SSC CGL Tier 2 Quant Previous Year Paper (Held On: 1 December 2016)

Shortcut Trick To find the order of x, √x, and x2 when 0 < x < 1, pick a convenient value like x = 0.25 (or 1/4). Step 1: x = 0.25 Step 2: x2 = 0.25 × 0.25 = 0.0625 Step 3: √x = √0.25 = 0.5 Comparing the values: 0.0625 < 0.25 < 0.5 ⇒ x2 < x < √x. ∴ The correct answer is x2 < x < √x.

Alternate Method Given: 0 < x < 1 (x is a proper fraction). Property Used: For any positive number x < 1, increasing the power decreases the value, and taking a root increases the value.

⇒ Let us take x = 0.36. ⇒ Square of x: x2 = 0.36 × 0.36 = 0.1296. ⇒ Square root of x: √x = √0.36 = 0.6. ⇒ Comparing 0.1296, 0.36, and 0.6. ⇒ Since 0.1296 < 0.36 < 0.6, the order is x2 < x < √x. ∴ The correct answer is x2 < x < √x.

Additional Information Inequality for x > 1 When x is greater than 1, the order reverses such that √x < x < x2. For example, if x = 4, then √4 = 2 and 42 = 16. Powers of Proper Fractions For any x in the range (0, 1), xn+1 < xn for all n > 0. Higher powers yield smaller results because multiplying a fraction by a value less than 1 reduces it. Roots of Proper Fractions For any x in the range (0, 1), x1/n > x for all n > 1. Taking a root of a proper fraction always results in a larger value than the original fraction.

Shortcut Trick

For recurring events like bells ringing, the common interval is the LCM of the individual intervals.

LCM (36, 40, 48) = 720 seconds.

Convert seconds to minutes: 720 ÷ 60 = 12 minutes.

∴ The correct answer is 12 minutes.

Alternate Method

Given intervals: 36 seconds, 40 seconds, and 48 seconds.

Formula: Time when they ring together = LCM of individual time intervals.

⇒ Prime factorization of intervals:

⇒ 36 = 2 × 2 × 3 × 3 = 2² × 3²

⇒ 40 = 2 × 2 × 2 × 5 = 2³ × 5¹

⇒ 48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3¹

⇒ LCM = 2⁴ × 3² × 5¹

⇒ LCM = 16 × 9 × 5 = 720

⇒ Time in minutes = 720 ÷ 60 = 12 minutes.

∴ The correct answer is 12 minutes.

Additional Information

Least Common Multiple (LCM)

LCM is the smallest positive integer that is divisible by each of the given numbers. In timing problems involving bells or lights, it represents the first common moment of occurrence.

Unit Conversion

To convert seconds into minutes, divide the total seconds by 60 (1 minute = 60 seconds).

HCF and LCM Relation

For two numbers a and b, Product of numbers = HCF(a, b) × LCM(a, b). This property is useful for finding one value when others are known.

Shortcut Trick

Test with a simple three-digit number: Let the number be 100.

Sum of digits of 100 = 1 + 0 + 0 = 1.

Subtracting the sum from the number: 100 − 1 = 99.

99 is divisible by both 3 and 9, but not by 6.

Test with another number: 123. Sum = 1+2+3 = 6. Difference = 123 − 6 = 117 (Divisible by 9 and 3).

∴ The correct answer is both 3 and 9.

Alternate Method

Let the three-digit number be represented as 100a + 10b + c, where a, b, and c are the digits.

The sum of the digits is given by: a + b + c.

⇒ Original Number = 100a + 10b + c

⇒ Sum of Digits = a + b + c

⇒ Required Difference = (100a + 10b + c) − (a + b + c)

⇒ Difference = 100a − a + 10b − b + c − c

⇒ Difference = 99a + 9b

⇒ Difference = 9 × (11a + b)

⇒ Since the result is a multiple of 9, it must be divisible by 9 and all its factors (1, 3, 9).

∴ The correct answer is both 3 and 9.

Additional Information

Divisibility Rule of 9

A number is divisible by 9 if the sum of its digits is a multiple of 9. Consequently, any number that is a multiple of 9 is also divisible by 3.

General Property of Digits

For any n-digit number N, the expression (N − sum of digits of N) is always divisible by 9. This forms the basis of the casting out nines method.

Divisibility by 6

A number is divisible by 6 if it is divisible by both 2 (even number) and 3. In this question, the difference is not always even, so it is not always divisible by 6.

Given: Fractions: $\frac{2}{3}, \frac{3}{5}, \frac{11}{15}$ Formula used: Common denominator method (LCM) Calculation: LCM of 3, 5, and 15 = 15 ⇒ $\frac{2}{3} = \frac{2 \times 5}{15} = \frac{10}{15}$ ⇒ $\frac{3}{5} = \frac{3 \times 3}{15} = \frac{9}{15}$ ⇒ $\frac{11}{15} = \frac{11}{15}$ Comparing the numerators: ⇒ 9 < 10 < 11 ⇒ $\frac{9}{15} < \frac{10}{15} < \frac{11}{15}$ ⇒ 3/5 < 2/3 < 11/15 ∴ The correct answer is option (2).

Shortcut Trick

The difference between (Sum) and (Difference) of two numbers is always twice the smaller number.

2 × Smaller Number = 30

⇒ Smaller Number = 15

Product = Greater Number × Smaller Number = 900

⇒ Greater Number = 900 ÷ 15 = 60

∴ The correct answer is 60.

Alternate Method

Given:

Product of two numbers (x × y) = 900

Sum exceeds Difference by 30 ⇒ (x + y) − (x − y) = 30

Formula Used:

(A + B) − (A − B) = 2B

Calculations:

⇒ (x + y) − (x − y) = 30

⇒ x + y − x + y = 30

⇒ 2y = 30

⇒ y = 15 (Smaller Number)

⇒ x × y = 900

⇒ x × 15 = 900

⇒ x = 900 ÷ 15

⇒ x = 60

∴ The correct answer is 60.

Additional Information

Basic Algebraic Identities

1. (x + y) + (x − y) = 2x (where x is the greater number).
   
2. (x + y) − (x − y) = 2y (where y is the smaller number).
   
3. (x + y)² − (x − y)² = 4xy.

Sum and Product of Roots

If S is the sum and P is the product of two numbers, the numbers are the roots of the quadratic equation t² − St + P = 0.

Given:

Numbers: $2\frac{1}{2}, 3\frac{1}{3}, 4\frac{1}{4}, 5\frac{1}{5}$

Formula used:

Sum = Integer parts + Fractional parts

Calculation:

Sum = $2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5}$

⇒ Sum = $(2 + 3 + 4 + 5) + (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5})$

⇒ Sum = $14 + (\frac{30 + 20 + 15 + 12}{60})$

⇒ Sum = $14 + \frac{77}{60}$

⇒ Sum = $14 + 1 + \frac{17}{60}$

⇒ Sum = $15 + \frac{17}{60}$

To make the result a whole number, add fraction x:

⇒ $\frac{17}{60} + x = 1$

⇒ $x = 1 - \frac{17}{60}$

⇒ $x = \frac{43}{60}$

∴ The correct answer is option (4).

Given:

Expression = $\sqrt[3]{-13824}$

Formula used:

$\sqrt[3]{-x} = -\sqrt[3]{x}$

Calculation:

⇒ $\sqrt[3]{-13824} = -\sqrt[3]{13824}$

⇒ 13824 = 24 × 24 × 24

⇒ 13824 = 24³

⇒ $-\sqrt[3]{24^3}$ = -24

∴ The correct answer is option (4).

Shortcut Trick

Given a + b + c = 18 and a + b = c, we get 2c = 18 → c = 9.

Product abc = 162 ⇒ ab × 9 = 162 ⇒ ab = 18.

Sum a + b = 9 and product ab = 18 implies the numbers are 3 and 6 (since 3 + 6 = 9 and 3 × 6 = 18).

Numbers are 3, 6, and 9.

Sum of squares = 3² + 6² + 9² = 9 + 36 + 81 = 126.

∴ The correct answer is 126.

Alternate Method

Let the three positive numbers be a, b, and c.

⇒ Given: a + b + c = 18

⇒ a + b = c

⇒ c + c = 18 ⇒ 2c = 18 ⇒ c = 9

⇒ Product: a × b × c = 162

⇒ a × b × 9 = 162 ⇒ ab = 18

⇒ Also, a + b = 9

⇒ Using (a + b)² = a² + b² + 2ab

⇒ 9² = a² + b² + 2(18)

⇒ 81 = a² + b² + 36 ⇒ a² + b² = 45

⇒ Sum of squares = a² + b² + c²

⇒ Sum of squares = 45 + 9² = 45 + 81 = 126

∴ The correct answer is 126.

Additional Information

Algebraic Identity for Three Variables

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca). This relates the sum, the sum of squares, and the sum of products taken two at a time.

Quadratic Formation

If the sum (S) and product (P) of two numbers are known, the numbers are roots of the equation x² − Sx + P = 0.

Arithmetic Progression Property

In this specific problem, since 3, 6, 9 differ by a constant 3, they are in Arithmetic Progression (AP).

