SSC CGL Tier 2 Quant Previous Year Paper (Held On: 12 Jan 2017)

Shortcut Trick

Check options to satisfy: x² − 21x = 100

Try Option (1), x = 25:

25² − 21 × 25 = 625 − 525 = 100

Since the condition is satisfied, 25 is the correct number.

∴ The correct answer is 25.

Alternate Method

Given: Let the positive number be x.

According to the question, 21 times the number is 100 less than its square.

⇒ 21x = x² − 100

⇒ x² − 21x − 100 = 0

⇒ x² − 25x + 4x − 100 = 0

⇒ x(x − 25) + 4(x − 25) = 0

⇒ (x − 25)(x + 4) = 0

⇒ x = 25 or x = −4

Since the problem specifies a positive number, we take x = 25.

∴ The correct answer is 25.

Additional Information

Quadratic Roots Formula

Roots of a quadratic equation ax² + bx + c = 0 can be found using x = (−b ± √(b² − 4ac)) ÷ 2a.

Sum and Product of Roots

For a quadratic equation, Sum of roots = −b/a and Product of roots = c/a.

Nature of Roots

The term (b² − 4ac) is called the discriminant (D). If D > 0, the roots are real and distinct; if D = 0, roots are real and equal.

Shortcut Trick

To find the greatest equal piece size without leaving any extra pipe, calculate the Highest Common Factor (HCF) of their lengths.

Convert lengths to centimeters for easy integer calculation: 1.5 m = 150 cm and 1.2 m = 120 cm.

The HCF of 150 and 120 is clearly 30.

Convert the result back to meters: 30 cm = 0.3 m.

∴ The correct answer is 0.3 m.

Alternate Method

Given:

Length of the first pipe = 1.5 m

Length of the second pipe = 1.2 m

Formula Used:

The greatest possible length of equal pieces cut from both pipes = HCF of their lengths.

HCF of Fractions = (HCF of Numerators) ÷ (LCM of Denominators)

Pipe 1: 1.5 m

0.3 m 0.3 m 0.3 m 0.3 m 0.3 m

Pipe 2: 1.2 m

0.3 m 0.3 m 0.3 m 0.3 m

Maximum equal piece cut = HCF(1.5, 1.2) = 0.3 m

Calculations:

⇒ First, express the given decimal lengths as fractions to use the formula.

⇒ 1.5 = 15/10 and 1.2 = 12/10

⇒ HCF of numerators (15, 12) = 3

⇒ LCM of denominators (10, 10) = 10

⇒ HCF of fractions = 3 / 10 = 0.3 m

∴ The correct answer is 0.3 m.

Additional Information

LCM of Fractions

LCM of Fractions = (LCM of Numerators) ÷ (HCF of Denominators)

HCF of Decimals

Make the number of decimal places equal in all given numbers. Find the HCF treating them as regular integers, then place the decimal point backward accordingly.

Application of HCF

HCF is applied when a practical problem asks to split things into smaller identical sections, equally distribute items, or find the maximum possible unit size without leaving a remainder.

Given: Total army men = 36562 Formula used: Unused men = Total army men - Side2 Calculation: ⇒ 1912 = 36481 ⇒ 1922 = 36864 ⇒ 1912 ≤ 36562 < 1922 ⇒ Unused men = 36562 - 36481 ⇒ Unused men = 81 ∴ The correct answer is option (3).

Shortcut Trick

Perform direct division of 756896 by 11 to find the remainder.

756896 ÷ 11 ⇒ Remainder = 8.

Smallest number to be added = Divisor − Remainder = 11 − 8 = 3.

∴ The correct answer is 3.

Alternate Method

Given Number: 756896

Formula: Divisibility rule of 11 states that (Sum of digits at odd places) − (Sum of digits at even places) must be 0 or a multiple of 11.

⇒ Sum of digits at odd positions (from right: 6, 8, 5) = 6 + 8 + 5 = 19

⇒ Sum of digits at even positions (from right: 9, 6, 7) = 9 + 6 + 7 = 22

⇒ Current Difference = 22 − 19 = 3

⇒ To make the number divisible by 11, we need to add a number x to the total sum such that the new difference becomes 0.

⇒ (19 + x) − 22 = 0 ⇒ x − 3 = 0 ⇒ x = 3.

∴ The correct answer is 3.

Additional Information

Divisibility Rule of 11

A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

Finding Number to Add/Subtract

To make N divisible by D: the smallest number to subtract is Remainder(N÷D), and the smallest number to add is [D − Remainder(N÷D)].

Divisibility Rule of 7

Double the last digit of the number and subtract it from the remaining number. If the result is divisible by 7, the original number is also divisible by 7.

Shortcut Trick

Correct decimal value = 1 − (0.25 + 0.20) = 0.55

Error in calculation = |0.55 − 0.45| = 0.10

Required Percentage Error = (Error ÷ Correct Value) × 100

⇒ (0.10 ÷ 0.55) × 100 = (2/11) × 100 = 200/11%

∴ The correct answer is (200/11)%.

Alternate Method

Given:

Unity = 1

Fractions to be summed = 1/4 and 1/5

Boy's Answer = 0.45

Formula Used:

Percentage Error = (Error ÷ True Value) × 100

Calculations:

⇒ Sum of fractions = 1/4 + 1/5

⇒ Sum = (5 + 4) ÷ 20 = 9/20

⇒ Correct value = 1 − 9/20

⇒ Correct value = 11/20 = 0.55

⇒ Error = Correct Value − Boy's Answer

⇒ Error = 0.55 − 0.45 = 0.10

⇒ Percentage Error = (0.10 ÷ 0.55) × 100

⇒ Percentage Error = (10 ÷ 55) × 100 = (2 ÷ 11) × 100

⇒ Percentage Error = 200/11%

∴ The correct answer is (200/11)%.

Additional Information

Percentage Error Formula

The percentage error is calculated as [|Approximate Value − Exact Value| ÷ Exact Value] × 100.

Unity in Mathematics

The term 'Unity' always refers to the number 1. In fractions, unity represents the whole part (100%).

Fraction to Decimal Conversion

To convert a fraction to a decimal, divide the numerator by the denominator. For example, 11/20 = (11 × 5) / (20 × 5) = 55/100 = 0.55.

Shortcut Trick

Identify values: Wings = 2, Fingers = 5, Tricycle wheels = 3.

Calculate first number: (2 + 2 × 5) ÷ 3 = (2 + 10) ÷ 3 = 4.

Calculate second number: 48 ÷ 4 = 12.

∴ The correct answer is 12.

Alternate Method

Given:

Product of two numbers = 48

First number expression: (Wings + 2 × Fingers) ÷ Tricycle wheels

Formula Used:

Number₁ × Number₂ = Product

Calculations:

⇒ Number of wings of a bird = 2

⇒ Number of fingers on a hand = 5

⇒ Number of wheels of a tricycle = 3

⇒ First number = (2 + 2 × 5) ÷ 3

⇒ First number = (2 + 10) ÷ 3 = 12 ÷ 3 = 4

⇒ Second number = Product ÷ First number

⇒ Second number = 48 ÷ 4 = 12

∴ The correct answer is 12.

Additional Information

BODMAS Rule

It specifies the order of operations: Brackets, Orders, Division/Multiplication, and Addition/Subtraction.

Multiplicative Property

The product of any two numbers x and y is given by x × y. If the product P and one number x are known, the other number y = P ÷ x.

Division Identity

Dividing a number by itself results in 1, and dividing any number by 1 results in the number itself.

Shortcut Trick

Check the second condition: if Natu gives 10 oranges to Buchku, they become equal.

This implies Natu has 20 more oranges than Buchku initially (Difference = 20).

Only Option (2) where Natu = 70 and Buchku = 50 has a difference of 20.

Verify condition 1: If Buchku gives 10 to Natu, Natu = 80 and Buchku = 40.

Since 80 is exactly twice 40, both conditions are perfectly satisfied.

∴ The correct answer is 70, 50.

Alternate Method

Given:

Let the number of oranges with Natu be N and with Buchku be B.

Formula Used:

Formation of linear equations in two variables: x + a = k(y − a).

Calculations:

According to the second condition (Natu gives 10 to Buchku):

⇒ N − 10 = B + 10

⇒ N − B = 20 ... (i)

According to the first condition (Buchku gives 10 to Natu):

⇒ N + 10 = 2(B − 10)

⇒ N + 10 = 2B − 20

⇒ N − 2B = −30 ... (ii)

Subtracting equation (ii) from equation (i):

⇒ (N − B) − (N − 2B) = 20 − (−30)

⇒ N − B − N + 2B = 20 + 30

⇒ B = 50

Substitute B = 50 in equation (i):

⇒ N − 50 = 20

⇒ N = 70

∴ The correct answer is 70, 50.

Additional Information

Linear Equations in Two Variables

Equations of the form ax + by + c = 0 require two independent conditions to find unique values for x and y.

Elimination Method

Subtracting or adding equations to eliminate one variable is often the fastest way to solve simultaneous equations.

Option Verification Technique

In competitive exams, substituting the given options into the logic of the problem can save significant time compared to algebraic derivation.

Shortcut Trick

Side of square (a) = √Area = √1127.6164 = 33.58 m

Total distance for one round = Perimeter = 4 × 33.58 = 134.32 m

Walking speed = 2.45 m/min

Time = 134.32 ÷ 2.45 ≈ 54.82 min

∴ The correct answer is 54.82 min.

Alternate Method

Given:

Area of square playground = 1127.6164 m²

Speed of the man = 2 + 9/20 = 2.45 m/min

Formula Used:

Area of Square = side²

Perimeter of Square = 4 × side

Time = Distance ÷ Speed

Calculations:

⇒ Area = a² = 1127.6164

⇒ a = √1127.6164 = 33.58 m

⇒ Perimeter (Distance) = 4 × a

⇒ Distance = 4 × 33.58 = 134.32 m

⇒ Speed = 2 + 0.45 = 2.45 m/min

⇒ Time = 134.32 ÷ 2.45

⇒ Time = 54.8244... min

⇒ Time ≈ 54.82 min

∴ The correct answer is 54.82 min.

Additional Information

Properties of a Square

A square is a quadrilateral where all sides are equal and all angles are 90°. Diagonal d = a√2.

Relationship between Speed, Distance and Time

Distance = Speed × Time. To convert km/h to m/s, multiply by 5/18. To convert m/min to m/s, divide by 60.

Square Root Extraction

For decimals, pair digits from the decimal point outwards. For 1127.6164, pairs are 11, 27 and 61, 64.

Shortcut Trick

LCM of 48, 72, and 108 determines the common interval.

LCM(48, 72, 108) = 432 seconds.

Convert 432 seconds to minutes ⇒ 432 ÷ 60 = 7 minutes 12 seconds.

Next beep time = 10:00:00 + 00:07:12 = 10:07:12 hrs.

∴ The correct answer is 10:07:12 hrs.

Alternate Method

Given:

Beep intervals: 48 sec, 72 sec, 108 sec.

Starting time: 10:00:00 a.m.

Formula Used:

Next simultaneous event = Initial time + LCM of intervals.

Calculations:

Find LCM using prime factorization:

⇒ 48 = 2⁴ × 3¹

⇒ 72 = 2³ × 3²

⇒ 108 = 2² × 3³

⇒ LCM = 2⁴ × 3³ = 16 × 27 = 432 seconds.

Convert 432 seconds into minutes:

⇒ 432 seconds = 420 seconds + 12 seconds

⇒ (420 ÷ 60) minutes + 12 seconds = 7 minutes 12 seconds.

Add the interval to the starting time:

⇒ 10:00:00 + 0:07:12 = 10:07:12 hrs.

∴ The correct answer is 10:07:12 hrs.

Additional Information

LCM for Recurring Events

LCM (Least Common Multiple) is used to find the first time multiple events with different periodic intervals will coincide again.

Time Unit Conversion

To convert seconds into minutes, divide by 60. The quotient represents minutes and the remainder represents the remaining seconds.

HCF vs LCM

Use LCM for "When will they meet next" or "Minimum distance to cover". Use HCF for "Maximum size of a container" or "Longest tape to measure".

Shortcut Trick

After the transfer, the oranges become equal in both baskets.

Total oranges = 640, so each basket finally has 640 ÷ 2 = 320 oranges.

In the first basket, 1/5^th was taken away, leaving 1 − 1/5 = 4/5^th of the oranges.

Let initial oranges in the first basket be X.

4/5 × X = 320 ⇒ X = 320 × 5 ÷ 4 = 400 oranges.

∴ The correct answer is 400 oranges.

Alternate Method

Given:

Total oranges = 640

Transfer amount = 1/5^th of the first basket

Formula Used:

Sum of items = Average × Number of items

Remaining Fraction = 1 − Given Fraction

Calculations:

Let the number of oranges in the first basket be x and in the second basket be y.

⇒ x + y = 640 ... (i)

According to the question, if 1/5^th of x is moved to y, they become equal:

⇒ x − x/5 = y + x/5

⇒ 4x/5 = y + x/5

⇒ y = 4x/5 − x/5

⇒ y = 3x/5

Substitute the value of y in equation (i):

⇒ x + 3x/5 = 640

⇒ (5x + 3x)/5 = 640

⇒ 8x/5 = 640

⇒ x = (640 × 5) ÷ 8

⇒ x = 80 × 5

⇒ x = 400

∴ The correct answer is 400 oranges.

Additional Information

Ratio and Proportion Method

From the equality condition 4x/5 = y + x/5, we get y = 3x/5, which means the ratio of oranges in Basket 1 to Basket 2 is x:y = 5:3.

Sum of Ratios

If the ratio is a:b and total is T, then the value of the first part is [a ÷ (a + b)] × T. Here, [5 ÷ (5 + 3)] × 640 = 400.

Percentage Application

A fraction of 1/5 is equivalent to 20%. The problem states that giving 20% of the first basket makes them equal.

