Quantitative Aptitude PYQs

Shortcut Trick

Speed ratio (New : Usual) = 5 : 8

Since distance is constant, Time ratio (New : Usual) = 8 : 5

Difference in time ratio parts = 8 − 5 = 3 units

Given, 3 units = 21 minutes ⇒ 1 unit = 7 minutes

Usual time = 5 units = 5 × 7 = 35 minutes

∴ The correct answer is 35 min.

Alternate Method

Given: New speed = 5/8 of Usual speed, Time delay = 21 minutes.

Formula Used: Distance = Speed × Time; If distance is constant, Speed ∝ 1/Time.

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

⇒ New speed = 5S/8

⇒ New time = 8T/5

⇒ Time difference = New time − Usual time = 21 min

⇒ 8T/5 − T = 21

⇒ 3T/5 = 21

⇒ T = (21 × 5) ÷ 3

⇒ T = 7 × 5 = 35 minutes

∴ The correct answer is 35 min.

Additional Information

Inverse Proportion Rule

When distance is constant, the ratio of speeds is inversely proportional to the ratio of time: S₁/S₂ = T₂/T₁.

Usual Time Formula

If a person walks at a/b of usual speed and is 't' minutes late/early, Usual Time = [a / (b − a)] × t.

Average Speed

For constant distance with two speeds x and y, Average Speed = 2xy ÷ (x + y).

Shortcut Trick

Given Selling Price (SP) = ₹3,600 and Profit% = 20%.

Using the multiplier method: SP = CP × (100 + P%) ÷ 100.

3,600 = CP × 120 ÷ 100 ⇒ 3,600 = CP × 1.2.

CP = 3,600 ÷ 1.2 = 3,000.

∴ The correct answer is ₹3,000.

Alternate Method

Given: Marked Price (MP) = ₹4,000, Selling Price (SP) = ₹3,600, Profit% = 20%.

Formulas: Profit = SP − CP | SP = CP × (100 + P%) ÷ 100 | Discount = MP − SP.

⇒ In this problem, the Marked Price and Discount are extra information because SP is already given.

⇒ Let Cost Price (CP) be x.

⇒ SP = CP + Profit ⇒ 3,600 = x + (20 ÷ 100) × x.

⇒ 3,600 = x + 0.2x ⇒ 3,600 = 1.2x.

⇒ x = 3,600 ÷ 1.2 = 3,000.

∴ The correct answer is ₹3,000.

Additional Information

Marked Price and Discount

Marked Price (MP) is the price printed on the product. Discount is always calculated on the MP: Discount = MP − SP.

Relation between CP and MP

The ratio of CP to MP can be expressed as: CP ÷ MP = (100 − D%) ÷ (100 + P%).

Shortcut Trick

Since the percentage score remains the same, the ratio of marks obtained to maximum marks is constant.

300 ÷ x = 200 ÷ 400

300 ÷ x = 1 ÷ 2

⇒ x = 300 × 2 = 600

∴ The correct answer is 600 marks.

Alternate Method

Given:

Marks secured in first test = 300

Maximum marks of first test = x

Marks secured in second test = 200

Maximum marks of second test = 400

Formula Used: Percentage = (Marks Obtained ÷ Maximum Marks) × 100

Calculations:

⇒ Percentage in second test = (200 ÷ 400) × 100

⇒ Percentage = 50%

⇒ According to the question, percentage in first test is also 50%.

⇒ 50% of x = 300

⇒ (50 ÷ 100) × x = 300

⇒ 0.5x = 300

⇒ x = 300 ÷ 0.5

⇒ x = 600

∴ The correct answer is 600 marks.

Additional Information

Percentage Comparison

When percentages are equal, the ratios of the parts to the whole are equal: a/b = c/d.

Percentage Change

Percentage change = [(New Value − Old Value) ÷ Old Value] × 100.

Conversion

To convert a fraction to a percentage, multiply by 100. To convert a percentage to a fraction, divide by 100.

Shortcut Trick

Apply the Chain Rule formula: (M₁ × D₁) ÷ W₁ = (M₂ × D₂) ÷ W₂

Given: M₁ = 21, D₁ = 32, W₁ = 1/4. Remaining work W₂ = 3/4. Remaining time D₂ = 74 − 32 = 42 days.

(21 × 32) ÷ (1/4) = (M₂ × 42) ÷ (3/4) ⇒ 21 × 32 = (M₂ × 42) ÷ 3

M₂ = (21 × 32 × 3) ÷ 42 = 48 men.

