Quantitative Aptitude PYQs

Shortcut Trick

Total deduction = 85 + 65 + 61 = ₹211

Remaining amount = 5575 − 211 = ₹5364

Divide ₹5364 in ratio 5 : 11 : 20 → Total parts = 36

Value of 1 part = 5364 ÷ 36 = ₹149

Difference (Y and Z) = (20 − 11) parts × 149 + (61 − 65) = 9 × 149 − 4 = 1341 − 4 = ₹1337

∴ The correct answer is ₹1337.

Alternate Method

Given: Total amount = ₹5,575. Deductions: X = ₹85, Y = ₹65, Z = ₹61. New ratio = 5 : 11 : 20.

⇒ Total Deductions = 85 + 65 + 61 = ₹211

⇒ Remaining Amount = 5575 − 211 = ₹5364

⇒ Sum of ratio parts = 5 + 11 + 20 = 36

⇒ 36 units = 5364 ⇒ 1 unit = 5364 ÷ 36 = 149

⇒ Reduced share of Y = 11 × 149 = ₹1639

⇒ Reduced share of Z = 20 × 149 = ₹2980

⇒ Original share of Y = 1639 + 65 = ₹1704

⇒ Original share of Z = 2980 + 61 = ₹3041

⇒ Difference = 3041 − 1704 = ₹1337

∴ The correct answer is ₹1337.

Additional Information

Ratio and Proportion Basics

If an amount A is divided in the ratio a:b:c, the shares are calculated as [a/(a+b+c)]×A, [b/(a+b+c)]×A, and [c/(a+b+c)]×A respectively.

Handling Deductions

When fixed amounts are subtracted from shares before applying a ratio, subtract the total deduction from the total sum before calculating unit values.

Difference Rule

To find the difference between original shares, calculate the difference between reduced shares and adjust by the difference in their respective deductions.

Shortcut Trick

Evaluate all the perfect square roots directly mentally:

⇒ √256 = 16, √169 = 13, √144 = 12, √25 = 5

⇒ Substitute these values directly: 16 − 13 + (12 − 5)²

⇒ 3 + (7)²

⇒ 3 + 49 = 52

∴ The correct answer is 52.

Alternate Method

Given:

Expression: √256 − √169 + (√144 − √25)²

Method Used:

Apply the BODMAS rule: evaluate Order (square roots), then Brackets, followed by Exponents, and finally Addition/Subtraction.

Calculations:

⇒ Find the values of individual square roots:

⇒ √256 = 16

⇒ √169 = 13

⇒ √144 = 12

⇒ √25 = 5

⇒ Substitute the values back into the main expression:

⇒ 16 − 13 + (12 − 5)²

⇒ Simplify the bracket first (12 − 5):

⇒ 16 − 13 + (7)²

⇒ Evaluate the square of 7:

⇒ 16 − 13 + 49

⇒ Perform subtraction and addition from left to right:

⇒ 3 + 49

⇒ 52

∴ The correct answer is 52.

Additional Information

BODMAS Rule

It defines the correct sequence for operations: Brackets, Order (Powers and Roots), Division × Multiplication (left-to-right), and Addition − Subtraction (left-to-right).

Perfect Squares up to 20

Memorising squares up to 30 significantly improves calculation speed in competitive exams. For example: 14² = 196, 15² = 225, 16² = 256, 17² = 289.

Basic Algebraic Identities

Although arithmetic simplification is best here, knowing (a − b)² = a² − 2ab + b² is essential for cases involving unknown variables.

Shortcut Trick

Fourth Proportional of 4, 8, 19 → (8 × 19) ÷ 4 = 38

Mean Proportional of 9, 4 → √(9 × 4) = 6

Required Ratio → 38 : 6 = 19 : 3

∴ The correct answer is 19 : 3.

Alternate Method

Given: Numbers for 4^th proportional (a=4, b=8, c=19) and mean proportional (m=9, n=4).