Shortcut Trick

For 3 consecutive even numbers, the Average is always the Middle number.

Sum = 3 × Average.

Given: Sum − Average = 28 ⇒ 2 × Average = 28.

Average (Middle Number) = 14.

Smallest even number = 14 − 2 = 12.

∴ The correct answer is 12.

Alternate Method

Given: Sum of three consecutive even numbers is 28 more than their average.

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

Let the three consecutive even numbers be x, (x + 2), and (x + 4).

⇒ Sum = x + (x + 2) + (x + 4) = 3x + 6

⇒ Average = (3x + 6) ÷ 3 = x + 2

According to the question:

⇒ 3x + 6 = (x + 2) + 28

⇒ 3x + 6 = x + 30

⇒ 3x − x = 30 − 6

⇒ 2x = 24

⇒ x = 12

∴ The correct answer is 12.

Additional Information

Average of Consecutive Terms

For any set of numbers in an Arithmetic Progression (AP) with an odd number of terms, the average is always the middle term.

Sum of Even Numbers

The sum of the first n even natural numbers is given by the formula n(n + 1). For consecutive even numbers starting from x, the difference between terms is always 2.

Average of First and Last Term

In any AP, Average = (First Term + Last Term) ÷ 2. In this case, (x + x + 4) ÷ 2 = x + 2.

Shortcut Trick

Given remainder r = 46.

Divisor (d) = 5 × r = 5 × 46 = 230.

Quotient (q) = d ÷ 10 = 230 ÷ 10 = 23.

Dividend = (d × q) + r = (230 × 23) + 46 = 5290 + 46 = 5336.

∴ The correct answer is 5336.

Alternate Method

Given:

Remainder (r) = 46

Divisor (d) = 10 × q (Quotient)

Divisor (d) = 5 × r (Remainder)

Formula Used:

Dividend = (Divisor × Quotient) + Remainder

Calculations:

⇒ d = 5 × 46 = 230

⇒ 10 × q = 230

⇒ q = 230 ÷ 10 = 23

⇒ Dividend = (230 × 23) + 46

⇒ Dividend = 5290 + 46

⇒ Dividend = 5336

∴ The correct answer is 5336.

Additional Information

Division Algorithm

For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

Divisibility Rule

A number is said to be completely divisible by the divisor if the remainder r = 0.

Properties of Remainder

The remainder is always non-negative and strictly less than the divisor. If the dividend is multiplied by k, the new remainder will be (r × k) mod d.

Shortcut Trick

Use the direct formula for individual time: Time = (T₁ × T₂) ÷ (T₁ − T₂)

Here, T₁ = 30 hours (Man) and T₂ = 20 hours (Man + Son)

Time taken by Son = (30 × 20) ÷ (30 − 20)

Time taken by Son = 600 ÷ 10 = 60 hours

∴ The correct answer is 60 hours.

Alternate Method

Given:

Time taken by Man = 30 hours

Time taken by Man + Son = 20 hours

Formula Used:

Total Work = LCM of individual times; Efficiency = Total Work ÷ Time

Calculations:

⇒ Efficiency of Man = 2 units/hour

⇒ Efficiency of Man + Son = 3 units/hour

⇒ Efficiency of Son = (Efficiency of Man + Son) − (Efficiency of Man)

⇒ Efficiency of Son = 3 − 2 = 1 unit/hour

⇒ Time taken by Son alone = Total Work ÷ Efficiency of Son

⇒ Time taken by Son = 60 ÷ 1 = 60 hours

∴ The correct answer is 60 hours.

Additional Information

Combined Work Principle

If A can do a work in x days and B can do it in y days, then together they can do it in (x × y) ÷ (x + y) days.

Work and Efficiency Relationship

Efficiency is inversely proportional to the time taken to complete the work. Total Work = Efficiency × Time.

Pipes and Cisterns Analog

This problem is similar to a tank being filled by two pipes; if one pipe is closed or acts as an outlet (negative efficiency), the net rate changes accordingly.

Shortcut Trick In 1 hour, the tap fills 1/p part of the tub. In 1 hour, the sink empties 1/q part of the tub. Since the tub fills in r hours, the net fill rate per hour is 1/r. Net Rate = Rate of Filling − Rate of Emptying. ∴ 1/r = 1/p − 1/q. ∴ The correct answer is 1/r = 1/p − 1/q.

Alternate Method Given: Time taken by tap to fill = p hours Time taken by sink to empty = q hours Combined time taken to fill = r hours Let the total capacity of the tub be the LCM of p and q, which is pq units.

Person/Pipe Time (hrs) Total Work Efficiency

Tap (Filling) p pq q

Sink (Emptying) q −p

Calculations: ⇒ Total Efficiency when both are open = q + (−p) = q − p ⇒ Combined time (r) = Total Work ÷ Net Efficiency ⇒ r = pq ÷ (q − p) ⇒ 1/r = (q − p) ÷ pq ⇒ 1/r = q/pq − p/pq ⇒ 1/r = 1/p − 1/q ∴ The correct answer is 1/r = 1/p − 1/q.

Additional Information Pipes and Cisterns Rule If a pipe can fill a tank in x hours and another can empty it in y hours (y > x), then the net part filled in 1 hour is (1/x − 1/y). Condition for Filling For a tank to be filled when both inlet and outlet are open, the efficiency of the inlet pipe must be greater than the outlet pipe, i.e., p < q. Work and Time Reciprocal Total time taken to complete a task is always the reciprocal of the total work done in one unit of time (Efficiency).

Shortcut Trick

John's full work time = 3 × 2 = 6 hours.

Remaining work after John = 1/2; Joe's full work time = 1 ÷ (1/4 × 1/2) = 8 hours.

Work remaining for George = 1 − (1/2 + 1/8) = 3/8; George's full work time = 5 ÷ (3/8) = 40/3 hours.

Combined efficiency with LCM 120: John = 20, Joe = 15, George = 9.

Total time = Total work ÷ Total efficiency = 120 ÷ (20 + 15 + 9) = 120 ÷ 44 = 30/11 = 2 8/11 hours.

∴ The correct answer is 2 8/11 hours.

Alternate Method

Given:

John: 1/2 work in 3 hours ⇒ 1 work in 6 hours.

Remaining work after John = 1 − 1/2 = 1/2.

Joe: 1/4 of 1/2 = 1/8 work in 1 hour ⇒ 1 work in 8 hours.

Remaining work after Joe = 1/2 − 1/8 = 3/8.

George: 3/8 work in 5 hours ⇒ 1 work in 5 × (8/3) = 40/3 hours.

Formula Used:

Total Work = LCM of individual times; Efficiency = Total Work ÷ Time; Time Together = Total Work ÷ Sum of Efficiencies.

Calculations:

⇒ Total Efficiency of John, Joe, and George = 20 + 15 + 9 = 44 units/hour.

⇒ Time taken by all three = Total Work ÷ Total Efficiency

⇒ Time = 120 ÷ 44 hours

⇒ Time = 30 ÷ 11 = 2 8/11 hours.

∴ The correct answer is 2 8/11 hours.

Additional Information

Work Rate Principle

The rate of work is the reciprocal of the time taken to complete the work (Rate = 1/Time). If multiple people work together, their combined rate is the sum of their individual rates.

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken. If a person is twice as efficient as another, they will take half the time to complete the same work.

LCM Method for Fractions

When time is in fractions (like 40/3), the LCM of the numerators is used as the total work units to simplify the efficiency calculations.

Shortcut Trick

A (full work) = 9 × 5/2 = 22.5 days

(A + B) (remaining 3/5 work) = 6 days → (A + B) (full work) = 6 × 5/3 = 10 days

Assume Total Work = LCM of 22.5 and 10 = 45 units

Efficiency: A = 2, A + B = 4.5 ∴ B = 4.5 − 2 = 2.5

B alone = 45 ÷ 2.5 = 18 days

∴ The correct answer is 18 days.

Alternate Method

Given:

A finishes 2/5 of work in 9 days.

Remaining work (1 − 2/5 = 3/5) is finished by A and B together in 6 days.

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Total time for A = 9 ÷ (2/5) = 45/2 = 22.5 days

⇒ Total time for (A + B) = 6 ÷ (3/5) = 30/3 = 10 days

⇒ Let Total Work = LCM(22.5, 10) = 45 units

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

⇒ Efficiency of B = 4.5 − 2 = 2.5 units/day

⇒ Time taken by B alone = Total Work ÷ Efficiency of B

⇒ Time = 45 ÷ 2.5 = 18 days

∴ The correct answer is 18 days.

Additional Information

Work and Efficiency

Efficiency is inversely proportional to the time taken. If a person takes T days to complete 1 unit of work, their efficiency is 1/T units per day.

Remaining Work Calculation

If a fraction x/y of work is completed, the remaining work is calculated as 1 − (x/y).

Combined Efficiency Rule

When two or more people work together, their combined efficiency is the sum of their individual efficiencies (E_total = E_A + E_B).

Shortcut Trick

Days taken by A = 63 ÷ 3.5 = 18 days.

Days taken by B = 75 ÷ 2.5 = 30 days.

Using LCM 90, efficiencies: A = 5, B = 3. Total efficiency = 8 units/day.