Additional men = 48 − 21 = 27 men.

∴ The correct answer is 27 men.

Alternate Method

Given: Total days = 74 days, Initial men (M₁) = 21, Days worked (D₁) = 32 days, Work done (W₁) = 1/4.

Formula Used: (M₁ × D₁) ÷ W₁ = (M₂ × D₂) ÷ W₂

Calculations:

⇒ Remaining Work (W₂) = 1 − 1/4 = 3/4

⇒ Remaining Days (D₂) = 74 − 32 = 42 days

⇒ (21 × 32) ÷ (1/4) = ((21 + x) × 42) ÷ (3/4)

⇒ 21 × 32 × 4 = (21 + x) × 42 × (4/3)

⇒ 21 × 32 = (21 + x) × 14

⇒ (21 × 32) ÷ 14 = 21 + x

⇒ 3 × 16 = 21 + x

⇒ 48 = 21 + x

⇒ x = 48 − 21 = 27

∴ The correct answer is 27 men.

Additional Information

Work Equivalence Formula

The general formula for Man-Days is (M₁×D₁×H₁)/W₁ = (M₂×D₂×H₂)/W₂, where M is men, D is days, H is hours, and W is work.

Efficiency and Time

Efficiency is inversely proportional to the time taken to complete a constant amount of work. (E₁ × T₁ = E₂ × T₂).

Shortcut Trick

Identify the Marked Price (MP) as 8,810 and Selling Price (SP) as 6,250.

The Discount Amount is the difference: 8,810 − 6,250 = 2,560.

Apply the multiplier logic: Discount% = (Discount ÷ MP) × 100.

Calculation: (2,560 ÷ 8,810) × 100 ≈ 29.0578%.

Rounding to two decimal places, we get 29.06%.

∴ The correct answer is 29.06%.

Alternate Method

Given: Marked Price (MP) = 8,810, Selling Price (SP) = 6,250.
Formulas: Discount = MP − SP | Discount% = (Discount ÷ MP) × 100

⇒ Discount = 8,810 − 6,250 = 2,560.
⇒ Discount Percentage = (2,560 ÷ 8,810) × 100.
⇒ Discount Percentage = 256,000 ÷ 8,810 ≈ 29.06%.

∴ The correct answer is 29.06%.

Additional Information

Marked Price & Discount

Marked Price is the price printed on the label. Discount is always calculated on the Marked Price unless stated otherwise.

Selling Price Calculation

Selling Price = Marked Price × (100 − Discount%) ÷ 100. This is the final amount paid by the customer.

Shortcut Trick

Total deduction = ₹(72 + 21 + 24) = ₹117.

Remaining amount to be divided = ₹5,177 − ₹117 = ₹5,060.

Ratio after deduction = 3 : 13 : 7 (Sum of parts = 23).

Value of 1 unit = 5,060 ÷ 23 = ₹220.

Difference between B and C after deduction = (13 − 7) units = 6 units.

Original difference = (Difference after deduction) + (Difference of deductions) = (6 × 220) + (21 − 24) = 1,320 − 3 = ₹1,317.

∴ The correct answer is ₹1,317.

Alternate Method

Given: Total sum = ₹5,177. Deductions for A, B, C are ₹72, ₹21, ₹24.

⇒ Sum of remaining shares = 5,177 − (72 + 21 + 24)

⇒ Sum of remaining shares = 5,177 − 117 = ₹5,060

⇒ Let the remaining shares be 3x, 13x, and 7x.

⇒ 3x + 13x + 7x = 5,060

⇒ 23x = 5,060

⇒ x = 5,060 ÷ 23 = 220

⇒ Share of B after deduction = 13 × 220 = ₹2,860

⇒ Share of C after deduction = 7 × 220 = ₹1,540

⇒ Original share of B = 2,860 + 21 = ₹2,881

⇒ Original share of C = 1,540 + 24 = ₹1,564

⇒ Required Difference = 2,881 − 1,564 = ₹1,317

∴ The correct answer is ₹1,317.

Additional Information

Ratio Division with Constraints

When fixed amounts are added or subtracted from shares, always perform the operation on the total sum first before applying the ratio division.

Compound Ratio

The compound ratio of (a : b) and (c : d) is calculated as (a × c) : (b × d).

Shortcut Trick

Total work (LCM of 2, 15, 10) = 30 units.

Combined efficiency of A + B + C = 15 + 2 + 3 = 20 units/day.

Twice the work = 30 × 2 = 60 units.

