Quantitative Aptitude PYQs

Shortcut Trick

Total work = LCM of days (24, 20, 8) = 120 units.

Efficiency of A = 120 ÷ 24 = 5 units/day.

Efficiency of (A + K) = 120 ÷ 20 = 6 units/day, and (A + K + S) = 120 ÷ 8 = 15 units/day.

Efficiency of S = 15 − 6 = 9 units/day.

Time taken by A and S together = Total work ÷ Combined Efficiency = 120 ÷ (5 + 9) = 120 ÷ 14 = 60/7 days.

∴ The correct answer is 60/7 days.

Alternate Method

Given:

Time taken by Anjani (A) = 24 days

Time taken by Anjani and Khushbu (A + K) = 20 days

Time taken by Anjani, Khushbu, and Sushmita (A + K + S) = 8 days

Calculations:

⇒ Efficiency of Sushmita (S) = Combined Efficiency of (A + K + S) − Combined Efficiency of (A + K)

⇒ Efficiency of S = 15 − 6 = 9 units/day

⇒ Combined Efficiency of Anjani and Sushmita (A + S) = Efficiency of A + Efficiency of S

⇒ Combined Efficiency of (A + S) = 5 + 9 = 14 units/day

⇒ Time taken by A and S together = Total Work ÷ Combined Efficiency of (A + S)

⇒ Time taken = 120 ÷ 14 = 60/7 days

∴ The correct answer is 60/7 days.

Additional Information

Work and Wages

Wages are distributed in proportion to the total work done by each individual. If they work for the same number of days, wages are distributed exactly in the ratio of their daily efficiencies.

Pipes and Cisterns

Follows the same logic as time and work concepts. A pipe filling a tank has a positive efficiency, while a leak or an emptying pipe has a negative efficiency. Net work equals the sum of these respective efficiencies.

Alternate Day Work

When individuals work on alternate days, calculate the total work completed in one full cycle (e.g., 2 days for 2 people). Then find the total number of full cycles needed to reach closest to the total work.

Shortcut Trick

Identify the base square roots first: √196 = 14, √324 = 18, and √676 = 26.

Adjust for decimals by halving the decimal places in the square root.

√0.0324 (4 decimal places) becomes 0.18 (2 decimal places).

√6.76 (2 decimal places) becomes 2.6 (1 decimal place).

Solve: 14 + 0.18 − 2.6 = 14.18 − 2.6 = 11.58.

∴ The correct answer is 11.58.

Alternate Method

Given:

The expression is: √196 + √0.0324 − √6.76

Method Used:

Convert decimal values into fractions to easily calculate their exact square roots.

Calculations:

⇒ √196 + √(324 ÷ 10000) − √(676 ÷ 100)

⇒ 14 + (18 ÷ 100) − (26 ÷ 10)

⇒ 14 + 0.18 − 2.6

⇒ 14.18 − 2.6

⇒ 11.58

∴ The correct answer is 11.58.

Additional Information

Decimal Square Roots

If a perfect square has 2n decimal places, its exact square root will have exactly n decimal places.

BODMAS Rule

Always follow the strict order of operations for simplifications: Bracket, Order (powers and roots), Division, Multiplication, Addition, and Subtraction.

Common Squares

Memorizing perfect squares up to 30 accelerates calculation speed. For example: 24² = 576, 25² = 625, and 26² = 676.

Shortcut Trick

If A + B = 90°, then sin²A + sin²B = 1, cos²A + cos²B = 1, and sec²A − cot²B = 1.

Applying these complementary identities directly to the given expression:

First term = (sin²40° + sin²50°) ÷ (cos²40° + cos²50°) = 1 ÷ 1 = 1

Second term = (sec²50° − cot²40°) ÷ (cos²20° + cos²70°) = 1 ÷ 1 = 1

Expression value = 1 − 1 = 0

∴ The correct answer is 0.