Formula Used: Fourth proportional x = (b × c) / a; Mean proportional y = √(m × n).

⇒ Fourth proportional (x) = (8 × 19) / 4

⇒ x = 2 × 19 = 38

⇒ Mean proportional (y) = √(9 × 4) = √36

⇒ y = 6

⇒ Ratio = x : y = 38 : 6

⇒ Ratio = 19 : 3

∴ The correct answer is 19 : 3.

Additional Information

Third Proportional

For two numbers a and b, the third proportional c is such that a : b :: b : c, giving c = b² / a.

Mean Proportional

The mean proportional between two numbers a and b is √(a × b), also known as the geometric mean.

Duplicate and Sub-duplicate Ratio

Duplicate ratio of a:b is a²:b² and sub-duplicate ratio is √a:√b.

Shortcut Trick

Identify the LCM of the given divisors: 12, 15, 18, and 20.

LCM (12, 15, 18, 20) = 180.

Required number = LCM − Increase Value

Required number = 180 − 10 = 170.

∴ The correct answer is 170.

Alternate Method

Given: Divisors are 12, 15, 18, and 20; Number is increased by 10.

Formula Used: Required Number = LCM(divisors) − k (where k is the increment).

⇒ Prime factors of 12 = 2² × 3

⇒ Prime factors of 15 = 3 × 5

⇒ Prime factors of 18 = 2 × 3²

⇒ Prime factors of 20 = 2² × 5

⇒ LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180

⇒ According to the condition: x + 10 = 180

⇒ x = 180 − 10 = 170

∴ The correct answer is 170.

Additional Information

LCM (Least Common Multiple)

It is the smallest positive integer that is evenly divisible by a given set of numbers.

Remainder and Increment Rules

If a number is increased by 'k', subtract 'k' from the LCM. If it leaves a remainder 'r', add 'r' to the LCM.

Common Remainder Case

If a number when divided by a, b, c leaves remainders x, y, z respectively such that (a−x) = (b−y) = (c−z) = K, then the number is LCM(a, b, c) − K.

Shortcut Trick

Find factors of Area (40) whose squares sum to 89.

Pairs: (1,40), (2,20), (4,10), (5,8).

⇒ 5² + 8² = 25 + 64 = 89 (Matches diagonal squared).

⇒ Perimeter = 2 × (8 + 5) = 2 × 13 = 26 cm.

∴ The correct answer is 26 cm.

Alternate Method

Given:

Diagonal (d) = √89 cm

Area (A) = l × b = 40 cm²

Formula Used:

⇒ d² = l² + b²

⇒ (l + b)² = l² + b² + 2lb

Calculations:

⇒ l² + b² = (√89)² = 89

⇒ lb = 40

⇒ (l + b)² = 89 + 2 × 40

⇒ (l + b)² = 89 + 80 = 169

⇒ l + b = √169 = 13 cm

⇒ Perimeter = 2(l + b) = 2 × 13 = 26 cm

∴ The correct answer is 26 cm.

Additional Information

Pythagorean Property of Rectangles

The diagonal of a rectangle forms a right-angled triangle with the length and breadth, satisfying l² + b² = d².

Algebraic Identity for Perimeter

Perimeter is related to diagonal and area by the identity: (Perimeter/2)² = d² + 2 × Area.

Area and Perimeter Relationship

For a fixed area, the perimeter of a rectangle is minimized when the rectangle is a square (l = b).

Shortcut Trick

Using the Man-Days relation: M₁ × D₁ = M₂ × D₂

20 × 12 = 80 × D₂

D₂ = 240 ÷ 80 = 3 days

∴ The correct answer is 3 days.

Alternate Method

Given:

Initial strength (M₁) = 20 pupils

Initial days (D₁) = 12 days

New strength (M₂) = 80 pupils

Formula Used:

M₁ × D₁ = M₂ × D₂

Calculations:

⇒ Total provisions = M₁ × D₁

⇒ Total provisions = 20 × 12 = 240 pupil-days

⇒ New days (D₂) = Total provisions ÷ M₂

⇒ D₂ = 240 ÷ 80

⇒ D₂ = 3 days

∴ The correct answer is 3 days.