Shortcut Trick Time for P = 10 × 4 = 40 days Time for Q = 40 × (100 ÷ 40) = 100 days Time for R = 13 × 3 = 39 days Comparison: R (39 days) < P (40 days) < Q (100 days) ∴ The correct answer is R.

Alternate Method Given: P can do 1/4 work in 10 days. Q can do 40% work in 40 days. R can do 1/3 work in 13 days.

Person Days Total Work Efficiency

P 40 7800 195

Q 100 78

R 39 200

⇒ P's total time = 10 ÷ (1/4) = 40 days. ⇒ Q's total time = 40 ÷ (40/100) = 40 × (100/40) = 100 days. ⇒ R's total time = 13 ÷ (1/3) = 39 days. ⇒ Comparing the time taken by each: 39 days < 40 days < 100 days. ⇒ R takes the least time, so R completes the work first. ∴ The correct answer is R.

Additional Information Relationship between Time and Efficiency Efficiency is inversely proportional to the time taken to complete a task (Efficiency = 1 ÷ Time). Higher efficiency means less time required. Total Time from Fractional Work If a person completes a fraction 'x/y' of work in 'D' days, then the total time 'T' to complete the whole work is T = D × (y/x). Work Comparison To determine who finishes first among several individuals, calculate the time each would take for 100% of the work and identify the minimum value.

Shortcut Trick

Apply the Chain Rule: M₁ × D₁ × H₁ = M₂ × D₂ × H₂

Substitute the values: 4 × 8 × 7 = M₂ × 4 × 8

Cancel out common factors (4 × 8) from both sides: M₂ = 7

∴ The correct answer is 7 men.

Alternate Method

Given:

Initial condition: M₁ = 4 men, D₁ = 8 days, H₁ = 7 hours/day

Final condition: D₂ = 4 days, H₂ = 8 hours/day, M₂ = ?

Formula used:

Product of workers, time, and days remains constant for the same work: M₁ × D₁ × H₁ = M₂ × D₂ × H₂

⇒ 4 × 8 × 7 = M₂ × 4 × 8

⇒ 224 = M₂ × 32

⇒ M₂ = 224 ÷ 32

⇒ M₂ = 7

∴ The correct answer is 7 men.

Additional Information

General Chain Rule Formula

The complete formula for work comparison is (M₁ × D₁ × H₁ × E₁) ÷ W₁ = (M₂ × D₂ × H₂ × E₂) ÷ W₂, where E is efficiency and W is work.

Inverse Proportion in Work

The number of men is inversely proportional to the number of days or hours required to complete a fixed amount of work.

Work and Wages

Wages are always distributed in the ratio of the actual work done by each person, which is the product of their efficiency and time spent.

Shortcut Trick

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

Remaining days = 60 − 32 = 28 days; Remaining work = 1 − 2/5 = 3/5

(35 × 32) / (2/5) = (M₂ × 28) / (3/5)

M₂ = (35 × 32 × 3) / (28 × 2) = 60 persons

Additional persons required = 60 − 35 = 25

∴ The correct answer is 25.

Alternate Method

Given:

Initial persons (M₁) = 35

Initial days (D₁) = 32

Initial work (W₁) = 2/5

Remaining days (D₂) = 60 − 32 = 28

Remaining work (W₂) = 1 − 2/5 = 3/5

Formula Used:

(M₁ × D₁) / W₁ = (M₂ × D₂) / W₂

Calculations:

⇒ (35 × 32) / (2/5) = (M₂ × 28) / (3/5)

⇒ (35 × 32) / 2 = (M₂ × 28) / 3

⇒ 17.5 × 32 = (M₂ × 28) / 3

⇒ 560 = (M₂ × 28) / 3

⇒ M₂ = (560 × 3) / 28

⇒ M₂ = 20 × 3 = 60 persons

⇒ Number of extra persons to be engaged = M₂ − M₁

⇒ Extra persons = 60 − 35 = 25 persons

∴ The correct answer is 25.

Additional Information

General MDH Rule

The relationship between Men (M), Days (D), Hours (H), Efficiency (E) and Work (W) is given by (M₁×D₁×H₁×E₁)/W₁ = (M₂×D₂×H₂×E₂)/W₂.

Inverse Proportionality

For a constant amount of work, the number of persons is inversely proportional to the time taken (M ∝ 1/D).

Work and Wages

Wages are always distributed in the ratio of the work done by each individual, which is the product of their efficiency and time (Work = E × T).

Shortcut Trick

Time for 5 children = 20 days ÷ 5 = 4 days.

Time for 4 men = 2 × 4 days = 8 days ⇒ 1 man takes 32 days; 1 woman takes 64 days.

Work rate of 12 men = 12/32 = 3/8; 10 children = 10/20 = 1/2; 8 women = 8/64 = 1/8.

Total work rate = 3/8 + 1/2 + 1/8 = 8/8 = 1 job/day.

∴ The correct answer is 1 day.

Alternate Method

Given:

Time for 1 child = 20 days.

Time for 4 men = 2 × (Time for 5 children).

Efficiency of 1 man = 2 × (Efficiency of 1 woman).

Calculations:

⇒ Total work (if 1 child does 1 unit/day) = 1 × 20 = 20 units.

⇒ Efficiency of 1 child = 1 unit/day.

⇒ Efficiency of 5 children = 5 × 1 = 5 units/day.

⇒ Time for 5 children = 20 ÷ 5 = 4 days.

⇒ Time for 4 men = 2 × 4 = 8 days.

⇒ Efficiency of 4 men = 20 ÷ 8 = 2.5 units/day.

⇒ Efficiency of 1 man = 2.5 ÷ 4 = 0.625 units/day.

⇒ Efficiency of 1 woman = 0.625 ÷ 2 = 0.3125 units/day.

⇒ Total efficiency (12M + 10C + 8W) = (12 × 0.625) + (10 × 1) + (8 × 0.3125).

⇒ Total efficiency = 7.5 + 10 + 2.5 = 20 units/day.

⇒ Total time = Total Work ÷ Total Efficiency = 20 ÷ 20 = 1 day.

∴ The correct answer is 1 day.

Additional Information

Efficiency and Time Relation

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

Work Equivalence Formula

M₁D₁H₁E₁ / W₁ = M₂D₂H₂E₂ / W₂, where M is persons, D is days, H is hours, E is efficiency, and W is work.

Combined Work Rate

When multiple groups work together, their individual work rates (1/Time) are added to find the combined work rate per unit time.

Shortcut Trick

Total work = LCM(8, 12, 40/3) = 120 units.

Total Efficiency (A+B+C) = 120 ÷ 8 = 15.

Efficiency (A+C) = 10 → B's Efficiency = 15 − 10 = 5.

Efficiency (A+B) = 9 → C's Efficiency = 15 − 9 = 6.

A's Efficiency = 15 − (5 + 6) = 4.

Ratio of money = Ratio of efficiency = 4 : 5 : 6.

∴ The correct answer is 4 : 5 : 6.

Alternate Method

Given:

Days for (A+B+C) = 8 days

Days for (A+C) = 12 days

Days for (A+B) = 13 1/3 = 40/3 days

Formula Used:

Efficiency = Total Work ÷ Time

Wages Ratio = Efficiency Ratio (when time worked is same for all)

Calculations:

⇒ Efficiency of B = Eff(A+B+C) − Eff(A+C)

⇒ Efficiency of B = 15 − 10 = 5 units/day

⇒ Efficiency of C = Eff(A+B+C) − Eff(A+B)

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

⇒ Efficiency of A = Eff(A+B+C) − (Eff of B + Eff of C)

⇒ Efficiency of A = 15 − (5 + 6) = 4 units/day

⇒ Ratio of shares A : B : C = 4 : 5 : 6

∴ The correct answer is 4 : 5 : 6.

Additional Information

Wages Distribution Rule

Wages are distributed in the ratio of the work done by each person. If they work for the same duration, the ratio of wages is equal to the ratio of their efficiencies.

Inverse Relation of Time and Efficiency

Efficiency is inversely proportional to the time taken to complete a task. Efficiency = 1 ÷ Time, provided the total work remains constant.

Combined Efficiency

When multiple people work together, their combined efficiency is the sum of their individual efficiencies: E_Total = E₁ + E₂ + ... + E_n.

Shortcut Trick

Remaining work fraction for (A+B) = (12 − 5) ÷ 12 = 7/12.

A completes this 7/12 part of the work in 14 days.

Time taken by A for the full work (1 unit) = 14 × (12 ÷ 7) = 24 days.

∴ The correct answer is 24 days.

Alternate Method

Given: Time for (A + B) = 12 days, Work duration together = 5 days, Remaining work by A = 14 days.

Formula: Total Work = LCM of given durations; Efficiency = Work ÷ Time.

⇒ Let Total Work = LCM of 12 and 14 = 84 units.

⇒ Efficiency of (A + B) = 84 ÷ 12 = 7 units/day.

⇒ Work completed by A and B in 5 days = 7 × 5 = 35 units.

⇒ Remaining work = 84 − 35 = 49 units.

⇒ Efficiency of A = Remaining work ÷ Time taken = 49 ÷ 14 = 3.5 units/day.

⇒ Time taken by A to complete full work alone = Total work ÷ Efficiency of A = 84 ÷ 3.5 = 24 days.

∴ The correct answer is 24 days.

Additional Information

Work and Wages

Wages are always distributed in the ratio of the efficiency of the workers or the ratio of the work done by them.

Pipes and Cisterns

This concept is similar to Time & Work, where an inlet pipe does positive work (+) and an outlet pipe does negative work (−).

Man-Days Formula

For different groups working on different tasks: (M₁×D₁×H₁)/W₁ = (M₂×D₂×H₂)/W₂, where M=Men, D=Days, H=Hours, W=Work.

Shortcut Trick

Apply successive discounts directly on the marked price using decimal multipliers.

Selling Price = Marked Price × (100 − d₁)% × (100 − d₂)%

Selling Price = 200 × 0.85 × 0.96

Selling Price = 170 × 0.96 = 163.2

∴ The correct answer is ₹163.2.

Alternate Method

Given: Marked Price (MP) = ₹200, First Discount = 15%, Second (Cash) Discount = 4%.

Formula: Final SP = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ First discount = 15% of ₹200 = (15 ÷ 100) × 200 = ₹30

⇒ Reduced price after first discount = 200 − 30 = ₹170

⇒ Second discount (on reduced price) = 4% of ₹170 = (4 ÷ 100) × 170 = ₹6.80

⇒ Final amount to be paid = 170 − 6.80 = ₹163.20

∴ The correct answer is ₹163.2.

Additional Information

Successive Discount Formula

Two successive discounts of x% and y% are equivalent to a single discount of [x + y − (xy ÷ 100)]%.

Net Change Method

Final Value = Initial Value × [1 ± (r₁ ÷ 100)] × [1 ± (r₂ ÷ 100)], where r is the percentage increase or decrease.

Cash Discount

A cash discount is often calculated on the net price after all trade discounts have been applied.

Shortcut Trick

CP : MP ratio = (100 − Discount%) : (100 + Profit%)

CP : MP = (100 − 10) : (100 + 20) = 90 : 120 = 3 : 4

Given CP = 3 units = ₹600 → 1 unit = ₹200

Marked Price (MP) = 4 units = 4 × ₹200 = ₹800

∴ The correct answer is ₹800.

Alternate Method

Given: Cost Price (CP) = ₹600, Profit = 20%, Discount = 10%

Formula Used: SP = CP × (100 + Profit%)/100 | MP = (SP × 100) / (100 − Discount%)

⇒ Profit = 20% of 600 = (20 ÷ 100) × 600 = ₹120

⇒ Selling Price (SP) = CP + Profit = 600 + 120 = ₹720

⇒ Let Marked Price be MP. Then, SP = MP × (100 − 10) ÷ 100

⇒ 720 = MP × 90 ÷ 100

⇒ MP = (720 × 100) ÷ 90 = 8 × 100 = ₹800

∴ The correct answer is ₹800.

Additional Information

CP and MP Relation

The direct ratio between Cost Price and Marked Price is CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%).

Markup Percentage

Markup % = [(MP − CP) ÷ CP] × 100. It can also be found using [(Profit% + Discount%) ÷ (100 − Discount%)] × 100.

Selling Price Formulas

SP = CP × (1 + P/100) when profit is given, and SP = MP × (1 − D/100) when discount is given.

Shortcut Trick

Price after first 15% discount on ₹1500 → 1500 − 225 = ₹1275.

The customer pays ₹1173, so second discount amount → 1275 − 1173 = ₹102.

Second discount rate → (102 ÷ 1275) × 100 = 8%.

∴ The correct answer is 8%.

Alternate Method

Given: Quoted Price (MP) = ₹1500, Final Selling Price (SP) = ₹1173, First Discount (d₁) = 15%.

Formula: SP = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ Price after first discount = 1500 − (15% of 1500)

⇒ Price after first discount = 1500 − 225 = ₹1275

⇒ Let the second discount be x%.

⇒ 1173 = 1275 × (100 − x) ÷ 100

⇒ 1173 ÷ 1275 = (100 − x) ÷ 100

⇒ 0.92 = (100 − x) ÷ 100

⇒ 92 = 100 − x

⇒ x = 100 − 92

⇒ x = 8%

∴ The correct answer is 8%.

Additional Information

Successive Discount Formula

The single equivalent discount for two successive discounts x% and y% is given by (x + y − (x × y)/100)%.

Profit and Loss Relation

Selling Price (SP) = Marked Price (MP) × (100 − Discount%)/100. Also, SP = Cost Price (CP) × (100 + Profit%)/100.

Marked Price (MP)

It is the price printed on the label, also known as list price or quoted price. Discounts are always calculated on the MP.

Shortcut Trick

When a dealer cheats x% while buying, he gets (100 + x) units for the price of 100.

When he cheats x% while selling, he gives (100 − x) units for the price of 100.

Total Gain % = [ (Goods Received − Goods Delivered) ÷ Goods Delivered ] × 100

Total Gain % = [ (100 + x − (100 − x)) ÷ (100 − x) ] × 100 = [ 2x ÷ (100 − x) ] × 100 %.