Alternate Method

Given:

Expression = (sin²40° + sin²50°) ÷ (cos²40° + cos²50°) − (sec²50° − cot²40°) ÷ (cos²20° + cos²70°)

Formulas Used:

⇒ sin(90° − θ) = cosθ

⇒ cos(90° − θ) = sinθ

⇒ cot(90° − θ) = tanθ

⇒ sin²θ + cos²θ = 1

⇒ sec²θ − tan²θ = 1

Calculations:

⇒ First term numerator = sin²40° + sin²(90° − 40°) = sin²40° + cos²40° = 1

⇒ First term denominator = cos²40° + cos²(90° − 40°) = cos²40° + sin²40° = 1

⇒ Second term numerator = sec²50° − cot²(90° − 50°) = sec²50° − tan²50° = 1

⇒ Second term denominator = cos²20° + cos²(90° − 20°) = cos²20° + sin²20° = 1

⇒ Substituting these values back into the expression:

⇒ Expression = (1 ÷ 1) − (1 ÷ 1)

⇒ Expression = 1 − 1 = 0

∴ The correct answer is 0.

Additional Information

Pythagorean Trigonometric Identities

The three core identities are: sin²θ + cos²θ = 1, sec²θ − tan²θ = 1, and cosec²θ − cot²θ = 1.

Complementary Angle Relations

For any acute angle θ: sin(90° − θ) = cosθ, tan(90° − θ) = cotθ, and sec(90° − θ) = cosecθ.

Shortcut Trick

Expand the brackets by applying the distributive property directly to save time.

⇒ 3x² − 50x + 28 − 3x² − 12x + 39

⇒ (3 − 3)x² − (50 + 12)x + (28 + 39)

⇒ −62x + 67

∴ The correct answer is −62x + 67.

Alternate Method

Given:

Expression: 2((3/2)x² − 25x + 14) − 3(x² + 4x − 13)

Method Used:

Simplify the algebraic expression by opening the brackets and combining like terms.

Calculations:

⇒ [2 × (3/2)x²] − [2 × 25x] + [2 × 14] − [3 × x²] − [3 × 4x] − [3 × (−13)]

⇒ 3x² − 50x + 28 − 3x² − 12x + 39

Group the terms with similar variables together:

⇒ (3x² − 3x²) + (−50x − 12x) + (28 + 39)

⇒ 0x² − 62x + 67

⇒ −62x + 67

∴ The correct answer is −62x + 67.

Additional Information

Distributive Law of Multiplication

It states that for any terms a, b, and c, multiplying the sum is equal to the sum of the multiplied terms: a(b ± c) = ab ± ac.

Like Terms

In algebra, like terms have exactly the same variables with the exact same exponents. They can be simplified into a single term through addition or subtraction.

BODMAS Rule

Calculations must always prioritize Brackets, followed by Order (powers/roots), Division, Multiplication, Addition, and Subtraction in sequence.

Shortcut Trick

Let the assumed overall average be 41 runs.

Deviation of the next 10 matches from assumed average = 10 × (15 − 41) = −260

Net effect on the overall average = −260 ÷ (23 + 10) = −260 ÷ 33 ≈ −7.88

Overall average = 41 − 7.88 = 33.12

∴ The correct answer is 33.12 runs.

Alternate Method

Given:

Average runs in the first 23 matches = 41

Average runs in the next 10 matches = 15

Formula Used:

Overall Average = Total Runs ÷ Total Matches

23 Matches
Avg: 41

10 Matches
Avg: 15

Overall Avg ≈ 33.12
Total Runs: 943
Total Runs: 150
Total Runs = 1093 | Total Matches = 33

Calculations:

⇒ Total matches = 23 + 10 = 33

⇒ Total runs in the first 23 matches = 23 × 41 = 943

⇒ Total runs in the next 10 matches = 10 × 15 = 150

⇒ Total runs in all 33 matches = 943 + 150 = 1093

⇒ Overall Average = 1093 ÷ 33

⇒ Overall Average = 33.1212... ≈ 33.12

∴ The correct answer is 33.12 runs.

Additional Information

Weighted Average

It is used to calculate the combined average of multiple groups with different sizes. Formula: A_w = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Deviation Method for Averages

Select an assumed average (A), find the total deviation of all elements from A, and distribute the net deviation equally among all items. Actual Average = Assumed Average + (Total Deviation ÷ Total Number of Items).

Combined Ratio Approach

When calculating weighted averages, only the ratio of the number of items (n₁ : n₂) matters, not their exact values. Here, the ratio of matches is directly taken as 23 : 10.

Shortcut Trick

Discount = Marked Price − Selling Price.

Discount = 500 − 425 = ₹75.

