Search Previous Year Questions

Shortcut Trick

Use the successive percentage formula: x + y + (xy / 100)

Here, x = 18% and y = 22%

Total Increase = 18 + 22 + (18 × 22 / 100)

Total Increase = 40 + 396 / 100 = 40 + 3.96 = 43.96%

∴ The correct answer is 43.96%.

Alternate Method

Given: Increase in 2021 = 18%; Increase in 2022 = 22%.

Assume the initial number of electric vehicles was 100.

⇒ Number after 2021 increase = 100 + (18% of 100) = 118

⇒ Number after 2022 increase = 118 + (22% of 118)

⇒ 22% of 118 = (22 × 118) / 100 = 25.96

⇒ Final number of vehicles = 118 + 25.96 = 143.96

⇒ Total percentage increase = 143.96 − 100 = 43.96%

∴ The correct answer is 43.96%.

Additional Information

Successive Percentage Formula

For two successive changes of a% and b%, the net change is (a + b + ab/100)%. Use positive values for increase and negative values for decrease.

Multiplying Factor Method

Final Value = Initial Value × (1 + r₁/100) × (1 + r₂/100). For 18% and 22% increases, the factors are 1.18 and 1.22 respectively.

Three Successive Changes

If there is a third change c%, calculate the net of a and b first, then use that result as a new 'a' with the third value 'c'.

Shortcut Trick

Decrease in population = 8,000 − 7,200 = 800

Fractional decrease = 800 ÷ 8,000 = 1/10

Percentage Decrease = 1/10 × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Initial Population of the town = 8,000

Final Population after one year = 7,200

Formula Used: Percentage Decrease = [(Original Value − New Value) ÷ Original Value] × 100

⇒ Total Decrease = 8,000 − 7,200 = 800

⇒ Percentage Decrease = (800 ÷ 8,000) × 100

⇒ Percentage Decrease = (1 ÷ 10) × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Percentage Change Rule

Percentage change is always calculated on the Initial Value (Base Value). Formula: [(Difference) ÷ (Initial Value)] × 100.

Successive Percentage Change

If a value changes by a% and then by b%, the net change is given by [a + b + (ab/100)]%.

Population Growth Formula

If population grows at a constant rate r% annually, Population after n years = P × (1 + r/100)ⁿ.

Given:

((22/5))¹⁸ × ((5/22))⁸ × ((22/5))⁶ = ((5/22))^9m+4

Formula used:

(a/b)ⁿ = (b/a)⁻ⁿ

a^x × a^y = a^x+y

Calculation:

To find the value of m, we first equate the bases on the left-hand side to match the right-hand side base of (5/22).

Using the reciprocal rule (a/b)ⁿ = (b/a)⁻ⁿ:

⇒ ((5/22))⁻¹⁸ × ((5/22))⁸ × ((5/22))⁻⁶ = ((5/22))^9m+4

Using the law of exponents (a^x × a^y = a^x+y) to combine the terms:

⇒ ((5/22))^{(−18 + 8 − 6)} = ((5/22))^9m+4

⇒ ((5/22))⁻¹⁶ = ((5/22))^9m+4

Since the bases are now equal, we compare the exponents:

⇒ −16 = 9m + 4

⇒ 9m = −16 − 4

⇒ 9m = −20

⇒ m = −20/9

∴ The correct answer is option (4).

Shortcut Trick

Efficiency ratio of X and Y = 15 : 20 = 3 : 4.

Total work = 20 × 3 = 60 units.

Work done by X and Y in 5 days = 5 × (3 + 4) = 35 units.

Remaining work = 60 − 35 = 25 units.

Time taken by Y to finish = 25 ÷ 4 = 6.25 days.

∴ The correct answer is 6.25 days.

Alternate Method

Given:

Time taken by X = 20 days

Time taken by Y = 15 days

Time worked together = 5 days

Formula used:

Efficiency = Total Work ÷ Time Taken

Calculations:

⇒ Total efficiency of X and Y = 3 + 4 = 7 units/day

⇒ Work completed by X and Y in 5 days = 7 × 5 = 35 units

⇒ Remaining work after X leaves = 60 − 35 = 25 units

⇒ Time taken by Y to complete 25 units = 25 ÷ 4

⇒ Time = 6.25 days

∴ The correct answer is 6.25 days.

Additional Information

Pipes and Cisterns

This concept follows the same principles as time and work, where an inlet pipe has positive efficiency and an outlet pipe has negative efficiency.