Time required = 60 ÷ 20 = 3 days.

∴ The correct answer is 3 days.

Alternate Method

Given: Time taken by A = 2 days, B = 15 days, C = 10 days.

Formula Used: Total Work = LCM of individual days; Efficiency = Total Work ÷ Time; Time = Work ÷ Efficiency.

⇒ Total Efficiency of (A + B + C) = 15 + 2 + 3 = 20 units/day

⇒ New Work to be completed = 2 × 30 = 60 units

⇒ Time taken together = 60 ÷ 20 = 3 days

∴ The correct answer is 3 days.

Additional Information

Combined Efficiency Rule

When multiple people work together, their individual efficiencies are added to find the total rate of work per day.

Work & Time Relationship

Time is inversely proportional to efficiency. If the work is doubled, the time required at the same efficiency also doubles.

Shortcut Trick

The average of the first n even natural numbers is always given by the direct formula: (n + 1).

Given n = 13 (the first 13 even numbers).

Average = 13 + 1 = 14.

∴ The correct answer is 14.

Alternate Method

Given: Total even numbers (n) = 13.

Formula Used: Average of an Arithmetic Progression (AP) = (First term + Last term) ÷ 2.

⇒ First even natural number (a₁) = 2

⇒ 13^th even natural number (a₁₃) = 2 × 13 = 26

⇒ Average = (2 + 26) ÷ 2

⇒ Average = 28 ÷ 2 = 14

∴ The correct answer is 14.

Additional Information

Sum of First n Even Numbers

The sum of the first n even natural numbers is calculated using the formula: S_n = n(n + 1).

Average of First n Odd Numbers

The average of the first n odd natural numbers is always equal to n.

Average of First n Natural Numbers

The average of the first n natural numbers (1, 2, 3, ..., n) is given by (n + 1) ÷ 2.

Shortcut Trick

Identify 2024 as a leap year, meaning it has 366 days.

Calculate total days: February (9) + March (31) + April (21) = 61 days.

Time period in years = 61 ÷ 366 = 1/6 year.

Simple Interest (SI) = 4000 × 6% × 1/6 = ₹40.

∴ The correct answer is ₹40.

Alternate Method

Given:

Principal (P) = ₹4,000

Rate (R) = 6% per annum

Duration: 20 February 2024 to 21 April 2024

Formula Used:

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

Calculations:

⇒ Year 2024 is a leap year (divisible by 4), so February has 29 days.

⇒ Days in February = 29 − 20 = 9 days

⇒ Days in March = 31 days

⇒ Days in April = 21 days

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

⇒ T (in years) = 61 ÷ 366 = 1/6 year

⇒ SI = (4000 × 6 × 1/6) ÷ 100

⇒ SI = (4000 × 1) ÷ 100 = 40

∴ The correct answer is ₹40.

Additional Information

Simple Interest Formula

Simple interest is calculated only on the principal amount: SI = (P × R × T) ÷ 100.

Leap Year Calculation

A year is a leap year if it is divisible by 4 (e.g., 2024, 2028). For such years, T is divided by 366 instead of 365.

Day Counting Rule

Standard practice excludes the day of deposit and includes the day of withdrawal to calculate the interest period.

Shortcut Trick

When the distance is constant for two speeds x and y, Average Speed = (2 × x × y) ÷ (x + y).

Average Speed = (2 × 45 × 36) ÷ (45 + 36)

Average Speed = (2 × 45 × 36) ÷ 81 = 3240 ÷ 81 = 40 km/h.

∴ The correct answer is 40 km/h.

Alternate Method

Given:

Speed to school (v₁) = 45 km/h

Speed back (v₂) = 36 km/h

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Let the distance between home and school be LCM(45, 36) = 180 km.

⇒ Time taken going to school (t₁) = 180 ÷ 45 = 4 hours.

⇒ Time taken coming back (t₂) = 180 ÷ 36 = 5 hours.

⇒ Total distance = 180 + 180 = 360 km.

⇒ Total time = 4 + 5 = 9 hours.

⇒ Average speed = 360 ÷ 9 = 40 km/h.

∴ The correct answer is 40 km/h.

Additional Information

Average Speed Formula

For two equal distances with speeds a and b, the average speed is the harmonic mean: 2ab/(a+b).

Relative Speed

In the same direction, subtract speeds (v₁ − v₂); in opposite directions, add speeds (v₁ + v₂).

Speed 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

Use the multiplying factor method for successive changes.

