Quantitative Aptitude PYQs

Shortcut Trick

Let the total salary be S.

Using the multiplying factor for successive deductions: S × 0.9 × 0.8 × 0.8 × 0.9 = 23328

S × (9/10) × (4/5) × (4/5) × (9/10) = 23328

S × (1296 / 2500) = 23328

S = 18 × 2500 = 45000

∴ The correct answer is ₹45,000.

Alternate Method

Given: Remaining amount = ₹23,328

⇒ Let Total Salary = 100x

⇒ After House Rent (10%): 100x − 10x = 90x

⇒ After Conveyance (20% of 90x): 90x − 18x = 72x

⇒ After Income Tax (20% of 72x): 72x − 14.4x = 57.6x

⇒ After Clothes (10% of 57.6x): 57.6x − 5.76x = 51.84x

⇒ 51.84x = 23328

⇒ x = 23328 ÷ 51.84 = 450

⇒ Total Salary = 100 × 450 = 45000

∴ The correct answer is ₹45,000.

Additional Information

Successive Percentage Change

When multiple deductions are made on the remaining balance, use the formula: Final Value = Initial Value × (1 − r₁/100) × (1 − r₂/100)...

Percentage to Fraction Conversion

10% = 1/10 (Remaining = 9/10) and 20% = 1/5 (Remaining = 4/5). These fractions simplify complex calculations significantly.

Shortcut Trick

Number of diagonals = n(n − 3) ÷ 2 = 9 ⇒ n(n − 3) = 18 ⇒ n = 6 (regular hexagon)

One exterior angle of a regular hexagon = 360° ÷ 6 = 60°

One interior angle of a regular hexagon = 180° − 60° = 120°

Difference between interior and exterior angle = 120° − 60° = 60°

∴ The correct answer is 60°.

Alternate Method

Given:

The number of diagonals of a regular polygon = 9

Formula Used:

1. Number of diagonals of an n-sided polygon = n(n − 3) ÷ 2
2. Each exterior angle of a regular polygon = 360° ÷ n
3. Each interior angle of a regular polygon = 180° − (Each exterior angle)

Calculations:

⇒ n(n − 3) ÷ 2 = 9

⇒ n(n − 3) = 18

⇒ n² − 3n − 18 = 0

⇒ n² − 6n + 3n − 18 = 0

⇒ n(n − 6) + 3(n − 6) = 0

⇒ (n − 6)(n + 3) = 0

⇒ n = 6 (since the number of sides n must be positive, n ≠ −3)

Therefore, the polygon is a regular hexagon with 6 sides.

⇒ One exterior angle = 360° ÷ 6 = 60°

⇒ One interior angle = 180° − 60° = 120°

⇒ Difference = 120° − 60° = 60°

Why other options are incorrect:

Option (1) 55° is incorrect because if the difference is 55°, the interior angle would be 117.5° and the exterior angle would be 62.5°, which does not correspond to a regular polygon with integer sides.

Option (2) 70° is incorrect because if the difference is 70°, the interior angle would be 125° and the exterior angle would be 55°, which is incorrect.

Option (4) 48° is incorrect because if the difference is 48°, the interior angle would be 114° and the exterior angle would be 66°, which is incorrect.

∴ The correct answer is 60°.

Additional Information

Sum of Interior Angles

The sum of all interior angles of an n-sided polygon is given by (n − 2) × 180°.

Sum of Exterior Angles

The sum of all exterior angles of any polygon is always 360°.

Relation between Interior and Exterior Angle

At any vertex of a polygon, the interior angle and the exterior angle are supplementary, meaning their sum is 180°.

Shortcut Trick

Assume all 269 children are girls, then the total amount = 269 × 7 = ₹1883.

The difference between the actual amount and this assumed amount = 2239 − 1883 = ₹356.

This difference arises because boys get ₹2 more than girls (9 − 7 = ₹2).

Number of boys = Total Difference ÷ Extra share per boy = 356 ÷ 2 = 178.