Time taken together = 90 ÷ 8 = 11.25 days.

Total cost = Total daily wage × Total days = (3.5 + 2.5) × 11.25 = ₹67.50.

∴ The correct answer is ₹67.50.

Alternate Method

Given:

Daily wage of A = ₹3.50

Daily wage of B = ₹2.50

Total wage A = ₹63, Total wage B = ₹75

Formula Used:

Number of days = Total Wage ÷ Daily Wage

Combined Time (T) = (A × B) ÷ (A + B)

Total cost together = (Wage A + Wage B) × Combined Time

Calculations:

⇒ Days taken by A = 63 ÷ 3.5 = 18 days.

⇒ Days taken by B = 75 ÷ 2.5 = 30 days.

⇒ Total efficiency = 5 + 3 = 8 units per day.

⇒ Time taken together = 90 ÷ 8 = 11.25 days.

⇒ Combined daily wage = 3.50 + 2.50 = ₹6.00.

⇒ Total cost of the work = 11.25 × 6 = ₹67.50.

∴ The correct answer is ₹67.50.

Additional Information

Work and Wages

Wages are generally distributed in the ratio of work done. If time is constant, wages are directly proportional to efficiency.

Combined Efficiency

When multiple people work together, their daily efficiencies are added. Combined Time = Total Work ÷ (Sum of efficiencies).

Relationship between Time and Work

Time taken is inversely proportional to efficiency. Efficiency = Work ÷ Time.

Shortcut Trick

Efficiency Ratio (Man : Boy) = 2 : 1

Total Work = (10 Men) × 8 days = (10 × 2) × 8 = 160 units

Combined Efficiency (3 Men + 4 Boys) = (3 × 2) + (4 × 1) = 10 units/day

Time Taken = Total Work ÷ Combined Efficiency = 160 ÷ 10 = 16 days

∴ The correct answer is 16 days.

Alternate Method

Given:

Efficiency: 1 Man = 2 Boys

Time for 10 Men = 8 days

Formula Used:

Total Work = Total Efficiency × Time

Calculations:

⇒ Let efficiency of 1 boy = 1 unit/day

⇒ Efficiency of 1 man = 2 units/day

⇒ Total work = (10 men × 2 units/day) × 8 days = 160 units

⇒ Efficiency of 3 men and 4 boys = (3 × 2) + (4 × 1) = 6 + 4 = 10 units/day

⇒ Required time = 160 ÷ 10 = 16 days

∴ The correct answer is 16 days.

Additional Information

Work and Efficiency Relationship

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

Chain Rule (M-D-H Formula)

The general formula for work comparison is (M₁ × D₁ × H₁) ÷ W₁ = (M₂ × D₂ × H₂) ÷ W₂, where M is persons, D is days, H is hours, and W is work units.

Group Efficiency Conversion

When multiple types of workers (men, women, children) are involved, always convert them to a single common unit (e.g., all to men or all to boys) using their efficiency ratios before solving.

Shortcut Trick

Apply successive percentage formula: Net Change = a + b + (ab/100)

Here, Markup (a) = +30% and Discount (b) = −10%

Gain % = 30 − 10 + [30 × (−10) / 100]

Gain % = 20 − 3 = 17%

∴ The correct answer is 17%.

Alternate Method

Given:

Markup Percentage = 30% higher than Cost Price

Discount Percentage = 10% on Marked Price

Let the Cost Price (CP) be ₹100.

⇒ Marked Price (MP) = CP + 30% of CP

⇒ MP = 100 + (30/100) × 100 = ₹130

⇒ Discount = 10% of MP

⇒ Discount = (10/100) × 130 = ₹13

⇒ Selling Price (SP) = MP − Discount

⇒ SP = 130 − 13 = ₹117

⇒ Gain = SP − CP = 117 − 100 = 17

⇒ Gain % = (Gain/CP) × 100 = (17/100) × 100 = 17%

∴ The correct answer is 17%.

Additional Information

Relation between CP and MP

The ratio of Marked Price to Cost Price is given by: MP/CP = (100 + Profit%)/(100 − Discount%)

Markup Percentage

Markup % = (Marked Price − Cost Price) / Cost Price × 100

Effective Discount

For two successive discounts d₁% and d₂%, the single equivalent discount is: d₁ + d₂ − (d₁ × d₂)/100

Shortcut Trick

Net profit % = Markup % − Discount % − (Markup × Discount) ÷ 100

Net profit % = 40 − 20 − (40 × 20) ÷ 100 = 20 − 8 = 12%

Given, 12% of CP = ₹48

100% of CP = (48 ÷ 12) × 100 = 4 × 100 = ₹400

∴ The correct answer is ₹400.

Alternate Method

Given:

Markup % = 40% of production cost (CP)

Discount % = 20% on marked price (MP)

Actual Profit = ₹48

Calculations:

⇒ Let Production Cost (CP) = 100 units

⇒ Marked Price (MP) = 100 + 40% of 100 = 140 units

⇒ Discount = 20% of 140 = (20 ÷ 100) × 140 = 28 units

⇒ Selling Price (SP) = MP − Discount = 140 − 28 = 112 units

⇒ Profit = SP − CP = 112 − 100 = 12 units

⇒ According to the question, 12 units = ₹48

⇒ 1 unit = ₹48 ÷ 12 = ₹4

⇒ Production Cost (100 units) = 100 × 4 = ₹400

∴ The correct answer is ₹400.

Additional Information

Markup Percentage

It is the percentage by which the cost price is increased to reach the marked price. Formula: Markup % = [(MP − CP) ÷ CP] × 100.

Successive Discounts

When two discounts d₁ and d₂ are given, the equivalent discount is d₁ + d₂ − (d₁ × d₂ ÷ 100).

Relation between CP and MP

CP ÷ MP = (100 − Discount %) ÷ (100 + Profit %). This ratio is extremely useful for solving markup problems directly.

Shortcut Trick

Total Cost Price (CP) = 500 + 10% of 500 = ₹550

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

CP : MP = (100 − 25) : (100 + 20) = 75 : 120 = 5 : 8

5 units = ₹550 → 1 unit = ₹110 → 8 units = 8 × 110 = ₹880

∴ The correct answer is ₹880.

Alternate Method

Given:

Cost Abroad = ₹500 | Customs Duty = 10% | Profit = 20% | Discount = 25%

Formula Used:

Total CP = Cost + Duty | SP = CP × (100 + P)/100 | MP = (SP × 100) / (100 − D)

Calculations:

⇒ Customs Duty = 10% of 500 = (10/100) × 500 = ₹50

⇒ Effective CP = 500 + 50 = ₹550

⇒ Selling Price (SP) = 550 + (20% of 550) = 550 + 110 = ₹660

⇒ Let Marked Price be MP. After 25% discount, SP = 75% of MP

⇒ 0.75 × MP = 660

⇒ MP = 660 ÷ 0.75 = 660 × (100/75)

⇒ MP = 660 × (4/3) = 220 × 4 = ₹880

∴ The correct answer is ₹880.

Additional Information

Direct Relation between CP and MP

When both profit and discount are known, the ratio is CP/MP = (100 − D%)/(100 + P%).

Markup Percentage

Markup% = (MP − CP)/CP × 100. It represents how much the price is raised above the cost price.

Effective Discount

For two successive discounts of a% and b%, the single equivalent discount = (a + b − ab/100)%.

Shortcut Trick

Profit 25% = ₹6000 ⇒ Cost Price (CP) = 6000 × 4 = ₹24000.

Selling Price (SP) = CP + Profit = 24000 + 6000 = ₹30000.

Discount = 20% ⇒ SP is 80% of Advertised Price (MP).

Advertised Price = 30000 ÷ 0.8 = ₹37500.

∴ The correct answer is ₹37500.

Alternate Method

Given:

Discount = 20%, Profit Percentage = 25%, Profit Amount = ₹6000.

Formula Used:

Profit = CP × Profit%

Selling Price (SP) = CP + Profit = MP × (100 − Discount)%

Calculations:

⇒ Profit = 25% of CP

⇒ 6000 = 0.25 × CP

⇒ CP = 6000 ÷ 0.25 = ₹24000

⇒ SP = CP + Profit = 24000 + 6000 = ₹30000

⇒ Let Advertised Price (MP) be x

⇒ SP = MP − 20% of MP = 80% of MP

⇒ 30000 = 0.80 × x

⇒ x = 30000 ÷ 0.8 = ₹37500

∴ The correct answer is ₹37500.

Additional Information

Relation between CP and MP

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

Markup Percentage

Markup% = [(MP − CP) ÷ CP] × 100. It represents how much the price is increased above the cost price before offering a discount.

Effective Discount

If two successive discounts d₁ and d₂ are given, the single equivalent discount is (d₁ + d₂ − d₁×d₂/100)%.

Shortcut Trick

Combine the ratios: A : B = 5 : 4 and B : C = 9 : 10.

A : B : C = (5 × 9) : (4 × 9) : (4 × 10) = 45 : 36 : 40.

Total Ratio Units = 45 + 36 + 40 = 121 units.

C's share = (Value of C unit ÷ Total units) × Total Amount

C's share = (40 ÷ 121) × 2420 = 40 × 20 = 800.