Net Factor = (1 − 0.35) × (1 − 0.55) × (1 − 0.35) × (1 + 0.65)

Net Factor = 0.65 × 0.45 × 0.65 × 1.65 ≈ 0.3137

Percentage change = (0.3137 − 1) × 100 = −68.63%

∴ The correct answer is Decreases by 68.63%.

Alternate Method

Given: Successive changes of −35%, −55%, −35% and +65%.

Formula: Final Price = Initial Price × (100 ± r₁)/100 × (100 ± r₂)/100 ...

⇒ Let the original price be 100 units.

⇒ Price after 1^st month = 100 × (65/100) = 65

⇒ Price after 2^nd month = 65 × (45/100) = 29.25

⇒ Price after 3^rd month = 29.25 × (65/100) = 19.0125

⇒ Price after 4^th month = 19.0125 × (165/100) = 31.370625

⇒ Net change = 31.370625 − 100 = −68.629375

⇒ Percentage decrease ≈ 68.63%

∴ The correct answer is Decreases by 68.63%.

Additional Information

Successive Percentage Formula

For two successive changes a% and b%, the net change is (a + b + ab/100)%.

Percentage Increase/Decrease

Percentage change = (Final Value − Initial Value) / Initial Value × 100.

Multiplying Factor (MF)

For an x% increase, MF = (1 + x/100). For an x% decrease, MF = (1 − x/100).

Shortcut Trick

Larger number = (Sum + Difference) ÷ 2 = (74 + 12) ÷ 2 = 43

Smaller number = (Sum − Difference) ÷ 2 = (74 − 12) ÷ 2 = 31

Required Ratio (Smaller : Larger) = 31 : 43

∴ The correct answer is 31 : 43.

Alternate Method

Given: Sum of numbers = 74, Difference = 12

Formula: If x + y = S and x − y = D, then x = (S+D)/2 and y = (S−D)/2

⇒ Let the numbers be x (larger) and y (smaller).

⇒ x + y = 74

⇒ x − y = 12

⇒ Adding both equations: 2x = 86 ⇒ x = 43

⇒ Subtracting both equations: 2y = 62 ⇒ y = 31

⇒ Ratio y : x = 31 : 43

∴ The correct answer is 31 : 43.

Additional Information

Linear Equations in Two Variables

When the sum (S) and difference (D) are known, the values are (S+D)/2 and (S−D)/2.

Ratio and Proportion

A ratio compares two quantities by division; always ensure the order (smaller to larger) matches the question.

Shortcut Trick

For a cyclic quadrilateral with sides a, b, c, and d, the area is given by Brahmagupta's Formula:

Area = √((s − a)(s − b)(s − c)(s − d))

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

Given: a = 10 cm, b = 14 cm, c = 8 cm, d = 12 cm

⇒ s = (10 + 14 + 8 + 12) ÷ 2 = 22 cm

⇒ Area = √((22 − 10)(22 − 14)(22 − 8)(22 − 12)) = √(12 × 8 × 14 × 10) = √(13440) = 8√210 cm²

∴ The correct answer is 8√210 cm².

Alternate Method

Given:

Sides of the cyclic quadrilateral ABCD:

⇒ AB = a = 10 cm

⇒ BC = b = 14 cm

⇒ CD = c = 8 cm

⇒ DA = d = 12 cm

Formula Used:

Area of a cyclic quadrilateral = √((s − a)(s − b)(s − c)(s − d))

where s = (a + b + c + d) ÷ 2

Calculations:

Step 1: Find the semi-perimeter (s) of the cyclic quadrilateral:

⇒ s = (10 + 14 + 8 + 12) ÷ 2

⇒ s = 44 ÷ 2

⇒ s = 22 cm

Step 2: Substitute the values of s, a, b, c, and d into Brahmagupta's formula:

⇒ Area = √((22 − 10) × (22 − 14) × (22 − 8) × (22 − 12))

⇒ Area = √(12 × 8 × 14 × 10)

⇒ Area = √(96 × 140)

⇒ Area = √(13440)

Step 3: Simplify the square root:

⇒ Area = √(64 × 210)

⇒ Area = 8√210 cm²

Verification of Options:

⇒ Option (1): 8√222 cm² = √(64 × 222) = √(14208) ≠ √(13440) (Incorrect)

⇒ Option (2): 10√210 cm² = √(100 × 210) = √(21000) ≠ √(13440) (Incorrect)

⇒ Option (3): 8√210 cm² = √(64 × 210) = √(13440) (Correct)

⇒ Option (4): 10√222 cm² = √(100 × 222) = √(22200) ≠ √(13440) (Incorrect)

∴ The correct answer is 8√210 cm².

