Quantitative Aptitude PYQs

Shortcut Trick

Find the factors of the product 45 that add up to the sum 14:

Factors of 45: 9 × 5 = 45

Sum of these factors: 9 + 5 = 14

Therefore, the two positive numbers are 9 and 5.

Their positive difference is: 9 − 5 = 4

∴ The correct answer is 4.

Alternate Method

Given:

Sum of two numbers (a + b) = 14

Product of two numbers (ab) = 45

A = 9

Shortcut Trick

The time difference of 42 minutes occurs only during the 98 km segment of the journey.

Let the original speed be s km/hr. Equation: (98 ÷ s) − (98 ÷ (s + 7)) = 42/60 = 0.7 hours.

Testing the provided answer 28 km/hr from the options:

(98 ÷ 28) − (98 ÷ 35) = 3.5 − 2.8 = 0.7 hours.

The condition is satisfied, thus the original speed is 28 km/hr.

∴ The correct answer is 28 km/hr.

Alternate Method

Given:

Total distance from Town A to Town B = 182 km

First segment distance = 84 km

Second segment distance = 98 km

Time saved = 42 minutes = 42 ÷ 60 hours = 0.7 hours

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Let the original speed of the cyclist be s km/hr.

⇒ Hypothetical time to cover the entire distance at original speed = 182 ÷ s

⇒ Actual time taken = (84 ÷ s) + (98 ÷ (s + 7))

⇒ According to the question: (182 ÷ s) − [(84 ÷ s) + (98 ÷ (s + 7))] = 0.7

⇒ (182 − 84) ÷ s − 98 ÷ (s + 7) = 0.7

⇒ 98 ÷ s − 98 ÷ (s + 7) = 0.7

⇒ 98 [ (s + 7 − s) ÷ (s(s + 7)) ] = 0.7

⇒ (98 × 7) ÷ (s² + 7s) = 0.7

⇒ 686 ÷ 0.7 = s² + 7s

⇒ s² + 7s − 980 = 0

⇒ s² + 35s − 28s − 980 = 0

⇒ s(s + 35) − 28(s + 35) = 0

⇒ (s − 28)(s + 35) = 0

⇒ Since speed cannot be negative, s = 28.

∴ The correct answer is 28 km/hr.

Additional Information

Speed-Distance-Time Relation

The fundamental relation is Speed = Distance ÷ Time. For a constant distance, speed is inversely proportional to time.

Average Speed

For any journey, Average Speed is calculated as the Total Distance divided by the Total Time taken for all segments.

Quadratic Equations in Motion

Problems involving speed changes and time savings frequently result in quadratic equations of the form s² + as − k = 0, where s is the speed.

Shortcut Trick

Apply the divisibility rule of 9: the sum of the digits must be a multiple of 9.

Sum of digits of 95061 = 9 + 5 + 0 + 6 + 1 = 21.

The nearest multiples of 9 to 21 are 18 (21 − 3) and 27 (21 + 6).

Subtracting 3 from the number: 95061 − 3 = 95058.

Adding 6 to the number: 95061 + 6 = 95067. Since 3 < 6, 95058 is the nearest.

∴ The correct answer is 95058.

Alternate Method

Given: Target number = 95061, Divisor = 9.

⇒ Divide 95061 by 9 to find the remainder.

⇒ 95061 ÷ 9 = 10562 with a remainder of 3.

⇒ Lower multiple = 95061 − 3 = 95058.

⇒ Higher multiple = 95061 + (9 − 3) = 95061 + 6 = 95067.

⇒ Distance to 95058 = |95061 − 95058| = 3.

⇒ Distance to 95067 = |95067 − 95061| = 6.

⇒ Comparing distances: 3 < 6.

∴ The correct answer is 95058.

Additional Information

Divisibility Rule of 9

A number is divisible by 9 if and only if the sum of its digits is divisible by 9. For example, in 95058, 9+5+0+5+8 = 27, which is divisible by 9.

Nearest Number Logic

To find the nearest number divisible by n, calculate the remainder r when the number N is divided by n. Compare r and (n − r). The smaller value determines whether to subtract or add.

Divisibility Rule of 3

Similar to the rule for 9, a number is divisible by 3 if the sum of its digits is divisible by 3.