∴ The correct answer is ₹800.

Alternate Method

Given:

Total Amount = ₹2420

A : B = 5 : 4

B : C = 9 : 10

Formula Used:

Combined Ratio (A : B : C) and Unitary Method.

Calculations:

⇒ To make B common in both ratios, multiply (A : B) by 9 and (B : C) by 4.

⇒ A : B = (5 × 9) : (4 × 9) = 45 : 36

⇒ B : C = (9 × 4) : (10 × 4) = 36 : 40

⇒ Combined Ratio A : B : C = 45 : 36 : 40

⇒ Sum of ratio parts = 45 + 36 + 40 = 121

⇒ Total Amount = ₹2420

⇒ Value of 121 units = 2420

⇒ Value of 1 unit = 2420 ÷ 121 = 20

⇒ C's share = 40 units × 20 = 800

∴ The correct answer is ₹800.

Additional Information

Compound Ratio

The compound ratio of two ratios a:b and c:d is (a × c) : (b × d).

Proportion Rule

If a : b :: c : d, then the product of extremes equals the product of means (a × d = b × c).

Mean Proportional

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

Duplicate and Sub-duplicate Ratio

The duplicate ratio of a : b is a² : b², and the sub-duplicate ratio is √a : √b.

Shortcut Trick

Initial mixture 5 : 2 (Total 7 units) = 49 kg ⇒ 1 unit = 7 kg.

Assam tea = 35 kg, Darjeeling tea = 14 kg.

In new ratio 2 : 1, Assam tea stays constant at 35 kg.

So, 2 units = 35 kg ⇒ 1 unit (new Darjeeling) = 17.5 kg.

Quantity to be added = 17.5 − 14 = 3.5 kg.

∴ The correct answer is 3.5 kg.

Alternate Method

Given: Total mixture = 49 kg, Initial Ratio (A : D) = 5 : 2, Final Ratio (A : D) = 2 : 1.

⇒ Total initial units = 5 + 2 = 7 units

⇒ 7 units = 49 kg ⇒ 1 unit = 7 kg

⇒ Initial Assam = 5 × 7 = 35 kg

⇒ Initial Darjeeling = 2 × 7 = 14 kg

⇒ Let quantity of Darjeeling added be x kg.

⇒ New ratio of Assam to Darjeeling = 35 ÷ (14 + x) = 2 ÷ 1

⇒ 35 = 2 × (14 + x)

⇒ 35 = 28 + 2x

⇒ 2x = 35 − 28 = 7

⇒ x = 7 ÷ 2 = 3.5 kg

∴ The correct answer is 3.5 kg.

Additional Information

Constant Quantity Principle

In mixture problems where only one component is added or removed, the quantity of the other component remains unchanged. Use this constant value to find the new total or the added part.

Ratio Balancing Method

If ratio A:B changes to C:D by adding B, make the parts of A equal in both ratios by multiplying with suitable constants: (A:B) × C and (C:D) × A.

Alligation for Mixtures

This method can also be solved using alligation by considering the concentration of Darjeeling tea in the original mixture (2/7), the added portion (100% or 1/1), and the final mixture (1/3).

Shortcut Trick

Total ratio parts = 9 + 2 = 11 units.

11 units = 132 students ⇒ 1 unit = 12 students.

Initial Successful students = 9 × 12 = 108.

Initial Unsuccessful students = 2 × 12 = 24.

New Successful = 108 + 4 = 112; New Unsuccessful = 24 − 4 = 20.

New ratio = 112 : 20 = 28 : 5.

∴ The correct answer is 28 : 5.

Alternate Method

Given: Total examinees = 132, Initial Ratio (S : U) = 9 : 2.

Formula: Individual Value = (Ratio Part ÷ Sum of Ratio) × Total.

⇒ Sum of ratio parts = 9 + 2 = 11.

⇒ Successful students = (9/11) × 132 = 108.

⇒ Unsuccessful students = (2/11) × 132 = 24.

⇒ New count of successful students = 108 + 4 = 112.

⇒ New count of unsuccessful students = 24 − 4 = 20.

⇒ New Ratio = 112 : 20.

⇒ Dividing both sides by 4: 28 : 5.

∴ The correct answer is 28 : 5.

Additional Information

Ratio and Constant Sum

When elements are transferred from one group to another within the same total, the sum of the ratio parts multiplied by the constant factor remains unchanged (Total = x + y).

Simplifying Ratios

A ratio a : b can be simplified by dividing both terms by their Highest Common Factor (HCF). For 112 : 20, the HCF is 4, resulting in 28 : 5.

Proportional Division

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

Shortcut Trick

Initial Ratio (Officers : Soldiers) = 3 : 31

Final Ratio (Officers : Soldiers) = 1 : 13

Change: Officers martyred = 6, Soldiers martyred = 22

Apply Cross-Product: (3 × 13) − (31 × 1) = 8 units

Value Difference: (6 × 13) − (22 × 1) = 78 − 22 = 56

8 units = 56 ⇒ 1 unit = 7 ⇒ Initial Officers = 3 × 7 = 21 officers

∴ The correct answer is 21 officers.

Alternate Method

Given: Initial ratio 3 : 31, Martyred officers = 6, Martyred soldiers = 22, Final ratio 1 : 13.

Formula Used: Ratio = Initial Value / Total Units

⇒ Let the initial number of officers be 3x and soldiers be 31x.

⇒ (3x − 6) / (31x − 22) = 1 / 13

⇒ 13 × (3x − 6) = 1 × (31x − 22)

⇒ 39x − 78 = 31x − 22

⇒ 39x − 31x = 78 − 22

⇒ 8x = 56

⇒ x = 7

⇒ Number of officers before the battle = 3x = 3 × 7 = 21

∴ The correct answer is 21 officers.

Additional Information

Definition of Ratio

A ratio is a comparison of two quantities of the same kind by division. If a and b are two quantities, the ratio is written as a : b or a/b.

Proportion Property

If four quantities a, b, c, d are in proportion (a:b = c:d), then the product of extremes equals the product of means: a × d = b × c.

Cross-Multiplication Shortcut

For problems where a constant or different value is added/subtracted from ratio parts, (Initial Ratio 1 × Final Ratio 2) − (Initial Ratio 2 × Final Ratio 1) units = Cross product of changes.

Shortcut Trick

Sum of ratio parts: Container 1 (4+1=5), Container 2 (3+1=4), Container 3 (5+2=7). LCM(5, 4, 7) = 140.

Multiply given volumes 3 : 4 : 5 by 140 ⇒ 420, 560, and 700 units respectively.

Total Milk = (4/5 × 420) + (3/4 × 560) + (5/7 × 700) = 336 + 420 + 500 = 1256 units.

Total Water = (1/5 × 420) + (1/4 × 560) + (2/7 × 700) = 84 + 140 + 200 = 424 units.

Final Ratio (Milk : Water) = 1256 : 424 ⇒ 157 : 53.

∴ The correct answer is 157 : 53.

Alternate Method

Given: Volumes ratio = 3 : 4 : 5

Ratios (Milk : Water): C1 = 4 : 1, C2 = 3 : 1, C3 = 5 : 2

⇒ Total Milk = [3 × (4 ÷ 5)] + [4 × (3 ÷ 4)] + [5 × (5 ÷ 7)]

⇒ Total Milk = 12/5 + 3 + 25/7 = (84 + 105 + 125) ÷ 35 = 314 ÷ 35

⇒ Total Water = [3 × (1 ÷ 5)] + [4 × (1 ÷ 4)] + [5 × (2 ÷ 7)]

⇒ Total Water = 3/5 + 1 + 10/7 = (21 + 35 + 50) ÷ 35 = 106 ÷ 35

⇒ Required Ratio = (314/35) : (106/35) = 314 : 106

⇒ Final Ratio = 157 : 53

∴ The correct answer is 157 : 53.

Additional Information

LCM Method for Mixtures

When combining multiple mixtures, find the LCM of the sum of ratio parts to normalize quantities before multiplying by the volume ratio.

Fractional Addition

Alternatively, calculate the absolute amount of components (Milk/Water) in each container as fractions: (Volume × Component Ratio Part ÷ Total Ratio Parts) and sum them.

Compound Ratio

For problems involving ratios of ratios, ensure all parts are relative to the same base value to perform direct addition.

Shortcut Trick

Apply the rule of alligation directly to find the ratio.

Difference between Dearer price and Mean price = 12 − 8 = 4

Difference between Mean price and Cheaper price = 8 − 7 = 1

Ratio of Quantities = (Mean price − Cheaper price) : (Dearer price − Mean price) = 1 : 4

∴ The correct answer is 1 : 4.

Alternate Method

Given:

Price of Sugar 1 (Dearer) = ₹12 per kg

Price of Sugar 2 (Cheaper) = ₹7 per kg

Mean Price of Mixture = ₹8 per kg

Formula Used:

Quantity of Cheaper / Quantity of Dearer = (CP of Dearer − Mean Price) / (Mean Price − CP of Cheaper)

Calculations:

⇒ Let the quantity of Sugar 1 be Q₁ and Sugar 2 be Q₂.

