Quantitative Aptitude PYQs

Shortcut Trick

For a number to be divisible by 45, it must be divisible by both 5 and 9.

Divisibility by 5: The last digit b must be 0 or 5. Since b is a positive integer, b = 5.

Divisibility by 9: The sum of digits (7 + a + 3 + 5 + 5) = 20 + a must be a multiple of 9.

The nearest multiple of 9 greater than 20 is 27. Therefore, 20 + a = 27 ⇒ a = 7.

The value of (a + b) = 7 + 5 = 12.

∴ The correct answer is 12.

Alternate Method

Given: The 5-digit number is 7a35b and it is divisible by 45.

Conditions: a and b are positive integers (a, b ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9}).

Step 1: Divisibility by 5

⇒ For a number to be divisible by 5, its last digit must be 0 or 5.

⇒ Since b is a positive integer, b cannot be 0. Thus, b = 5.

Step 2: Divisibility by 9

⇒ Sum of digits = 7 + a + 3 + 5 + b

⇒ Sum = 15 + a + 5 = 20 + a

⇒ For 20 + a to be divisible by 9, the possible values for (20 + a) are 27, 36, ...

⇒ If 20 + a = 27, then a = 7.

⇒ If 20 + a = 36, then a = 16 (which is not a single-digit integer).

Step 3: Calculate (a + b)

⇒ a + b = 7 + 5 = 12.

∴ The correct answer is 12.

Additional Information

Divisibility Rule of 9

A number is divisible by 9 if the sum of its digits is divisible by 9.

Divisibility Rule of 5

A number is divisible by 5 if its last digit is either 0 or 5.

Composite Divisibility

To check divisibility by a composite number like 45, check divisibility by its coprime factors (5 and 9).

Shortcut Trick

Effective rate of Compound Interest (CI) for 2 years at 10% = 10 + 10 + (10 × 10 / 100) = 21%.

Effective rate of Simple Interest (SI) for 2 years at 10% = 10 × 2 = 20%.

According to the question: 21% = ₹147.

So, 1% = 147 ÷ 21 = ₹7.

Therefore, 20% (SI) = 20 × 7 = ₹140.

∴ The correct answer is ₹140.

Alternate Method

Given:

Compound Interest (CI) = ₹147

Rate (R) = 10% per annum

Time (T) = 2 years

Formula Used:

CI = P[(1 + R/100)^T − 1]

SI = (P × R × T) ÷ 100

Calculations:

⇒ 147 = P[(1 + 10/100)² − 1]

⇒ 147 = P[(11/10)² − 1]

⇒ 147 = P[121/100 − 1]

⇒ 147 = P[21/100]

⇒ P = (147 × 100) ÷ 21 = 7 × 100 = ₹700

⇒ Now, SI for 2 years = (700 × 10 × 2) ÷ 100

⇒ SI = 7 × 10 × 2 = ₹140

∴ The correct answer is ₹140.

Additional Information

Direct SI-CI Relation for 2 Years

For a period of 2 years, the relationship is given by: SI = CI × 200 ÷ (200 + R). In this case, 147 × 200 ÷ 210 = ₹140.

Difference between CI and SI

The difference (D) for 2 years is calculated as: D = CI − SI = P(R/100)². Here, D = 147 − 140 = ₹7.

Successive Rate Formula

Compound interest for 2 years can be treated as a successive percentage change: Effective Rate = x + y + (xy/100).

Shortcut Trick

Relative Speed = 180 / 12 = 15 m/s

Convert to km/hr: 15 × (18 / 5) = 54 km/hr

In opposite direction, Relative Speed = Speed_Train + Speed_Man

54 = Speed_Train + 4 ⇒ Speed_Train = 50 km/hr

∴ The correct answer is 50 km/hr.