Additional Information

Brahmagupta's Formula

This formula is a generalization of Heron's formula for the area of a triangle and is specifically used to find the area of any cyclic quadrilateral given only its side lengths.

Ptolemy's Theorem

For any cyclic quadrilateral ABCD, the product of the diagonals is equal to the sum of the products of the opposite sides: AC × BD = AB × CD + BC × DA.

Opposite Angles Property

In a cyclic quadrilateral, the sum of opposite angles is always supplementary, i.e., ∠A + ∠C = 180° and ∠B + ∠D = 180°.

Shortcut Trick

For two successive discounts of a% and b%, the equivalent single discount is given by the formula: (a + b − ab ÷ 100)%.

Here, a = 9% and b = 6%.

Equivalent discount = (9 + 6 − (9 × 6) ÷ 100)%.

Equivalent discount = (15 − 54 ÷ 100)% = (15 − 0.54)% = 14.46%.

∴ The correct answer is 14.46%.

Alternate Method

Given → First discount (d₁) = 9%, Second discount (d₂) = 6%.
Formulas: Single Equivalent Discount = 100 − [100 × (100 − d₁)/100 × (100 − d₂)/100]

⇒ Let the Marked Price (MP) be 100.
⇒ Price after 9% discount = 100 − 9 = 91.
⇒ Price after 6% discount on 91 = 91 − (6% of 91) = 91 − 5.46 = 85.54.
⇒ Total discount = 100 − 85.54 = 14.46.
⇒ Single equivalent discount % = (14.46 ÷ 100) × 100 = 14.46%.
∴ The correct answer is 14.46%.

Additional Information

Successive Discounts

When a series of discounts are allowed one after another, they are called successive discounts. The net discount is always less than the sum of individual discount rates.

Effective Discount for 3 values

For three successive discounts a%, b%, and c%, first calculate the equivalent of a and b, then use that result with c.

Shortcut Trick

Combined efficiency (P+Q) = 1/15, P efficiency = 1/60.

Efficiency of Q = (1/15) − (1/60) = (4 − 1)/60 = 3/60 = 1/20.

Time for Q to fill 1/4^th of cistern = (1/4) ÷ (1/20) = 20 ÷ 4 = 5 hours.

∴ The correct answer is 5 hours.

Alternate Method

⇒ Total capacity of cistern = LCM of 15 and 60 = 60 units.

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

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

⇒ One-fourth of cistern capacity = 60 × (1/4) = 15 units.

⇒ Time taken by Q = Work ÷ Efficiency = 15 ÷ 3 = 5 hours.

∴ The correct answer is 5 hours.

Additional Information

Combined Efficiency Rule

If two pipes fill a tank in T₁ and T₂ hours respectively, their combined work rate is 1/T₁ + 1/T₂.

Pipes and Cisterns Logic

Inlet pipes have positive efficiency, while outlet (leakage) pipes have negative efficiency. Total Work = Time × Efficiency.

Shortcut Trick

For Compound Interest at 6% for 2 years, the Effective Interest Rate = x + y + (xy/100)

Effective Rate = 6 + 6 + (6 × 6)/100 = 12.36%

Total Amount = 100% (Principal) + 12.36% (Interest) = 112.36% of Principal

Amount = 7500 × 112.36/100 = 75 × 112.36 = ₹8,427

∴ The correct answer is ₹8,427.

Alternate Method

Given: Principal (P) = ₹7,500, Rate (R) = 6% per annum, Time (n) = 2 years.

Formula Used: Amount (A) = P(1 + R/100)ⁿ

⇒ A = 7500 × (1 + 6/100)²

⇒ A = 7500 × (106/100) × (106/100)

⇒ A = 7500 × (1.06) × (1.06)

⇒ A = 7500 × 1.1236

⇒ A = ₹8,427

∴ The correct answer is ₹8,427.

Additional Information

CI vs SI Difference

The difference between Compound Interest and Simple Interest for 2 years is given by the formula: D = P(R/100)².

Half-Yearly Compounding

When interest is compounded half-yearly, the rate becomes R/2% and the time period becomes 2n.

Effective Rate for 3 Years

The effective rate for 3 years can be calculated by applying the successive formula [x + y + (xy/100)] twice or using 3R.3R²R³ shortcut.

Shortcut Trick

Total Cost Price (CP) of 24 kg = 3,600 ⇒ CP per kg = 3,600 ÷ 24 = 150.