⇒ By Alligation Rule: Q₁ / Q₂ = (Mean − Price 2) / (Price 1 − Mean)

⇒ Q₁ / Q₂ = (8 − 7) / (12 − 8)

⇒ Q₁ / Q₂ = 1 / 4

⇒ Q₁ : Q₂ = 1 : 4

∴ The correct answer is 1 : 4.

Additional Information

Rule of Alligation

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

Mean Price

The mean price is the cost price of a unit quantity of the mixture. It always lies between the values of the cheaper and the dearer components.

Application in Percentage

Alligation can also be used for percentages, like mixing two solutions of different concentrations or calculating overall profit/loss percentage from two parts.

Shortcut Trick

Ratio of number of theatres = 15 : (15 − 6) = 15 : 9 = 5 : 3.

Since the total attendance remains constant, the average attendance per theatre is inversely proportional to the number of theatres.

Ratio of average daily attendance = 3 : 5.

3 units = 600 customers ⇒ 1 unit = 200 customers.

New average = 5 units = 5 × 200 = 1000 customers.

∴ The correct answer is 1000 customers.

Alternate Method

Given:

Initial number of theatres = 15

Initial average attendance = 600 customers

Number of theatres closed = 6

Formula Used:

Total Attendance = Number of Theatres × Average Attendance

Calculations:

⇒ Total attendance = 15 × 600

⇒ Total attendance = 9000 customers

⇒ Number of remaining theatres = 15 − 6

⇒ Number of remaining theatres = 9

⇒ New average attendance = Total attendance ÷ Number of remaining theatres

⇒ New average attendance = 9000 ÷ 9

⇒ New average attendance = 1000 customers

∴ The correct answer is 1000 customers.

Additional Information

Concept of Average

Average is defined as the sum of all observations divided by the total number of observations. Average = Sum ÷ Count.

Inverse Variation in Averages

When the total sum of a dataset is constant, the average value is inversely proportional to the number of elements: Average₁ × Count₁ = Average₂ × Count₂.

Weighted Average

When combining two groups, the combined average is A_w = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where n represents counts and A represents averages.

Shortcut Trick

Weight of B = (Sum of overlapping groups) − (Total sum of all individuals)

Weight of B = (2 × Average of A and B) + (2 × Average of B and C) − (3 × Average of A, B and C)

Weight of B = (2 × 40) + (2 × 43) − (3 × 45)

Weight of B = 80 + 86 − 135 = 166 − 135 = 31 kg

∴ The correct answer is 31 kg.

Alternate Method

Given:

Average weight of (A + B + C) = 45 kg

Average weight of (A + B) = 40 kg

Average weight of (B + C) = 43 kg

Formula: Sum = Average × Number of persons

⇒ Total sum of weights of A, B and C = 45 × 3 = 135 kg --- (i)

⇒ Sum of weights of A and B = 40 × 2 = 80 kg --- (ii)

⇒ Sum of weights of B and C = 43 × 2 = 86 kg --- (iii)

⇒ Adding (ii) and (iii): A + B + B + C = 80 + 86 = 166 kg

⇒ (A + B + C) + B = 166 kg

⇒ 135 + B = 166 kg [From equation (i)]

⇒ B = 166 − 135 = 31 kg

∴ The correct answer is 31 kg.

Additional Information

General Average Formula

Average = (Sum of all observations) ÷ (Number of observations). Sum = Average × Count.

Included/Excluded Value

To find the weight of an excluded or included person: New Weight = (New Count × New Average) − (Old Count × Old Average).

Net Deviation Method

The sum of deviations of all observations from their average is always zero. This is useful for finding missing terms in large datasets.

Shortcut Trick

Sum of (Highest + Lowest) score = (Total Innings × New Average reduction) + (Old Average × Number of Excluded Innings)

H + L = (38 × 2) + (50 × 2) = 76 + 100 = 176

Given: H − L = 172

Highest Score (H) = (Sum + Difference) ÷ 2 = (176 + 172) ÷ 2 = 174

∴ The correct answer is 174 runs.

Alternate Method

Given:

Average of 40 innings = 50 runs

Average of 38 innings = 48 runs

Difference between Highest (H) and Lowest (L) = 172 runs

Formula Used:

Total Sum = Average × Number of innings

Calculations:

⇒ Sum of 40 innings = 40 × 50 = 2000 runs

⇒ Sum of 38 innings = 38 × 48 = 1824 runs

⇒ (H + L) = 2000 − 1824 = 176 runs

⇒ H + L = 176 ....(i)

⇒ H − L = 172 ....(ii)

⇒ Adding (i) and (ii): 2H = 176 + 172

⇒ 2H = 348

⇒ H = 348 ÷ 2 = 174

∴ The correct answer is 174 runs.

Additional Information

Sum and Difference Rule

If the sum (S) and difference (D) of two numbers are known, the larger number is (S + D) ÷ 2 and the smaller number is (S − D) ÷ 2.

Average Change Concept

When certain items are excluded, the total sum decreases by (Old Avg × Count of excluded items) + (Reduction in Avg × Count of remaining items).

Batting Average

Batting Average = Total Runs Scored ÷ Number of Innings in which the player was out (for standard problems, total innings is usually used).

Shortcut Trick

For an odd number of consecutive terms (n = 7), the average is exactly the middle term.

Middle term (4^th term) = 20.

Consecutive numbers increase by 1, so the largest number (7^th term) = Middle term + 3.

Largest number = 20 + 3 = 23.

∴ The correct answer is 23.

Alternate Method

Given: Average of 7 consecutive numbers = 20.

Formula: Average = Sum of terms ÷ Number of terms.

⇒ Let the seven consecutive numbers be x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5), and (x + 6).

⇒ Sum of these numbers = x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 7x + 21

⇒ Average = Sum ÷ 7 = (7x + 21) ÷ 7

⇒ 20 = x + 3

⇒ x = 20 − 3 = 17

⇒ The largest number = x + 6 = 17 + 6 = 23

∴ The correct answer is 23.

Additional Information

Average of an Arithmetic Progression

For any series in AP (like consecutive numbers), Average = (First term + Last term) ÷ 2.

Middle Term Property

If the number of terms (n) is odd, the middle term is exactly at the position (n + 1) ÷ 2.

Sum of Consecutive Integers

The sum of n consecutive integers is given by n × Average. Here, Sum = 7 × 20 = 140.

Shortcut Trick

Let Pankaj = 100 units.

Soham = 150 units (50% more than Pankaj).

Mukesh = 300 units (Twice of Soham).

Ratio P : S : M = 100 : 150 : 300 = 2 : 3 : 6. Total = 11 units.

Average = 110 → Total money = 110 × 3 = 330.

Mukesh's share = (6 ÷ 11) × 330 = 180.

∴ The correct answer is ₹180.

Alternate Method

Given:

Money with Mukesh = 2 × Money with Soham

Money with Soham = 1.5 × Money with Pankaj (as 50% more means 150% of Pankaj)

Average money (Mukesh, Soham, Pankaj) = ₹110

Formula Used:

Average = Sum of values ÷ Number of values

Calculations:

⇒ Let money with Pankaj be P. Then Soham S = 1.5P and Mukesh M = 2S = 2(1.5P) = 3P.

⇒ Ratio of P : S : M = P : 1.5P : 3P = 1 : 1.5 : 3

⇒ Multiply by 2 to remove decimal: Ratio P : S : M = 2 : 3 : 6

⇒ Let the amounts be 2x, 3x, and 6x respectively.

⇒ Average = (2x + 3x + 6x) ÷ 3 = 110

⇒ 11x ÷ 3 = 110

⇒ 11x = 330

⇒ x = 30

⇒ Money with Mukesh = 6x = 6 × 30 = 180.

∴ The correct answer is ₹180.

Additional Information

Average of Quantities

The sum of n quantities is equal to (Average × n). Here, Sum = 110 × 3 = 330.

Ratio Balancing

If A:B = x:y and B:C = p:q, then to find A:B:C, the common term B must be made equal by multiplying the ratios by suitable constants.

Percentage to Ratio

A value being 50% more than another means it is 150/100 or 3/2 of that value. A:B = 3:2.

Shortcut Trick

Let the average income of a woman be x. Then average income of a man is (x + 10) and a boy is (x − 10).

Net deviation from the woman's average = (7 × 10) + (11 × 0) + (2 × −10) = 70 − 20 = 50.

Average deviation per person = 50 ÷ 20 = 2.5.

Average income of a woman = 257.50 − 2.5 = 255.

Average income of a man = 255 + 10 = 265.

∴ The correct answer is Rs. 265.

Alternate Method

Given:

Total persons = 7 men + 11 women + 2 boys = 20 persons

Average daily income of all = Rs. 257.50

Man's average = Woman's average + 10

Woman's average = Boy's average + 10

Calculations:

Let the average daily income of a woman be x.

⇒ Average daily income of a man = x + 10

⇒ Average daily income of a boy = x − 10

⇒ Total daily income = (7 × (x + 10)) + (11 × x) + (2 × (x − 10))

⇒ Total daily income = 7x + 70 + 11x + 2x − 20

⇒ Total daily income = 20x + 50

Also, Total daily income = 20 × 257.50 = 5150

⇒ 20x + 50 = 5150

⇒ 20x = 5100

⇒ x = 5100 ÷ 20 = 255

⇒ Average daily income of a man = x + 10 = 255 + 10 = 265

∴ The correct answer is Rs. 265.