Alternate Method

Given:

Speed of walking man = 4 km/hr

Length of the train (Distance) = 180 m

Crossing time = 12 seconds

Formula Used:

Relative Speed = Total Distance / Total Time

Speed (km/hr) = Speed (m/s) × (18/5)

Calculations:

⇒ Relative Speed = 180 / 12 = 15 m/s

⇒ Relative Speed in km/hr = 15 × (18 / 5)

⇒ Relative Speed = 3 × 18 = 54 km/hr

⇒ Since they are moving in the opposite direction:

⇒ Relative Speed = Speed of Train + Speed of Man

⇒ 54 = Speed of Train + 4

⇒ Speed of Train = 54 − 4 = 50 km/hr

∴ The correct answer is 50 km/hr.

Additional Information

Relative Speed Concept

When two objects move in opposite directions, their relative speed is the sum of their individual speeds (u + v). If they move in the same direction, it is the difference (u − v).

Speed Conversion

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

Crossing a Stationary Point/Man

When a train crosses a man, a pole, or a point object, the distance traveled is equal to the length of the train itself.

Shortcut Trick

Let the number of ₹20 notes be x. Then ₹10 notes = x + 16 and ₹5 notes = 2x.

Subtract the value of extra 16 notes of ₹10: 1000 − (16 × 10) = ₹840.

Effective value per unit: ₹10 (from x) + ₹20 (from x) + ₹10 (from 2x of ₹5) = ₹40.

Number of units (x) = 840 ÷ 40 = 21.

Total notes = 4x + 16 = 4(21) + 16 = 100.

∴ The correct answer is 100 notes.

Alternate Method

Given:

Total amount with Ram = ₹1000

Denominations are ₹10, ₹20, and ₹5.

Let the number of ₹20 notes be x.

Number of ₹10 notes = x + 16

Number of ₹5 notes = twice of ₹20 notes = 2x

Calculations:

⇒ Total Value = (Value of ₹10 notes) + (Value of ₹20 notes) + (Value of ₹5 notes)

⇒ 1000 = 10 × (x + 16) + 20 × (x) + 5 × (2x)

⇒ 1000 = 10x + 160 + 20x + 10x

⇒ 1000 − 160 = 40x

⇒ 840 = 40x

⇒ x = 840 ÷ 40 = 21

⇒ Number of ₹20 notes = 21

⇒ Number of ₹10 notes = 21 + 16 = 37

⇒ Number of ₹5 notes = 2 × 21 = 42

⇒ Total number of notes = 37 + 21 + 42 = 100

∴ The correct answer is 100 notes.

Additional Information

Currency Denomination Formula

The total value of money is calculated as: Total Value = ∑(Face Value of Note × Number of Notes).

Linear Equation Application

When multiple variables are related to a single base (like ₹20 notes here), express all terms in that base variable to solve the equation easily.

Verification Technique

Always verify the final count by multiplying back: (37 × 10) + (21 × 20) + (42 × 5) = 370 + 420 + 210 = 1000.

Shortcut Trick

Net filling rate per hour = (1/4 + 1/8 − 1/4) = 1/8 of the cistern.

Time to fill the full cistern = 1 ÷ (1/8) = 8 hours.

Time to fill one-fourth of the cistern = 8 × 1/4 = 2 hours.

∴ The correct answer is 2 hr.

Alternate Method

Given: Filling time of Tap A = 4 hr, Tap B = 8 hr. Emptying time of Tap C = 4 hr.

Formula Used: Efficiency = Total Work ÷ Time; Time = Work ÷ Total Efficiency.

⇒ Total Efficiency when all taps are open = 2 + 1 − 2 = 1 unit/hr.

⇒ One-fourth capacity of the cistern = 8 × 1/4 = 2 units.

⇒ Required Time = 2 units ÷ 1 unit/hr = 2 hours.

∴ The correct answer is 2 hr.

Additional Information

Inlet and Outlet Pipes

An Inlet pipe fills the cistern (positive efficiency) while an Outlet pipe empties it (negative efficiency).

Net Efficiency Rule

If multiple pipes operate, the net rate of filling is the algebraic sum of their individual efficiencies (Rates).

Work & Wages in Pipes

If workers or pipes are paid based on work, the wage ratio is equal to the ratio of their individual efficiencies.