Since this expression is not present in the options, the result is None of these.

∴ The correct answer is None of these.

Alternate Method

Given:

Cheating percentage in buying = x%

Cheating percentage in selling = x%

Formula Used:

Gain % = [ (True Weight − False Weight) ÷ False Weight ] × 100

Calculations:

Let the cost price (CP) of 1 unit of goods be ₹1.

⇒ While buying, the dealer pays ₹100 but receives (100 + x) units.

⇒ While selling, the dealer charges for 100 units (₹100) but only gives (100 − x) units.

⇒ Total SP of (100 − x) units = ₹100

⇒ SP of 1 unit = 100 ÷ (100 − x)

⇒ Total units available with the dealer = (100 + x)

⇒ Total SP of all goods = (100 + x) × [ 100 ÷ (100 − x) ]

⇒ Total Profit = [ 100(100 + x) ÷ (100 − x) ] − 100

⇒ Total Profit = 100 × [ (100 + x − (100 − x)) ÷ (100 − x) ]

⇒ Total Profit = 100 × [ 2x ÷ (100 − x) ]

⇒ Gain % = (Total Profit ÷ Total CP) × 100

⇒ Gain % = [ (200x ÷ (100 − x)) ÷ 100 ] × 100

⇒ Gain % = [ 200x ÷ (100 − x) ] %

Since none of the options (1) 2x%, (2) (10/x + x²)%, or (4) (x + x²/100)% match this result:

∴ The correct answer is None of these.

Additional Information

Faulty Weight Formula

Profit % = [ (Error ÷ False Value) × 100 ]%. Here, False Value is what is actually delivered to the customer.

Successive Profit Myth

While some texts use [ x + x + (x × x ÷ 100) ]%, this applies only if the cheating is on the price/base, not the weight.

Dishonest Dealer with Markup

If a dealer markups price by m% and uses a weight of w₁ instead of w₂, then Gain % = [ ((100 + m)/100) × (w₂/w₁) − 1 ] × 100.

Shortcut Trick

Initial Boys = 48 × (5/8) = 30 and Initial Girls = 48 × (3/8) = 18.

The new ratio of Boys to Girls is given as 6 : 5.

Since the number of boys remains unchanged, 6 ratio units = 30 boys ⇒ 1 ratio unit = 5.

New number of girls = 5 units = 5 × 5 = 25.

Number of girls added = 25 − 18 = 7.

∴ The correct answer is 7.

Alternate Method

Initial State (Total = 48 students)

Boys = 30 (5 parts) Girls = 18 (3 parts)

Final State (New Ratio 6 : 5)

Boys = 30 (6 parts) Girls = 25 (5 parts)

• 7 Girls added

⇒ Total number of students = 48

⇒ Initial ratio of boys to girls = 5 : 3

⇒ Let the number of boys be 5x and the number of girls be 3x.

⇒ 5x + 3x = 48

⇒ 8x = 48 ⇒ x = 6

⇒ Number of boys = 5 × 6 = 30

⇒ Number of girls = 3 × 6 = 18

⇒ Let the number of girls added be y.

⇒ New number of girls = 18 + y

⇒ The number of boys remains unchanged at 30.

⇒ New ratio of boys to girls = 30 / (18 + y) = 6 / 5

⇒ 150 = 6 × (18 + y)

⇒ 150 = 108 + 6y

⇒ 6y = 150 − 108

⇒ 6y = 42

⇒ y = 7

∴ The correct answer is 7.

Additional Information

Compound Ratio

When two or more ratios are multiplied together term by term, the resulting ratio is called a compound ratio. For a:b and c:d, the compound ratio is ac:bd.

Duplicate Ratio

The duplicate ratio of a given ratio a:b is the ratio of their squares, which is a²:b². Similarly, the sub-duplicate ratio is √a:√b.

Proportion Rule

When two ratios are equal, they are in proportion. For a:b = c:d, the product of means equals the product of extremes, meaning ad = bc.

Shortcut Trick

Bottle 1 (10L, 2:3) → Syrup = 4L, Water = 6L.

Bottle 2 (21L, 3:4) → Syrup = 9L, Water = 12L.

Sum (B1 + B2) → Syrup = 13L, Water = 18L. Deficit of Syrup = 5L.

In Bottle 3 (7:5), Syrup surplus = 7 − 5 = 2 units per 12 units total.

To balance 5L deficit, quantity = 5 × (12 / 2) = 30 Litres.

∴ The correct answer is 30 Litres.

Alternate Method

Given:

Mixture 1 = 10 Litres (Ratio 2 : 3)

Mixture 2 = 21 Litres (Ratio 3 : 4)

Mixture 3 = x Litres (Ratio 7 : 5)

Final Ratio = 1 : 1

Formula Used:

Quantity of component = Total quantity × (Component ratio part / Total ratio parts)

Calculations:

⇒ Syrup in 1st bottle = 10 × (2 / 5) = 4 Litres

⇒ Water in 1st bottle = 10 − 4 = 6 Litres

⇒ Syrup in 2nd bottle = 21 × (3 / 7) = 9 Litres

⇒ Water in 2nd bottle = 21 − 9 = 12 Litres

⇒ Let quantity of 3rd bottle be x Litres.

⇒ Syrup in 3rd bottle = (7 / 12) × x

⇒ Water in 3rd bottle = (5 / 12) × x

⇒ In final mixture, Syrup = Water

⇒ 4 + 9 + 7x / 12 = 6 + 12 + 5x / 12

⇒ 13 + 7x / 12 = 18 + 5x / 12

⇒ 7x / 12 − 5x / 12 = 18 − 13

⇒ 2x / 12 = 5

⇒ x / 6 = 5

⇒ x = 30 Litres

∴ The correct answer is 30 Litres.

Additional Information

Rule of Alligation

It is a rule that enables us to find the ratio in which two or more ingredients at the given price/ratio must be mixed to produce a mixture of a desired price/ratio.

Weighted Average of Ratio

When mixing multiple quantities, the final concentration C_avg = (Q₁C₁ + Q₂C₂ + ... + Q_nC_n) ÷ (Q₁ + Q₂ + ... + Q_n).

Concentration in Mixtures

In a ratio A : B, the concentration of A is A ÷ (A + B). For a 1 : 1 final ratio, the concentration of each component must be 1/2 or 50%.

Shortcut Trick

Total ratio parts: Whole picture = 4 + 3 = 7; Upper half = 2 + 3 = 5.

To make the Whole (2 units) divisible by the Half (1 unit), find LCM(7, 5) = 35. Let Half = 35 units, then Whole = 70 units.

Whole Picture B : Y ⇒ 4 : 3 → (4 × 10) : (3 × 10) = 40 : 30.

Upper Half B : Y ⇒ 2 : 3 → (2 × 7) : (3 × 7) = 14 : 21.

Lower Half B = 40 − 14 = 26; Lower Half Y = 30 − 21 = 9.

∴ The correct answer is 26 : 9.

Alternate Method

Given:

Ratio of Blue (B) to Yellow (Y) in whole picture = 4 : 3

Ratio of B : Y in upper half = 2 : 3

Formula Used:

Total Quantity = Quantity in Upper Half + Quantity in Lower Half

Calculations:

Let the total quantity of the picture be 70 units (LCM of 7 and 5 is 35, and we double it for two halves).

⇒ Total Blue in picture = (4/7) × 70 = 40 units

⇒ Total Yellow in picture = (3/7) × 70 = 30 units

⇒ Quantity of Upper Half = 70 ÷ 2 = 35 units

⇒ Blue in Upper Half = (2/5) × 35 = 14 units

⇒ Yellow in Upper Half = (3/5) × 35 = 21 units

⇒ Blue in Lower Half = Total Blue − Upper Blue = 40 − 14 = 26 units

⇒ Yellow in Lower Half = Total Yellow − Upper Yellow = 30 − 21 = 9 units

⇒ Ratio in Lower Half = 26 : 9

∴ The correct answer is 26 : 9.

Additional Information

Ratio Balancing

When comparing parts of a whole, ensure the sum of ratio parts for the whole is consistent with the sum of ratio parts for its components by using a common multiplier.

Compounded Ratio

The compounded ratio of (a : b) and (c : d) is (ac : bd). This is often used in partnership or multi-step mixture problems.

Inverse Ratio

The inverse ratio of a : b is 1/a : 1/b, which simplifies to b : a. This is crucial in speed-time or work-efficiency relationships.

Shortcut Trick

Ratio of Profit sharing = (Capital × Time) for each partner.

A's units = (4000 × 8) + (6000 × 4) = 32000 + 24000 = 56000

B's units = (5000 × 9) + (3000 × 3) = 45000 + 9000 = 54000

Ratio A : B = 56 : 54 = 28 : 27 | Total units = 55

Net profit for distribution = 6700 − (100 × 12) = 5500

B's share = (27 ÷ 55) × 5500 = 2700

∴ The correct answer is ₹2700.

Alternate Method

Given:

A's initial investment = ₹4000 for 8 months, increased to ₹6000 for 4 months.

B's initial investment = ₹5000 for 9 months, decreased to ₹3000 for 3 months.

Allowance for A = ₹100 per month. Total Profit = ₹6700.

Calculations:

⇒ Total Allowance for A = 100 × 12 = ₹1200

⇒ Remaining profit to be shared = 6700 − 1200 = ₹5500

⇒ A's investment × time = (4000 × 8) + (6000 × 4) = 32000 + 24000 = 56000

⇒ B's investment × time = (5000 × 9) + (3000 × 3) = 45000 + 9000 = 54000

⇒ Ratio of Profit (A : B) = 56000 : 54000 = 28 : 27

⇒ Sum of ratio parts = 28 + 27 = 55 units

⇒ Value of 55 units = ₹5500

⇒ Value of 1 unit = 5500 ÷ 55 = ₹100

⇒ B's share = 27 units × 100 = ₹2700

∴ The correct answer is ₹2700.

Additional Information

Active and Sleeping Partners

An active partner manages the business and usually receives a salary or allowance from the profit before it is distributed. A sleeping partner only invests capital.

Profit Distribution Formula

Profit Ratio = (C₁ × T₁) : (C₂ × T₂), where C is the capital and T is the time period for which the capital is invested.

Effective Capital

When investments change over time, effective capital is the sum of (Investment × Duration) for each period during the partnership term.

Shortcut Trick

Sum of reductions = 22 + 35 + 48 = 105.

Net sum to divide = 15525 − 105 = 15420.

Total ratio parts = 7 + 10 + 13 = 30 → 1 unit = 15420 ÷ 30 = 514.

New shares = (7 × 514 + 22 + 16) : (10 × 514 + 35 + 77) : (13 × 514 + 48 + 37).

Simplifying: 3636 : 5252 : 6767 = 36 : 52 : 67 (divided by 101).

∴ The correct answer is 36 : 52 : 67.

Alternate Method

Given:

Total Sum = ₹15525

Reductions: Sunil = ₹22, Anil = ₹35, Jamil = ₹48

Remaining Ratio = 7 : 10 : 13

Additions: Sunil = ₹16, Anil = ₹77, Jamil = ₹37

Calculations:

⇒ Sum of reductions = 22 + 35 + 48 = ₹105

⇒ Remaining amount = 15525 − 105 = ₹15420

⇒ Ratio sum = 7 + 10 + 13 = 30 units

⇒ 30 units = 15420

⇒ 1 unit = 514

⇒ Sunil's reduced share = 7 × 514 = ₹3598

⇒ Anil's reduced share = 10 × 514 = ₹5140

⇒ Jamil's reduced share = 13 × 514 = ₹6682

⇒ Sunil's original share = 3598 + 22 = ₹3620

⇒ Anil's original share = 5140 + 35 = ₹5175

⇒ Jamil's original share = 6682 + 48 = ₹6730

⇒ New shares after adding 16, 77, 37:

⇒ Sunil = 3620 + 16 = 3636

⇒ Anil = 5175 + 77 = 5252

⇒ Jamil = 6730 + 37 = 6767

⇒ New Ratio = 3636 : 5252 : 6767

⇒ Divide each by 101: 36 : 52 : 67

∴ The correct answer is 36 : 52 : 67.

Additional Information

Ratio Division Rule

To divide a quantity S in the ratio a : b : c, the shares are [a/(a+b+c)]×S, [b/(a+b+c)]×S, and [c/(a+b+c)]×S.

Ratio Simplification

A ratio a : b : c can be simplified by dividing all terms by their Highest Common Factor (HCF).

Changes in Ratio

If specific amounts are added or subtracted from the original parts of a ratio, the new ratio must be recalculated based on the absolute values.

Shortcut Trick

Let B income = x, then A = x + 140.

Given B : D = 1 : 2 ⇒ D = 2x, so C = 2x + 80.

A : C = 2 : 3 ⇒ (x + 140) / (2x + 80) = 2 / 3.

Cross multiply: 3x + 420 = 4x + 160 ⇒ x = 260.

Incomes: A = 400, B = 260, C = 600, D = 520.

∴ The correct answer is ₹400, ₹260, ₹600 and ₹520.

Alternate Method

Given:

A − B = 140

C − D = 80

A : C = 2 : 3

B : D = 1 : 2

Let incomes of B and D be 1k and 2k respectively.

⇒ B = k, D = 2k

⇒ A = B + 140 = k + 140

⇒ C = D + 80 = 2k + 80

Since A / C = 2 / 3:

⇒ (k + 140) / (2k + 80) = 2 / 3

⇒ 3(k + 140) = 2(2k + 80)

⇒ 3k + 420 = 4k + 160

⇒ 4k − 3k = 420 − 160

⇒ k = 260

Now, calculate each income:

⇒ B = 260

⇒ A = 260 + 140 = 400

⇒ D = 2 × 260 = 520

⇒ C = 520 + 80 = 600

∴ The correct answer is ₹400, ₹260, ₹600 and ₹520.