Additional Information

Average Formula

Average = (Sum of all observations) ÷ (Total number of observations). It represents the central value of a data set.

Weighted Average

When groups have different sizes, the combined average is (n₁a₁ + n₂a₂ + ...) ÷ (n₁ + n₂ + ...), where n is count and a is average.

Net Deviation Rule

The sum of deviations of all observations from the arithmetic mean is always zero. This property is used to simplify complex average problems.

The correct answer is Rs. 390

Key Points

• To find the Cost Price (CP), we can use the basic formulas for Profit and Loss based on the condition given in the question.
• According to the question, Profit at Selling Price 1 (SP₁ = Rs. 425) is equal to the Loss at Selling Price 2 (SP₂ = Rs. 355).
• We know that: Profit = Selling Price − Cost Price and Loss = Cost Price − Selling Price.
• Let the Cost Price be CP.
• Equation: SP₁ − CP = CP − SP₂
• Substitute the given values: 425 − CP = CP − 355
• Rearranging the equation to bring CP terms to one side: 425 + 355 = CP + CP
• 780 = 2CP
• CP = 780 ÷ 2 = 390.
• Thus, the Cost Price of the article is Rs. 390.

Additional Information

• A quick shortcut for such problems: When the numerical value of Profit earned at one price is equal to the Loss incurred at another price, the Cost Price is simply the Average of the two Selling Prices.
• Shortcut Formula: CP = (SP₁ + SP₂) ÷ 2
• Calculation: CP = (425 + 355) ÷ 2 = 780 ÷ 2 = 390.
• In accounting and commerce, Profit Margin is calculated based on the difference between the revenue and the cost of goods sold.

The correct answer is Rs. 750

Key Points

• Let the profit share of A be 'A' and the profit share of B be 'B'.
• The total profit is given as Rs. 1650. Therefore, the first equation is: A + B = 1650.
• According to the problem, 1/3 of A’s profit is equal to 2/5 of B’s profit. This can be written as: (1/3) × A = (2/5) × B.
• To find the ratio of their profits, we rearrange the equation: A / B = (2/5) / (1/3).
• A / B = (2/5) × (3/1) = 6 / 5. So, the ratio A : B = 6 : 5.
• The total ratio units are 6 + 5 = 11 units.
• Since 11 units correspond to the total profit of Rs. 1650, the value of 1 unit is: 1650 ÷ 11 = 150.
• B’s share is 5 units. Therefore, B’s profit = 5 × 150 = Rs. 750.
• Checking A’s share: A = 6 × 150 = Rs. 900. Summing them up: 900 + 750 = 1650, which satisfies the condition.

Additional Information

• Ratio and Proportion is a fundamental concept in arithmetic used to compare two or more quantities. In partnership problems, profits are generally shared in the ratio of the investment multiplied by the time period for which the investment was made.
• If the time period is constant for all partners, the profit is shared directly in the ratio of their capital investments.
• In this specific question, the relationship between shares was explicitly provided as a fractional equality, which allows us to determine the ratio directly without knowing individual investments.

The correct answer is 4 : 5 Key Points

According to the first part of the question, 4% of the selling price (SP) of an article is equal to 5% of its cost price (CP). This can be written mathematically as: (4 / 100) × SP = (5 / 100) × CP. By canceling out 100 from both denominators, we get: 4 × SP = 5 × CP. To find the ratio of cost price to selling price (CP : SP), rearrange the equation: CP / SP = 4 / 5. The ratio of CP & SP is 4 : 5. We can verify this with the second condition: 20% of SP is Rs.120 more than 22% of CP. Let CP = 4k and SP = 5k. 0.20(5k) − 0.22(4k) = 120 1k − 0.88k = 120 0.12k = 120 ⇒ k = 120 / 0.12 = 1000. Substituting k = 1000, Cost Price = 4000 and Selling Price = 5000. The ratio 4000 : 5000 simplifies to 4 : 5.

Additional Information

Cost Price (CP): The price at which an item is purchased by a trader or shopkeeper. Selling Price (SP): The price at which an item is eventually sold to a customer. Profit: Occurs when the Selling Price is greater than the Cost Price (SP > CP). Profit = SP − CP. Loss: Occurs when the Cost Price is greater than the Selling Price (CP > SP). Loss = CP − SP. The ratio is a comparison of two quantities of the same units, indicating how many times one number contains another.

Shortcut Trick

Reduction in price = 25% = 1/4.

If price reduces by 1/x, consumption increases by 1/(x−1) to keep expenditure same.

Here, price reduces by 1/4 → Consumption increases by 1/(4−1) = 1/3 of original.

Given, 1/3 of original consumption = 2 dozen ⇒ Original consumption = 6 dozen.

Original Rate = Total Investment ÷ Original Quantity = 162 ÷ 6 = ₹27 per dozen.

∴ The correct answer is ₹27.

Alternate Method

Given:

Total amount invested = ₹162

Reduction in rate = 25%

Extra eggs bought = 2 dozen

Calculations:

⇒ Price reduction = 25% = 1/4

⇒ Price ratio (Old : New) = 4 : 3

⇒ Since Expenditure is constant (₹162), Quantity ratio is inverse of Price ratio.

⇒ Quantity ratio (Old : New) = 3 : 4

⇒ Difference in units = 4 − 3 = 1 unit

⇒ 1 unit = 2 dozen (given)

⇒ Original Quantity = 3 units = 3 × 2 dozen = 6 dozen

⇒ Original Rate = Total Amount ÷ Original Quantity

⇒ Original Rate = 162 ÷ 6 = ₹27

∴ The correct answer is ₹27.

Additional Information

Price-Consumption Relationship

When expenditure is constant, the quantity consumed is inversely proportional to the price. If price changes by x/y, the quantity changes in the ratio y/x.

Direct Formula for New Rate

Reduced Rate per unit = (Percentage Reduction × Total Amount) ÷ Extra Quantity bought. In this case: (25% × 162) ÷ 2 = ₹20.25.

Calculating Original Rate from New Rate

Original Rate = New Rate ÷ (1 − reduction decimal). Here: 20.25 ÷ (1 − 0.25) = 20.25 ÷ 0.75 = ₹27.

Shortcut Trick

When SP is same for two items, one at x% profit and another at x% loss, there is always a net loss.

Net Loss % = (x ÷ 10)² = (25 ÷ 10)² = 6.25%.

Total SP = 2 × 20,000 = ₹40,000.

Net Loss Amount = (Total SP × Loss %) ÷ (100 − Loss %) = (40,000 × 6.25) ÷ 93.75 = ₹2666.67.

Since ₹2666.67 > ₹2000, the result is a loss of more than ₹2000.

∴ The correct answer is ₹2666.67 loss.

Alternate Method

Given:

Selling Price of each painting (SP₁ = SP₂) = ₹20,000

Profit on 1^st painting = 25% | Loss on 2^nd painting = 25%

Formula Used:

Cost Price (CP) = SP × 100 ÷ (100 ± Profit/Loss %)

Calculations:

⇒ CP of 1^st painting = 20,000 × 100 ÷ 125

⇒ CP₁ = 20,000 × 4 ÷ 5 = ₹16,000

⇒ CP of 2^nd painting = 20,000 × 100 ÷ 75

⇒ CP₂ = 20,000 × 4 ÷ 3 = ₹26,666.67

⇒ Total CP = 16,000 + 26,666.67 = ₹42,666.67

⇒ Total SP = 20,000 + 20,000 = ₹40,000

⇒ Net Loss = Total CP − Total SP

⇒ Net Loss = 42,666.67 − 40,000 = ₹2666.67

⇒ Comparison: ₹2666.67 > ₹2000

∴ The correct answer is ₹2666.67 loss.

Additional Information

Equal SP Profit and Loss Rule

When two articles are sold at the same price, one at x% profit and the other at x% loss, the transaction always results in a loss of (x²/100)%.

Effective Percentage Formula

For two successive percentage changes a% and b%, the net change is given by (a + b + ab/100)%. For loss, use a negative sign.

Ratio Method for Profit/Loss

A 25% profit means CP:SP = 4:5. A 25% loss means CP:SP = 4:3. To equate SP, multiply the first ratio by 3 and the second by 5, giving SP = 15 units for both.

Shortcut Trick

Ratio of SP to CP due to price markup = 110 : 100 = 11 : 10

Ratio of effective quantity due to weight cheating = 100 : 70 = 10 : 7

Net Profit Ratio (SP : CP) = (11 ÷ 10) × (10 ÷ 7) = 11 : 7

Profit % = [(11 − 7) ÷ 7] × 100 = 400 ÷ 7 = 57¹/₇%

∴ The correct answer is 57¹/₇%.

Alternate Method

Given:

Profit on selling price = 10%

Weight cheating = 30% less than actual measure

Calculations:

⇒ Let the cost price (CP) of 1000g of rice be ₹1000.

⇒ The shopkeeper uses 30% less weight, so actual weight delivered = 1000g − 30% of 1000g = 700g.