Additional Information

Inverse Proportion

When the number of people increases, the food provisions will last for fewer days. This is a case of inverse variation, represented as x × y = k (constant).

Chain Rule Formula

For work and provisions, the general formula is (M₁ × D₁ × H₁) / W₁ = (M₂ × D₂ × H₂) / W₂, where M is men, D is days, H is hours, and W is work.

Shortcut Trick

Use the Successive Percentage method for 2 years: Net Rate = x + y + (xy/100)

Net Interest Rate = 15 + 15 + (15 × 15)/100 = 30 + 2.25 = 32.25%

Total Amount = 132.25% of Principal = 4800 × 1.3225

Amount = 48 × 132.25 = ₹6,348

∴ The correct answer is ₹6,348.

Alternate Method

Given:

Principal (P) = ₹4,800

Rate (R) = 15% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

⇒ A = 4800 × (1 + 15/100)²

⇒ A = 4800 × (1 + 3/20)²

⇒ A = 4800 × (23/20) × (23/20)

⇒ A = 4800 × (529/400)

⇒ A = (4800 ÷ 400) × 529

⇒ A = 12 × 529

⇒ A = 6348

∴ The correct answer is ₹6,348.

Additional Information

Difference between CI and SI

For 2 years, the difference (D) between Compound Interest and Simple Interest is given by: D = P × (R/100)².

Compound Interest Formula

Interest (CI) = Amount − Principal = P[(1 + R/100)ⁿ − 1].

Compounding Half-Yearly

If interest is compounded half-yearly, the Rate becomes R/2 and the Time (n) becomes 2n.

Shortcut Trick

When the distance for both journeys is the same (from A to B and back to A), use the harmonic mean formula.

Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂)

Average Speed = (2 × 70 × 111) ÷ (70 + 111) = 15540 ÷ 181

Average Speed ≈ 85.8563...

∴ The correct answer is 85.86 km/hr.

Alternate Method

Given:

Distance (A to B) = 933 km

Speed (A to B), S₁ = 70 km/hr

Speed (B to A), S₂ = 111 km/hr

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Total Distance = 933 + 933 = 1866 km

⇒ Time taken from A to B (T₁) = 933 ÷ 70 = 13.3285 hr

⇒ Time taken from B to A (T₂) = 933 ÷ 111 = 8.4054 hr

⇒ Total Time = 13.3285 + 8.4054 = 21.7339 hr

⇒ Average Speed = 1866 ÷ 21.7339 ≈ 85.856 km/hr

∴ The correct answer is 85.86 km/hr.

Additional Information

Average Speed (Equal Distances)

When an object covers equal distances at different speeds, the average speed is independent of the actual distance and is calculated as the harmonic mean of speeds.

Relative Speed

Relative speed in same direction is (S₁ − S₂) and in opposite direction is (S₁ + S₂).

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Shortcut Trick

Ratio of ages of Ram and Lalith = 6 : 1

Product of ratio parts = 6 × 1 = 6 units²

Given product = 216 ⇒ 1 unit² = 216 ÷ 6 = 36 ⇒ 1 unit = √36 = 6 years

Difference in ratio = 6 − 1 = 5 units

Difference in ages = 5 × 6 = 30 years

∴ The correct answer is 30 years.

Alternate Method

Given:

Product of Ram and Lalith ages = 216 years

Ratio of their ages = 6 : 1

Let the age of Ram be 6x and Lalith be x.

⇒ 6x × x = 216

⇒ 6x² = 216

⇒ x² = 216 ÷ 6

⇒ x² = 36

⇒ x = √36 = 6

⇒ Age of Ram = 6 × 6 = 36 years

⇒ Age of Lalith = 1 × 6 = 6 years

⇒ Difference = 36 − 6 = 30 years

∴ The correct answer is 30 years.