∴ The correct answer is 178.

Alternate Method

Given:

Total children = 269

Total amount = ₹2,239

Amount per boy = ₹9 | Amount per girl = ₹7

Formula Used:

Total Amount = (Number of boys × Share per boy) + (Number of girls × Share per girl)

Let the number of boys be x.

Then, the number of girls = 269 − x.

⇒ 9 × x + 7 × (269 − x) = 2239

⇒ 9x + 1883 − 7x = 2239

⇒ 2x + 1883 = 2239

⇒ 2x = 2239 − 1883

⇒ 2x = 356

⇒ x = 356 ÷ 2

⇒ x = 178

∴ The correct answer is 178.

Additional Information

Alligation Rule

This problem can be solved using the alligation method where the mean value is 2239 ÷ 269. The ratio of boys to girls is (Mean − Girl's share) : (Boy's share − Mean).

Weighted Average

The total sum is the weighted average of individual shares: Total = (n₁w₁ + n₂w₂), where n is the count and w is the weight/share.

Linear Equations

For any two-variable problem (x + y = A and ax + by = B), substituting one variable in terms of the other is the most direct algebraic approach.

Shortcut Trick

Net Deviation Method: Assume the Average is 15.

Deviations: (6.5 − 15) + (10.7 − 15) + (20.2 − 15) + (a − 15) = 0

−8.5 − 4.3 + 5.2 + (a − 15) = 0

−7.6 + (a − 15) = 0 ⇒ a − 15 = 7.6

a = 15 + 7.6 = 22.6

∴ The correct answer is 22.6.

Alternate Method

Given:

Numbers = 6.5, 10.7, 20.2 and a

Total count of numbers (n) = 4

Average = 15

Formula Used:

Average = (Sum of all observations) ÷ (Total number of observations)

Calculations:

⇒ 15 = (6.5 + 10.7 + 20.2 + a) ÷ 4

⇒ 15 × 4 = 37.4 + a

⇒ 60 = 37.4 + a

⇒ a = 60 − 37.4

⇒ a = 22.6

∴ The correct answer is 22.6.

Additional Information

Sum of Observations

The total sum can always be calculated as Total Sum = Average × Number of terms.

Net Deviation Rule

The algebraic sum of deviations of all observations from their arithmetic mean is always zero.

Change in Average

If a constant k is added to or subtracted from every term, the new average also increases or decreases by k.

Shortcut Trick

Total amount to be deducted = 16 + 15 + 75 = ₹106.

Remaining amount = 5,398 − 106 = ₹5,292.

Ratio of remaining shares = 2 : 21 : 4 (Total units = 27).

Difference between Y and Z units = 21 − 4 = 17 units.

Difference between original shares = (Remaining difference) + (Difference in deductions) = (17/27 × 5,292) + (15 − 75) = 3,332 − 60 = ₹3,272.

∴ The correct answer is ₹3,272.

Alternate Method

Given: Total sum = ₹5,398. Deductions: X = ₹16, Y = ₹15, Z = ₹75. Ratio = 2 : 21 : 4.

⇒ Total deduction = 16 + 15 + 75 = ₹106.

⇒ Remaining sum = 5,398 − 106 = ₹5,292.

⇒ Total ratio parts = 2 + 21 + 4 = 27.

⇒ Value of 1 unit = 5,292 ÷ 27 = ₹196.

⇒ Y’s remaining share = 21 × 196 = ₹4,116.

⇒ Y’s original share = 4,116 + 15 = ₹4,131.

⇒ Z’s remaining share = 4 × 196 = ₹784.

⇒ Z’s original share = 784 + 75 = ₹859.

⇒ Required difference = 4,131 − 859 = ₹3,272.

∴ The correct answer is ₹3,272.

Additional Information

Ratio and Proportion

Ratio is a comparison of two or more quantities of the same kind. If a sum S is divided in ratio a:b:c, the shares are (a/Σratio)×S, (b/Σratio)×S, etc.