⇒ The Cost Price (CP) for the shopkeeper is for 700g = ₹700.

⇒ The shopkeeper claims to sell 1000g at 10% profit.

⇒ Selling Price (SP) = ₹1000 + 10% of 1000 = ₹1100.

⇒ Total Profit = SP − CP = 1100 − 700 = ₹400.

⇒ Profit Percentage = (Total Profit ÷ Actual CP) × 100

⇒ Profit % = (400 ÷ 700) × 100

⇒ Profit % = 400 ÷ 7 = 57¹/₇%

∴ The correct answer is 57¹/₇%.

Additional Information

Dishonest Dealer Formula

When a dealer sells at cost price but uses a false weight, Profit % = [Error ÷ (True Value − Error)] × 100.

Net Profit with Price Markup

If a dealer marks up price by x% and also cheats on weight by y%, then Net Profit % = [(x + y) ÷ (100 − y)] × 100.

Multiplication Factor Method

Net SP/CP = (Claimed Weight ÷ Actual Weight) × (SP ÷ CP). If this ratio is 11/7, profit is (11−7)/7 = 4/7.

Given:

Value = 30 minutes

Total time = 1 day

Formula used:

Percentage = $(\frac{\text{Value}}{\text{Total}}) \times 100$

1 day = 24 hours

1 hour = 60 minutes

Calculation:

Total minutes in a day = 24 × 60 = 1440 minutes

Required percentage = $(\frac{30}{1440}) \times 100$

⇒ $(\frac{1}{48}) \times 100$

⇒ $\frac{25}{12}$

⇒ 2.0833...

⇒ 2.083%

∴ The correct answer is option (2).

Shortcut Trick

Successive Multiplier for 2 years (Cycle): (1 + 25/100) × (1 − 4/100) = 5/4 × 24/25 = 6/5 = 1.2

Total time = 5 years = 2 full cycles (4 years) + Year 5 (+25%)

Total Multiplier = 1.2 × 1.2 × 1.25 = 1.44 × 1.25 = 1.8

Present Earning = 72000 ÷ 1.8 = 40000

∴ The correct answer is ₹40,000.

Alternate Method

Given:

Earnings after 5 years = ₹72,000

Pattern: Year 1, 3, 5 (+25%) and Year 2, 4 (−4%)

Formula Used:

Final Value = Present Earning × (1 + r₁/100) × (1 − r₂/100) × ...

Calculations:

⇒ Let Present Earning be P.

⇒ 72000 = P × (125/100) × (96/100) × (125/100) × (96/100) × (125/100)

⇒ 72000 = P × (5/4) × (24/25) × (5/4) × (24/25) × (5/4)

⇒ 72000 = P × (5/4 × 24/25) × (5/4 × 24/25) × 5/4

⇒ 72000 = P × (6/5) × (6/5) × 5/4

⇒ 72000 = P × (36/25) × 5/4

⇒ 72000 = P × (9/5)

⇒ P = (72000 × 5) ÷ 9

⇒ P = 8000 × 5

⇒ P = 40000

∴ The correct answer is ₹40,000.

Additional Information

Successive Percentage Change

When a value changes by a% and then b%, the net percentage change is given by the formula: [a + b + (a × b)/100]%.

Fraction Multiplier Method

Convert percentage increase/decrease into fractions (e.g., +25% = 5/4, −4% = 24/25) and multiply them sequentially for the total factor.

Compounded Growth

Earning patterns like these follow the principle of Compound Interest where the base changes after every increment or decrement.

Shortcut Trick

Total % passed in at least one subject = %Passed(QA) + %Passed(GA) − %Passed(Both)

Total Passed % = 73% + 70% − 64% = 79%

Total Failed % = 100% − 79% = 21%

Since 21% of total students = 6300

Total examinees = (6300 ÷ 21) × 100 = 30000

∴ The correct answer is 30000 examinees.

Alternate Method

Given:

Candidates passed in QA = 73%

Candidates passed in GA = 70%

Candidates passed in both = 64%

Number of candidates failed in both = 6300

Calculations:

⇒ Candidates passed in QA only = 73% − 64% = 9%

⇒ Candidates passed in GA only = 70% − 64% = 6%

⇒ Total candidates who passed in at least one subject = (QA only) + (GA only) + (Both)

⇒ Total Passed % = 9% + 6% + 64% = 79%

⇒ Candidates who failed in both subjects = 100% − 79% = 21%

⇒ Let total number of examinees be X.

⇒ 21% of X = 6300

⇒ (21 ÷ 100) × X = 6300

⇒ X = (6300 × 100) ÷ 21

⇒ X = 300 × 100

⇒ X = 30000

∴ The correct answer is 30000 examinees.

Additional Information

Inclusion-Exclusion Principle (2 Sets)

For two sets A and B, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). This represents the union of two groups.

Venn Diagram for Failed Candidates

Total Failed = 100 − (Passed in at least one). Alternatively, you can solve using fail percentages directly: %Fail(Both) = 100 − [%Pass(A) + %Pass(B) − %Pass(A ∩ B)].

Case of 3 Sets

For three sets A, B, and C: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C).

Shortcut Trick

Let initial Income = 100, then Expenditure = 75 and Savings = 25.

Ratio of Expenditure : Savings = 75 : 25 = 3 : 1.

Using the weighted average method for Net Income Increase:

(3 × 10%) + (1 × x%) = 4 × 20%

30 + x = 80 ⇒ x = 50%

∴ The correct answer is 50%.

Alternate Method

Given:

Expenditure = 75% of Income

Income Increase = 20%

Expenditure Increase = 10%

Formula used: Savings = Income − Expenditure

⇒ Let initial Income = ₹100

⇒ Initial Expenditure = 75% of 100 = ₹75

⇒ Initial Savings = 100 − 75 = ₹25

⇒ New Income = 100 + (20% of 100) = ₹120

⇒ New Expenditure = 75 + (10% of 75) = 75 + 7.5 = ₹82.5

⇒ New Savings = 120 − 82.5 = ₹37.5

⇒ Increase in Savings = 37.5 − 25 = ₹12.5

⇒ Percentage increase in savings = (12.5 ÷ 25) × 100 = 50%

∴ The correct answer is 50%.

Additional Information

Income Relation

The fundamental relation is Income = Expenditure + Savings. Any change in two variables directly impacts the third.

Percentage Change Formula

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

Base Change Rule

When calculating percentage increase/decrease, always ensure the denominator is the original (initial) value of that specific component.

Shortcut Trick

Q is the midpoint ⇒ Round trip time for PR is exactly twice the round trip time for PQ.

Total time for (P → R) + (R → P) = 2 × 12 hours = 24 hours.

Given: Time taken for P → R = 16 hours 40 minutes.

Time for R → P = Total round trip time − Time for P → R.

Time for R → P = 24 hr − 16 hr 40 min = 7 hr 20 min = 7¹/₃ hr.

∴ The correct answer is 7¹/₃ hr.

Alternate Method

Given:

Time (P → Q + Q → P) = 12 hr

Time (P → R) = 16 hr 40 min = 16²/₃ hr

Q is the midpoint of PR ⇒ PQ = QR = 1/2 PR

Since the speed is constant for the same direction:

⇒ Time (P → Q) = 1/2 × Time (P → R)

⇒ Time (P → Q) = 1/2 × 16 hr 40 min = 8 hr 20 min

⇒ From (P → Q + Q → P) = 12 hr:

⇒ Time (Q → P) = 12 hr − 8 hr 20 min = 3 hr 40 min

⇒ Since Q is the midpoint, Time (R → P) = 2 × Time (Q → P)

⇒ Time (R → P) = 2 × 3 hr 40 min = 6 hr 80 min

⇒ Time (R → P) = 7 hr 20 min = 7²⁰/₆₀ hr = 7¹/₃ hr

∴ The correct answer is 7¹/₃ hr.

Additional Information

Upstream and Downstream Speed

Downstream speed (a) = u + v and Upstream speed (b) = u − v, where u is boat speed in still water and v is stream speed.

Midpoint Time Relation

For a constant speed, the time taken to travel half the distance is exactly half of the total time: t_half = t_total ÷ 2.

Average Speed Formula

For a round trip with speeds x and y, Average Speed = (2 × x × y) ÷ (x + y).

Shortcut Trick

When distance is constant, speed is inversely proportional to time.

Ratio of Time (T₁ : T₂) = 10 : 7

Ratio of Speed (S₁ : S₂) = 7 : 10

Given S₁ = 42 kmph ⇒ 7 units = 42 kmph ⇒ 1 unit = 6 kmph

Increase in speed = 10 − 7 = 3 units = 3 × 6 = 18 kmph

∴ The correct answer is 18 kmph.

Alternate Method

Given:

Initial Speed (S₁) = 42 kmph

Initial Time (T₁) = 10 hours

New Time (T₂) = 7 hours

Formula Used:

Distance = Speed × Time

Calculations:

⇒ Total Distance = S₁ × T₁

⇒ Total Distance = 42 × 10 = 420 km

⇒ New Speed (S₂) = Total Distance ÷ T₂

⇒ S₂ = 420 ÷ 7 = 60 kmph

⇒ Increase in speed = S₂ − S₁

⇒ Increase = 60 − 42 = 18 kmph

∴ The correct answer is 18 kmph.