Additional Information

Ratio and Product

When the product of two quantities in ratio a : b is given as P, always use a constant x such that (ax) × (bx) = P, leading to x = √(P / ab).

Age Difference Constant

The difference between the ages of two persons remains constant throughout their lives, regardless of the number of years passed.

Ratio Difference

To find the difference directly from ratio a : b, calculate (a − b) × x, where x is the value of one ratio unit.

Shortcut Trick

Assume savings = 100. Then expenditure = 100 + 200 = 300.

Income = 300 + 100 = 400.

New expenditure = 300 − 10% of 300 = 270. New savings = 100 + 34% of 100 = 134.

New income = 270 + 134 = 404.

% Increase in income = (4 / 400) × 100 = 1%.

∴ The correct answer is 1%.

Alternate Method

Given that Expenditure is 200% more than Savings.

⇒ Let Savings = S and Expenditure = E.

⇒ E = S + 2S = 3S.

⇒ Income (I) = E + S = 3S + S = 4S.

⇒ If S = 100, then E = 300 and I = 400.

⇒ New E = 300 × (90/100) = 270.

⇒ New S = 100 × (134/100) = 134.

⇒ New I = 270 + 134 = 404.

⇒ Increase in Income = 404 − 400 = 4.

⇒ Percentage increase = (4 / 400) × 100 = 1%.

∴ The correct answer is 1%.

Additional Information

Income-Expenditure Relationship

The fundamental formula is Income = Expenditure + Savings. Any change in two variables directly impacts the third.

Percentage Change in Income

It can also be calculated using weighted averages: %ΔIncome = (%ΔE × E + %ΔS × S) / (E + S).

X% More Than Concept

If A is x% more than B, then A = B × (1 + x/100). For 200% more, the multiplier is (1 + 2) = 3.

Shortcut Trick

Let actual cost of a pen be x and a pencil be y.

We get two equations: 2x + 19y = 118 and 5x + 8y = 137

Using cross-multiplication to solve directly for y:

y = (2 × 137 − 118 × 5) ÷ (2 × 8 − 19 × 5) = −316 ÷ −79 = 4

x = (118 − 19 × 4) ÷ 2 = 42 ÷ 2 = 21

Cost of 14 pens and 6 pencils = 14 × 21 + 6 × 4 = 294 + 24 = 318

∴ The correct answer is ₹318.

Alternate Method

Given:

Cost of 2 pens and 19 pencils = ₹118

When the cost of a pen decreases by ₹5 and a pencil increases by ₹4, the cost of 5 pens and 8 pencils becomes ₹144.

Formula Used:

Standard elimination method for a system of linear equations in two variables.

Calculations:

⇒ Let the actual cost of a pen be ₹x and a pencil be ₹y.

⇒ 2x + 19y = 118   --- (Equation 1)

⇒ The new cost of a pen is ₹(x − 5) and a pencil is ₹(y + 4).

⇒ 5(x − 5) + 8(y + 4) = 144

⇒ 5x − 25 + 8y + 32 = 144

⇒ 5x + 8y + 7 = 144

⇒ 5x + 8y = 137   --- (Equation 2)

⇒ Multiplying Equation 1 by 5 and Equation 2 by 2, we get:

⇒ 10x + 95y = 590

⇒ 10x + 16y = 274

⇒ Subtracting the second modified equation from the first:

⇒ 79y = 316

⇒ y = 4

⇒ Substituting the value of y in Equation 1:

⇒ 2x + 19(4) = 118

⇒ 2x + 76 = 118

⇒ 2x = 42

⇒ x = 21

⇒ The actual cost of a pen is ₹21 and a pencil is ₹4.

⇒ Cost of 14 pens and 6 pencils = 14x + 6y

⇒ 14(21) + 6(4) = 294 + 24 = 318

∴ The correct answer is ₹318.