Handling Deductions

When specific amounts are deducted from shares before a ratio is applied, always subtract the total of these deductions from the total sum first.

Unitary Method

Find the value of a single 'ratio unit' by dividing the modified total amount by the sum of all ratio parts.

Shortcut Trick

Let the length, breadth, and height be l, b, and h.

Adjacent face areas: lb = x, bh = y, hl = z.

Multiplying these: (lb) × (bh) × (hl) = x × y × z

l²b²h² = xyz ⇒ (lbh)² = xyz

Since Volume (V) = lbh, then V² = xyz ⇒ V = √xyz.

∴ The correct answer is √xyz cubic units.

Alternate Method

Given:

Area of first face (x) = length (l) × breadth (b)

Area of second face (y) = breadth (b) × height (h)

Area of third face (z) = height (h) × length (l)

Formula Used:

Volume of Cuboid (V) = l × b × h

Calculations:

⇒ x × y × z = (l × b) × (b × h) × (h × l)

⇒ xyz = l² × b² × h²

⇒ xyz = (lbh)²

⇒ lbh = √xyz

⇒ V = √xyz

∴ The correct answer is √xyz cubic units.

Additional Information

Total Surface Area of Cuboid

The total surface area is given by 2(lb + bh + hl) or 2(x + y + z) square units.

Diagonal of a Cuboid

The length of the longest rod that can be placed inside is √(l² + b² + h²) units.

Lateral Surface Area

The area of the four walls is 2h(l + b) square units.

Identify twin primes (a, b) between 2 and 50: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43).

Total count n = 6.

Least prime α = 2 and greatest prime β = 47 (both < 50).

Value = 6 × (2 + 47) = 6 × 49 = 294.

∴ The correct answer is 294.

Alternate Method

Given: 2 < a < b < 50, b − a = 2. α and β are least and greatest primes < 50.

Step 1: Find twin primes (a, b) where both are prime and differ by 2.

⇒ Pairs are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43).

⇒ Total count n = 6.

Step 2: Find α (least prime number < 50) and β (greatest prime number < 50).

⇒ α = 2

⇒ β = 47

Step 3: Calculate the expression n(α + β).

⇒ Value = 6 × (2 + 47)

⇒ Value = 6 × 49 = 294.

∴ The correct answer is 294.

Additional Information

Twin Prime Numbers

Pairs of prime numbers that have a difference of 2 are called twin primes (e.g., 5 and 7).

Prime Numbers < 50

The primes are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. Total 15 primes.

Co-prime Numbers

Two numbers are co-prime if their HCF is 1. They do not necessarily have to be prime themselves.

Shortcut Trick

The total sum of 4 numbers with an average of 18 is 18 × 4 = 72.

The sum of the three known numbers is 6.5 + 10.7 + 20.2 = 37.4.

Value of a = 72 − 37.4 = 34.6.

∴ The correct answer is 34.6.

Alternate Method

Given:

Observations: 6.5, 10.7, 20.2, and a

Number of observations (n) = 4

Average (Arithmetic Mean) = 18

Formula Used:

Average = (Sum of all observations) ÷ (Total number of observations)

Calculations:

⇒ 18 = (6.5 + 10.7 + 20.2 + a) ÷ 4

⇒ 18 × 4 = 37.4 + a

⇒ 72 = 37.4 + a

⇒ a = 72 − 37.4

⇒ a = 34.6

∴ The correct answer is 34.6.

Additional Information

Arithmetic Mean (Average)

It represents the central value of a set of data and is calculated as the sum of all values divided by the total number of values.

Property of Deviations

The sum of the deviations of all observations from their arithmetic mean is always zero. This can be used to quickly find missing values.

Weighted Average

When different groups have different averages, the combined average is (n₁x₁ + n₂x₂) ÷ (n₁ + n₂).

Shortcut Trick

Net rate of filling = 1/2 − 1/4 = 1/4 of the tank per minute.

Time required to fill the full tank = 4 minutes.

Time required to fill one-half of the tank = 4 ÷ 2 = 2 minutes.