Given:

$\dfrac{(3^2 \times 9)^3}{27^2 \times 3^4}$ = ?

Formula used:

1. (a^m)ⁿ = a^{m × n}
2. a^m × aⁿ = a^{m + n}
3. a^m / aⁿ = a^{m − n}

Calculations:

To Simplify the expression, convert all terms to Base 3:

9 = 3²

27 = 3³

Substituting these values in the Numerator and Denominator:

⇒ $\dfrac{(3^2 \times 3^2)^3}{(3^3)^2 \times 3^4}$

Using the Laws of Exponents:

⇒ $\dfrac{(3^{2+2})^3}{3^{3 \times 2} \times 3^4}$

⇒ $\dfrac{(3^4)^3}{3^6 \times 3^4}$

⇒ $\dfrac{3^{4 \times 3}}{3^{6+4}}$

⇒ $\dfrac{3^{12}}{3^{10}}$

⇒ 3^{12 − 10}

⇒ 3²

⇒ 9

∴ The correct answer is option (1).

Shortcut Trick

If every number in a data set is decreased by a constant value k, then the average of the set also decreases by the same value k.

Given Old Average = 43 and Decrease (k) = 7.

New Average = Old Average − k

⇒ 43 − 7 = 36

∴ The correct answer is 36.

Alternate Method

Given:

Total numbers = 8

Original average = 43

Decrease in each number = 7

Formula Used:

Average = Sum ÷ Total Numbers

New Average = (Original Sum − Total Decrease) ÷ Total Numbers

Calculations:

⇒ Original Sum = 8 × 43 = 344

⇒ Total Decrease = 8 × 7 = 56

⇒ New Sum = 344 − 56 = 288

⇒ New Average = 288 ÷ 8 = 36

∴ The correct answer is 36.

Additional Information

Multiplication and Division Rule

If each observation is multiplied or divided by a non-zero constant a, the new average becomes the old average multiplied or divided by a.

Addition Rule

If a constant value k is added to every number in a series, the average of that series also increases exactly by k.

Average of Consecutive Numbers

For any arithmetic progression, the average is simply the middle term or (First term + Last term) ÷ 2.

Shortcut Trick

Effective rate for compounded half-yearly is 8% ÷ 2 = 4% per half-year.

Ratio of Principal to Amount for 4% (1/25) for 2 half-years = 25² : 26² = 625 : 676.

Given 676 units = ₹10,816 ⇒ 1 unit = 10816 ÷ 676 = ₹16.

Principal = 625 units = 625 × 16 = ₹10,000.

∴ The correct answer is ₹10,000.

Alternate Method

Given:

Total Amount (A) = ₹10,816

Annual Rate of Interest (R) = 8% per annum

Time (T) = 1 year

Compounding: Half-yearly

Formula Used: A = P(1 + r/100)ⁿ

Where r = Annual Rate ÷ 2 and n = Time in years × 2.

Calculations:

⇒ Half-yearly rate (r) = 8 ÷ 2 = 4%

⇒ Number of half-years (n) = 1 × 2 = 2

⇒ 10816 = P(1 + 4/100)²

⇒ 10816 = P(1 + 1/25)²

⇒ 10816 = P(26/25)²

⇒ 10816 = P × (676 ÷ 625)

⇒ P = (10816 × 625) ÷ 676

⇒ P = 16 × 625

⇒ P = ₹10,000

∴ The correct answer is ₹10,000.

Additional Information

Effective Rate of Interest

For two successive growth periods of r%, the effective rate R_eff = r + r + (r × r)/100. In this case, 4 + 4 + (4 × 4)/100 = 8.16%.

Quarterly Compounding

If interest is compounded quarterly, the rate becomes R/4 and the number of periods becomes T × 4.

Compound Interest vs Simple Interest

For 2 periods, the difference between CI and SI is P(r/100)², where r is the rate per period.

Shortcut Trick

Present sum of ages = Sum after 6 years − (Number of people × Years passed)

Present sum = 60 − (2 × 6) = 48 years

Ratio of A : B is 5 : 7, so total parts = 12

12 parts = 48 ⇒ 1 part = 4

A’s present age = 5 parts = 5 × 4 = 20 years

∴ The correct answer is 20 years.