Additional Information

Linear Equation in Two Variables

An equation of the form ax + by + c = 0, where a and b are non-zero constants, representing a straight line on a Cartesian plane.

Methods of Solving Simultaneous Equations

Commonly solved using substitution (expressing one variable in terms of another), elimination (making coefficients equal to cancel out a variable), or cross-multiplication.

Conditions for Solvability

For a₁x + b₁y = c₁ and a₂x + b₂y = c₂: A unique solution exists if a₁ ÷ a₂ ≠ b₁ ÷ b₂.

Shortcut Trick

The product of two numbers is equal to the product of their HCM and LCM.

⇒ 1^st Number × 2^nd Number = HCM × LCM

⇒ 182 × N₂ = 91 × 546

⇒ N₂ = (91 × 546) ÷ 182 = 546 ÷ 2

⇒ N₂ = 273

∴ The correct answer is 273.

Alternate Method

Given:

HCM (Highest Common Measure) = 91

LCM (Lowest Common Multiple) = 546

First Number = 182

Formula Used:

Product of the two numbers = Product of their HCM and LCM

Num 1 × Num 2
182 × ?

=

HCM × LCM
91 × 546

Calculations:

⇒ Let the other number be x.

⇒ 182 × x = 91 × 546

⇒ x = (91 × 546) ÷ 182

⇒ We know that 182 is exactly twice of 91 (91 × 2 = 182).

⇒ x = 546 ÷ 2

⇒ x = 273

∴ The correct answer is 273.

Additional Information

Product of Two Numbers Rule

For any two positive integers a and b, it is always true that a × b = HCF(a, b) × LCM(a, b). Note that this specific rule applies strictly to TWO numbers only and does not scale directly to three or more numbers.

Co-prime Numbers

Two numbers are said to be co-prime if their HCF is 1. In such instances, their LCM is exactly equal to the product of the two numbers.

HCF and LCM of Fractions

HCF of fractions = (HCF of numerators) ÷ (LCM of denominators). Conversely, the LCM of fractions = (LCM of numerators) ÷ (HCF of denominators).

Shortcut Trick

60% can be written as the fraction 3/5.

Let the original number be 5 units. The increased number will be 5 + 3 = 8 units.

8 units = 3550 ⇒ 1 unit = 3550 ÷ 8 = 443.75

Original number = 5 units = 5 × 443.75 = 2218.75

∴ The correct answer is 2218.75.

Alternate Method

Given:

Number after a 60% increase = 3550

Formula Used:

Final Value = Initial Value × (1 + Percentage Increase ÷ 100)

Original (100%)

• 60%

Increased Number (160%) = 3550

Calculations:

Let the original number be x.

⇒ x + 60% of x = 3550

⇒ x + 0.60x = 3550

⇒ 1.60x = 3550

⇒ x = 3550 ÷ 1.60

⇒ x = 35500 ÷ 16

⇒ x = 2218.75

∴ The correct answer is 2218.75.

Additional Information

Percentage Increase Formula

New Value = Original Value × (1 + Percentage Increase ÷ 100). It is often faster to convert the percentage into a fraction and use the ratio method to solve proportionately.

Percentage Decrease Formula

New Value = Original Value × (1 − Percentage Decrease ÷ 100).

Successive Percentage Change

When a number is changed successively by a% and b%, the net percentage change is calculated using the standard formula: (a + b + ab ÷ 100)%.

Shortcut Trick

Let 100g of rice cost 100 units.

Selling Price (SP) for 100g stated value = 100 + 39% profit = 139 units.

Actual Cost Price (CP) for weight given (17% less) = 100 − 17 = 83 units.

Profit % = [(SP − CP) ÷ CP] × 100 = [(139 − 83) ÷ 83] × 100.

Profit % = (56 ÷ 83) × 100 ≈ 67.47% ≈ 67%.

∴ The correct answer is 67%.