Shortcut Trick

Ratio of speeds after meeting: S_A / S_B = √(Time_B / Time_A)

60 / S_B = √(9 / 16) = 3 / 4

S_B = (60 × 4) / 3 = 80 km/h

Speed of Train B in m/s = 80 × (5 / 18) = 400 / 18 m/s

Length of Train B = Speed × Crossing Time = (400 / 18) × 18 = 400 m

∴ The correct answer is 400 m.

Alternate Method

Given:

Speed of Train A (S_A) = 60 km/h

Time taken by Train A to reach Mumbai after meeting (t_A) = 16 hr

Time taken by Train B to reach Delhi after meeting (t_B) = 9 hr

Time taken by Train B to cross a pole = 18 sec

Formula Used:

1. S_A / S_B = √(t_B / t_A)
2. Distance = Speed × Time
3. km/h to m/s conversion: Speed × (5 / 18)

Calculations:

⇒ 60 / S_B = √(9 / 16)

⇒ 60 / S_B = 3 / 4

⇒ 3 × S_B = 60 × 4

⇒ S_B = 240 / 3 = 80 km/h

⇒ Speed of Train B in m/s = 80 × (5 / 18) = 400 / 18 m/s

⇒ Length of Train B = Speed × Time to cross pole

⇒ Length of Train B = (400 / 18) × 18

⇒ Length of Train B = 400 m

∴ The correct answer is 400 m.

Additional Information

Meeting Time Formula

If two objects start at the same time and meet at a point, the time they take to meet is T = √(t₁ × t₂), where t₁ and t₂ are times taken to reach destinations after meeting.

Train Crossing a Pole

When a train crosses a stationary object of negligible length (pole, man, tree), the distance traveled is equal to the length of the train itself.

Unit Conversion

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

Shortcut Trick

Total surface area (TSA) of one solid hemisphere = 3 units (πr² factor).

TSA of two separate hemispheres = 3 + 3 = 6 units.

Surface area of the resulting sphere = 4 units.

Reduction = 6 − 4 = 2 units.

Percentage Reduction = (2 ÷ 6) × 100 = 33.33%.

∴ The correct answer is 33.33%.

Alternate Method

Given: Two identical solid hemispheres are joined at their flat bases to form a sphere of the same radius r.

Formula Used:

⇒ Total Surface Area of a Solid Hemisphere = 3πr²

⇒ Surface Area of a Sphere = 4πr²

Calculations:

⇒ Combined TSA of two separate hemispheres = 2 × 3πr² = 6πr²

⇒ Surface Area of the single sphere formed = 4πr²

⇒ Reduction in Surface Area = 6πr² − 4πr² = 2πr²

⇒ Percentage Reduction = (Reduction ÷ Original Area) × 100

⇒ Percentage Reduction = (2πr² ÷ 6πr²) × 100

⇒ Percentage Reduction = (1 ÷ 3) × 100 ≈ 33.33%

∴ The correct answer is 33.33%.

Additional Information

Curved Surface Area (CSA) of Hemisphere

The area of only the curved part of a hemisphere is 2πr². The flat base has an area of πr².

Volume Conservation

When solids are joined or melted, the total volume remains constant. Volume of sphere = (4/3)πr³.

Total Surface Area of Solid Hemisphere

TSA = CSA + Base Area = 2πr² + πr² = 3πr².

Shortcut Trick

Let initial Cost Price (CP) = 100, then initial Selling Price (SP) = 125.

New CP = 100 × 1.20 = 120.

New SP = 125 × 0.90 = 112.5.

Net Loss = 120 − 112.5 = 7.5.

Loss Percentage = (7.5 ÷ 120) × 100 = 6.25%.

∴ The correct answer is 6.25%.

Alternate Method

Given: Initial profit = 25%, CP increase = 20%, SP reduction = 10%.

Formula: Profit/Loss % = (SP − CP) ÷ CP × 100.

⇒ Let the initial Cost Price be ₹100.

⇒ Initial Selling Price = 100 + 25% of 100 = ₹125.

⇒ New Cost Price = 100 + 20% of 100 = ₹120.

⇒ New Selling Price = 125 − 10% of 125 = 125 − 12.5 = ₹112.5.

⇒ Loss = New CP − New SP = 120 − 112.5 = ₹7.5.