Alternate Method

Given: Present ratio of ages of A and B = 5 : 7. Sum of ages after 6 years = 60 years.

Let the present ages of A and B be 5x and 7x respectively.

⇒ Age of A after 6 years = 5x + 6

⇒ Age of B after 6 years = 7x + 6

⇒ (5x + 6) + (7x + 6) = 60

⇒ 12x + 12 = 60

⇒ 12x = 60 − 12

⇒ 12x = 48

⇒ x = 4

⇒ A’s present age = 5x = 5 × 4 = 20

∴ The correct answer is 20 years.

Additional Information

Sum of Ages in the Future/Past

If the sum of ages of 'n' persons is S now, then after 't' years, the sum will be S + (n × t). Similarly, 't' years ago, it was S − (n × t).

Ratio of Ages

If the ages are in ratio a : b, their actual values can be taken as ax and bx. The difference between their ages (b−a)x remains constant regardless of time.

Proportional Division

To find A's share from total sum S with ratio A:B = a:b, use the formula: A = [a ÷ (a + b)] × S.

Shortcut Trick

Let the total amount be T. The sum of money remains constant regardless of transfers.

Condition 1: Juhi has 2 parts out of 3 (total 2+1=3) ⇒ Juhi's amount = (2/3)T + 30 (since she gave 30).

Condition 2: Juhi has 4 parts out of 5 (total 4+1=5) ⇒ Juhi's amount = (4/5)T − 10 (since she received 10).

Equating Juhi's original amount: (4/5)T − 10 = (2/3)T + 30 ⇒ (4/5 − 2/3)T = 40.

(2/15)T = 40 ⇒ T = 20 × 15 = 300.

∴ The correct answer is ₹300.

Alternate Method

Given: Juhi (J) and Mamta (M) exchange money in two scenarios. Scenario 1: J gives ₹30 to M. New Ratio J:M = 2:1. Scenario 2: M gives ₹10 to J. New Ratio J:M = 4:1.

Step-by-step Calculation:

⇒ From scenario 1: (J − 30) = 2(M + 30)

⇒ J − 30 = 2M + 60 ⇒ J − 2M = 90 —— (Equation 1)

⇒ From scenario 2: (J + 10) = 4(M − 10)

⇒ J + 10 = 4M − 40 ⇒ J − 4M = −50 —— (Equation 2)

⇒ Subtracting (Equation 2) from (Equation 1): (J − 2M) − (J − 4M) = 90 − (−50)

⇒ 2M = 140 ⇒ M = 70

⇒ Substituting M = 70 in Equation 1: J − 2(70) = 90 ⇒ J − 140 = 90 ⇒ J = 230

⇒ Total initial amount = J + M = 230 + 70 = 300

∴ The correct answer is ₹300.

Additional Information

Simultaneous Linear Equations

Used to find the values of two variables when two distinct conditions are provided. The method of elimination or substitution is typically employed.

Constant Sum Property

When money is transferred internally between two individuals, the sum of their assets (total money) remains unchanged. This allows for quick ratio comparisons.

Ratio Balancing

If the total units in two ratios differ (e.g., 3 units and 5 units), they can be equated by multiplying each ratio by the total of the other to find a common denominator.

Shortcut Trick

The ratio of height to shadow is given as √3 : 1.

In trigonometry, the tangent of the angle of elevation (θ) is the ratio of height (perpendicular) to shadow (base).

⇒ tan θ = √3 / 1 = √3.

Since we know that tan 60° = √3, the value of θ must be 60°.

∴ The correct answer is 60°.

Alternate Method

Given: Ratio of the height of the tower to the length of its shadow = √3 : 1.

Formula Used: tan θ = Perpendicular (Height) ÷ Base (Shadow).

⇒ Let the height of the tower (AB) be √3x.

⇒ Let the length of the shadow (BC) be 1x.

⇒ Let the angle of elevation of the sun be θ.

⇒ In right-angled triangle ABC, tan θ = AB ÷ BC

⇒ tan θ = √3x ÷ 1x

⇒ tan θ = √3

⇒ tan θ = tan 60°

⇒ θ = 60°

∴ The correct answer is 60°.