Alternate Method

Given:

⇒ Claimed profit on sale = 39%

⇒ Less weight used by the trader = 17%

Formula Used:

⇒ Profit % = [(SP − CP) ÷ CP] × 100

Actual CP = 83(83g given)Actual SP = 139(100g claimed)Profit = 56Dishonest Dealer ConceptProfit % = (56 ÷ 83) × 100 ≈ 67%

Calculations:

⇒ Let the stated value be 100g and its cost price be 100 units.

⇒ The trader claims a profit of 39%, so the Selling Price (SP) of 100g = 100 + 39 = 139 units.

⇒ The trader uses 17% less weight, so the actual weight given = 100 − 17 = 83g.

⇒ The Cost Price (CP) incurred by the trader is only for 83g = 83 units.

⇒ Total Profit = SP − CP = 139 − 83 = 56 units.

⇒ Profit % = (Profit ÷ Actual CP) × 100

⇒ Profit % = (56 ÷ 83) × 100

⇒ Profit % = 5600 ÷ 83 ≈ 67.469%

⇒ Rounding to the nearest whole number, the profit percentage is 67%.

∴ The correct answer is 67%.

Additional Information

Dishonest Dealer Selling at Cost Price

If a trader claims to sell goods at cost price but uses a false weight, the profit percentage is calculated purely on the weight discrepancy: Profit % = [Error ÷ (True Value − Error)] × 100.

Multiplying Factor Method

For dishonest dealer problems involving multiple markups or discounts, the final profit ratio is the product of individual multipliers: (Final SP ÷ Final CP) = (Claimed SP ÷ True CP) × (True Weight ÷ False Weight).

Marked Price and Discounts

When a dealer marks up the price and offers a discount, the effective Selling Price is determined by successive percentage changes on the Cost Price: SP = CP × (1 + Markup %) × (1 − Discount %).

Shortcut Trick

Evaluate the square roots of the given perfect squares directly.

√400 = 20

√0.0225 = 0.15

√9 = 3

Expression value = 20 + 0.15 − 3 = 17.15

∴ The correct answer is 17.15.

Alternate Method

Given:

Expression: √400 + √0.0225 − √9

Concept Used:

To easily find the square root of a decimal number, convert it into a fraction first.

Calculation:

⇒ √400 = 20

⇒ √0.0225 = √(225 ÷ 10000) = 15 ÷ 100 = 0.15

⇒ √9 = 3

Substitute these values back into the expression:

⇒ 20 + 0.15 − 3

⇒ 20.15 − 3

⇒ 17.15

∴ The correct answer is 17.15.

Additional Information

Square Root of Decimals

Count the number of decimal places in the perfect square. The square root will always have exactly half as many decimal places.

BODMAS Rule

Always solve mathematical expressions in the order of Brackets, Orders (powers and roots), Division, Multiplication, Addition, and Subtraction.

Perfect Squares

A number that can be expressed as the product of an integer with itself. Memorising perfect squares up to 30 speeds up simplification significantly.

Shortcut Trick

Subtract the respective remainders from the given numbers to find the exactly divisible numbers.

527 − 7 = 520 and 632 − 8 = 624.

The required largest number is the Highest Common Factor (HCF) of 520 and 624.

Using the difference trick for HCF: 624 − 520 = 104.

Since 104 completely divides both 520 (104 × 5) and 624 (104 × 6), the HCF is 104.

∴ The correct answer is 104.

Alternate Method

Given:

Numbers = 527 and 632

Remainders = 7 and 8

Formula Used:

Required largest number = HCF of (N₁ − R₁) and (N₂ − R₂)

Calculations:

⇒ First divisible number = 527 − 7 = 520

⇒ Second divisible number = 632 − 8 = 624

⇒ Find the prime factorization of 520 and 624.

⇒ Factors of 520 = 2³ × 5 × 13

⇒ Factors of 624 = 2⁴ × 3 × 13

⇒ The common prime factors with the lowest power are 2³ and 13.