∴ The correct answer is 2 minutes.

Alternate Method

Given:

Time taken by filling pipe = 2 minutes

Time taken by emptying pipe = 4 minutes

Formula Used:

Total Capacity = LCM of individual time durations

Efficiency = Total Capacity ÷ Time Taken

Calculations:

⇒ Net efficiency when both pipes are opened together = +2 − 1 = +1 unit/minute

⇒ Capacity of one-half of the tank = 4 ÷ 2 = 2 units

⇒ Time taken to fill one-half of the tank = 2 units ÷ 1 unit/minute = 2 minutes

∴ The correct answer is 2 minutes.

Additional Information

Inlet and Outlet Pipes

An inlet pipe performs positive work by filling a tank, whereas an outlet pipe performs negative work by emptying the tank.

Net Work Formula

If an inlet pipe fills a tank in x hours and an outlet pipe empties it in y hours (where y > x), then the net part filled in 1 hour when both are open is (1/x − 1/y).

Shortcut Trick

Rate of interest = 5% = 1/20 in fraction form.

For 2 years, the ratio of Principal to Amount is (20)² : (21)² = 400 : 441.

Given 400 units = ₹2,800 ⇒ 1 unit = ₹7.

Total Amount = 441 × 7 = ₹3,087.

∴ The correct answer is ₹3087.

Alternate Method

Given:

Principal (P) = ₹2,800

Rate (R) = 5% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

⇒ A = 2800 × (1 + 5/100)²

⇒ A = 2800 × (1 + 1/20)²

⇒ A = 2800 × (21/20)²

⇒ A = 2800 × (441 / 400)

⇒ A = 7 × 441

⇒ A = ₹3,087

∴ The correct answer is ₹3087.

Additional Information

Compound Interest Formula

The interest is calculated on the principal and also on the accumulated interest of previous periods: CI = P[(1 + R/100)ⁿ − 1].

Effective Rate of Interest

For two years at rate R%, the effective compound interest rate is (x + y + xy/100)%. For 5%, it is 5 + 5 + (5 × 5)/100 = 10.25%.

Difference between SI and CI

The difference for 2 years is given by the formula D = P(R/100)².

Shortcut Trick
Scheme A: Net discount = 28 + 28 − (28 × 28) ÷ 100 = 56 − 7.84 = 48.16%.
Scheme B: Buy 1 get 4 means 5 items for the price of 1. Discount = 4 ÷ 5 × 100 = 80%.
Scheme C: Single flat discount = 35%.
Scheme D: Net discount = 27 + 26 − (27 × 26) ÷ 100 = 53 − 7.02 = 45.98%.
Comparing all: 80% > 48.16% > 45.98% > 35%.
∴ The correct answer is B.

Alternate Method
Given → Scheme A: 28% & 28% | Scheme B: Buy 1 Get 4 | Scheme C: 35% | Scheme D: 27% & 26%.
Formulas: Successive = a + b − ab ÷ 100 | Buy X Get Y Discount% = [Y ÷ (X + Y)] × 100.

⇒ For Scheme B: Buy 1, Get 4 implies the customer gets 1 (paid) + 4 (free) = 5 total items.
⇒ Discount = (Free Items ÷ Total Items) × 100 = (4 ÷ 5) × 100 = 80%.
⇒ The best scheme for a customer is the one offering the highest discount percentage.
⇒ Since 80% is significantly higher than all other options, Scheme B is the best.
∴ The correct answer is B.

Additional Information
Successive Discounts
When two discounts x% and y% are given, the total effective discount is always less than the sum (x+y) due to the reduction applied to the already discounted price.
Buy X Get Y Free
In 'Buy X Get Y' schemes, the discount percentage is calculated as [Y ÷ (X+Y)] × 100.

Note:

'Buy 1 Get 4' often implies 4 are free, whereas 'Buy 1 Get 4 total' means 3 are free.

Shortcut Trick

For the first n natural numbers, the formula for the average of squares is (n + 1)(2n + 1) ÷ 6.