⇒ Loss % = (7.5 ÷ 120) × 100 = (1 ÷ 16) × 100 = 6.25%.

∴ The correct answer is 6.25%.

Additional Information

Profit and Loss Percentage

Profit % = (Profit ÷ CP) × 100 and Loss % = (Loss ÷ CP) × 100. The denominator is always the Cost Price unless specified otherwise.

Successive Changes

When multiple changes occur, always update the base. For example, a 10% reduction on 125 is 125 × 0.90, not 125 − 10.

Base Change Rule

If CP changes, the new profit or loss must be calculated on the revised CP to find the effective percentage.

Shortcut Trick

Sum of ratio parts = 5 + 2 + 3 = 10 units.

10 units = 100 g (Total weight).

1 unit = 10 g.

Chemical Y share = 2 units = 2 × 10 = 20 g.

∴ The correct answer is 20 g.

Alternate Method

Given:

Ratio of Chemical X : Chemical Y : Chemical Z = 5 : 2 : 3

Total weight of the mixture = 100 g

Calculations:

Let the weight of Chemical X be 5k, Chemical Y be 2k, and Chemical Z be 3k.

⇒ Total weight = 5k + 2k + 3k = 10k

⇒ 10k = 100

⇒ k = 100 ÷ 10

⇒ k = 10 g

⇒ Weight of Chemical Y = 2k = 2 × 10

⇒ Weight of Chemical Y = 20 g

∴ The correct answer is 20 g.

Additional Information

Ratio and Proportions

A ratio indicates how many times one number contains another. To divide a quantity in a given ratio, the total quantity is divided by the sum of the ratio terms to find the value of one unit.

Percentage in Mixtures

The percentage of any component can be found by (Component Ratio / Total Ratio) × 100. In this case, Chemical Y is (2/10) × 100 = 20% of the total mixture.

Shortcut Trick

Total sum of interior angles of a hexagon (n = 6) = (6 − 2) × 180° = 720°.

Sum of 5 given angles = 140° + 80° + 145° + 155° + 100° = 620°.

Unknown angle = 720° − 620° = 100°.

∴ The correct answer is 100°.

Alternate Method

Given: Angles of an irregular hexagon are 140°, 80°, 145°, 155°, and 100°.

Formula Used: Sum of interior angles of a polygon = (n − 2) × 180°, where n is the number of sides.

For a hexagon, n = 6:

⇒ Sum = (6 − 2) × 180°

⇒ Sum = 4 × 180°

⇒ Sum = 720°

Let the unknown 6^th angle be x.

⇒ 140° + 80° + 145° + 155° + 100° + x = 720°

⇒ 620° + x = 720°

⇒ x = 720° − 620°

⇒ x = 100°

∴ The correct answer is 100°.

Additional Information

Polygon Interior Angles Theorem

The sum of the interior angles of any convex polygon is given by (n − 2) × 180°, where n represents the number of sides.

Exterior Angle Sum

For any convex polygon, the sum of the exterior angles (one at each vertex) is always equal to 360° regardless of the number of sides.

Regular Hexagon

In a regular hexagon, all interior angles are equal, and each angle measures 720° ÷ 6 = 120°.

Shortcut Trick

Let the total capacity of the cistern be the LCM of (10, 20, 12) = 60 units.

Combined Efficiency: (A+B) = 6, (B+C) = 3, (C+A) = 5.

Summing them: 2(A+B+C) = 6 + 3 + 5 = 14 ⇒ A+B+C = 7 units/hr.

Efficiency of Pipe B = (A+B+C) − (C+A) = 7 − 5 = 2 units/hr.

Time taken by B = 60 ÷ 2 = 30 hr.

∴ The correct answer is 30 hr.

Alternate Method

Given:

Time taken by (Pipe A + Pipe B) = 10 hr

Time taken by (Pipe B + Pipe C) = 20 hr

Time taken by (Pipe C + Pipe A) = 12 hr

Formula Used:

Efficiency = Total Work ÷ Time taken

Calculations:

⇒ Adding efficiencies: (A + B) + (B + C) + (C + A) = 6 + 3 + 5

⇒ 2(A + B + C) = 14

⇒ A + B + C = 14 ÷ 2 = 7 units/hr

⇒ We know efficiency of (C + A) = 5 units/hr

⇒ Efficiency of B = (A + B + C) − (C + A) = 7 − 5 = 2 units/hr

⇒ Time taken by Pipe B alone = Total Work ÷ Efficiency of B

⇒ Time = 60 ÷ 2 = 30 hr

∴ The correct answer is 30 hr.