Discount % = (Discount ÷ Marked Price) × 100.

Discount % = (75 ÷ 500) × 100 = 15%.

∴ The correct answer is 15%.

Alternate Method

Given:

Marked Price (MP) = ₹500

Selling Price (SP) = ₹425

Formula Used:

Discount = MP − SP

Discount % = (Discount ÷ MP) × 100

Marked Price = ₹500

Selling Price = ₹425

₹75
Off

Discount % = (75 ÷ 500) × 100 = 15%

Calculations:

⇒ Discount = 500 − 425 = ₹75

⇒ Discount % = (75 ÷ 500) × 100

⇒ Discount % = 15%

∴ The correct answer is 15%.

Additional Information

Successive Discounts

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

Relation between CP and MP

When Profit % (P) and Discount % (D) are both given, the ratio of Marked Price to Cost Price is MP ÷ CP = (100 + P) ÷ (100 − D).

Mark Up Percentage

Mark up percentage is calculated over the Cost Price. Mark Up % = [(MP − CP) ÷ CP] × 100.

Shortcut Trick

Let the total capacity of the tank be the LCM of given hours: LCM(15, 27, 10) = 270 units.

Combined efficiency of A, B, and C = (270 ÷ 15) + (270 ÷ 27) + (270 ÷ 10) = 18 + 10 + 27 = 55 units/hour.

Time taken to fill the tank together = Total Capacity ÷ Combined Efficiency = 270 ÷ 55 hours.

On simplifying: 270 ÷ 55 = 54 ÷ 11 = 4 ¹⁰⁄₁₁ hours.

∴ The correct answer is 4 ¹⁰⁄₁₁ hours.

Alternate Method

Given:

Time taken by pipe A = 15 hours

Time taken by pipe B = 27 hours

Time taken by pipe C = 10 hours

Formula Used:

Total Work = LCM of time taken

Efficiency = Total Work ÷ Time Taken

Time Taken Together = Total Work ÷ Total Efficiency

Calculations:

⇒ Total efficiency of pipes (A + B + C) = 18 + 10 + 27 = 55 units/hour

⇒ Time taken when operating together = Total Work ÷ Total efficiency

⇒ Time = 270 ÷ 55

⇒ Time = 54 ÷ 11 = 4 ¹⁰⁄₁₁ hours

∴ The correct answer is 4 ¹⁰⁄₁₁ hours.

Additional Information

Emptying Pipes (Negative Efficiency)

If a pipe empties a full tank in 'y' hours, its efficiency is taken as negative. When combining with a filling pipe, the net efficiency = (1 ÷ x) − (1 ÷ y).

Alternate Working of Pipes

When pipes are opened in alternate hours, calculate the combined work done in 1 cycle (e.g., 2 hours for 2 pipes). Multiply by the closest whole number factor before dealing with the remainder.

Actual Capacity of Tank

If the filling or emptying rate is given (e.g., liters per minute), the exact volume can be found by: Capacity = Time required to fill the tank × Rate of flow.

Shortcut Trick

Let the cost price of 100g of wheat be ₹100.

Claimed selling price at 20% loss = ₹100 − 20% = ₹80.

Actual weight given by vendor = 100g − 35% = 65g.

Actual cost price for 65g = ₹65.

Profit % = [(SP − Actual CP) ÷ Actual CP] × 100 = [(80 − 65) ÷ 65] × 100 = 23.08%

∴ The correct answer is 23.08%.

Alternate Method

Given:

⇒ Claimed loss percentage = 20%

⇒ Weight reduction by cheating = 35%

Formula Used:

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

⇒ Profit = Selling Price − Cost Price

Actual CP = ₹65
SP = ₹80
Stated CP = ₹100

(65g weight)
(100g weight)

Actual Profit = ₹15

20% Loss Claimed

35% Less Weight Used (100g → 65g)

Calculations:

⇒ Let the true weight be 100g and Cost Price (CP) of 1g be ₹1.

⇒ Therefore, Cost Price of 100g = ₹100.

⇒ Since the vendor claims to sell at a 20% loss, Selling Price (SP) = ₹100 − 20% of ₹100 = ₹80.

⇒ The vendor uses weights that are 35% less than the stated weight.

⇒ Actual weight given = 100g − 35g = 65g.