Given n = 5.

Average = (5 + 1)(2 × 5 + 1) ÷ 6

Average = (6 × 11) ÷ 6 = 11

∴ The correct answer is 11.

Alternate Method

Given: First 5 natural numbers are 1, 2, 3, 4, 5.

Formula Used: Average = (Sum of all observations) ÷ (Total number of observations)

⇒ First 5 natural numbers: 1, 2, 3, 4, 5

⇒ Squares of these numbers: 1², 2², 3², 4², 5²

⇒ Values of squares: 1, 4, 9, 16, 25

⇒ Sum of squares: 1 + 4 + 9 + 16 + 25 = 55

⇒ Total count: 5

⇒ Average: 55 ÷ 5 = 11

∴ The correct answer is 11.

Additional Information

Sum of First n Natural Numbers

The sum is given by the formula S_n = [n(n + 1)] ÷ 2.

Sum of Squares of First n Natural Numbers

The sum is given by the formula S_n² = [n(n + 1)(2n + 1)] ÷ 6.

Average of First n Natural Numbers

The average is simply the middle term for consecutive numbers: (n + 1) ÷ 2.

Sum of Cubes of First n Natural Numbers

The sum of cubes is the square of the sum of the first n natural numbers: [n(n + 1) ÷ 2]².

Shortcut Trick

Total Work = 10 Skilled × 8 = 80 units = 10 Semi-skilled × 12 = 120 units.

Efficiency: 1 Skilled (S) = 1/80; 1 Semi-skilled (SS) = 1/120 per person-day.

Required Efficiency = 4 × (1/80) + 6 × (1/120) = 1/20 + 1/20 = 2/20 = 1/10 units/day.

Time Required = 1 ÷ (1/10) = 10 days.

∴ The correct answer is 10 days.

Alternate Method

Given:

10 Skilled persons (S) = 8 days

10 Semi-skilled persons (SS) = 12 days

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ 10S × 8 = 10SS × 12

⇒ 80S = 120SS

⇒ 2S = 3SS ⇒ S/SS = 3/2 (Efficiency ratio)

⇒ Total Work = 10 × 3 × 8 = 240 units

⇒ Combined Efficiency of 4S and 6SS = 4(3) + 6(2)

⇒ Efficiency = 12 + 12 = 24 units/day

⇒ Time = 240 ÷ 24 = 10 days

∴ The correct answer is 10 days.

Additional Information

Chain Rule Formula

Used for work comparison: M₁D₁H₁/W₁ = M₂D₂H₂/W₂, where M is persons, D is days, H is hours, and W is work.

Work and Wages

Wages are always distributed in the ratio of the actual work done or the efficiency of the individuals if they work for the same time.

Combined Efficiency

When different groups work together, the total efficiency is the sum of their individual efficiencies (Total E = E₁ + E₂ + ... + E_n).

Shortcut Trick

Volume relationship: π × R₁² × H₁ × 0.95 = π × R₂² × H₂

7² × 10 × 0.95 = 5² × H₂

49 × 9.5 = 25 × H₂

H₂ = 465.5 ÷ 25 = 18.62 m

∴ The correct answer is 18.62 m.

Alternate Method

Given: Radius of first tank (r₁) = 7 m, Height (h₁) = 10 m, Leakage = 5%, Radius of second tank (r₂) = 5 m.

Formula Used: Volume of Cylinder = πr²h

⇒ Total Volume of first tank = π × 7 × 7 × 10 = 490π m³

⇒ Water remaining after 5% leakage = 95% of 490π

⇒ Remaining Volume = 0.95 × 490π = 465.5π m³

⇒ Volume in second tank = π × 5 × 5 × H

⇒ 465.5π = 25π × H

⇒ H = 465.5 ÷ 25

⇒ H = 18.62 m

∴ The correct answer is 18.62 m.

Additional Information

Volume of Cylinder

The space occupied by a cylindrical object is calculated as Area of base × Height = πr²h.