Additional Information

Combined Efficiency Rule

When pipes work in pairs, the sum of their individual efficiencies is half the sum of their pairwise efficiencies: (A+B+C) = [(A+B) + (B+C) + (C+A)] ÷ 2.

Pipes and Cisterns Basics

Inlet pipes perform positive work (filling), while outlet pipes perform negative work (emptying). Total work is the volume of the tank/cistern.

Time and Work Relation

Time taken and efficiency are inversely proportional to each other. Time = 1 ÷ Efficiency (for constant work).

Shortcut Trick

Approximate Effective Discount for each Scheme:

(A) 21 + 8 − (21 × 8 / 100) = 27.32%

(B) Successive 38%, 42%, 30% → clearly much higher than 50% (≈ 74.8%)

(C) Weighted Average = [(8 × 17) + (2 × 20)] ÷ 10 = 17.6%

(D) Buy 10 Get 7 Free = [7 ÷ (10 + 7)] × 100 ≈ 41.18%

Comparing the values, 17.6% is the lowest discount, making it the least beneficial.

∴ The correct answer is C.

Alternate Method

Given:

⇒ Total Toys to buy = 10

⇒ Let Marked Price (MP) of 1 toy = ₹100

⇒ Total Marked Price for 10 toys = ₹1000

Formula Used:

⇒ Successive Discount = a + b − (ab/100)

⇒ Buy X Get Y Free Discount % = [Y ÷ (X + Y)] × 100

Calculations:

Scheme (A): Successive 21% and 8%

⇒ Effective Discount = 21 + 8 − (21 × 8 / 100) = 29 − 1.68 = 27.32%

Scheme (B): Successive 38%, 42%, 30%

⇒ First two: 38 + 42 − (38 × 42 / 100) = 80 − 15.96 = 64.04%

⇒ Final: 64.04 + 30 − (64.04 × 30 / 100) = 94.04 − 19.21 = 74.83%

Scheme (C): 17% on first 8, 20% on next 2

⇒ Discount on 8 toys = 8 × 17 = ₹136

⇒ Discount on 2 toys = 2 × 20 = ₹40

⇒ Total Discount = 136 + 40 = ₹176

⇒ Effective % = (176 ÷ 1000) × 100 = 17.6%

Scheme (D): Buy 10 Get 7 Free

⇒ Effective Discount % = [7 ÷ (10 + 7)] × 100 = (7 ÷ 17) × 100 ≈ 41.18%

⇒ Comparing all schemes, 17.6% is the minimum discount.

∴ The correct answer is C.

Additional Information

Successive Discounts

When multiple discounts are applied one after another, the total discount is always less than the sum of individual percentages. Formula: a + b − (ab/100).

Buy X Get Y Free

The effective discount percentage in such offers is calculated as: [Free Items ÷ Total Items (Bought + Free)] × 100.

Least Beneficial vs Most Beneficial

For a customer, the least beneficial scheme is the one offering the lowest effective discount percentage, while the most beneficial offers the highest.

Shortcut Trick

Convert decimals to fractions for faster calculation: 0.25 = 1/4 and 0.5 = 1/2.

Numerator: (1/4) × 1.6 = 0.4

Denominator: 0.04 × (1/2) = 0.02

Result = 0.4 ÷ 0.02 = 40 ÷ 2 = 20

∴ The correct answer is 20.

Alternate Method

Given: (0.25 × 1.6) ÷ (0.04 × 0.5)

Method Used: Apply the BODMAS rule (perform operations inside Brackets first).

⇒ Step 1 (First bracket): 0.25 × 1.6 = 0.400 = 0.4

⇒ Step 2 (Second bracket): 0.04 × 0.5 = 0.020 = 0.02

⇒ Step 3 (Division): 0.4 ÷ 0.02

⇒ To remove the decimal from the divisor, multiply both numerator and denominator by 100.