Work and Wages

Wages are distributed among workers in the ratio of the total work done by each, which is (Efficiency × Time).

Alternate Day Work

In such problems, calculate the work done in one full cycle (e.g., 2 days) and then find the total cycles needed to approach the total work.

Shortcut Trick

For a number to be divisible by 11, the difference between the sum of digits at odd positions and even positions must be 0 or a multiple of 11.

Sum of digits at odd places (7+2+3+2) = 14

Sum of digits at even places (9+4+p+5) = 18 + p

Difference = (18 + p) − 14 = p + 4

For divisibility, p + 4 = 11 ⇒ p = 7.

∴ The correct answer is 7.

Alternate Method

Given: The 8-digit number is 79243p25 and it is divisible by 11.

Formula Used: A number is divisible by 11 if |(Sum of digits at odd places) − (Sum of digits at even places)| is 0, 11, 22, etc.

Calculations:

⇒ Sum of digits at odd places = 7 + 2 + 3 + 2 = 14

⇒ Sum of digits at even places = 9 + 4 + p + 5 = 18 + p

⇒ Difference = (Sum of even places) − (Sum of odd places)

⇒ Difference = (18 + p) − 14 = p + 4

⇒ For the number to be divisible by 11, (p + 4) must be a multiple of 11.

⇒ Since p is a single digit (0−9), the only possible multiple is 11.

⇒ p + 4 = 11

⇒ p = 11 − 4 = 7

∴ The correct answer is 7.

Additional Information

Divisibility by 3 and 9

A number is divisible by 3 (or 9) if the sum of all its digits is divisible by 3 (or 9).

Divisibility by 8

A number is divisible by 8 if its last three digits are divisible by 8.

Divisibility by 11 (Alternative)

Subtract the last digit from the remaining number. If the result is divisible by 11, the original number is too.

Shortcut Trick

Net change in sum = New value − Old value = 20 − 10 = +10

Net change in mean = Net change in sum ÷ Total count = 10 ÷ 5 = +2

Since the result is positive, the mean increases by the calculated value.

∴ The correct answer is It increases by 2.

Alternate Method

Given: Total numbers (n) = 5, Initial Arithmetic Mean = 20.

Formula Used: Total Sum = Mean × Number of observations

⇒ Initial total sum = 20 × 5 = 100

⇒ One number changes from 10 to 20.

⇒ New total sum = 100 − 10 + 20 = 110

⇒ New Arithmetic Mean = 110 ÷ 5 = 22

⇒ Difference in mean = 22 − 20 = +2

∴ The correct answer is It increases by 2.

Additional Information

Properties of Arithmetic Mean

If every observation in a data set is increased or decreased by a constant value k, the mean also increases or decreases by k.

Effect of Replacement

When a value x is replaced by y, the change in the mean is given by (y − x) ÷ n, where n is the total number of observations.

Weighted Average

The combined mean of two groups with sizes n₁, n₂ and means x₁, x₂ is (n₁x₁ + n₂x₂) ÷ (n₁ + n₂).

Shortcut Trick

Since the filling rate of the second tap and the emptying rate of the third tap are identical (both 17 hr), they cancel each other out.

Net filling rate = 1/4 + 1/17 − 1/17 = 1/4 of the cistern per hour.

Time to fill the full cistern = 4 hours.

Time to fill one-fourth of the cistern = 1/4 × 4 = 1 hour.

∴ The correct answer is 1 hr.

Alternate Method

Given: Filling time for Tap A = 4 hr, Tap B = 17 hr, and emptying time for Tap C = 17 hr.

Total capacity (Work) = LCM of (4, 17, 17) = 68 units.

⇒ Total net efficiency = 17 + 4 − 4 = 17 units/hr.

⇒ Required work (one-fourth of cistern) = 68 ÷ 4 = 17 units.

⇒ Time required = Required Work ÷ Net Efficiency

⇒ Time = 17 ÷ 17 = 1 hour.

∴ The correct answer is 1 hr.

Additional Information

Net Efficiency in Pipes and Cisterns

The net rate of filling is the sum of the rates of inlet pipes minus the sum of the rates of outlet pipes.

Relationship between Time and Work

Total Work = Efficiency × Time. For a cistern, work is represented by the volume or capacity, often calculated as the LCM of time intervals.