Total Surface Area

The total surface area of a closed cylinder includes the curved surface and two circular bases: 2πrh + 2πr².

Conversion of Volume

When a liquid is transferred from one container to another, its volume remains constant unless there is a loss (leakage).

Shortcut Trick

Convert percentages and fractions into a single multiplier: 60% = 3/5 and 80% = 4/5.

Multiplier = (3/5) × (4/5) × (3/5) = 36/125.

Given: (36/125) × Number = 216.

Number = (216 × 125) ÷ 36 = 6 × 125 = 750.

∴ The correct answer is 750.

Alternate Method

Let the unknown number be x.

Given: 60% of 80% of (3/5) of x = 216.

⇒ (60/100) × (80/100) × (3/5) × x = 216

⇒ (3/5) × (4/5) × (3/5) × x = 216

⇒ (36/125) × x = 216

⇒ x = (216 × 125) ÷ 36

⇒ x = 6 × 125

⇒ x = 750

∴ The correct answer is 750.

Additional Information

Percentage to Fraction Conversion

To simplify calculations, remember common fractions: 20% = 1/5, 40% = 2/5, 60% = 3/5, and 80% = 4/5.

Successive Percentages

A% of B% of C is equivalent to (A × B ÷ 100)% of C. Multiplying the individual fractions directly is the fastest method.

Base Change Rule

If a value is reduced by x%, to restore it to the original value, it must be increased by [x/(100 − x)] × 100%.

Shortcut Trick

The relationship between Selling Price (SP) and Marked Price (MP) is given by: SP = MP × (100 − Discount%) ÷ 100.

Ratio method: 20% discount ⇒ 1/5, where 5 is MP and 1 is discount, so SP = 4 units.

Given 4 units = 660, then 1 unit = 165.

Marked Price (5 units) = 165 × 5 = 825.

∴ The correct answer is ₹825.

Alternate Method

Given: Selling Price (SP) = ₹660, Discount% = 20%.
Formula: Marked Price (MP) = (SP × 100) ÷ (100 − Discount%)

⇒ MP = (660 × 100) ÷ (100 − 20)
⇒ MP = (660 × 100) ÷ 80
⇒ MP = 66000 ÷ 80
⇒ MP = 825

∴ The correct answer is ₹825.

Additional Information

Marked Price & Discount

Marked Price is the price printed on the label. Discount is always calculated on the Marked Price. SP = MP − Discount.

Discount Percentage

Discount% = (Discount ÷ Marked Price) × 100. It represents the reduction offered on the tagged price.

Shortcut Trick

Speed and Time are inversely proportional when Distance is constant.

Speed ratio (S₁ : S₂) = 42 : 63 = 2 : 3

Time ratio (T₁ : T₂) = 3 : 2

Given T₁ = 12 hours (3 units → 12), then 1 unit = 4.

Required time T₂ = 2 units = 2 × 4 = 8 hours.

∴ The correct answer is 8 hr.

Alternate Method

Given: Initial Speed (S₁) = 42 km/hour, Initial Time (T₁) = 12 hours, New Speed (S₂) = 63 km/hour.

Formula Used: Distance = Speed × Time

⇒ Total Distance = S₁ × T₁

⇒ Total Distance = 42 × 12 = 504 km

⇒ New Time (T₂) = Total Distance ÷ S₂

⇒ T₂ = 504 ÷ 63

⇒ T₂ = 8 hours

∴ The correct answer is 8 hr.

Additional Information

Inverse Proportion in Speed

When the distance is fixed, speed is inversely proportional to time: S₁T₁ = S₂T₂.

Average Speed for Same Distance

If a journey involves two equal distances covered at different speeds 'x' and 'y', the average speed is (2xy) ÷ (x + y).

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

Total amount = ₹12,405. Deductions = 53 + 88 + 76 = ₹217.

Remaining amount = 12,405 − 217 = ₹12,188.

Remaining ratio P : Q : R = 15 : 20 : 9. Total ratio units = 15 + 20 + 9 = 44 units.