Additional Information

Ratio and Proportion

A ratio compares two quantities. If A:B = x:y, then A = xk and B = yk for some constant k.

Income-Expenditure Relation

Income = Expenditure + Savings. While not directly used here, it is a common extension of income ratio problems.

Linear Equations in Ratios

When differences between terms are given along with ratios, converting the ratios into terms of a single variable (like k or x) is the most efficient algebraic approach.

Shortcut Trick

New Average = New Score − (Previous Innings × Increase in Average)

New Average = 96 − (12 × 5)

New Average = 96 − 60 = 36 runs.

∴ The correct answer is 36 runs.

Alternate Method

Given:

Number of innings = 12

13th inning score = 96 runs

Average increase = 5 runs

Formula Used:

Average = Total Runs ÷ Total Innings

Calculations:

⇒ Let the average for 12 innings be x.

⇒ Total runs in 12 innings = 12 × x = 12x.

⇒ Runs after 13th inning = 12x + 96.

⇒ New Average = x + 5.

⇒ (12x + 96) ÷ 13 = x + 5

⇒ 12x + 96 = 13 × (x + 5)

⇒ 12x + 96 = 13x + 65

⇒ 13x − 12x = 96 − 65

⇒ x = 31

⇒ New Average = x + 5 = 31 + 5 = 36 runs.

∴ The correct answer is 36 runs.

Additional Information

Batting Average

It is the ratio of total runs scored to the number of innings played (excluding not-out innings). Formula: Total runs ÷ Innings.

Bowling Average

It is the ratio of total runs conceded to the total number of wickets taken. Formula: Runs Conceded ÷ Wickets Taken.

Effect of New Entry

When a new value is added, the new average = Old Average + (Change in Average × New Count of items).

Shortcut Trick

Hypothetical total score (if he scored 92) = 84 × 8 = 672 points.

Difference in scores = 92 − 85 = 7 points.

Actual total score = Hypothetical total − Difference = 672 − 7 = 665 points.

∴ The correct answer is 665 points.

Alternate Method

Given:

Number of persons (n) = 8

Actual score of best marksman = 85

Hypothetical score of best marksman = 92

Hypothetical average score = 84

Formula Used:

Total Score = Average Score × Number of persons

⇒ Hypothetical Total Score = 84 × 8

⇒ Hypothetical Total Score = 672

⇒ Score increase applied = 92 − 85 = 7

⇒ Actual Total Score = Hypothetical Total Score − Score Increase

⇒ Actual Total Score = 672 − 7

⇒ Actual Total Score = 665

∴ The correct answer is 665 points.

Additional Information

Definition of Average

Average = (Sum of all observations) ÷ (Total number of observations). It is also known as the arithmetic mean.

Effect of Change in a Value

If one value in a data set is changed, the new average = Old Average + (Difference in value ÷ n), where n is the number of terms.

Total Sum Property

The sum of all observations is always equal to the product of the Average and the total count: Sum = Average × n.

Shortcut Trick

Extra cost spent for 4 books = ₹336.

Total reduction in value for all 64 books (60 + 4) = 64 × ₹1 = ₹64.

Total value of these 4 books at the original average price = 336 + 64 = ₹400.

Previous average price per book = 400 ÷ 4 = ₹100.

∴ The correct answer is ₹100.

Alternate Method

Given:

Initial number of books = 60

Extra books purchased = 4

Additional expenditure = ₹336

Reduction in average price = ₹1

Formula Used:

Average Price = Total Expenditure ÷ Total Number of Books

Calculations:

⇒ Let the previous average price be ₹x.

⇒ Total cost of 60 books = 60 × x = ₹60x

⇒ Total cost of 64 books (60 + 4) = 60x + 336

⇒ New average price = (60x + 336) ÷ 64

⇒ According to the problem: New Average = x − 1

⇒ (60x + 336) ÷ 64 = x − 1

⇒ 60x + 336 = 64 × (x − 1)

⇒ 60x + 336 = 64x − 64

⇒ 64x − 60x = 336 + 64

⇒ 4x = 400

⇒ x = 100

∴ The correct answer is ₹100.

Additional Information

Weighted Average Formula

The combined average of two groups is given by A = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂), where n is the quantity and a is the average.

Net Change in Average

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

Inclusion Property

When a new item is added, New Sum = (Old Average × Old Count) + New Item Value.

Shortcut Trick

Total subjects = 7. Group 1 (4 sub) average = 50. Group 2 (4 sub) average = 70. Overall average = 58.

Deviation of Group 1 = 4 × (50 − 58) = −32.

Deviation of Group 2 = 4 × (70 − 58) = +48.

Maths score = Overall average + Net deviation = 58 + (−32 + 48) = 58 + 16 = 74.

∴ The correct answer is 74 marks.

Alternate Method

Given:

Average of English, Maths, Hindi, Drawing (4 subjects) = 50

Average of Maths, Science, Social Studies, Craft (4 subjects) = 70

Average of all 7 distinct subjects = 58

Formula Used:

Total Sum = Average × Number of subjects

Overlapping term (Maths) = (Sum of Group 1 + Sum of Group 2) − Total Sum of distinct items

⇒ Sum of Group 1 subjects = 4 × 50 = 200

⇒ Sum of Group 2 subjects = 4 × 70 = 280

⇒ Total Sum including overlapping Maths = 200 + 280 = 480

⇒ Actual total sum of 7 subjects = 7 × 58 = 406

⇒ Marks in Maths = 480 − 406

⇒ Marks in Maths = 74

∴ The correct answer is 74 marks.

Additional Information

General Average Formula

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

Overlapping Elements in Sets

When one element is common between two sets of sizes n₁ and n₂ that form a total set of N elements, the common element value is (Sum₁ + Sum₂) − Sum_Total, where N = n₁ + n₂ − 1.

Weighted Average

The combined average of two groups is given by (n₁a₁ + n₂a₂) ÷ (n₁ + n₂), where a₁ and a₂ are the respective averages.

Shortcut Trick

Weight of D = New total − Old total = (4 × 80) − (3 × 84) = 320 − 252 = 68 Kg.

Weight of E = D + 3 = 68 + 3 = 71 Kg.

Difference between Sum(A,B,C,D) and Sum(B,C,D,E) = 4 × (80 − 79) = 4 Kg.

Since B, C, and D are common, A − E = 4 ⇒ A = 71 + 4 = 75 Kg.

∴ The correct answer is 75 Kg.

Alternate Method

Given:

Average weight of A, B, C = 84 Kg

Average weight of A, B, C, D = 80 Kg

Weight of E = Weight of D + 3 Kg

Average weight of B, C, D, E = 79 Kg

Formula:

Total Weight = Average Weight × Number of persons

⇒ Total weight of A, B, C = 3 × 84 = 252 Kg

⇒ Total weight of A, B, C, D = 4 × 80 = 320 Kg

⇒ Weight of D = 320 − 252 = 68 Kg

⇒ Weight of E = 68 + 3 = 71 Kg

⇒ Total weight of B, C, D, E = 4 × 79 = 316 Kg

⇒ Difference: (A + B + C + D) − (B + C + D + E) = 320 − 316

⇒ A − E = 4 Kg

⇒ A = E + 4 = 71 + 4 = 75 Kg

∴ The correct answer is 75 Kg.

Additional Information

Replacement Concept

When a person of weight X is replaced by a person of weight Y, the change in average is (Y − X) ÷ n, where n is the total number of people.

Weighted Average

The average of two or more groups combined: Avg_comb = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Deviation Method

If a new person joins and the average decreases, the new person weight = Old Average − (Decrease × New Total People).

Shortcut Trick

Using Alligation Rule to find the ratio of Officers to Remaining employees:

Ratio (Officers : Remaining) = (Overall Avg − Remaining Avg) : (Officer Avg − Overall Avg)

Ratio = (8840 − 8000) : (15000 − 8840) = 840 : 6160

Simplifying the ratio: 84 : 616 = 3 : 22

Percentage of Officers = [3 ÷ (3 + 22)] × 100 = (3 ÷ 25) × 100 = 12%

∴ The correct answer is 12%.

Alternate Method

Given:

Average salary of Officers = ₹15000

Average salary of Remaining employees = ₹8000

Average salary of All employees = ₹8840

⇒ Let the number of Officers be n₁ and Remaining employees be n₂.

⇒ Total Salary = (n₁ + n₂) × 8840 = 15000n₁ + 8000n₂

⇒ 8840n₁ + 8840n₂ = 15000n₁ + 8000n₂

⇒ 8840n₂ − 8000n₂ = 15000n₁ − 8840n₁

⇒ 840n₂ = 6160n₁

⇒ n₁ ÷ n₂ = 840 ÷ 6160 = 3 ÷ 22

⇒ Percentage of Officers = [n₁ ÷ (n₁ + n₂)] × 100

⇒ Percentage of Officers = [3 ÷ (3 + 22)] × 100

⇒ Percentage of Officers = (3 ÷ 25) × 100 = 12%

∴ The correct answer is 12%.

Additional Information

Weighted Average Formula

Average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where n is the number of items and A is the individual average.

Alligation Rule

It is a faster way to find the ratio in which two quantities at different prices (or averages) are mixed to produce a mixture of a certain mean price: Ratio = (Dearer − Mean) : (Mean − Cheaper).

Percentage Composition

If the ratio of two parts is a : b, then the percentage of the first part is [a ÷ (a + b)] × 100.

Shortcut Trick

Ratio of Cost Price (CP) : Selling Price (SP) = 20 : 21

Profit = 21 − 20 = 1 unit

Gain % = (Profit ÷ CP) × 100

Gain % = (1 ÷ 20) × 100 = 5%

∴ The correct answer is 5%.

Alternate Method

Given:

CP : SP = 20 : 21

Formula Used:

Gain % = [(SP − CP) ÷ CP] × 100

⇒ Let CP = 20x and SP = 21x

⇒ Gain = SP − CP

⇒ Gain = 21x − 20x = x

⇒ Gain % = (x ÷ 20x) × 100

⇒ Gain % = (1 ÷ 20) × 100

⇒ Gain % = 5%

∴ The correct answer is 5%.

Additional Information

Profit and Loss Percentage

Profit % and Loss % are always calculated on the Cost Price (CP) unless stated otherwise. Formula: Loss % = (Loss ÷ CP) × 100.

Relationship between SP and CP

Selling Price can be directly found using CP and Gain%: SP = CP × (100 + Gain%) ÷ 100.

Selling Price for Loss

When there is a loss: SP = CP × (100 − Loss%) ÷ 100.

Shortcut Trick

Ratio of CP : SP = 25 : 26

Profit = 26 − 25 = 1 unit

Profit % = (Profit ÷ CP) × 100

Profit % = (1 ÷ 25) × 100 = 4%

∴ The correct answer is 4%.

Alternate Method

Given:

Ratio of Cost Price (CP) and Selling Price (SP) = 25 : 26

Formula Used:

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

Calculations:

⇒ Let Cost Price (CP) = 25x

⇒ Let Selling Price (SP) = 26x

⇒ Profit = SP − CP

⇒ Profit = 26x − 25x = x

⇒ Profit % = (x ÷ 25x) × 100

⇒ Profit % = 100 ÷ 25

⇒ Profit % = 4%

∴ The correct answer is 4%.

Additional Information

Loss Percentage

When CP is greater than SP, Loss = CP − SP. Loss % is calculated as (Loss ÷ CP) × 100.

Marked Price (MP) and Discount

Discount is the reduction offered on the Marked Price. Selling Price (SP) = MP − Discount. Discount % = (Discount ÷ MP) × 100.

Successive Discounts

If two discounts d₁% and d₂% are given successively, the equivalent single discount is [d₁ + d₂ − (d₁ × d₂ ÷ 100)]%.

Shortcut Trick

Required SP per kg = [CP per kg × (100 + Profit%)] ÷ (100 − Damage%)

Required SP = [150 × 120] ÷ 85

Required SP = 18000 ÷ 85 = 3600 ÷ 17

Required SP = ₹211 13/17

∴ The correct answer is ₹211 13/17.

Alternate Method

Given: CP = ₹150/kg, Damage = 15%, Desired Profit = 20%

Formula: Total SP = Total CP × (1 + Profit/100)

Assume Total Quantity = 100 kg

⇒ Total CP = 100 × 150 = ₹15,000

⇒ Target Total SP = 15,000 × 1.20 = ₹18,000

⇒ Usable Quantity = 100 − 15% of 100 = 85 kg

⇒ Selling Price per kg = Total SP ÷ Usable Quantity

⇒ SP per kg = 18,000 ÷ 85

⇒ SP per kg = 3600 ÷ 17

⇒ SP per kg = ₹211 13/17

∴ The correct answer is ₹211 13/17.

Additional Information

Profit and Loss with Wastage

When goods are damaged or lost, the cost of the lost items must be recovered from the sale of the remaining items. The effective CP increases as the sellable quantity decreases.

Marked Price and Discount

SP = MP × (100 − Discount%)/100. If no discount is given, MP = SP. Relationship: CP × (100 + P%) = MP × (100 − D%).

Dishonest Dealer Concept

Profit% = [Error ÷ (True Value − Error)] × 100. This is used when a seller uses false weights to gain profit.

Shortcut Trick

When the Cost Price (CP) of two articles is the same, the overall profit or loss percentage is the simple arithmetic mean of the individual percentages.

Net Profit or Loss % = (Profit₁% + Profit₂%) ÷ 2

Here, Profit₁ = +6% and Profit₂ = −4% (loss is taken as negative).

Overall % = (6 − 4) ÷ 2 = 2 ÷ 2 = 1%.

Since the result is positive, it is a net gain of 1%.

∴ The correct answer is 1% gain.