⇒ Highest Common Factor (HCF) = 2³ × 13

⇒ HCF = 8 × 13 = 104

∴ The correct answer is 104.

Additional Information

Highest Common Factor (HCF)

The HCF of two or more numbers is the greatest number that completely divides each of them without leaving a remainder. It is also known as the Greatest Common Divisor (GCD).

Same Remainder Case

The largest number that divides a, b, and c leaving the same remainder 'r' in each case is given by the HCF of (a − r), (b − r), and (c − r).

Unknown Same Remainder Case

The largest number that divides a, b, and c leaving the same unknown remainder in each case is given by the HCF of |a − b|, |b − c|, and |c − a|.

Product of Two Numbers

For any two numbers 'a' and 'b', the product of the numbers is equal to the product of their HCF and LCM: a × b = HCF(a, b) × LCM(a, b).

Shortcut Trick

If xA = yB = zC, the ratio A : B : C is directly found by taking (yz) : (xz) : (xy).

For 9A = 3B = 7C, the ratio A : B : C = (3 × 7) : (9 × 7) : (9 × 3) = 21 : 63 : 27.

Simplifying by dividing by 3, we get A : B : C = 7 : 21 : 9.

Total units = 7 + 21 + 9 = 37 units.

37 units = ₹74,000 ⇒ 1 unit = ₹2,000

Share of A = 7 units × ₹2,000 = ₹14,000.

∴ The correct answer is ₹14,000.

Alternate Method

Given:

⇒ Total amount to be divided = ₹74,000

⇒ Relationship between shares: 9 × A = 3 × B = 7 × C

Total = 37 parts = ₹74,000

A (7x)
B (21x)
C (9x)

A = ₹14,000

Calculations:

⇒ Let 9A = 3B = 7C = k

⇒ A = k/9, B = k/3, C = k/7

⇒ Ratio of their shares, A : B : C = 1/9 : 1/3 : 1/7

⇒ To simplify, multiply the entire ratio by the LCM of denominators (9, 3, 7) which is 63.

⇒ A : B : C = (63 × 1/9) : (63 × 1/3) : (63 × 1/7)

⇒ A : B : C = 7 : 21 : 9

⇒ Sum of ratio parts = 7 + 21 + 9 = 37 parts

⇒ Value of 37 parts = ₹74,000

⇒ Value of 1 part = 74,000 ÷ 37 = ₹2,000

⇒ Share of A = 7 parts = 7 × 2,000 = ₹14,000

∴ The correct answer is ₹14,000.

Additional Information

Inverse Ratio Concept

If pA = qB = rC, then the simplest ratio form of A : B : C is calculated as (1/p) : (1/q) : (1/r).

Dividing a Quantity

To divide a total quantity Q into a ratio a : b : c, the actual value of the first part is (a ÷ (a + b + c)) × Q.

Combining Multiple Ratios

If given two sets of ratios such as A:B = m:n and B:C = p:q, the combined ratio A:B:C becomes (m × p) : (n × p) : (n × q).

Shortcut Trick

Directly expand the brackets and combine the like terms mentally:

⇒ 6x² − 3x + 5x² − 20 + 18

⇒ (6 + 5)x² − 3x + (18 − 20)

⇒ 11x² − 3x − 2

∴ The correct answer is 11x² − 3x − 2.

Alternate Method

Given:

Expression: x(6x − 3) + 5(x² − 4) + 18

Calculation:

⇒ First, apply the distributive property to multiply out the brackets.

⇒ Multiply x by (6x − 3): x × 6x − x × 3 = 6x² − 3x

⇒ Multiply 5 by (x² − 4): 5 × x² − 5 × 4 = 5x² − 20

⇒ Substitute these expanded forms back into the original expression:

⇒ 6x² − 3x + 5x² − 20 + 18

⇒ Now, group the like terms together (terms with x², terms with x, and constants):

⇒ (6x² + 5x²) − 3x + (−20 + 18)

⇒ Add the coefficients of the grouped terms:

⇒ 11x² − 3x − 2

∴ The correct answer is 11x² − 3x − 2.