⇒ Actual Cost Price (CP) incurred by the vendor = CP of 65g = ₹65.

⇒ Actual Profit = SP − Actual CP = 80 − 65 = ₹15.

⇒ Profit Percentage = (15 ÷ 65) × 100

⇒ Profit Percentage = (3 ÷ 13) × 100 ≈ 23.0769...

⇒ Rounding off to two decimal places, Profit = 23.08%

∴ The correct answer is 23.08%.

Additional Information

Dishonest Dealer (Selling at CP)

If a trader professes to sell his goods at cost price, but uses false weights, then Profit % = [Error ÷ (True Value − Error)] × 100.

Dishonest Dealer with Markup or Loss

When a dealer claims to sell at x% profit/loss but also uses false weights, compute the final SP based on the claim and final CP based on the actual weight delivered.

Shortcut Trick

Check if the given sides form a Pythagorean triplet:

9² + 12² = 81 + 144 = 225, and 15² = 225.

Since 9² + 12² = 15², the given triangle is a right-angled triangle.

Area of a right-angled triangle = (1 ÷ 2) × Base × Height

Area = (1 ÷ 2) × 12 × 9 = 54 cm².

∴ The correct answer is 54 cm².

Alternate Method

12 cm
9 cm
15 cm
Area = ?

Given:

Sides of the triangle: a = 9 cm, b = 12 cm, c = 15 cm.

Formula Used:

Heron's Formula: Area = √(s × (s − a) × (s − b) × (s − c))

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

Calculations:

⇒ s = (9 + 12 + 15) ÷ 2

⇒ s = 36 ÷ 2 = 18 cm

⇒ Area = √(18 × (18 − 9) × (18 − 12) × (18 − 15))

⇒ Area = √(18 × 9 × 6 × 3)

⇒ Area = √(18 × 9 × 18)

⇒ Area = √(18² × 3²)

⇒ Area = 18 × 3 = 54 cm²

∴ The correct answer is 54 cm².

Additional Information

Pythagorean Triplets

Sets of three integers (a, b, c) that satisfy a² + b² = c² always form a right-angled triangle. Common base triplets include (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Area of an Equilateral Triangle

If all three sides of a triangle are equal (side = a), the area is given by (√3 ÷ 4) × a².

Area of an Isosceles Triangle

For a triangle with two equal sides 'a' and a base 'b', the Area = (b ÷ 4) × √(4a² − b²).

Shortcut Trick

Evaluate coefficients of x², x, and constant term independently.

Coefficient of x² : 2 × (6 ÷ 2) − 6 × 1 = 6 − 6 = 0

Coefficient of x : 2 × (−22) − 6 × 5 = −44 − 30 = −74

Constant term : 2 × 15 − 6 × (−18) = 30 + 108 = 138

Combining the terms, we directly get −74x + 138.

∴ The correct answer is −74x + 138.

Alternate Method

Given:

2((6 ÷ 2)x² − 22x + 15) − 6(x² + 5x − 18)

Calculations:

⇒ Simplify the fraction inside the first bracket: 6 ÷ 2 = 3

⇒ The expression becomes: 2(3x² − 22x + 15) − 6(x² + 5x − 18)

⇒ Expand the first bracket by multiplying each term by 2:

⇒ (2 × 3x²) − (2 × 22x) + (2 × 15) = 6x² − 44x + 30

⇒ Expand the second bracket by multiplying each term by −6:

⇒ (−6 × x²) − (6 × 5x) − (−6 × −18) = −6x² − 30x + 108

⇒ Combine both expanded parts together:

⇒ 6x² − 44x + 30 − 6x² − 30x + 108

⇒ Group the like terms:

⇒ (6x² − 6x²) + (−44x − 30x) + (30 + 108)

⇒ 0x² − 74x + 138

⇒ −74x + 138

∴ The correct answer is −74x + 138.

Additional Information

BODMAS Rule

It dictates the correct order of mathematical operations: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction to ensure calculation consistency.

Distributive Property

Multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products together: a(b + c) = ab + ac.

Like Terms

Terms in an algebraic expression that have the exact same variable raised to the same power.

Only like terms can be simplified by addition or subtraction.

Shortcut Trick

Total remaining bulbs = 37 − 19 = 18.