⇒ (0.4 × 100) ÷ (0.02 × 100) = 40 ÷ 2

⇒ Result = 20

∴ The correct answer is 20.

Additional Information

BODMAS Rule

It stands for Brackets, Of, Division, Multiplication, Addition, and Subtraction. Operations must be performed in this specific order to get the correct result.

Multiplying Decimals

Multiply the numbers as if they were whole numbers, then place the decimal point in the product so that it has the same number of decimal places as the sum of the decimal places in the factors.

Dividing by Decimals

Move the decimal point in the divisor to the right until it is a whole number, and move the decimal point in the dividend the same number of places to the right before dividing.

Shortcut Trick

Average Speed = Total Distance ÷ Total Time

Total Distance = 100 + 150 + 50 = 300 km

Total Time = (100/50) + (150/60) + (50/40) = 2 + 2.5 + 1.25 = 5.75 hours

Average Speed = 300 ÷ 5.75 ≈ 52.17 km/h

∴ The correct answer is 52.17 km/h.

Alternate Method

Given:

Distance₁ = 100 km, Speed₁ = 50 km/h

Distance₂ = 150 km, Speed₂ = 60 km/h

Distance₃ = 50 km, Speed₃ = 40 km/h

Formula Used:

Time = Distance ÷ Speed

Average Speed = (Total Distance) ÷ (Total Time)

Calculations:

⇒ Time for first part (t₁) = 100 ÷ 50 = 2 hours

⇒ Time for second part (t₂) = 150 ÷ 60 = 2.5 hours

⇒ Time for third part (t₃) = 50 ÷ 40 = 1.25 hours

⇒ Total Time (T) = 2 + 2.5 + 1.25 = 5.75 hours

⇒ Average Speed = 300 ÷ 5.75

⇒ Average Speed = 30000 ÷ 575 = 1200 ÷ 23

⇒ Average Speed ≈ 52.1739... km/h

∴ The correct answer is 52.17 km/h.

Additional Information

Average Speed for Equal Distances

When a journey is divided into two equal distances with speeds x and y, the Average Speed is the harmonic mean: (2 × x × y) ÷ (x + y).

Average Speed for Equal Time Intervals

If an object travels at speed x for time T and then at speed y for the same time T, the Average Speed is the arithmetic mean: (x + y) ÷ 2.

Unit Conversion

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

Shortcut Trick

Assume the total quantity is 12 units (LCM of 3 and 4).

Net profit/loss units = (1/3 × 12 × 20) + (1/4 × 12 × (−10)) + (5/12 × 12 × 0)

Net effect = 80 − 30 + 0 = 50 units.

Overall percentage = 50 ÷ 12 = 25/6 = 4 ¹/₆%.

∴ The correct answer is 4 ¹/₆% profit.

Alternate Method

Given: Total Cost Price (CP) = ₹2400

Formula Used: Profit % = (Total Profit ÷ Total CP) × 100

⇒ Profit on first part = (1/3) × 2400 × 20% = 800 × 0.20 = ₹160

⇒ Loss on second part = (1/4) × 2400 × 10% = 600 × 0.10 = ₹60

⇒ Remaining goods = 1 − (1/3 + 1/4) = 1 − 7/12 = 5/12

⇒ Profit/Loss on remaining part = (5/12) × 2400 × 0 = ₹0

⇒ Total Net Profit = 160 − 60 + 0 = ₹100

⇒ Overall Profit Percentage = (100 ÷ 2400) × 100 = 100/24 = 25/6 = 4 ¹/₆%

∴ The correct answer is 4 ¹/₆% profit.

Additional Information

Weighted Average Profit/Loss

When goods are sold in parts, the net profit/loss percentage is the weighted average: % = (w₁p₁ + w₂p₂ + ...) ÷ (w₁ + w₂ + ...).

Cost Price Selling

Selling an item at its Cost Price (CP) implies there is no profit and no loss (0% change).

Fractional Parts Calculation

Always ensure the sum of all fractional parts equals 1 (e.g., 1/3 + 1/4 + 5/12 = 12/12 = 1).