First 12 kg (half) gain: 12 × 150 × 20% = +360.

Remaining 12 kg (half) loss: 12 × 150 × 10% = −180.

Net Profit/Loss = 360 − 180 = 180 (Positive value indicates Gain).

∴ The correct answer is Gain ₹180.

Alternate Method

Given → Total CP = 3,600 | 1^st half gain = 20% | 2^nd half loss = 10%.
Formulas: Profit = SP − CP | Profit% = (Profit ÷ CP) × 100 | SP = CP × (100 + P%) ÷ 100.

⇒ Total Cost Price = 3,600. CP of 12 kg = 1,800.
⇒ SP of first 12 kg = 1,800 × (120/100) = 2,160.
⇒ SP of remaining 12 kg = 1,800 × (90/100) = 1,620.
⇒ Total Selling Price = 2,160 + 1,620 = 3,780.
⇒ Net Gain = Total SP − Total CP = 3,780 − 3,600 = 180.

∴ The correct answer is Gain ₹180.

Additional Information

Net Gain/Loss on Equal Parts

When two equal parts are sold at x% gain and y% loss, the net percentage change is (x − y) ÷ 2.

Selling Price Calculation

SP = CP × (100 ± Change%) ÷ 100. This is the basic multiplier for profit or loss adjustments.

Shortcut Trick

Let the total votes be 100%.

Dinesh got 46.5%, so Rizwan got 100% − 46.5% = 53.5%.

Difference in percentage = 53.5% − 46.5% = 7%.

Given 7% ≡ 3150 votes.

Rizwan's votes (53.5%) = (3150 ÷ 7) × 53.5 = 450 × 53.5 = 24075.

∴ The correct answer is 24,075 votes.

Alternate Method

Given: Votes for Dinesh = 46.5% | Margin = 3150 votes | All votes are valid.

⇒ Total Votes = 100%

⇒ Votes obtained by Rizwan = 100% − 46.5% = 53.5%

⇒ Difference between Rizwan and Dinesh = 53.5% − 46.5% = 7%

⇒ According to the question, 7% of Total Votes = 3150

⇒ 1% of Total Votes = 3150 ÷ 7 = 450

⇒ Rizwan's share = 53.5 × 450

⇒ Rizwan's share = 24075

∴ The correct answer is 24,075 votes.

Additional Information

Percentage Change

Percentage change is calculated as (Difference ÷ Original Value) × 100.

Valid and Invalid Votes

Total Cast Votes = Valid Votes + Invalid Votes. Winners and losers are usually calculated based on Valid Votes.

Winning Margin

Margin = Votes of Winner − Votes of Loser. It represents the number of votes by which the winner defeated the opponent.

Shortcut Trick

For third proportional to a and b, the formula is c = b²/a.

Given: a = 50, b = Z, and c = 18.

⇒ 18 = Z² ÷ 50

⇒ Z² = 18 × 50 = 900

⇒ Z = √900 = 30

∴ The correct answer is 30.

Alternate Method

Given:

First number (a) = 50

Second number (b) = Z

Third proportional (c) = 18

Formula Used:

In continued proportion a, b, c ⇒ a/b = b/c

Calculations:

⇒ 50 / Z = Z / 18

⇒ Z × Z = 50 × 18

⇒ Z² = 900

⇒ Z = √900

⇒ Z = 30

∴ The correct answer is 30.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is √(a × b).

Fourth Proportional

If a : b :: c : d, then d is the fourth proportional and d = (b × c) ÷ a.

Continued Proportion

Three quantities a, b, c are in continued proportion if the ratio of first to second equals second to third (a/b = b/c).

Shortcut Trick

Average Speed = Total Distance ÷ Total Time

Total Distance = 300 + 300 = 600 m

Total Time = 100 + 50 = 150 sec

Average Speed = 600 ÷ 150 = 4 m/sec

∴ The correct answer is 4 m/sec.

Alternate Method

Given:

Distance of first lap = 300 m, Time taken = 100 sec

Distance of second lap = 300 m, Time taken = 50 sec

Formula Used:

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

Calculations:

⇒ Total Distance = 300 + 300 = 600 m

⇒ Total Time = 100 + 50 = 150 sec

⇒ Average Speed = 600 ÷ 150

⇒ Average Speed = 4 m/sec

∴ The correct answer is 4 m/sec.

Additional Information

Average Speed (Equal Distances)

When two equal distances are covered at speeds x and y, the average speed is 2xy ÷ (x + y).

Unit Conversion

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