Total Selling Price (SP) = 18 × ₹3.90 = ₹70.20.

Total Cost Price (CP) = ₹37.

Profit = SP − CP = 70.20 − 37 = 33.20.

∴ The correct answer is ₹33.20.

Alternate Method

Given:

Total light bulbs purchased = 37

Total Cost Price (CP) = ₹37

Number of broken bulbs = 19

Selling Price (SP) per remaining bulb = ₹3.90

Formula Used:

Profit = Total Selling Price − Total Cost Price

CP = ₹37.00
Profit = ₹33.20

Total Selling Price (SP) = 18 × ₹3.90 = ₹70.20

Cost of all 37 bulbs
Extra amount earned

Calculations:

⇒ Remaining light bulbs = 37 − 19 = 18

⇒ Total Selling Price (SP) = 18 × 3.90

⇒ Total SP = ₹70.20

⇒ Profit = Total SP − Total CP

⇒ Profit = 70.20 − 37

⇒ Profit = ₹33.20

∴ The correct answer is ₹33.20.

Additional Information

Profit and Loss Basic Formula

Profit occurs when Selling Price (SP) > Cost Price (CP). The fundamental formula is: Profit = SP − CP.

Profit Percentage

Profit percentage is always calculated on the base Cost Price (CP) unless specifically mentioned otherwise. Formula: Profit % = (Profit ÷ CP) × 100.

Effective Price of Remaining Items

When certain items are damaged or lost in transit, the overall Cost Price remains unchanged but it must be absorbed entirely by the remaining good items, effectively increasing the Cost Price per unit for the rest.

Shortcut Trick

The difference between the ages of two people always remains constant over time.

After 7 years: 6 × (P − Q) = 30 ⇒ P − Q = 5.

Given present condition: 3P − Q = 23.

Rewrite as: 2P + (P − Q) = 23 ⇒ 2P + 5 = 23 ⇒ 2P = 18 ⇒ P = 9.

Thus, Q's present age = P − 5 = 9 − 5 = 4.

∴ The correct answer is 4 years.

Alternate Method

Given:

Three times P's present age exceeds Q's present age by 23 years.

After 7 years, 6 times Q's age is 30 less than 6 times P's age.

Formula Used:

Let the present age of P be P and the present age of Q be Q.

Present Ages
P & Q

• 7 Years

After 7 Years
P + 7 & Q + 7

Age difference remains constant over time

Calculations:

⇒ From the first condition: 3P − Q = 23 . . . . (Equation 1)

⇒ From the second condition (after 7 years):

⇒ 6 × (Q + 7) = 6 × (P + 7) − 30

⇒ 6Q + 42 = 6P + 42 − 30

⇒ 6Q = 6P − 30

⇒ 6P − 6Q = 30

⇒ P − Q = 5 . . . . (Equation 2)

⇒ From Equation 2, we get: Q = P − 5

⇒ Substitute the value of Q in Equation 1:

⇒ 3P − (P − 5) = 23

⇒ 3P − P + 5 = 23

⇒ 2P = 18

⇒ P = 9 years

⇒ Q = P − 5 = 9 − 5 = 4 years.

∴ The correct answer is 4 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals remains perfectly constant at any point in time (past, present, or future).

Present, Past and Future Ages

If the present age is assumed as x, then the age 'n' years ago is calculated as (x − n) and the age after 'n' years is given by (x + n).

Ratio of Ages Trick

When the ratio of ages at two different timelines is provided, equate the unit difference of the ratio parts to the chronological time gap to determine the exact age multipliers.

Shortcut Trick

In right-angled ▵BDC, the remaining acute angle is ∠C = 90° − 44° = 46°.

By applying the exterior angle theorem on ▵AEC, the exterior angle ∠AEB equals the sum of the two opposite interior angles.

∠AEB = ∠CAE + ∠C

∠AEB = 23° + 46° = 69°.

∴ The correct answer is 69°.

Alternate Method

A
B
C
D
E

23°

44°

?

Given:

⇒ In ▵ABC, BD ⊥ AC at D.

⇒ ∠DBC = 44°

⇒ ∠CAE = 23°

⇒ Point E is on the line segment BC.

Calculations:

⇒ In the right-angled ▵BDC, the sum of internal angles is 180°.