1 unit = 12,188 ÷ 44 = ₹277.

Difference between Q and R (shares) = (20 − 9) units + (88 − 76) = 11 × 277 + 12 = 3047 + 12 = ₹3,059.

∴ The correct answer is ₹3,059.

Alternate Method

Given: Total = ₹12,405; Deductions = ₹53, ₹88, ₹76; Ratio = 15 : 20 : 9.

Formula: Value = (Ratio Share × Adjusted Total) ÷ Sum of Ratios.

⇒ Sum of remaining shares = 12,405 − (53 + 88 + 76) = 12,188

⇒ 15x + 20x + 9x = 12,188

⇒ 44x = 12,188

⇒ x = 277

⇒ Q's original share = 20 × 277 + 88 = 5540 + 88 = 5628

⇒ R's original share = 9 × 277 + 76 = 2493 + 76 = 2569

⇒ Difference = 5628 − 2569 = 3059

∴ The correct answer is ₹3,059.

Additional Information

Ratio Distribution with Constant Change

When fixed amounts are added or subtracted before dividing in a ratio, always adjust the Total Sum first by adding or subtracting the net change before applying the ratio parts.

Ratio & Proportion Basics

If a sum S is divided in ratio a:b:c, the share of b is [b ÷ (a+b+c)] × S. For original shares, remember to add back any deducted amounts.

Calculation Trick

To divide by 44, you can divide by 4 and then by 11 sequentially to simplify mental math during competitive exams.

Shortcut TrickFor a right-angled triangle, tanθ = Perpendicular (P) / Base (B) = 3 / 4.Using Pythagorean triplet (3, 4, 5), the Hypotenuse (H) = 5.cosecθ = Hypotenuse (H) / Perpendicular (P) = 5 / 3.∴ The correct answer is 5/3.Alternate MethodGiven: tanθ = 3/4Formula Used: tanθ = P/B, H2 = P2 + B2, cosecθ = H/P⇒ Let Perpendicular (P) = 3 and Base (B) = 4.⇒ H2 = 32 + 42⇒ H2 = 9 + 16 = 25⇒ H = √25 = 5⇒ cosecθ = H / P = 5 / 3∴ The correct answer is 5/3.Additional InformationTrigonometric RatiosReciprocal relationships: cosecθ = 1/sinθ, secθ = 1/cosθ, and cotθ = 1/tanθ.Pythagorean Identity1 + cot2θ = cosec2θ. Since cotθ = 4/3, cosec2θ = 1 + (4/3)2 = 1 + 16/9 = 25/9 → cosecθ = 5/3.Acute Angle PropertyFor an acute angle (0° < θ < 90°), all trigonometric ratios are positive.

Shortcut Trick

The HCF cannot be greater than the minimum difference between any two numbers.

Difference between 172 and 148 = 24

Difference between 198 and 172 = 26

The factors of the smallest difference (24) are 24, 12, 8, 6, 4, 3, 2, 1.

Testing the options: 148 is not divisible by 6 or 3 (sum of digits of 148 = 13). Hence, only 2 divides all three numbers.

∴ The correct answer is 2.

Alternate Method

Given: Numbers are 148, 172, and 198.

Formula Used: Prime Factorisation Method

Write the prime factorization of each number:

⇒ 148 = 2 × 2 × 37 = 2² × 37

⇒ 172 = 2 × 2 × 43 = 2² × 43

⇒ 198 = 2 × 3 × 3 × 11 = 2¹ × 3² × 11

To find the HCF, we take the lowest power of the common prime factors:

⇒ Common prime factor is 2, with the lowest power being 2¹.

⇒ HCF = 2¹ = 2

∴ The correct answer is 2.

Additional Information

Highest Common Factor (HCF)

The HCF of two or more numbers is the greatest number that divides each of them exactly without leaving any remainder.

Relationship between HCF and LCM

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).

HCF of Fractions

HCF of fractions = (HCF of Numerators) ÷ (LCM of Denominators).