Additional Information

Angle of Elevation

It is the angle formed by the line of sight with the horizontal when the object being viewed is above the horizontal level.

Standard Trigonometric Ratios

For height-shadow problems, remember: tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3.

Sun Altitude Relation

As the angle of elevation of the sun increases (from 0° to 90°), the length of the shadow of a fixed object decreases.

Shortcut Trick

The difference between the direct profit percentage and the first profit percentage equals Q's profit relative to P's Cost Price.

Difference in profit = 25% − 12% = 13%

This 13% of CP is equal to Q's profit of ₹26.

13% of Cost Price = ₹26 ⇒ 1% = ₹2

100% (Cost Price) = 2 × 100 = ₹200

∴ The correct answer is ₹200.

Alternate Method

Given:

Profit by P selling to Q = 12%

Profit by Q selling to R = ₹26

Direct profit if P sells to R at same price = 25%

Calculations:

⇒ Let the Cost Price (CP) of the item for P be ₹x.

⇒ Selling Price for P (which is CP for Q) = x + 12% of x = 1.12x

⇒ Selling Price for Q (which is the price R pays) = 1.12x + 26

⇒ According to the condition, if P sells to R at 1.12x + 26, profit = 25%.

⇒ 1.25x = 1.12x + 26

⇒ 1.25x − 1.12x = 26

⇒ 0.13x = 26

⇒ x = 26 ÷ 0.13 = 2600 ÷ 13

⇒ x = 200

∴ The correct answer is ₹200.

Additional Information

Successive Profit/Loss

When an item is sold through multiple hands, the effective profit is calculated by multiplying the factors: Final Price = CP × (1 + r₁/100) × (1 + r₂/100)...

Profit Percentage Formula

Profit % = [(Selling Price − Cost Price) ÷ Cost Price] × 100.

Difference between Two Profits

If SP₁ and SP₂ are selling prices at profit % p₁ and p₂ respectively, then SP₂ − SP₁ = (p₂ − p₁)% of Cost Price.

Shortcut Trick

Total deduction = ₹25 + ₹52 + ₹18 = ₹95

Remaining amount = ₹1,472 − ₹95 = ₹1,377

Vikas's share ratio part = 17 ÷ (15 + 19 + 17) = 17/51 = 1/3

Vikas's amount after deduction = 1/3 × ₹1,377 = ₹459

∴ The correct answer is ₹459.

Alternate Method

Given: Total amount = ₹1,472, Deductions = ₹25, ₹52, ₹18, Ratio after deduction = 15 : 19 : 17

⇒ Total Deductions = 25 + 52 + 18 = ₹95

⇒ Amount left to be divided = 1472 − 95 = ₹1,377

⇒ Let the shares of Rohit, Maya, and Vikas after deduction be 15x, 19x, and 17x respectively.

⇒ 15x + 19x + 17x = 1377

⇒ 51x = 1377

⇒ x = 1377 ÷ 51 = 27

⇒ Vikas's amount after deduction = 17 × 27 = ₹459

∴ The correct answer is ₹459.

Additional Information

Ratio and Proportion

To divide a quantity S into parts a:b:c, the parts are calculated as [a/(a+b+c)] × S, [b/(a+b+c)] × S, and [c/(a+b+c)] × S.

Original Amount Calculation

If the original share is required, add the deducted amount back to the calculated share: Original Share = Share after deduction + Deducted amount.

Sum of Ratio Parts

The sum of all individual ratio shares must always equal the total amount being distributed.

Shortcut Trick

The mode is the value that appears most frequently in the given data set.

Checking frequencies: 7 occurs 2 times, 8 occurs 1 time, 9 occurs 1 time, and 10 occurs 1 time.

The number with the highest frequency is 7.

∴ The correct answer is 7.

Alternate Method

Given: Data set = {7, 8, 7, 9, 10}

Formula Used: Mode = Value with the maximum frequency in the observation set.

⇒ Count of number 7 = 2

⇒ Count of number 8 = 1

⇒ Count of number 9 = 1

⇒ Count of number 10 = 1

⇒ Maximum frequency = 2 (for number 7)

⇒ Mode = 7

∴ The correct answer is 7.