Alternate Method

Given: Cost Price (CP) of each cycle = ₹750

Profit on 1st cycle = 6% | Loss on 2nd cycle = 4%

⇒ Total CP = 750 + 750 = ₹1500

⇒ Selling Price (SP) of 1st cycle = 750 × (106 ÷ 100) = 7.5 × 106 = ₹795

⇒ Selling Price (SP) of 2nd cycle = 750 × (96 ÷ 100) = 7.5 × 96 = ₹720

⇒ Total SP = 795 + 720 = ₹1515

⇒ Net Profit = Total SP − Total CP = 1515 − 1500 = ₹15

⇒ Gain Percent = (Net Profit ÷ Total CP) × 100

⇒ Gain Percent = (15 ÷ 1500) × 100 = 1%

∴ The correct answer is 1% gain.

Additional Information

Same Selling Price Rule

If two articles are sold at the same price, one at x% gain and another at x% loss, there is always a net loss of (x ÷ 10)² %.

Successive Percentage Change

For two successive changes of a% and b%, the net effective change is given by [a + b + (a × b) ÷ 100]%.

Profit and Loss Formulas

SP = CP × (100 + Gain%) ÷ 100 or CP × (100 − Loss%) ÷ 100. Profit always depends on the base CP unless stated otherwise.

Shortcut Trick

Total percentage difference = Profit % − (Loss %) = 8% − (−8%) = 16%

According to the question, this 16% difference is equal to ₹28.

Cost Price (100%) = (Difference ÷ Percentage Difference) × 100

CP = (28 ÷ 16) × 100 = 1.75 × 100 = ₹175.

∴ The correct answer is ₹175.

Alternate Method

Given:

Profit percentage = 8%

Loss percentage = 8%

Difference in selling prices = ₹28

Formula Used:

Selling Price (SP) = CP × (100 + Profit%)/100 or CP × (100 − Loss%)/100

Calculations:

Let the Cost Price (CP) be ₹x.

⇒ SP at 8% profit (SP₁) = x × (108/100) = 1.08x

⇒ SP at 8% loss (SP₂) = x × (92/100) = 0.92x

⇒ According to the question: SP₁ − SP₂ = 28

⇒ 1.08x − 0.92x = 28

⇒ 0.16x = 28

⇒ x = 28 ÷ 0.16

⇒ x = 2800 ÷ 16

⇒ x = 175

∴ The correct answer is ₹175.

Additional Information

Direct CP Formula for SP Difference

When one transaction is a profit (P%) and other is a loss (L%), CP = (Difference in SP × 100) ÷ (P + L). If both are profits or both are losses, use the difference of percentages in the denominator.

Marked Price and Discount

Relationship between CP and MP: CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%). This is useful for problems involving marked price and discounts.

Successive Percentage Change

Net change = a + b + (a × b) ÷ 100. Used for successive discounts or profit/loss scenarios.

Shortcut Trick

Since the quantities sold are equal (50% and 50%), use the average profit formula.

Let the profit percent on the second half be x%.

Average Profit % = (Profit₁ + Profit₂) ÷ 2

10% = (5% + x%) ÷ 2

⇒ 20 = 5 + x ⇒ x = 15%

∴ The correct answer is 15%.

Alternate Method

Given:

Total quantity = 500 m

Cost price per metre = 50 paise = ₹0.50

Total Gain required = 10%

Calculations:

⇒ Total CP = 500 × 0.50 = ₹250

⇒ Total Profit required = 10% of 250 = (10 ÷ 100) × 250 = ₹25

⇒ CP of 50% wire = 250 × 0.50 = ₹125

⇒ Profit from first part = 5% of 125 = (5 ÷ 100) × 125 = ₹6.25

⇒ Profit needed from second part = Total Profit − Profit from first part

⇒ Profit needed from second part = 25 − 6.25 = ₹18.75

⇒ Required Profit % = (Profit ÷ CP of part) × 100

⇒ Required Profit % = (18.75 ÷ 125) × 100 = 15%

∴ The correct answer is 15%.

Additional Information

Weighted Average Profit

Overall Profit % = (n₁p₁ + n₂p₂) ÷ (n₁ + n₂), where n represents the quantity or cost price part and p represents the profit percentage.

Profit/Loss on Equal Cost Price

If two items are bought at the same CP, and one is sold at x% profit and another at y% loss, the net result is (x − y) ÷ 2 %.

Successive Profit/Loss

When an item is sold successively at a% and b% profit, the effective profit is given by (a + b + (a × b) ÷ 100)%.

Shortcut Trick

Error % = (Error ÷ True Value) × 100

Error = 1.55 − 1.5 = 0.05 m

Error % = (0.05 ÷ 1.5) × 100 = (5 ÷ 150) × 100

Error % = 100 ÷ 30 = 10 ÷ 3 = 3¹/₃%

∴ The correct answer is 3¹/₃%.

Alternate Method

Given:

True Length = 1.5 m

Measured Length = 1.55 m

True Length = 1.5 m

1.5 m +0.05

Error

Measured Length = 1.55 m

Formula Used:

Error % = (Absolute Error ÷ True Value) × 100

Calculations:

⇒ Error = Measured Length − True Length

⇒ Error = 1.55 − 1.5 = 0.05 m

⇒ Error % = (0.05 ÷ 1.5) × 100

⇒ Error % = (5 ÷ 150) × 100

⇒ Error % = (1 ÷ 30) × 100

⇒ Error % = 10 ÷ 3

⇒ Error % = 3¹/₃%

∴ The correct answer is 3¹/₃%.

Additional Information

Percentage Error

Percentage error measures the inaccuracy of a measurement compared to its true value. Formula: (Absolute Error ÷ True Value) × 100.

Percentage Change

Used to find the increase or decrease in a value over time. Formula: [(New Value − Old Value) ÷ Old Value] × 100.

Successive Percentage

When a value changes consecutively, the net percentage change is calculated as: Net Change % = a + b + (a × b ÷ 100).

Shortcut Trick

Total Duty = (10% of 2,10,000) + (8% of 1,00,000) + (5% of 1,50,000)

Duty on Laptops = 21,000

Duty on Mobiles = 8,000

Duty on TV sets = 7,500

Total Duty = 21,000 + 8,000 + 7,500 = 36,500

∴ The correct answer is ₹36500.

Alternate Method

Given values and duty rates:

Value of Laptops = ₹2,10,000 | Duty Rate = 10%

Value of Mobile Phones = ₹1,00,000 | Duty Rate = 8%

Value of Television Sets = ₹1,50,000 | Duty Rate = 5%

Formula used: Duty Amount = (Duty Rate ÷ 100) × Worth of Item

⇒ Duty on Laptops = (10 ÷ 100) × 2,10,000 = ₹21,000

⇒ Duty on Mobile phones = (8 ÷ 100) × 1,00,000 = ₹8,000

⇒ Duty on Television sets = (5 ÷ 100) × 1,50,000 = ₹7,500

⇒ Total Duty = 21,000 + 8,000 + 7,500

⇒ Total Duty = ₹36,500

∴ The correct answer is ₹36500.

Additional Information

Percentage Calculation

To calculate x% of a value V, use the formula (x ÷ 100) × V. For simple values like 10%, just shift the decimal one place to the left.

Weighted Average Percentage

The overall percentage of duty is calculated as [(Rate₁ × Value₁ + Rate₂ × Value₂ + ...) ÷ Total Value] × 100.

Effective Tax Rate

Effective Duty % = (Total Duty ÷ Total Worth) × 100. In this case, 36500 ÷ 460000 × 100 ≈ 7.93%.

Shortcut Trick

Convert percentages to fractions: 7.5% = 3/40 and 75% = 3/4.

Remaining fraction after 1st spend = 1 − 3/40 = 37/40.

Remaining fraction after 2nd spend = 37/40 × (1 − 3/4) = 37/40 × 1/4 = 37/160.

Now, 37 units = ₹370 ⇒ 1 unit = ₹10.

Total money (160 units) = 160 × 10 = ₹1600.

∴ The correct answer is ₹1600.

Alternate Method

Given:

Initial spending = 7.5% of total money.

Second spending = 75% of remaining money.

Final balance = ₹370.

Let the total money be ₹x.

⇒ Remaining after 7.5% spend = x − (7.5/100)x = 92.5x/100 = 37x/40.

⇒ Remaining after spending 75% of (37x/40) = 25% of (37x/40).

⇒ (25/100) × (37x/40) = 370.

⇒ (1/4) × (37x/40) = 370.

⇒ 37x/160 = 370.

⇒ x = (370 × 160) ÷ 37.

⇒ x = 10 × 160.

⇒ x = 1600.

∴ The correct answer is ₹1600.

Additional Information

Successive Percentage Change

When a value is changed by a% and then the remaining by b%, the net remaining fraction is (1 − a/100) × (1 − b/100).

Fraction Equivalents

Common percentages to fractions: 7.5% = 3/40, 12.5% = 1/8, 25% = 1/4, 75% = 3/4.

Percentage Base Change

Always identify if the second percentage is of the original total or the remaining amount before applying formulas.

Shortcut Trick

Initial success rate = 60%, Success rate of new matches (all lost) = 0%.

Target overall success rate = 30%.

Using Alligation: Ratio of (Initial matches) : (New matches) = (30 − 0) : (60 − 30) = 30 : 30 = 1 : 1.

Given new matches = 30 (1 part), so initial matches = 30 (1 part).

Total matches played = 30 + 30 = 60.

∴ The correct answer is 60 ODIs.

Alternate Method

Given:

Initial success rate = 60%

Number of consecutive losses = 30

Final success rate = 30%

Formula Used:

Success Rate = (Total Wins ÷ Total Matches) × 100

Let the number of initial matches be x.

⇒ Number of wins = 60% of x = 0.6x

⇒ New total matches = x + 30

⇒ New success rate = (0.6x ÷ (x + 30)) × 100 = 30

⇒ 0.6x ÷ (x + 30) = 0.3

⇒ 0.6x = 0.3(x + 30)

⇒ 0.6x = 0.3x + 9

⇒ 0.3x = 9

⇒ x = 9 ÷ 0.3 = 30

⇒ Total number of ODIs = x + 30 = 30 + 30 = 60

∴ The correct answer is 60 ODIs.

Additional Information

Success Rate Formula

It is defined as (Total Wins ÷ Total Events) × 100. Adding losses increases the denominator without changing the numerator, thus decreasing the percentage.

Rule of Alligation

It provides the ratio in which two quantities at different concentrations or rates must be mixed to produce a desired average rate: Q₁/Q₂ = (R₂ − R_avg)/(R_avg − R₁).

Weighted Average Concept

The overall success rate is the weighted average of individual success rates: Total Success % = (n₁p₁ + n₂p₂) ÷ (n₁ + n₂).

Shortcut Trick

Relative Speed (opposite direction) = Speed 1 + Speed 2

Relative Speed = 3 + 2 = 5 m/sec

Time required = Total Distance ÷ Relative Speed

Time = 400 ÷ 5 = 80 sec

∴ The correct answer is 80 sec.

Alternate Method

Given:

Distance (D) = 400 m

Speed of first donkey (s₁) = 3 m/sec

Speed of second donkey (s₂) = 2 m/sec

Formula Used:

Time = Distance ÷ Relative Speed

Calculations:

⇒ When two objects move towards each other, relative speed is the sum of individual speeds.

⇒ Relative Speed = s₁ + s₂

⇒ Relative Speed = 3 + 2 = 5 m/sec

⇒ Time = Distance ÷ Relative Speed

⇒ Time = 400 ÷ 5

⇒ Time = 80 sec

∴ The correct answer is 80 sec.

Additional Information

Relative Speed in Same Direction

When two objects move in the same direction, Relative Speed = |Speed₁ − Speed₂|.

Unit Conversion

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

Average Speed

For the same distance covered at two different speeds x and y, the average speed is given by 2xy ÷ (x + y).

Shortcut Trick

Use the direct formula for distance when reaching late and early:

Distance = [ (Product of Speeds) ÷ (Difference of Speeds) ] × Total Time Difference

Time Difference = 5 min late + 5 min early = 10 minutes = 10/60 hours = 1/6 hr

Distance = [ (3 × 4) ÷ (4 − 3) ] × (1/6)

Distance = [ 12 ÷ 1 ] × (1/6) = 12 × (1/6) = 2 km

∴ The correct answer is 2 km.

Alternate Method

Given:

Speed 1 (s₁) = 3 km/hr | Speed 2 (s₂) = 4 km/hr

Time 1 = Actual Time + 5 min | Time 2 = Actual Time − 5 min

Since Distance is constant, Speed (s) is inversely proportional to Time (t):

⇒ Ratio of speeds = s₁ : s₂ = 3 : 4

⇒ Ratio of times = t₁ : t₂ = 4 : 3

⇒ Difference in time units = 4 − 3 = 1 unit

⇒ Actual time difference = 5 min late − (−5 min early) = 10 minutes

⇒ 1 unit = 10 minutes = 10/60 hr = 1/6 hr

⇒ Time taken at 3 km/hr (t₁) = 4 units = 4 × (1/6) = 2/3 hr

⇒ Distance = Speed × Time

⇒ Distance = 3 km/hr × (2/3) hr = 2 km

∴ The correct answer is 2 km.

Additional Information

Time Difference Rule

If both cases are late or both are early, subtract the times. If one is late and one is early, add the times to find the total difference.

Average Speed

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

Speed-Time Relationship

For a fixed distance, speed and time are inversely proportional (s₁/s₂ = t₂/t₁).

Shortcut Trick

Total time taken = 2 hours 20 minutes = 2 + 20/60 = 2 + 1/3 = 7/3 hours.

Shortcut formula for distance when total time for a round trip is known: Distance = (S₁ × S₂) / (S₁ + S₂) × Total Time

Distance = (12 × 9) / (12 + 9) × (7/3)

Distance = (108 / 21) × (7/3) = (36 / 7) × (7/3) = 12 km.

∴ The correct answer is 12 km.

Alternate Method

Given:

Speed while going (S₁) = 12 km/hr

Speed while returning (S₂) = 9 km/hr

Total time (T) = 2 hours 20 minutes = 7/3 hours

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Let the distance between the two points be 'd' km.