⇒ ∠BDC + ∠DBC + ∠C = 180°

⇒ 90° + 44° + ∠C = 180°

⇒ 134° + ∠C = 180°

⇒ ∠C = 180° − 134° = 46°

⇒ Now, consider ▵AEC. Since point E lies on the line segment BC, extending side CE to point B forms the exterior angle ∠AEB at vertex E.

⇒ According to the exterior angle theorem, the measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior opposite angles.

⇒ ∠AEB = ∠CAE + ∠C

⇒ ∠AEB = 23° + 46°

⇒ ∠AEB = 69°

∴ The correct answer is 69°.

Additional Information

Exterior Angle Theorem

The exterior angle of a triangle is exactly equal to the sum of the two opposite interior angles. Formula: ∠Exterior = ∠Int₁ + ∠Int₂.

Angle Sum Property

The sum of all interior angles of a triangle is always 180°. This fundamental property applies uniformly across all types of triangles in planar geometry.

Right-Angled Triangle

In any right-angled triangle, the sum of the two acute angles is strictly 90°, as the right angle itself accounts for 90° out of the total 180° limit.

Shortcut Trick

Average of the ratio parts = (3 + 5 + 7) ÷ 3 = 15 ÷ 3 = 5 parts.

Given the average price is ₹5,300, which corresponds to the average of the parts.

⇒ 5 parts = ₹5,300

⇒ 1 part = 5300 ÷ 5 = ₹1,060

Price of the cheapest item (3 parts) = 3 × 1060 = ₹3,180.

∴ The correct answer is ₹3,180.

Alternate Method

Given:

Average price of 3 items = ₹5,300

Ratio of their prices = 3 : 5 : 7

Formula Used:

Total Sum = Average × Number of items

Total Sum = ₹5300 × 3 = ₹15,900
Ratio = 3 : 5 : 7 (Total 15 parts)

3x

5x

7x

Cheapest
3 × 1060
₹3180

15x = ₹15,900
⇒ x = ₹1060

Calculations:

⇒ Total price of 3 items = 5300 × 3 = ₹15,900

Let the individual prices of the three items be 3x, 5x, and 7x respectively.

⇒ Total price = 3x + 5x + 7x = 15x

⇒ 15x = 15900

⇒ x = 15900 ÷ 15 = 1060

The cheapest item corresponds to the smallest ratio part, which is 3x.

⇒ Price of the cheapest item = 3 × 1060 = ₹3,180

∴ The correct answer is ₹3,180.

Additional Information

Average Concept

The average of a set of numbers is the total sum divided by the number of observations: Average = Sum of Observations ÷ Total Number of Observations.

Dividing a Quantity in a Ratio

When a total quantity T is divided in the ratio a : b : c, the individual share of the first part is calculated directly as [a ÷ (a + b + c)] × T.

Weighted Average

If Group 1 has n₁ items with an average of A₁, and Group 2 has n₂ items with an average of A₂, then Combined Average = (n₁×A₁ + n₂×A₂) ÷ (n₁ + n₂).

Shortcut Trick

Eliminate options using basic divisibility tests before proceeding to long division.

Option 87 is a multiple of 3 (87 = 29 × 3).

Sum of digits of 541096943 = 5 + 4 + 1 + 0 + 9 + 6 + 9 + 4 + 3 = 41.

Since 41 is not divisible by 3, 541096943 is not divisible by 87.

Testing the remaining primes by direct division: 541096943 ÷ 97 = 5578319 (Remainder 0).

∴ The correct answer is 97.

Alternate Method

Given:

Number = 541096943

Method Used:

Long division and divisibility rules.

Calculations:

⇒ First, check for composite numbers in the options to quickly eliminate them: 87 = 29 × 3.

⇒ A number is exactly divisible by 3 if the sum of its digits is divisible by 3.

⇒ Sum of digits = 5 + 4 + 1 + 0 + 9 + 6 + 9 + 4 + 3 = 41.

⇒ Since 41 is not a multiple of 3, the number cannot be divided by 87.

⇒ Next, perform direct division for the first prime option, which is 97.

⇒ 541096943 ÷ 97 = 5578319

⇒ Since the division leaves a remainder of 0, the number is perfectly divisible by 97.

∴ The correct answer is 97.