Additional Information

Empirical Relationship

For a moderately skewed distribution, the relationship between mean, median, and mode is: Mode = 3 × Median − 2 × Mean.

Types of Mode

A data set with one mode is called unimodal, two modes is bimodal, and more than two is multimodal.

No Mode Case

If all observations in a data set occur with the same frequency, the data set is said to have no mode.

Shortcut Trick

Ratio of radii of smaller to larger sphere (r : R) = 1 : 5.

Ratio of surface areas = (1)² : (5)² = 1 : 25.

Total surface area ratio units = 1 + 25 = 26 units.

Given, 4π × (Sum of r²) = 12584π ⇒ 26r² = 12584 ÷ 4 = 3146.

r² = 3146 ÷ 26 = 121 ⇒ r = 11 cm.

Diameter of larger sphere = 2 × (5 × 11) = 110 cm.

∴ The correct answer is 110 cm.

Alternate Method

Given:

Sum of surface areas of two spheres = 12584π cm².

Radius of larger sphere (R) = 5 × Radius of smaller sphere (r).

Formula Used:

Surface area of a sphere = 4πr²

Calculations:

⇒ 4πR² + 4πr² = 12584π

⇒ 4π(R² + r²) = 12584π

⇒ R² + r² = 3146

Substitute R = 5r:

⇒ (5r)² + r² = 3146

⇒ 25r² + r² = 3146

⇒ 26r² = 3146

⇒ r² = 121

⇒ r = √121 = 11 cm

⇒ Radius of larger sphere (R) = 5 × 11 = 55 cm

⇒ Diameter of larger sphere = 2 × R = 2 × 55 = 110 cm

∴ The correct answer is 110 cm.

Additional Information

Volume of a Sphere

The volume of a sphere is given by the formula V = (4/3)πr³, where r is the radius.

Hemisphere Properties

For a hemisphere, the Curved Surface Area (CSA) is 2πr² and the Total Surface Area (TSA) is 3πr².

Ratio of Attributes

If the ratio of radii of two spheres is a:b, then the ratio of their surface areas is a²:b² and the ratio of their volumes is a³:b³.

Shortcut Trick

Find the average (mid-point) of the two numbers: 36 ÷ 2 = 18.

The consecutive prime numbers immediately smaller and larger than 18 are 17 and 19.

Verify the sum: 17 + 19 = 36.

Calculate the product: 17 × 19 = (18 − 1) × (18 + 1) = 18² − 1² = 324 − 1 = 323.

∴ The correct answer is 323.

Alternate Method

Given: The sum of two distinct consecutive prime numbers is 36.

Method: Test pairs of prime numbers that add up to 36.

⇒ Let the primes be p₁ and p₂.

⇒ Prime numbers list: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31...

⇒ Testing pairs: 5 + 31 = 36 (Not consecutive), 7 + 29 = 36 (Not consecutive), 13 + 23 = 36 (Not consecutive).

⇒ Testing 17 and 19: They are consecutive primes.

⇒ Sum = 17 + 19 = 36 (Satisfied).

⇒ Product = 17 × 19 = 323.

∴ The correct answer is 323.

Additional Information

Prime Numbers

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Twin Primes

A twin prime is a prime number that is either 2 less or 2 more than another prime number (e.g., 17 and 19 are twin primes).

Consecutive Primes

Consecutive primes are prime numbers that follow one another in the sequence of prime numbers without any other primes in between.

Product of Primes

The product of two odd primes is always an odd number. If p₁ + p₂ is even, both must be odd (except when one is 2).

Shortcut Trick

Rate = 20%, Time = 3 years ⇒ Total Interest = 60%

Total Amount = 100% (Principal) + 60% (Interest) = 160%

Given, 160% = ₹5,040

Required SI for 2 years = 20% × 2 = 40%

Required SI = (5,040 ÷ 160) × 40 = 5,040 ÷ 4 = ₹1,260

∴ The correct answer is ₹1,260.

Alternate Method

Given: Amount after 3 years (A) = ₹5,040, Rate (R) = 20% per annum, Time (T₁) = 3 years.