⇒ Time taken to travel from A to B = d/12 hours

⇒ Time taken to travel from B to A = d/9 hours

⇒ According to the question: d/12 + d/9 = 7/3

⇒ Taking LCM of 12 and 9 (which is 36): (3d + 4d) / 36 = 7/3

⇒ 7d / 36 = 7/3

⇒ d = (7/3) × (36/7)

⇒ d = 12 km

∴ The correct answer is 12 km.

Additional Information

Average Speed for Same Distance

When a person covers equal distances at speeds x and y, the average speed for the whole journey is (2xy) / (x + y).

Distance Formula (Time Difference)

If the difference in time taken at two different speeds is ΔT, then Distance = [ (S₁ × S₂) / |S₁ − S₂| ] × ΔT.

Unit Conversion

To convert speed from km/hr to m/s, multiply by 5/18. To convert from m/s to km/hr, multiply by 18/5.

Shortcut Trick

Sum of speeds (opposite direction) = 15 ÷ 0.5 = 30 km/hr

Difference of speeds (same direction) = 15 ÷ 2.5 = 6 km/hr

Speed of faster person = (Sum + Difference) ÷ 2

Faster speed = (30 + 6) ÷ 2 = 36 ÷ 2 = 18 km/hr

∴ The correct answer is 18 km/hr.

Alternate Method

Given:

Distance (D) = 15 km

Time for opposite direction (t₁) = 0.5 hr

Time for same direction (t₂) = 2.5 hr

Formula Used:

Relative Speed (Opposite) = S₁ + S₂ = Distance ÷ Time₁

Relative Speed (Same) = S₁ − S₂ = Distance ÷ Time₂

Calculations:

⇒ Let the speed of the faster person be x km/hr and the slower be y km/hr.

⇒ x + y = 15 ÷ 0.5

⇒ x + y = 30 ---- (Eq. 1)

⇒ x − y = 15 ÷ 2.5

⇒ x − y = 6 ---- (Eq. 2)

⇒ Adding (Eq. 1) and (Eq. 2):

⇒ (x + y) + (x − y) = 30 + 6

⇒ 2x = 36

⇒ x = 18 km/hr

⇒ Substituting x in (Eq. 1): 18 + y = 30 → y = 12 km/hr

∴ The correct answer is 18 km/hr.

Additional Information

Relative Speed Concept

When two bodies move in the same direction, relative speed is the difference (S₁ − S₂). When they move in opposite directions, it is the sum (S₁ + S₂).

Conversion of Units

To convert speed from km/hr to m/s, multiply by 5/18. To convert from m/s to km/hr, multiply by 18/5.

Average Speed

If a person travels equal distances at speeds x and y, then the average speed for the whole journey is (2 × x × y) ÷ (x + y).

Shortcut Trick

Effective rate of Compound Interest for 2 years = x + y + (x × y)/100

Effective rate = 15 + 15 + (15 × 15)/100 = 30 + 2.25 = 32.25%

Given Compound Interest = ₹3225

∴ 32.25% of Principal = 3225

Principal = (3225 ÷ 32.25) × 100 = 100 × 100 = 10000

∴ The correct answer is ₹10000.

Alternate Method

Given:

Compound Interest (CI) = ₹3225

Rate (R) = 15% per annum

Time (T) = 2 years

Formula Used:

CI = P[(1 + R/100)^T − 1]

Calculations:

⇒ 3225 = P[(1 + 15/100)² − 1]

⇒ 3225 = P[(115/100)² − 1]

⇒ 3225 = P[(23/20)² − 1]

⇒ 3225 = P[529/400 − 1]

⇒ 3225 = P[(529 − 400)/400]

⇒ 3225 = P[129/400]

⇒ P = (3225 × 400) ÷ 129

⇒ P = 25 × 400

⇒ P = 10000

∴ The correct answer is ₹10000.

Additional Information

Effective Rate of Interest

For two successive growth rates x% and y%, the effective rate is calculated as [x + y + (xy/100)]%.

Difference between CI and SI

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

Compound Interest Amount Formula

The total amount (A) after n years is A = P(1 + R/100)ⁿ, where P is the principal and R is the annual interest rate.

Shortcut Trick

For Compound Interest, the ratio of Amount to Principal for 3 years is (1 + R/100)³.

Ratio A : P = 3993 : 3000

Divide both by 3 → 1331 : 1000

Take cube root for 3 years: √³(1331) : √³(1000) = 11 : 10

Rate x = [(11 − 10) / 10] × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given:

Principal (P) = ₹3000, Amount (A) = ₹3993, Time (n) = 3 years, Rate = x%

Formula Used:

A = P(1 + r/100)ⁿ

⇒ 3993 = 3000(1 + x/100)³

⇒ 3993 / 3000 = (1 + x/100)³

⇒ 1331 / 1000 = (1 + x/100)³

⇒ (11/10)³ = (1 + x/100)³

⇒ 11/10 = 1 + x/100

⇒ 1.1 = 1 + x/100

⇒ x/100 = 0.1

⇒ x = 10

∴ The correct answer is 10%.

Additional Information

Compound Interest vs Simple Interest

In CI, interest is calculated on the principal plus accumulated interest. For 1 year, SI and CI are equal if compounded annually.

Half-Yearly Compounding

If interest is compounded half-yearly, the rate is halved (R/2) and the time period is doubled (2n).

Effective Rate of Interest

For two successive years with rate r%, the effective rate is [r + r + (r × r)/100]%.

Shortcut Trick

Rate 5% = 1/20 → Present Value factor = 20/21

PV of 1st Installment = 3150 × (20/21) = ₹3000

PV of 2nd Installment = 4410 × (20/21)² = 4410 × 400/441 = ₹4000

Total Principal (Sum) = 3000 + 4000 = ₹7000

∴ The correct answer is ₹7000.

Alternate Method

Given:

1st Year Payment (I₁) = ₹3150

2nd Year Payment (I₂) = ₹4410

Rate of Interest (r) = 5% per annum

Formula Used:

Principal (P) = I₁/(1 + r/100)¹ + I₂/(1 + r/100)²

Calculations:

⇒ P = 3150 / (1 + 5/100) + 4410 / (1 + 5/100)²

⇒ P = 3150 / (1.05) + 4410 / (1.05)²

⇒ P = 3150 / (21/20) + 4410 / (21/20)²

⇒ P = (3150 × 20) / 21 + (4410 × 400) / 441

⇒ P = 150 × 20 + 10 × 400

⇒ P = 3000 + 4000

⇒ P = 7000

∴ The correct answer is ₹7000.

Additional Information

Installment Plan in Compound Interest

When a loan is repaid in installments, the principal is the sum of the present values of all future payments. P = ∑ [I_n / (1 + r/100)ⁿ].

Equal Installments Formula

If all installments are equal (I), for 2 years: P = I × [(1 + r/100) + 1] / (1 + r/100)², or more generally using GP sum formulas.

Relationship between SI and CI for 2 Years

The difference between Compound Interest and Simple Interest for two years is given by D = P × (r/100)².

Shortcut Trick

Rate 4% = 1/25 ⇒ Multiplying factor = 26/25.

Time difference = 6 − 3 = 3 years.

Ratio of shares (Ram : Shyam) = (1.04)³ : 1 = (26/25)³ : 1 = 17576 : 15625.

Total units = 17576 + 15625 = 33201 units → ₹260200.

Ram share = (17576 ÷ 33201) × 260200 ≈ ₹137745.

∴ The correct answer is ₹137745.

Alternate Method

Given: Total sum = ₹260200, Rate (r) = 4%, Time for Ram (n₁) = 3 years, Time for Shyam (n₂) = 6 years.

Formula: Amount (A) = P(1 + r/100)ⁿ

⇒ Let Ram share be R and Shyam share be S.

⇒ R(1 + 4/100)³ = S(1 + 4/100)⁶

⇒ R / S = (1 + 0.04)⁶ / (1 + 0.04)³

⇒ R / S = (1.04)⁶⁻³ = (1.04)³

⇒ R / S = (104/100)³ = (26/25)³

⇒ R / S = 17576 / 15625

⇒ Ram share = [17576 / (17576 + 15625)] × 260200

⇒ Ram share = (17576 / 33201) × 260200

⇒ Ram share ≈ 0.52938 × 260200 ≈ ₹137745.1

∴ The correct answer is ₹137745.

Additional Information

Division of Sum for Equal Amounts

When a sum is divided into parts P₁ and P₂ such that amounts after t₁ and t₂ years at rate r% are equal, the ratio is P₁/P₂ = (1 + r/100)^{(t2 − t1)}.

Compound Interest Formula

The amount A accumulated on principal P at rate r% per annum for n years is A = P(1 + r/100)ⁿ.

Successive Percentage Shortcut

For a 4% rate over 3 years, the effective CI rate can be calculated using successive % formula: x + y + (xy/100).

Shortcut Trick

Volume of a cylinder is proportional to (Radius)² × Height.

Ratio of volumes = (Ratio of radii)² × (Ratio of heights)

Ratio of volumes = (2/3)² × (5/3) = (4/9) × (5/3)

Ratio of volumes = 20 : 27

∴ The correct answer is 20 : 27.

Alternate Method

Given:

Ratio of radii (r₁ : r₂) = 2 : 3

Ratio of heights (h₁ : h₂) = 5 : 3

Formula used: Volume of cylinder (V) = πr²h

⇒ Let radii be 2r and 3r, and heights be 5h and 3h.

⇒ Volume of first cylinder (V₁) = π × (2r)² × 5h = 20πr²h

⇒ Volume of second cylinder (V₂) = π × (3r)² × 3h = 27πr²h

⇒ Ratio V₁ : V₂ = 20πr²h : 27πr²h

⇒ Ratio = 20 : 27

∴ The correct answer is 20 : 27.

Additional Information

Curved Surface Area (CSA) of a Cylinder

The area of the curved surface is given by the formula CSA = 2πrh, where r is the radius and h is the height.

Total Surface Area (TSA) of a Cylinder

The total surface area includes the curved surface and the two circular bases: TSA = 2πr(h + r).

Volume of a Cone

If a cone has the same radius and height as a cylinder, its volume is exactly one-third of the cylinder: Volume = (1/3)πr²h.

Shortcut Trick

Total Volume = sum of individual volumes of the three cubes.

Volume of new cube = 6³ + 8³ + 10³

Volume = 216 + 512 + 1000 = 1728 cm³.

Edge of new cube = ³√1728 = 12 cm.

∴ The correct answer is 12 cm.

Alternate Method

Given:

Edge of first cube (a₁) = 6 cm

Edge of second cube (a₂) = 8 cm

Edge of third cube (a₃) = 10 cm

Formula used:

Volume of a cube = (Edge)³

Calculations:

⇒ Volume of 1^st cube = 6 × 6 × 6 = 216 cm³

⇒ Volume of 2^nd cube = 8 × 8 × 8 = 512 cm³

⇒ Volume of 3^rd cube = 10 × 10 × 10 = 1000 cm³

⇒ Let the edge of the new cube be A.

⇒ Total Volume = Volume of new cube

⇒ 216 + 512 + 1000 = A³

⇒ 1728 = A³

⇒ A = ³√1728

⇒ A = 12 cm

∴ The correct answer is 12 cm.

Additional Information

Volume and Surface Area of Cube

For a cube with edge 'a', Volume = a³ and Total Surface Area = 6a².

Melting and Recasting Rule

When solid shapes are melted to form a new shape, the total volume remains constant (Sum of old volumes = New volume).

Diagonal of a Cube

The longest rod that can be placed in a cube is its diagonal, given by d = a√3.

Shortcut Trick

Area of a ring/annulus = π(R² − r²)

Using the identity a² − b² = (a + b)(a − b):

Area = π(68 + 22)(68 − 22)

Area = π × 90 × 46 = 4140π

∴ The correct answer is 4140π sq. cm.

Alternate Method

Given:

Radius of outer circle (R) = 68 cm

Radius of inner circle (r) = 22 cm

Formula Used:

Area of region between two concentric circles = Area of outer circle − Area of inner circle

Area = πR² − πr² = π(R² − r²)

Calculations:

⇒ Area = π × (68² − 22²)

⇒ Area = π × (68 + 22) × (68 − 22)

⇒ Area = π × 90 × 46

⇒ Area = 4140π sq. cm.

∴ The correct answer is 4140π sq. cm.

Additional Information

Circumference of a Circle

The boundary length of a circle is given by C = 2πr, where r is the radius.

Area of a Sector

The area of a portion of a circle subtending an angle θ at the center is (θ/360) × πr².

Concentric Circles

Circles that share the same center point but have different radii are called concentric circles.

Shortcut Trick

When a solid is melted to form another shape, their volumes remain equal.

Length of wire (h) = (4 ÷ 3) × R³ ÷ r²

h = (4 ÷ 3) × 6 × 6 × 6 ÷ (0.2 × 0.2)

h = (4 × 2 × 36) ÷ 0.04 = 288 ÷ 0.04 = 7200 cm

Length in meters = 7200 ÷ 100 = 72 m.

∴ The correct answer is 72 m.

Alternate Method

Given:

Radius of sphere (R) = 6 cm

Radius of wire (r) = 0.2 cm

Formula Used:

Volume of Sphere = (4/3) × π × R³

Volume of Cylinder (Wire) = π × r² × h

Calculations:

⇒ Volume of sphere = Volume of wire

⇒ (4/3) × π × (6)³ = π × (0.2)² × h

⇒ (4/3) × 216 = 0.04 × h

⇒ 4 × 72 = 0.04 × h

⇒ 288 = 0.04 × h

⇒ h = 288 ÷ 0.04

⇒ h = 7200 cm

⇒ h = 7200 ÷ 100 m

⇒ h = 72 m

∴ The correct answer is 72 m.

Additional Information

Volume of Cylinder

The volume of a cylinder with radius r and height h is given by V = πr²h.

Volume of Sphere

The volume of a sphere with radius R is given by V = (4/3)πR³.

Conservation of Volume

When a solid is reshaped by melting or recasting, the total volume before and after remains the same.