Concept of Negative Work

An emptying tap performs negative work. If a tap empties a tank in 'n' hours, its efficiency is taken as −1/n per hour.

Shortcut Trick

LCM of numbers = (LCM of ratio parts) × HCF

Ratio of numbers = 3 : 2 : 11

LCM of ratio parts (3, 2, 11) = 3 × 2 × 11 = 66 (since they are co-prime)

66 units = 3564

1 unit (HCF) = 3564 ÷ 66 = 54

∴ The correct answer is 54.

Alternate Method

Given:

Ratio of three numbers = 3 : 2 : 11

LCM of the numbers = 3564

Formula Used:

If the numbers are ax, bx, and cx, where x is the HCF, then their LCM = x × LCM(a, b, c).

Calculations:

Let the three numbers be 3x, 2x, and 11x, where x is the HCF.

⇒ LCM(3x, 2x, 11x) = 3564

⇒ x × LCM(3, 2, 11) = 3564

⇒ x × (3 × 2 × 11) = 3564

⇒ 66x = 3564

⇒ x = 3564 ÷ 66

⇒ x = 54

∴ The correct answer is 54.

Additional Information

HCF and LCM Relation

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 × LCM.

Co-prime Numbers

If a set of numbers are co-prime (meaning they have no common factors other than 1), their HCF is 1 and their LCM is the product of the numbers.

HCF and LCM Properties

The HCF of a group of numbers is always a factor of their LCM. HCF is the largest number that divides all given numbers exactly.

Shortcut Trick

Combine the ratios by equating the common term B: A : B : C = (15 × 14) : (17 × 14) : (17 × 19) = 210 : 238 : 323.

Total units = 210 + 238 + 323 = 771 units.

Given 771 units = 1542 runs ⇒ 1 unit = 2 runs.

A's score = 210 × 2 = 420.

∴ The correct answer is 420.

Alternate Method

Given:

Ratio A : B = 15 : 17

Ratio B : C = 14 : 19

Total runs scored = 1542

Formula Used:

If A : B = x : y and B : C = p : q, then A : B : C = (x × p) : (y × p) : (y × q)

Calculations:

⇒ To merge the ratios, find the LCM of B's values (17 and 14), which is 238.

⇒ A : B = (15 × 14) : (17 × 14) = 210 : 238

⇒ B : C = (14 × 17) : (19 × 17) = 238 : 323

⇒ Combined Ratio A : B : C = 210 : 238 : 323

⇒ Sum of parts = 210 + 238 + 323 = 771

⇒ Value of 1 part = 1542 ÷ 771 = 2

⇒ Runs scored by A = 210 × 2 = 420

∴ The correct answer is 420.

Additional Information

Merging Ratios

To combine ratios like A:B and B:C, the common variable (B) must have the same value in both. Multiply each ratio by the value of B from the other ratio.

Unitary Method in Ratios

Once the total number of parts is known, the value of one part is calculated as (Total Value ÷ Total Parts). Individual shares are then (Individual Parts × Value of 1 part).

Inverse Ratio

The inverse ratio of a : b is 1/a : 1/b, which simplifies to b : a.

Shortcut Trick

Check the sum of the bases: 22 + (−16) + (−6) = 22 − 22 = 0.

Apply the algebraic identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Result = 3 × 22 × (−16) × (−6) = 66 × 96 = 6336.

∴ The correct answer is 6336.

Alternate Method

Given: The expression to evaluate is 22³ + (−16)³ + (−6)³.

Formula Used: If a + b + c = 0, then the identity a³ + b³ + c³ = 3abc holds true.

⇒ Assign values: a = 22, b = −16, and c = −6.

⇒ Verify the condition: a + b + c = 22 + (−16) + (−6) = 22 − 22 = 0.

⇒ Substitute into the identity: 22³ + (−16)³ + (−6)³ = 3 × 22 × (−16) × (−6).

⇒ Multiply the constants: 3 × 22 = 66 and (−16) × (−6) = 96.

⇒ Final calculation: 66 × 96 = 66 × (100 − 4) = 6600 − 264 = 6336.

∴ The correct answer is 6336.

Additional Information

General Algebraic Identity

The complete identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Special Conditions

The expression a³ + b³ + c³ = 3abc occurs when either a + b + c = 0 or a = b = c.

Negative Base Powers

Since the exponent 3 is an odd number, the cube of a negative number remains negative, i.e., (−x)³ = −x³.