Formula Used: Amount = P + (P × R × T) ÷ 100; SI = (P × R × T) ÷ 100

⇒ Amount = P[1 + (R × T₁) ÷ 100]

⇒ 5,040 = P[1 + (20 × 3) ÷ 100]

⇒ 5,040 = P[1 + 0.6] ⇒ 5,040 = 1.6P

⇒ Principal (P) = 5,040 ÷ 1.6 = ₹3,150

⇒ Simple Interest for 2 years (SI₂) = (P × R × T₂) ÷ 100

⇒ SI₂ = (3,150 × 20 × 2) ÷ 100

⇒ SI₂ = 3,150 × 0.4 = ₹1,260

∴ The correct answer is ₹1,260.

Additional Information

Simple Interest Formula

The basic formula is SI = (P × R × T) ÷ 100, where P is Principal, R is Rate per annum, and T is Time in years.

Relationship between Amount and Principal

Amount (A) = Principal (P) + Simple Interest (SI). In percentage terms, if P is 100%, A is (100 + R × T)%.

Constant Interest Property

In Simple Interest, the interest earned every year remains the same because it is always calculated on the original Principal.

Shortcut Trick

To find the lowest selling price, compare the product of the remaining percentages (100% − discount%).

Shop I factor: 0.11 × 0.32 = 0.0352

Shop II factor: 0.50 × 0.49 = 0.2450

Shop III factor: 0.75 × 0.89 = 0.6675

Shop IV factor: 0.06 × 0.51 × 0.79 = 0.024174

Since 0.024174 is the smallest multiplier, Shop IV offers the lowest price.

∴ The correct answer is IV.

Alternate Method

Assume the Marked Price (MP) of the cooker is ₹100 at all shops.

Formula: Selling Price (SP) = MP × (1 − d₁/100) × (1 − d₂/100) × ...

Shop I SP:

⇒ 100 × (1 − 0.89) × (1 − 0.68)

⇒ 100 × 0.11 × 0.32 = ₹3.52

Shop II SP:

⇒ 100 × (1 − 0.50) × (1 − 0.51)

⇒ 100 × 0.50 × 0.49 = ₹24.50

Shop III SP:

⇒ 100 × (1 − 0.25) × (1 − 0.11)

⇒ 100 × 0.75 × 0.89 = ₹66.75

Shop IV SP:

⇒ 100 × (1 − 0.94) × (1 − 0.49) × (1 − 0.21)

⇒ 100 × 0.06 × 0.51 × 0.79 = ₹2.4174

Comparing all values: ₹2.4174 < ₹3.52 < ₹24.50 < ₹66.75.

∴ The correct answer is IV.

Additional Information

Successive Discount Formula

For two discounts x% and y%, the effective discount is (x + y − xy/100)%. For more than two, apply the formula repeatedly or use the multiplier method.

Multiplying Factor Method

Final SP = MP × Product of (1 − d_i/100). The shop with the smallest product provides the highest discount and the lowest price.

Marked Price and Selling Price

Selling Price = Marked Price − Discount. Discount is always calculated on the Marked Price unless stated otherwise.

Shortcut Trick

Present ratio of Father and Son = 12 : 7 (Difference = 5 units)

Future ratio (after 4 years) = 5 : 3 (Difference = 2 units)

Equalize differences: Multiply present by 2 and future by 5 → Present = 24 : 14 | Future = 25 : 15

Ratio increase = 25 − 24 = 1 unit ⇒ 1 unit = 4 years

Sum of present ages = (24 + 14) units = 38 units = 38 × 4 = 152 years

∴ The correct answer is 152 years.

Alternate Method

Given: Present ratio = 12:7 and Ratio after 4 years = 5:3.

Let the present ages of Father and Son be 12x and 7x respectively.

⇒ According to the question, (12x + 4) ÷ (7x + 4) = 5 ÷ 3

⇒ 3 × (12x + 4) = 5 × (7x + 4)

⇒ 36x + 12 = 35x + 20

⇒ 36x − 35x = 20 − 12

⇒ x = 8

⇒ Father's present age = 12 × 8 = 96 years

⇒ Son's present age = 7 × 8 = 56 years

⇒ Sum of their present ages = 96 + 56 = 152 years

∴ The correct answer is 152 years.

Additional Information

Concept of Age Difference

The difference between the ages of two persons always remains constant, regardless of how many years pass.

