴ The correct answer is ₹44,800.

Additional Information

Effective Rate of Compound Interest

For 2 years at R% per annum, the effective compound interest rate is (2R + R²/100)% of the principal.

Difference between CI and SI

For 2 years, the difference between Compound Interest and Simple Interest is exactly the interest earned on the 1^st year's interest, which equals P(R/100)².

Multiplicative Factor Method

Amount after n years under CI is calculated as A = P(1 + R/100)ⁿ. The factor (1 + R/100) multiplies the previous year's amount directly.

Shortcut Trick

Check if the given sides form a Pythagorean triplet: 80² + 18² = 6400 + 324 = 6724.

Since 82² = 6724, it satisfies the Pythagorean theorem, meaning it is a right-angled triangle.

The hypotenuse is the longest side (82 cm), making the base and height 18 cm and 80 cm.

Area = (1 ÷ 2) × Base × Height

Area = (1 ÷ 2) × 18 × 80 = 720 cm².

∴ The correct answer is 720 cm².

Alternate Method

Given:

The sides of the triangle are a = 82 cm, b = 18 cm, and c = 80 cm.

80 cm
18 cm
82 cm

Formula Used:

Heron's formula for the area of a triangle: Area = √(s × (s − a) × (s − b) × (s − c))

Where 's' is the semi-perimeter: s = (a + b + c) ÷ 2

Calculations:

⇒ s = (82 + 18 + 80) ÷ 2

⇒ s = 180 ÷ 2 = 90 cm

⇒ Area = √(90 × (90 − 82) × (90 − 18) × (90 − 80))

⇒ Area = √(90 × 8 × 72 × 10)

⇒ Area = √(720 × 720)

⇒ Area = 720 cm²

∴ The correct answer is 720 cm².

Additional Information

Right-Angled Triangle

A triangle in which one angle is exactly 90°. The sides follow the Pythagorean theorem: a² + b² = c², and its Area = (1 ÷ 2) × Base × Height.

Heron's Formula

Used to find the area of any general triangle when all three sides are known. Area = √(s(s − a)(s − b)(s − c)), where s is the semi-perimeter.

Equilateral Triangle

A triangle with all three sides equal. Its Area = (√3 ÷ 4) × a², where a is the side length.