Unit Conversion

1 meter (m) = 100 centimeters (cm) | 1 cm² = 100 mm².

Shortcut Trick

For a wire (cylinder), volume V = πr²l. If volume remains constant, l ∝ 1/r².

Given the radius is decreased to 1/3, the new radius factor is 1/3.

The square of the radius factor is (1/3)² = 1/9.

Since length is inversely proportional to r², the length must be multiplied by the reciprocal of 1/9, which is 9.

∴ The correct answer is 9 times.

Alternate Method

Given:

Original radius = r

New radius (r′) = r/3

Original volume (V₁) = New volume (V₂)

Formula Used:

Volume of a cylinder (wire) = πr²l, where r is the radius and l is the length.

Calculations:

⇒ Let the original length be l and the new length be l′.

⇒ V₁ = π × r² × l

⇒ V₂ = π × (r/3)² × l′

⇒ V₂ = π × (r²/9) × l′

⇒ Since V₁ = V₂:

⇒ π × r² × l = π × (r²/9) × l′

⇒ l = l′ / 9

⇒ l′ = 9 × l

∴ The correct answer is 9 times.

Additional Information

Volume of a Cylinder

The volume is given by V = πr²h, where r is the base radius and h is the height or length.

Relationship between Dimensions

When volume is constant, the ratio of lengths is inversely proportional to the ratio of the squares of the radii: l₁ / l₂ = (r₂ / r₁)².

Curved Surface Area (CSA)

The surface area of the side of the wire is 2πrl. In this problem, the CSA changes by a factor of (1/3) × 9 = 3 times.

Shortcut Trick

In right-angled triangle ADC, using Pythagorean triplet (9, 40, 41):

Height (AD) = √(AC² − CD²) = √(41² − 40²) = √81 = 9 cm.

Area of trapezium = 1/2 × (Sum of parallel sides) × Height

Area = 1/2 × (15 + 40) × 9 = 27.5 × 9 = 247.5 cm².

∴ The correct answer is 247.5 cm².

Alternate Method

Given:

Parallel sides AB = 15 cm and CD = 40 cm.

Diagonal AC = 41 cm and ∠ADC = 90°.

Formula Used:

Area of Trapezium = 1/2 × (a + b) × h

Pythagoras Theorem: h² + base² = hypotenuse²

Calculations:

⇒ In ΔADC, ∠D = 90°

⇒ AD² + CD² = AC²

⇒ AD² + 40² = 41²

⇒ AD² = 1681 − 1600

⇒ AD² = 81

⇒ AD = √81 = 9 cm

⇒ Since AB || CD and ∠ADC = 90°, AD is the height (h) of the trapezium.

⇒ Area = 1/2 × (AB + CD) × AD

⇒ Area = 1/2 × (15 + 40) × 9

⇒ Area = 1/2 × 55 × 9

⇒ Area = 27.5 × 9 = 247.5 cm²

∴ The correct answer is 247.5 cm².

Additional Information

Properties of a Right Trapezium

A right trapezium has at least two adjacent right angles. The height is equal to the length of the non-parallel side perpendicular to the bases.

Pythagorean Triplets

Common triplets like (9, 40, 41), (3, 4, 5), and (8, 15, 17) help solve geometry problems quickly without full calculation of squares.

Median of a Trapezium

The median (mid-segment) connects the midpoints of non-parallel sides and its length is the average of the parallel sides: m = (a + b) / 2.

Shortcut Trick

In a rhombus, diagonals bisect each other at 90° forming right-angled triangles.

Half of the given diagonal = 12 ÷ 2 = 6 cm and Side = 10 cm.

Using Pythagorean triplet (6, 8, 10), the other half-diagonal must be 8 cm.

Second diagonal (d₂) = 8 × 2 = 16 cm.

Area = 1/2 × d₁ × d₂ = 1/2 × 12 × 16 = 96 cm².

∴ The correct answer is 96 cm².

Alternate Method

Given:

Side (a) = 10 cm

One diagonal (d₁) = 12 cm

Formula Used:

Area of Rhombus = 1/2 × d₁ × d₂

Relation between side and diagonals: a² = (d₁/2)² + (d₂/2)²

Calculations:

⇒ 10² = (12/2)² + (d₂/2)²

⇒ 100 = 6² + (d₂/2)²

⇒ 100 = 36 + (d₂/2)²

⇒ (d₂/2)² = 100 − 36 = 64

⇒ d₂/2 = √64 = 8 cm

⇒ d₂ = 8 × 2 = 16 cm

⇒ Area = 1/2 × 12 × 16

⇒ Area = 6 × 16 = 96 cm²

∴ The correct answer is 96 cm².

Additional Information

Diagonal and Side Relationship

The relationship between the side (a) and the two diagonals (d₁ and d₂) is given by 4a² = d₁² + d₂².

Properties of Rhombus

All four sides are equal, and the diagonals bisect each other at right angles (90°). Diagonals also bisect the interior angles of the rhombus.

Perimeter of Rhombus

The perimeter of a rhombus is calculated as 4 × side (a).

Shortcut Trick

Area of field = Total levelling cost ÷ Rate = 7700 ÷ 0.50 = 15400 m².

Standard Circle Property: If Area = 154, Radius = 7 and Circumference = 44.

Since Area = 154 × 100, then Radius = 7 × 10 = 70 m and Circumference = 44 × 10 = 440 m.

Total Fencing Cost = Circumference × Rate = 440 × 1.20 = ₹528.

∴ The correct answer is ₹528.

Alternate Method

Given:

Cost of levelling = ₹7700

Rate of levelling = 50 Paise per m² = ₹0.50 per m²

Rate of fencing = ₹1.20 per metre

Formula Used:

Area of a circle = πr²

Circumference of a circle = 2πr

Calculations:

⇒ Area of the field = Total levelling cost ÷ Levelling rate

⇒ Area = 7700 ÷ 0.50 = 15400 m²

⇒ πr² = 15400

⇒ (22/7) × r² = 15400

⇒ r² = (15400 × 7) ÷ 22

⇒ r² = 700 × 7 = 4900

⇒ r = √4900 = 70 m

⇒ Circumference of the field = 2 × π × r

⇒ Circumference = 2 × (22/7) × 70

⇒ Circumference = 2 × 22 × 10 = 440 m

⇒ Cost of fencing = Circumference × Fencing rate

⇒ Cost of fencing = 440 × 1.20 = ₹528

∴ The correct answer is ₹528.

Additional Information

Standard Circle Values (for π = 22/7)

If r = 7, then Circumference = 44 and Area = 154. If radius is multiplied by 'k', circumference is multiplied by 'k' and area by 'k'².

Relationship between Cost and Dimension

Cost of Levelling = Area × Rate per unit area. Cost of Fencing = Perimeter (Circumference) × Rate per unit length.

Area in terms of Circumference

Area = C² ÷ (4π), where C is the circumference of the circle.

Shortcut Trick

When squares of side 's' are cut from corners of a sheet (L × W), the box dimensions are (L−2s), (W−2s), and 's'.

Length of box = 25 − 2(2) = 21 cm

Width of box = 20 − 2(2) = 16 cm

Height of box = 2 cm

Volume = 21 × 16 × 2 = 672 cm³.

∴ The correct answer is 672 cm³.

Alternate Method

Given: Dimensions of rectangular sheet = 25 cm × 20 cm. Side of the square cut from each corner = 2 cm.

Formula: Volume of a cuboid = Length × Breadth × Height

⇒ Length of the box (l) = 25 − (2 + 2) = 21 cm

⇒ Breadth of the box (b) = 20 − (2 + 2) = 16 cm

⇒ Height of the box (h) = Side of the square cut = 2 cm

⇒ Volume of the box = l × b × h

⇒ Volume = 21 × 16 × 2

⇒ Volume = 336 × 2 = 672 cm³

∴ The correct answer is 672 cm³.

Additional Information

Total Surface Area of Cuboid

The total surface area of a closed cuboid is given by 2(lb + bh + hl), where l, b, and h are length, breadth, and height.

Surface Area of an Open Box

For an open box, the surface area is the area of the base plus the area of the four side walls: lb + 2h(l + b).

Diagonal of a Cuboid

The length of the longest rod that can be placed in a cuboidal box is the diagonal, calculated as √(l² + b² + h²).

Shortcut Trick

For a right circular cylinder with height h and Total Surface Area TSA, the radius r can be directly calculated as:

r = √((h ÷ 2)² + TSA ÷ 2π) − (h ÷ 2)

Given h = 4 cm and TSA = 8π sq.cm.

r = √((4 ÷ 2)² + 8π ÷ 2π) − (4 ÷ 2) = √(4 + 4) − 2 = √8 − 2 = 2√2 − 2 cm.

∴ The correct answer is (2√2 − 2) cm.

Alternate Method

Given:

Height of cylinder (h) = 4 cm

Total Surface Area (TSA) = 8π sq.cm.

Formula Used:

Total Surface Area of Cylinder = 2πr(r + h)

Calculations:

⇒ 2 × π × r(r + 4) = 8 × π

⇒ r(r + 4) = 4

⇒ r² + 4r − 4 = 0

⇒ Using quadratic formula r = [−b ± √(b² − 4ac)] ÷ 2a

⇒ r = [−4 ± √(4² − 4 × 1 × (−4))] ÷ (2 × 1)

⇒ r = [−4 ± √(16 + 16)] ÷ 2

⇒ r = [−4 ± √32] ÷ 2

⇒ r = [−4 ± 4√2] ÷ 2

⇒ r = −2 ± 2√2

⇒ Since radius cannot be negative, r = 2√2 − 2 cm

∴ The correct answer is (2√2 − 2) cm.

Additional Information

Curved Surface Area of Cylinder

The area of the curved side of a cylinder excluding the top and bottom bases is given by CSA = 2πrh.

Volume of Cylinder

The total space occupied by the cylinder is calculated using the formula V = πr²h.

Relation between TSA and CSA

Total Surface Area (TSA) is the sum of Curved Surface Area (CSA) and the area of the two circular bases: TSA = CSA + 2πr².

Shortcut Trick

Ratio given: r = h ÷ 2 ⇒ h = 2r.

Inner surface area (open top) = πr² + 2πrh = 5πr².

5 × (22/7) × r² = 616 ⇒ r² = 39.2.

r = √39.2 = 14 ÷ √5 = 14 ÷ 2.23 ≈ 6.28 cm.

Volume = πr²h = (22/7) × 39.2 × (2 × 6.28) ≈ 1547 cm³ ≈ 1.547 litres.

∴ The correct answer is 1.53 litres.

Alternate Method

Given:

Radius (r) = h ÷ 2 ⇒ Height (h) = 2r

Total Surface Area (Inner) = 616 cm²

π = 22/7, √5 = 2.23

Formula Used:

Inner TSA of open cylinder = Area of base + Curved Surface Area = πr² + 2πrh

Volume of cylinder = πr²h

Calculations:

⇒ TSA = πr² + 2πr(2r) = πr² + 4πr² = 5πr²

⇒ 5 × (22/7) × r² = 616

⇒ (110/7) × r² = 616

⇒ r² = (616 × 7) ÷ 110 = 5.6 × 7 = 39.2

⇒ r = √39.2 = √(196 ÷ 5) = 14 ÷ √5

⇒ r = 14 ÷ 2.23 ≈ 6.278 cm

⇒ h = 2 × 6.278 = 12.556 cm

⇒ Volume (V) = πr²h = (22/7) × 39.2 × 12.556

⇒ V = 123.2 × 12.556 = 1546.89 cm³

⇒ Capacity in litres = 1546.89 ÷ 1000 ≈ 1.547 litres

⇒ Closest option is 1.53 litres.

∴ The correct answer is 1.53 litres.

Additional Information

Volume of Cylinder

The volume of a cylinder is given by V = πr²h, where r is the radius and h is the height.

Surface Area of Cylinder

Curved Surface Area (CSA) = 2πrh. Total Surface Area (TSA) of a closed cylinder = 2πr(r + h).

Capacity Conversion

1 litre = 1000 cm³ or 1000 millilitres. For calculations in litres, divide the volume in cm³ by 1000.

Open vs Closed Containers

For an open-top container, the surface area is base area + curved area (πr² + 2πrh), whereas a closed one includes two circular bases (2πr² + 2πrh).

Shortcut Trick

Volume of sphere = Volume of cylinder ⇒ (4/3) × 21³ = r² × 7 (since 2.1 dm = 21 cm)

Solving for cylinder radius: r² = 4 × 21² ⇒ r = 42 cm

Ratio of TSA (Rod : Sphere) = 2πr(r + h) : 4πR²

Ratio = [2 × 42 × (42 + 7)] : [4 × 21 × 21] = [84 × 49] : [4 × 441]

Ratio = 4116 : 1764 = 7 : 3

∴ The correct answer is 7 : 3.

Alternate Method

Given:

Radius of sphere (R) = 2.1 dm = 21 cm

Length of rod (h) = 7 cm

Formula Used:

Volume of Sphere = (4/3)πR³

Volume of Cylinder = πr²h

Total Surface Area (TSA) of Sphere = 4πR²

TSA of Cylinder = 2πr(r + h)

Calculations:

⇒ Volume of sphere = Volume of cylinder

⇒ (4/3) × π × (21)³ = π × r² × 7

⇒ (4/3) × 21 × 21 × 21 = r² × 7

⇒ 4 × 7 × 21 × 21 = r² × 7

⇒ r² = 4 × 441 = 1764

⇒ r = √1764 = 42 cm

⇒ TSA of sphere = 4 × π × (21)² = 1764π

⇒ TSA of rod = 2 × π × 42 × (42 + 7) = 2 × π × 42 × 49 = 4116π

⇒ Ratio = TSA(rod) : TSA(sphere) = 4116π : 1764π

⇒ Ratio = 4116 / 1764 = 7 / 3

∴ The correct answer is 7 : 3.

Additional Information

Conversion of Solids

When one solid is melted and recast into another, the total volume remains constant (Volume₁ = Volume₂).