Ratio Balancing Method

To solve age problems quickly, multiply the ratios so that the difference between the terms of each ratio is the same. This makes the unit change per year consistent.

Cross-Multiplication Rule

For any ratio a:b becoming c:d after k years: x = [k × (d − c)] ÷ (ad − bc), where x is the common multiplier for the initial ratio.

Shortcut Trick

Equalize decimal places: 15.0, 10.0, 0.4.

Remove decimals: 150, 100, 4.

Find LCM of 150, 100, and 4: 150 = 2 × 3 × 5², 100 = 2² × 5², 4 = 2².

LCM = 2² × 3 × 5² = 4 × 3 × 25 = 300.

Re-adjust decimal: 300 ÷ 10 = 30.

∴ The correct answer is 30.

Alternate Method

Given: The numbers are 15, 10, and 0.4.

Formula Used: LCM of fractions = LCM of Numerators ÷ HCF of Denominators.

Convert the decimal and integers into simple fractions:

⇒ 15 = 15/1

⇒ 10 = 10/1

⇒ 0.4 = 4/10 = 2/5

⇒ Numerators = 15, 10, 2

⇒ Denominators = 1, 1, 5

⇒ LCM(15, 10, 2) = 30

⇒ HCF(1, 1, 5) = 1

⇒ Required LCM = 30 ÷ 1 = 30

∴ The correct answer is 30.

Additional Information

LCM of Fractions

LCM of (a/b, c/d, e/f) = LCM(a, c, e) ÷ HCF(b, d, f). Ensure all fractions are simplified before calculating.

HCF of Fractions

HCF of (a/b, c/d, e/f) = HCF(a, c, e) ÷ LCM(b, d, f).

Property of Two Numbers

The product of two numbers is equal to the product of their LCM and HCF (n₁ × n₂ = LCM × HCF).

Shortcut Trick

Total Work = LCM(12, 6) = 12 units.

Efficiency of A = 1 unit/day and efficiency of B = 2 units/day.

Work done by A and B in 2 days = (1 + 2) × 2 = 6 units.

Remaining work = 12 − 6 = 6 units.

Time taken by B alone to complete the rest = 6 ÷ 2 = 3 days.

Total time taken = 2 days (together) + 3 days (alone) = 5 days.

∴ The correct answer is 5 days.

Alternate Method

Given: Time taken by A = 12 days and time taken by B = 6 days.

Formula Used: Total Work = Efficiency × Time; Total Time = Initial days + Remaining days.

⇒ Let Total Work be the LCM of 12 and 6 = 12 units.

⇒ Combined efficiency of A and B = 1 + 2 = 3 units/day.

⇒ Work completed by both in 2 days = 3 × 2 = 6 units.

⇒ Remaining Work to be done by B = 12 − 6 = 6 units.

⇒ Time taken by B to finish remaining work = 6 ÷ 2 = 3 days.

⇒ Whole work completed in = 2 + 3 = 5 days.

∴ The correct answer is 5 days.

Additional Information

Efficiency and Time Relation

Efficiency is inversely proportional to the Time taken to complete a work. If the ratio of time taken is x:y, the ratio of their efficiency will be y:x.

Work and Wages

Wages are generally distributed in the ratio of the Work Done by each person or the ratio of their Efficiencies if they work for the same duration.

Pipes and Cisterns

Similar logic applies to pipes where the Inlet pipe has positive efficiency and the Outlet pipe has negative efficiency.

Given:

{(20/36) ÷ (20/5)} ÷ (5/11 × 22/15 + 3/12) + 5/5 ÷ 33/21 of 21/5

Formula used:

The order of operations follows VBODMAS: Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Calculation:

⇒ {(20/36) ÷ (20/5)} ÷ (5/11 × 22/15 + 3/12) + 5/5 ÷ (33/21 × 21/5)

⇒ (20/36 × 5/20) ÷ (2/3 + 1/4) + 1 ÷ 33/5

⇒ (5/36) ÷ (11/12) + (1 × 5/33)

⇒ (5/36 × 12/11) + 5/33

⇒ 5/33 + 5/33

⇒ 10/33

∴ The correct answer is option (1).