Additional Information

BODMAS Rule

BODMAS stands for Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction. It determines the correct sequence of operations in any mathematical expression.

Distributive Property

The distributive property states that a(b + c) = ab + ac. This is essential for expanding expressions with brackets before combining the terms.

Combining Like Terms

Terms with the exact same variables and powers are called like terms.

Only like terms can be added or subtracted from one another (e.g., ax² + bx² = (a + b)x²).

Shortcut Trick

List the odd numbers in the given range: 467, 469, 471, 473, and 475.

Eliminate numbers divisible by 3 and 5: 471 (sum of digits = 12) and 475 (ends in 5).

Check the rest using prime factors up to √476 ≈ 21.

469 = 7 × 67, and 473 = 11 × 43. Both are non-prime.

Only 467 has no prime factors up to 19. Hence, it is the only prime number.

∴ The correct answer is 1.

Alternate Method

Given:

Range of numbers: 466 to 476 (inclusive).

Concept Used:

A prime number has exactly two factors: 1 and itself.

To verify if a number N is prime, check for divisibility by all prime numbers up to √N.

Calculation:

⇒ Even numbers in the range (466, 468, 470, 472, 474, 476) are divisible by 2. Thus, they are not prime.

⇒ The remaining odd numbers to test are: 467, 469, 471, 473, 475.

⇒ 469 ÷ 7 = 67

⇒ 471 ÷ 3 = 157

⇒ 473 ÷ 11 = 43

⇒ 475 ÷ 5 = 95

⇒ The number 467 is not divisible by any prime numbers up to 19 (2, 3, 5, 7, 11, 13, 17, 19). Therefore, 467 is a prime number.

⇒ The only prime number present in this given range is 467.

∴ The correct answer is 1.

Additional Information

Prime Number Definition

A natural number greater than 1 that has only two factors: 1 and itself.

Divisibility Test for Primes

To check if a number N is prime, verify its divisibility by all prime numbers less than or equal to √N. If none divide it evenly, N is prime.

Co-prime Numbers

Two integers are said to be co-prime or mutually prime if their highest common factor (HCF) is exactly 1.

Twin Primes

Twin primes are a pair of prime numbers that have a difference of exactly 2, such as (11, 13) or (41, 43).

Shortcut Trick

Total Selling Price = Remaining plates × Selling price per plate

Total Selling Price = (79 − 13) × 5.90 = 66 × 5.90 = ₹389.40

Total Cost Price = ₹79

Profit = Total SP − Total CP = 389.40 − 79 = ₹310.40

∴ The correct answer is ₹310.40.

Alternate Method

Given:

Total cost price of 79 plates = ₹79

Number of broken plates = 13

Selling price per plate = ₹5.90

Formula Used:

Profit = Total Selling Price − Total Cost Price

Total CP
₹79
(79 plates)

Total SP
₹389.40
(66 plates)

Remaining: 66
Selling Rate:
₹5.90 / plate

Profit = SP − CP = ₹310.40

Calculations:

⇒ Number of remaining plates = 79 − 13 = 66

⇒ Total Selling Price = 66 × 5.90

⇒ Total Selling Price = ₹389.40

⇒ Total Cost Price = ₹79

⇒ Profit = Total Selling Price − Total Cost Price

⇒ Profit = 389.40 − 79 = ₹310.40

∴ The correct answer is ₹310.40.

Additional Information

Profit & Loss Formula

Profit occurs when the Selling Price (SP) is strictly greater than the Cost Price (CP). The formula used is: Profit = SP − CP.

Effective Cost Price

In scenarios involving damaged or lost items, the total initial investment remains the effective Cost Price. The revenue generated solely from the remaining items dictates the final actual Selling Price.

Profit Percentage

If required to calculate profit percentage, it is universally calculated over the effective Cost Price. Formula: Profit % = (Profit ÷ CP) × 100.