Shortcut Trick

Let the HCF of the three numbers be x.

The numbers are then 3x, 7x, and 2x.

LCM of (3x, 7x, 2x) = x × LCM(3, 7, 2) = x × 42.

Given, 42x = 8820.

x = 8820 ÷ 42 = 210.

∴ The correct answer is 210.

Alternate Method

Given:

Ratio of three numbers = 3 : 7 : 2

LCM of the numbers = 8820

Formula Used:

If numbers are in the ratio a : b : c, then the numbers can be represented as ax, bx, and cx, where x is their HCF.

The LCM of these numbers = x × LCM(a, b, c).

Calculations:

⇒ Let the numbers be 3x, 7x, and 2x.

⇒ LCM(3, 7, 2) = 3 × 7 × 2 = 42 (as they are co-prime).

⇒ LCM of (3x, 7x, 2x) = 42x.

⇒ 42x = 8820.

⇒ x = 8820 ÷ 42.

⇒ x = 210.

∴ The correct answer is 210.

Additional Information

HCF (Highest Common Factor)

The greatest number that divides two or more numbers without leaving a remainder.

LCM (Least Common Multiple)

The smallest non-zero number that is a multiple of two or more numbers.

Product Rule for Two Numbers

For any two numbers A and B: A × B = HCF(A, B) × LCM(A, B).

Co-prime Numbers

Two or more numbers are co-prime if their HCF is 1. Their LCM is simply the product of the numbers.

Shortcut Trick

Effective Interest Rate for 2 years at 4% = 4 + 4 + (4 × 4)/100 = 8.16%.

Given that 8.16% of the Principal (y) = ₹765.

The Total Amount (A) percentage = 100% + 8.16% = 108.16%.

⇒ Amount = (765 ÷ 8.16) × 108.16

⇒ Amount = 93.75 × 108.16 = ₹10,140.

∴ The correct answer is ₹10,140.

Alternate Method

Given: Principal (P) = ₹y, Rate (R) = 4% p.a., Time (n) = 2 years, Compound Interest (CI) = ₹765.

Formula Used: CI = P[(1 + R/100)ⁿ − 1] and Amount (A) = P + CI

⇒ 765 = y[(1 + 4/100)² − 1]

⇒ 765 = y[(1 + 1/25)² − 1]

⇒ 765 = y[(26/25)² − 1]

⇒ 765 = y[676/625 − 1]

⇒ 765 = y[(676 − 625)/625]

⇒ 765 = y[51/625]

⇒ y = (765 × 625) ÷ 51

⇒ y = 15 × 625 = ₹9,375

⇒ Amount (A) = Principal (y) + CI

⇒ A = 9,375 + 765 = ₹10,140

∴ The correct answer is ₹10,140.

Additional Information

Compound Interest Formula

The amount A is calculated as A = P(1 + r/100)ⁿ, where P is principal, r is rate, and n is time.

Effective Rate for 2 Years

The net percentage interest for two years is given by x + y + (xy/100), where x and y are annual rates.

Difference between CI and SI

For 2 years, the difference between Compound Interest and Simple Interest is P(R/100)².

Shortcut Trick

Net percentage change = a + b + (ab ÷ 100)

Here, a = +12% and b = −10%

Net change = 12 − 10 + (12 × (−10) ÷ 100) = 2 − 1.2 = 0.8%

Given, 0.8% increase = ₹1,584

Original Salary (100%) = (1,584 ÷ 0.8) × 100 = 1,980 × 100 = ₹1,98,000

∴ The correct answer is ₹1,98,000.

Alternate Method

Given:

First year revision = increased by 12%

Second year revision = decreased by 10%

Increase in original monthly salary = ₹1,584

Formula Used:

Final Value = Initial Value × (1 + r₁/100) × (1 + r₂/100)

Calculations:

Let Sangeeta's salary be ₹x.