Additional Information

Divisibility Rule of 3

A number is divisible by 3 if the sum of all its individual digits is a multiple of 3.

Checking Divisibility for Large Primes

For large prime divisors like 97, 103, or 107, standard direct division is typically the fastest and most reliable method during an exam, unless specific osculator rules are heavily practiced.

Osculator Rule for 97

The negative osculator for 97 is 29. Multiply the unit digit of the number by 29 and subtract it from the remaining left-over number. Repeat this process until a manageable two-digit or three-digit number is reached.

Shortcut Trick

Since the rate is given per month and time is given in months, we can directly apply the simple interest formula without converting units.

SI = (Principal × Rate × Time) ÷ 100

⇒ SI = (1200 × 4 × 12) ÷ 100

⇒ SI = 12 × 48 = 576

∴ The correct answer is ₹576.

Alternate Method

Given:

Principal (P) = ₹1200

Rate of Interest (R) = 4% per month

Time (T) = 12 months

P = ₹1200

R = 4% per month
T = 12 months

SI = ₹576

Formula Used:

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

Calculation:

⇒ SI = (1200 × 4 × 12) ÷ 100

⇒ SI = 12 × 4 × 12

⇒ SI = 12 × 48

⇒ SI = 576

∴ The correct answer is ₹576.

Additional Information

Total Amount (A)

The total amount at the end of the specified term is the sum of the principal and the accrued interest. The formula is A = P + SI = P(1 + RT/100).

Time & Rate Consistency

Always ensure that the unit of Time (T) matches the frequency of the Interest Rate (R). If the rate is per month, the time must be in months. If the rate is per annum, the time must be in years.

Compound Interest (CI)

Unlike Simple Interest, Compound Interest is calculated on the principal plus any accumulated interest from previous periods. The formula used is A = P(1 + R/100)^T.

Shortcut Trick

List the odd numbers from 393 to 401: 393, 395, 397, 399, 401.

Eliminate 395 (ends in 5) and multiples of 3 (393 and 399, as their digit sum is divisible by 3).

The only remaining possible primes are 397 and 401.

Check their divisibility by primes up to 19 (since √401 ≈ 20).

Neither 397 nor 401 is divisible by 7, 11, 13, 17, or 19, meaning both are prime numbers.

∴ The correct answer is 2.

Alternate Method

Given:

Numbers lying between 393 and 401 (inclusive).

Method Used:

Filter out the composite numbers using standard divisibility rules (by 2, 3, and 5) and verify the remaining numbers using the prime divisibility test up to √N.

Calculations:

⇒ The list of numbers to evaluate: 393, 394, 395, 396, 397, 398, 399, 400, 401

⇒ Eliminate the even numbers (divisible by 2): 394, 396, 398, 400

⇒ Eliminate the numbers ending in 5 (divisible by 5): 395

⇒ Eliminate the numbers whose digit sum is a multiple of 3: 393 (sum = 15) and 399 (sum = 21)

⇒ The only numbers left to verify are 397 and 401.

⇒ Check 397: √397 ≈ 19.9. Prime numbers ≤ 19 are 2, 3, 5, 7, 11, 13, 17, 19.

⇒ 397 is not exactly divisible by any of these primes, making 397 a prime number.

⇒ Check 401: √401 ≈ 20.0. Prime numbers ≤ 19 are 2, 3, 5, 7, 11, 13, 17, 19.

⇒ 401 is not exactly divisible by any of these primes, making 401 a prime number.

⇒ The prime numbers in the given range are exactly two (397 and 401).

∴ The correct answer is 2.

Additional Information

Prime Number Test

To check if a number N is prime, find the approximate square root of N. Test its divisibility by all prime numbers less than or equal to √N. If none divide it completely, then N is a prime number.

Twin Primes

Twin primes are a pair of prime numbers that differ by exactly 2. Common examples include (3, 5), (11, 13), and (41, 43).

Co-Prime Numbers

Two numbers are called co-prime (or mutually prime) if they do not share any common factor other than 1. Their Highest Common Factor (HCF) is always exactly 1.

Shortcut Trick
The product of two consecutive natural numbers lies exactly between their squares.
We know that 212 = 441 and 222 = 484.
The given product 462 lies exactly between 441 and 484.
⇒ The two consecutive numbers are 21 and 22.
⇒ 21 × 22 = 462, which satisfies the given condition.
⇒ Therefore, the greater number of the two is 22.
∴ The correct answer is 22.