Unit Conversion

1 decimetre (dm) = 10 centimetres (cm) | 1 metre = 10 dm = 100 cm.

Cylinder and Sphere Formulas

Curved Surface Area (CSA) of cylinder = 2πrh; Surface Area of sphere = 4πR².

Shortcut Trick

Sum of sides (L + B) = 6 ⇒ (L + B)² = 36 ⇒ L² + B² + 2LB = 36.

Ratio of Area_square : Area_rect = (L² + B²) : LB = 5 : 2.

Let L² + B² = 5x and LB = 2x.

Substitute in expansion: 5x + 2(2x) = 36 ⇒ 9x = 36 ⇒ x = 4.

Area of square = L² + B² = 5x = 5 × 4 = 20 cm².

∴ The correct answer is 20 cm².

Alternate Method

Given:

Sum of length (L) and breadth (B) = 6 cm

Side of square (s) = Diagonal of rectangle (d)

Ratio of Area of square to Area of rectangle = 5 : 2

Formula Used:

Diagonal of rectangle (d) = √(L² + B²)

Area of square = s² = d² = L² + B²

Area of rectangle = L × B

Calculations:

⇒ L + B = 6

⇒ (L + B)² = 6² = 36

⇒ L² + B² + 2LB = 36 --- (Equation 1)

⇒ Area of Square / Area of Rectangle = (L² + B²) / LB = 5 / 2

⇒ 2(L² + B²) = 5LB

⇒ LB = 0.4(L² + B²)

⇒ Substitute LB in Equation 1:

⇒ (L² + B²) + 2(0.4)(L² + B²) = 36

⇒ (L² + B²) + 0.8(L² + B²) = 36

⇒ 1.8(L² + B²) = 36

⇒ L² + B² = 36 ÷ 1.8

⇒ L² + B² = 20

⇒ Area of square = s² = d² = L² + B² = 20 cm²

∴ The correct answer is 20 cm².

Additional Information

Properties of Rectangles and Squares

For a rectangle with sides L and B, the diagonal d = √(L² + B²) and perimeter P = 2(L + B).

Relation between Sum and Squares

The algebraic identity (a + b)² = a² + b² + 2ab connects the sum of dimensions, the area of the rectangle, and the square of the diagonal.

Area of Square via Diagonal

Area of a square can be expressed as (Side)² or (Diagonal)² ÷ 2 if the diagonal of the square itself is given.

Shortcut Trick

For an equilateral triangle of side a, the area between the circumcircle and incircle is always π × (a/2)².

Given side a = 8 cm.

Area = (22/7) × (8/2)²

Area = (22/7) × 4² = (22/7) × 16 = 352/7

Area = 50 2/7 cm².

∴ The correct answer is 50 2/7 cm².

Alternate Method

Given:

Side of equilateral triangle (a) = 8 cm

π = 22/7

Formula Used:

Circumradius (R) = a / √3

Inradius (r) = a / (2√3)

Required Area = Area of circumcircle − Area of incircle = πR² − πr² = π(R² − r²)

⇒ Circumradius (R) = 8 / √3 cm

⇒ Inradius (r) = 8 / (2√3) = 4 / √3 cm

⇒ Area = π × [(8/√3)² − (4/√3)²]

⇒ Area = π × [64/3 − 16/3]

⇒ Area = π × [48/3]

⇒ Area = π × 16

⇒ Area = (22/7) × 16 = 352/7

⇒ Area = 50 2/7 cm²

∴ The correct answer is 50 2/7 cm².

Additional Information

Equilateral Triangle Radius Relation

For any equilateral triangle, the circumradius (R) is always double the inradius (r), i.e., R = 2r.

Area of Circles Ratio

The ratio of the area of the circumcircle to the area of the incircle in an equilateral triangle is always 4 : 1.

Height of Equilateral Triangle

The height (h) is the sum of the circumradius and inradius: h = R + r = (√3/2) × a.

Area of Equilateral Triangle

The area of the triangle itself is given by (√3/4) × a².

Shortcut Trick

Volume of Sphere = Volume of Hollow Cylinder

4/3 × 3³ = 4 × (5² − r²)

36 = 4 × (25 − r²) ⇒ 9 = 25 − r²

r² = 16 ⇒ r = 4 cm

Thickness = External Radius (R) − Internal Radius (r) = 5 − 4 = 1 cm

∴ The correct answer is 1 cm.

Alternate Method

Given:

Radius of solid sphere (r_s) = 3 cm

Length of hollow cylinder (h) = 4 cm

External radius of cylinder (R) = 5 cm

Formula Used:

Volume of sphere = 4/3 × π × r³

Volume of hollow cylinder = π × h × (R² − r²)

Calculations:

⇒ Volume of Sphere = 4/3 × π × (3)³

⇒ Volume of Sphere = 4/3 × π × 27 = 36π cm³

⇒ Volume of Hollow Cylinder = π × 4 × (5² − r²)

⇒ 36π = 4π × (25 − r²)

⇒ 36 / 4 = 25 − r²

⇒ 9 = 25 − r²

⇒ r² = 25 − 9 = 16

⇒ r = √16 = 4 cm

⇒ Thickness of tube = External radius (R) − Internal radius (r)

⇒ Thickness = 5 cm − 4 cm = 1 cm

∴ The correct answer is 1 cm.

Additional Information

Volume of Solid Objects

For a solid sphere, Volume = 4/3 πr³. When an object is melted and reshaped, its total volume remains constant.

Hollow Cylinder Properties

Volume = πh(R² − r²), where R is external radius and r is internal radius. Thickness = R − r.

Surface Area of Sphere

The total surface area of a solid sphere is given by 4πr².

Shortcut Trick If x2 + 1/x2 = a, then x + 1/x = √(a + 2). Given x2 + 1/x2 = 98, so x + 1/x = √(98 + 2) = 10 (since x > 0). If x + 1/x = k, then x3 + 1/x3 = k3 − 3k. Required value = 103 − 3(10) = 1000 − 30 = 970. ∴ The correct answer is 970. Alternate Method Given: x2 + 1/x2 = 98 and x > 0 Formula used: (x + 1/x)2 = x2 + 1/x2 + 2 and x3 + 1/x3 = (x + 1/x)3 − 3(x + 1/x) ⇒ (x + 1/x)2 = 98 + 2 ⇒ (x + 1/x)2 = 100 ⇒ x + 1/x = √100 ⇒ x + 1/x = 10 ⇒ x3 + 1/x3 = (10)3 − 3 × (10) ⇒ x3 + 1/x3 = 1000 − 30 ⇒ x3 + 1/x3 = 970 ∴ The correct answer is 970. Additional Information Square Reciprocal Identity For any positive x, x2 + 1/x2 = (x + 1/x)2 − 2 or (x − 1/x)2 + 2. Cubic Reciprocal Identity The sum of cubes is derived as x3 + 1/x3 = (x + 1/x)3 − 3(x + 1/x) and the difference as x3 − 1/x3 = (x − 1/x)3 + 3(x − 1/x). Successive Higher Powers To find x4 + 1/x4, square the value of x2 + 1/x2 and subtract 2.

Shortcut Trick

Use the substitution method by picking a value for 'a' that satisfies the conditions.

Let a = 1/2.

Substitute in first eq: 1/2 + 1/b = 1 ⇒ 1/b = 1/2 ⇒ b = 2.

Substitute in second eq: 2 + 1/c = 1 ⇒ 1/c = −1 ⇒ c = −1.

Calculate: c + 1/a = −1 + 1/(1/2) = −1 + 2 = 1.

∴ The correct answer is 1.

Alternate Method

Given: a + 1/b = 1 and b + 1/c = 1

Formula: Express c and 1/a in terms of the same variable to find the sum.

⇒ From the first equation: a + 1/b = 1

⇒ 1/b = 1 − a

⇒ b = 1 / (1 − a)

⇒ From the second equation: b + 1/c = 1

⇒ 1/c = 1 − b

⇒ 1/c = 1 − [1 / (1 − a)]

⇒ 1/c = (1 − a − 1) / (1 − a) = −a / (1 − a)

⇒ c = (1 − a) / −a = −1/a + 1

⇒ c + 1/a = 1

∴ The correct answer is 1.

Additional Information

Cyclic Symmetric Expressions

In cyclic equations of the form x + 1/y = k and y + 1/z = k, the third expression z + 1/x usually results in the same constant k.

Reciprocal Algebra

If x + 1/y = 1, then y = 1 / (1 − x). This transformation is useful for solving nested fraction problems.

Product Rule

Multiplying such cyclic equations can often simplify the expression to abc + 1 = 0 or similar identities depending on the constant.

Shortcut Trick

Use substitution: Let y = 1 and z = 1.

Since x = y + z, then x = 1 + 1 = 2.

Calculate x³ − y³ − z³ ⇒ 2³ − 1³ − 1³ = 8 − 1 − 1 = 6.

Check options: 3xyz = 3 × 2 × 1 × 1 = 6, which matches the result.

∴ The correct answer is 3xyz.

Alternate Method

Given: x = y + z

Formula used: (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ x = y + z

Cubing both sides:

⇒ x³ = (y + z)³

⇒ x³ = y³ + z³ + 3yz(y + z)

Substitute (y + z) = x back into the equation:

⇒ x³ = y³ + z³ + 3yz(x)

⇒ x³ = y³ + z³ + 3xyz

Rearranging the terms:

⇒ x³ − y³ − z³ = 3xyz

∴ The correct answer is 3xyz.

Additional Information

Conditional Identity

If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Trinomial Cube Factorisation

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

Binomial Cube (Subtraction)

(a − b)³ = a³ − b³ − 3ab(a − b).

Shortcut Trick

Assume convenient values for a, b, c, and d such that their sum is 4 and no denominator becomes zero.

Let a = 4, b = 0, c = 0, and d = 0 (since 4 + 0 + 0 + 0 = 4).

Substitute these values: 1/((1 − 4)(1 − 0)(1 − 0)) + 1/((1 − 0)(1 − 0)(1 − 0)) + 1/((1 − 0)(1 − 0)(1 − 4)) + 1/((1 − 0)(1 − 4)(1 − 0))

⇒ 1/(−3) + 1/1 + 1/(−3) + 1/(−3) = −1/3 + 1 − 1/3 − 1/3 = 1 − 1 = 0.

∴ The correct answer is 0.

Alternate Method

Given: a + b + c + d = 4

Formula Used: Algebraic transformation and common denominator method.

Let x = (1 − a), y = (1 − b), z = (1 − c), and w = (1 − d).

⇒ x + y + z + w = (1 − a) + (1 − b) + (1 − c) + (1 − d)

⇒ x + y + z + w = 4 − (a + b + c + d)

⇒ x + y + z + w = 4 − 4 = 0

⇒ The given expression is: 1/(xyz) + 1/(yzw) + 1/(zwx) + 1/(wxy)

⇒ Taking the LCM of denominators: (w + x + y + z) / (xyzw)

⇒ Since x + y + z + w = 0, the expression becomes: 0 / (xyzw) = 0

∴ The correct answer is 0.

Additional Information

Variable Substitution

In competitive math, substituting variables that satisfy the given condition (like x + y + z + w = 0) is the most efficient way to simplify complex rational expressions.

Symmetric Expressions

When terms in an expression are cyclic or symmetric, the sum often simplifies to zero or a constant based on the total sum of the individual variables.

BODMAS in Algebra

Always follow the order of operations when simplifying denominators before combining terms using a least common multiple (LCM).

Shortcut Trick

Simplify denominators: √12 − √18 = √6(√2 − √3) = −√6(√3 − √2).

First term: (√3 − √2) ÷ [−√6(√3 − √2)] = −1/√6.

Second term: (1/3) × √27 = (1/3) × 3√3 = √3.

Third term: (1/2) × ³√27 = (1/2) × 3 = 1.5 = 1 + 1/2.

Total = −1/√6 − √3 − 1 − 1/2 = − [ √3 + 1 + 1/2 + 1/√6 ].

∴ The correct answer is − [ √3 + 1 + 1/2 + 1/√6 ].

Alternate Method

Given:

Expression = [(√3 − √2) / (√12 − √18)] − [(1/3) × √27] − [(1/2) × ³√27]

Formula/Properties Used:

√(a²b) = a√b | ³√a³ = a | Rationalization of surds.

Calculations:

⇒ Step 1: Simplify √12, √18, √27, and ³√27

⇒ √12 = √(4 × 3) = 2√3

⇒ √18 = √(9 × 2) = 3√2

⇒ √27 = √(9 × 3) = 3√3

⇒ ³√27 = ³√3³ = 3

⇒ Step 2: Substitute these values into the expression

⇒ [(√3 − √2) / (2√3 − 3√2)] − [(1/3) × 3√3] − [(1/2) × 3]

⇒ Step 3: Rationalize the first term

⇒ [(√3 − √2)(2√3 + 3√2)] / [(2√3 − 3√2)(2√3 + 3√2)]

⇒ [2(3) + 3√6 − 2√6 − 3(2)] / [4(3) − 9(2)]

⇒ [6 + √6 − 6] / [12 − 18] = √6 / −6 = −1/√6

⇒ Step 4: Simplify the remaining terms

⇒ Term 2 = −√3

⇒ Term 3 = −3/2 = −(1 + 1/2)

⇒ Step 5: Combine all terms

⇒ −1/√6 − √3 − 1 − 1/2

⇒ − [ √3 + 1 + 1/2 + 1/√6 ]

∴ The correct answer is − [ √3 + 1 + 1/2 + 1/√6 ].

Additional Information

Rationalization of Surds

To rationalize a denominator of the form (a − b), multiply both the numerator and denominator by its conjugate (a + b) to eliminate roots using (a − b)(a + b) = a² − b².

Laws of Radicals

Properties like √ab = √a × √b and ⁿ√aⁿ = a are fundamental for simplifying complex expressions involving roots and powers.

BODMAS Rule

Operations must follow the order: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction for accurate results.