⇒ Salary after 1^st year = x × (112 ÷ 100) = 1.12x

⇒ Salary after 2^nd year = 1.12x × (90 ÷ 100) = 1.008x

⇒ Total increase in salary = 1.008x − x = 0.008x

⇒ 0.008x = 1,584

⇒ x = 1,584 ÷ 0.008

⇒ x = 1,584,000 ÷ 8

⇒ x = 1,98,000

∴ The correct answer is ₹1,98,000.

Additional Information

Successive Percentage Change

When a value is changed by a% and then by b%, the net effective change is (a + b + ab/100)%.

Multiplying Factor Method

For an increase of r%, the factor is (1 + r/100). For a decrease, the factor is (1 − r/100). Final value = Initial value × Factor₁ × Factor₂.

Base Value Property

Percentage changes are always calculated on the preceding value in a series of successive changes, not on the original base value each time.

Shortcut Trick

Convert science-to-total ratios into decimals to compare proportions:

Batch A: 5 ÷ (5 + 8) = 5/13 ≈ 0.3846

Batch B: 7 ÷ (7 + 11) = 7/18 ≈ 0.3888

Batch C: 3 ÷ (3 + 5) = 3/8 = 0.3750

Since 0.3888 > 0.3846 > 0.3750, Batch B has the highest proportion.

∴ The correct answer is Batch B.

Alternate Method

Given:

Ratio of Science to Arts students:

Batch A = 5 : 8 | Batch B = 7 : 11 | Batch C = 3 : 5

Formula Used:

Proportion of Science students = Science ÷ (Science + Arts)

Calculations:

⇒ Batch A proportion = 5 / (5 + 8) = 5/13

⇒ Batch B proportion = 7 / (7 + 11) = 7/18

⇒ Batch C proportion = 3 / (3 + 5) = 3/8

⇒ Comparing 5/13 and 7/18 by cross-multiplication: 5 × 18 = 90 and 7 × 13 = 91. Since 91 > 90, Batch B > Batch A.

⇒ Comparing 7/18 and 3/8 by cross-multiplication: 7 × 8 = 56 and 3 × 18 = 54. Since 56 > 54, Batch B > Batch C.

⇒ Batch B has the highest proportion of science students.

∴ The correct answer is Batch B.

Additional Information

Ratio to Proportion Conversion

If the ratio of two components is a:b, the proportion of the first component in the total is given by a/(a+b).

Comparing Fractions Shortcut

To compare a/b and c/d without division, compare the products (a × d) and (b × c). If ad > bc, then a c/d.

Percentage Share

The percentage share of a component is (Value / Total) × 100, which is useful for comparing proportions across different group sizes.

Shortcut Trick

Let Marked Price (MP) = 100 units.

Cost Price (CP) = 100 − 20% = 80 units.

Selling Price (SP) = 80 + 40% of 80 = 112 units.

Difference (SP − MP) = 112 − 100 = 12 units → ₹10.

Marked Price = (10 ÷ 12) × 100 = 1000 ÷ 12 = 83.33.

∴ The correct answer is ₹83.33.

Alternate Method

Given:

Discount on Marked Price = 20%

Selling Price (SP) = Marked Price (MP) + ₹10

Profit Percentage = 40%

⇒ Let the Marked Price be x.

⇒ Cost Price (CP) = x − 20% of x = 0.8x.

⇒ Selling Price (SP) = x + 10.

⇒ Profit = SP − CP = (x + 10) − 0.8x = 0.2x + 10.

⇒ Profit Percentage = (Profit ÷ CP) × 100

⇒ 40 = [(0.2x + 10) ÷ 0.8x] × 100

⇒ 0.40 = (0.2x + 10) ÷ 0.8x

⇒ 0.32x = 0.2x + 10

⇒ 0.12x = 10

⇒ x = 10 ÷ 0.12 = 83.333...

∴ The correct answer is ₹83.33.

Additional Information

Relationship between CP and MP

The direct ratio between Cost Price and Marked Price is given by: CP/MP = (100 − Discount%) / (100 + Profit%).

Successive Discounts

Two successive discounts of a% and b% are equivalent to a single discount of [a + b − (a × b)/100]%.

Mark-up Percentage

Mark-up % = [(MP − CP) ÷ CP] × 100. It represents how much more the Marked Price is compared to the Cost Price.