Alternate Method
Given:
The product of two consecutive natural numbers is 462.
Method Used:
Formulate a quadratic equation and solve for its roots using factorisation.
Calculations:
⇒ Let the smaller natural number be x.
⇒ The consecutive greater natural number will be (x + 1).
⇒ According to the question: x × (x + 1) = 462
⇒ x2 + x = 462
⇒ x2 + x − 462 = 0
⇒ x2 + 22x − 21x − 462 = 0
⇒ x(x + 22) − 21(x + 22) = 0
⇒ (x − 21)(x + 22) = 0
⇒ The roots of the equation are x = 21 and x = −22.
⇒ Since x must be a natural number, we reject −22 and consider x = 21.
⇒ The greater number = x + 1 = 21 + 1 = 22.
∴ The correct answer is 22.

Additional Information
Consecutive Numbers
Numbers that follow each other continuously in order from smallest to largest, with a constant common difference of 1 (e.g., n, n+1, n+2).
Product of Consecutive Numbers
The product of two consecutive integers n and (n+1) is always an even number, because one of the numbers in the pair is strictly even.
Perfect Square Check
The product of two consecutive positive integers always lies strictly between the squares of the two integers. That is, n2 < n(n+1) < (n+1)2.

Shortcut Trick

Discount = MP − SP = 1500 − 1200 = ₹300

Discount % = (Discount ÷ MP) × 100

Discount % = (300 ÷ 1500) × 100

Discount % = (1 ÷ 5) × 100 = 20%

∴ The correct answer is 20%.

Alternate Method

Marked Price (MP) = ₹1,500

Selling Price (SP) = ₹1,200

Discount
₹300

Given:

Marked Price (MP) = ₹1,500

Selling Price (SP) = ₹1,200

Formula Used:

Discount = MP − SP

Discount % = (Discount ÷ MP) × 100

Calculations:

⇒ Discount = 1500 − 1200

⇒ Discount = ₹300

⇒ Discount % = (300 ÷ 1500) × 100

⇒ Discount % = (1 ÷ 5) × 100

⇒ Discount % = 20%

∴ The correct answer is 20%.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP), while Profit or Loss is calculated on the Cost Price (CP).

Successive Discounts

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

Markup Price

It is the price written on the article to show a higher value before applying a discount. MP = CP + Markup Amount.

Shortcut Trick

Total days from 8 Feb to 9 Apr 2024 = 21 (Feb) + 31 (Mar) + 9 (Apr) = 61 days.

Since 2024 is a leap year, time T = 61 ÷ 366 = 1/6 years.

SI = (P × R × T) ÷ 100

SI = 4000 × 7.5 × (1/6) ÷ 100 = 300 ÷ 6 = 50

∴ The correct answer is ₹50.

Alternate Method

Given:

Principal (P) = ₹4,000

Rate of interest (R) = 7.5% per annum

Time period = 8 February 2024 to 9 April 2024

₹4,000
Principal (P)

R = 7.5%
T = 61 days

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

Formula Used:

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

Calculations:

⇒ The year 2024 is divisible by 4, hence it is a leap year with 366 days and February has 29 days.

⇒ Number of days in February (excluding the deposit day) = 29 − 8 = 21 days

⇒ Number of days in March = 31 days

⇒ Number of days in April (including the withdrawal day) = 9 days

⇒ Total time in days (T) = 21 + 31 + 9 = 61 days

⇒ Time in years (T) = 61 ÷ 366 = 1/6 years

⇒ SI = (4000 × 7.5 × 1/6) ÷ 100

⇒ SI = (40 × 7.5) ÷ 6

⇒ SI = 300 ÷ 6 = 50

∴ The correct answer is ₹50.

Additional Information

Simple Interest Rule

Simple Interest is always computed on the initial principal amount throughout the entire period. Formula: SI = (P × R × T) ÷ 100.

Leap Year Calculation

A year is a leap year if it is evenly divisible by 4 (and by 400 for century years). A leap year contains 366 days instead of 365 days.

Day Counting Convention

When computing interest between two specific dates, banks generally include either the date of deposit or the date of withdrawal, but not both.