Additional Information

Constant Distance Rule

If distance is constant, S₁ × T₁ = S₂ × T₂. This is used to find any one missing variable when the other three are known.

Relative Speed

For two objects moving in the same direction, relative speed = S₁ − S₂. For opposite directions, relative speed = S₁ + S₂.

Unit Conversion

To convert kmph to m/s, multiply by 5/18. To convert m/s to kmph, multiply by 18/5.

Shortcut Trick

Change in mode: 60 km less by train and 60 km more by car increases time by 20 mins.

To eliminate the train journey entirely, we shift the remaining 240 km from train to car.

Additional time for 240 km shift = (240 ÷ 60) × 20 mins = 80 mins.

Total time if the entire 450 km is covered by car = 8 hr 40 min + 80 min = 10 hr.

Speed of car = Total distance ÷ Total time = 450 ÷ 10 = 45 km/hr.

∴ The correct answer is 45 km/hr.

Alternate Method

Given: Total Distance = 450 km.

Case 1: Train = 240 km, Car = 210 km, Time = 8 hr 40 min = 26/3 hrs.

Case 2: Train = 180 km, Car = 270 km, Time = 9 hr.

Let speed of train be T and car be C.

⇒ 240/T + 210/C = 26/3 — (Eq 1)

⇒ 180/T + 270/C = 9 — (Eq 2)

Multiply Eq 1 by 3 and Eq 2 by 4 to eliminate T:

⇒ 720/T + 630/C = 26

⇒ 720/T + 1080/C = 36

Subtracting the two equations:

⇒ (1080 − 630) / C = 36 − 26

⇒ 450/C = 10

⇒ C = 45 km/hr.

∴ The correct answer is 45 km/hr.

Additional Information

Average Speed

Average speed is the ratio of total distance covered to the total time taken. Average Speed = Total Distance ÷ Total Time.

Speed 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.

Relative Speed

If two objects move in the same direction, relative speed = |u − v|. If they move in opposite directions, relative speed = u + v.

Shortcut Trick

Total distance to cover = Sum of lengths = 150 + 250 = 400 m = 0.4 km.

Time taken = 2 minutes = 2/60 = 1/30 hours.

Relative speed in same direction = Distance ÷ Time = 0.4 ÷ (1/30) = 12 kmph.

Relative speed = Faster speed − Slower speed ⇒ 12 = 100 − Speed of Train C.

Speed of Train C = 100 − 12 = 88 kmph.

∴ The correct answer is 88 kmph.

Alternate Method

Given:

Speed of Train B (V_B) = 100 kmph

Length of Train B (L_B) = 150 m

Length of Train C (L_C) = 250 m

Time (t) = 2 minutes = 120 seconds

Formula Used:

Relative Speed (Same direction) = (L₁ + L₂) ÷ Time

To convert m/s to kmph, multiply by 18/5.

Calculations:

⇒ Total distance to be covered = 150 + 250 = 400 m

⇒ Total time taken = 120 seconds

⇒ Relative speed (in m/s) = 400 ÷ 120 = 10/3 m/s

⇒ Relative speed (in kmph) = (10/3) × (18/5)

⇒ Relative speed = 2 × 6 = 12 kmph

⇒ Since both trains move in the same direction: V_B − V_C = 12

⇒ 100 − V_C = 12

⇒ V_C = 100 − 12 = 88 kmph

∴ The correct answer is 88 kmph.

Additional Information

Relative Speed Concepts

When two objects move in the same direction, Relative Speed = |S₁ − S₂|. When they move in opposite directions, Relative Speed = S₁ + S₂.

Crossing Rule

Whether moving in same or opposite directions, the distance covered while crossing each other is always the sum of their lengths (L₁ + L₂).

Unit Conversion

1 kmph = 5/18 m/s and 1 m/s = 18/5 kmph.

Shortcut Trick

Total Compound Interest percentage = (4347 ÷ 30000) × 100 = 14.49%.

Given annual rate (R) = 7%.

Effective interest for 2 years = R + R + (R × R) ÷ 100.

Effective rate = 7 + 7 + (49 ÷ 100) = 14.49%.

Since the calculated interest percentage matches the required effective rate, n = 2.

∴ The correct answer is 2 years.

Alternate Method

Given: Principal (P) = ₹30,000, Rate (R) = 7% p.a., Compound Interest (CI) = ₹4,347.

Formula used: Amount (A) = P(1 + R/100)ⁿ, where A = P + CI.

⇒ Amount (A) = 30000 + 4347 = ₹34,347

⇒ 34347 = 30000(1 + 7/100)ⁿ

⇒ 34347 ÷ 30000 = (107/100)ⁿ

⇒ 11449 ÷ 10000 = (1.07)ⁿ

⇒ (107 ÷ 100)² = (1.07)ⁿ

⇒ (1.07)² = (1.07)ⁿ

⇒ n = 2

∴ The correct answer is 2 years.

Additional Information

Effective Interest Rate

For a rate R, the effective compound interest rate for 2 years is calculated as R + R + (R² ÷ 100)%.

CI vs SI Difference

The difference between Compound Interest and Simple Interest for 2 years is given by P × (R ÷ 100)².

Amount in Compound Interest

The total amount A after n years is A = P(1 + r/100)ⁿ, where compounding is done annually.

Shortcut Trick

Equate the final amount for both persons A and B.

Amount = Principal + (Principal × Rate × Time) ÷ 100

P + (P × x × t) ÷ 100 = Q + (Q × y × t) ÷ 100

(Pxt − Qyt) ÷ 100 = Q − P

t (Px − Qy) = 100(Q − P)

∴ The correct answer is 100 ((Q − P) ÷ (Px − Qy)) years.

Alternate Method

Given:

Principal for A = ₹ P, Rate for A = x% per annum

Principal for B = ₹ Q, Rate for B = y% per annum

Let the time after which their debts become equal be 't' years.

Formula Used:

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

Amount (A) = Principal (P) + Simple Interest (SI)

Calculations:

⇒ Amount for A = P + (P × x × t) ÷ 100

⇒ Amount for B = Q + (Q × y × t) ÷ 100

According to the question, these amounts are equal:

⇒ P + (Pxt ÷ 100) = Q + (Qyt ÷ 100)

⇒ (Pxt ÷ 100) − (Qyt ÷ 100) = Q − P

⇒ (t ÷ 100) × (Px − Qy) = Q − P

⇒ t × (Px − Qy) = 100 × (Q − P)

⇒ t = 100 × (Q − P) ÷ (Px − Qy)

∴ The correct answer is 100 ((Q − P) ÷ (Px − Qy)) years.

Additional Information

Simple Interest Formula

SI = (P × R × T) ÷ 100, where P is Principal, R is Rate of Interest, and T is Time in years.

Amount in Simple Interest

The total debt or maturity value is A = P + SI = P(1 + RT ÷ 100).

Rate and Time Units

Ensure the Rate (R) is per annum and Time (T) is in years. If time is in months, divide by 12; if in days, divide by 365.

Shortcut Trick

Multiplying factor for 1 year = Amount after 3 years ÷ Amount after 2 years

Factor = 2662 ÷ 2420 = 1.1 = 11/10

We know, Amount after 2 years = Principal × (Multiplying factor)²

2420 = P × (11/10)²

P = 2420 × (100 / 121) = 2000

∴ The correct answer is Rs. 2000.

Alternate Method

Year 0 Year 2 Year 3

P = ? Rs. 2420 Rs. 2662

Amount compounded Interest = Rs. 242

Given:

Amount after 2 years (A₂) = Rs. 2420

Amount after 3 years (A₃) = Rs. 2662

Formula Used:

Amount = P(1 + R/100)^T

Calculations:

⇒ In compound interest, the amount of the previous year acts as the principal for the next year.

⇒ Interest for the 3rd year = A₃ − A₂

⇒ Interest = 2662 − 2420 = Rs. 242

⇒ Rate of Interest (R) = (Interest / Principal for 3rd year) × 100

⇒ R = (242 / 2420) × 100 = 10%

⇒ Now, substitute the rate into the formula for the amount after 2 years:

⇒ 2420 = P(1 + 10/100)²

⇒ 2420 = P(11/10)²

⇒ 2420 = P × (121 / 100)

⇒ P = (2420 × 100) / 121

⇒ P = 20 × 100 = 2000

∴ The correct answer is Rs. 2000.

Additional Information

Compound Interest Successive Method

Compound Interest works exactly like successive percentage increments. For a 10% rate, the value increases by 10% each year (Multiplier = 1.1).

Effective Rate of Interest

The effective interest rate for 2 years at R% p.a. is given by the formula (2R + R²/100)%. For R = 10%, the 2-year CI is 21%.

Difference between SI and CI

The difference between Compound Interest and Simple Interest for 2 years is calculated using the formula: Difference = P × (R/100)².

Shortcut Trick

Interest for (4.5 − 2) = 2.5 years = 5500 − 4000 = ₹1500.

Simple Interest for 1 year = 1500 ÷ 2.5 = ₹600.

Simple Interest for 2 years = 600 × 2 = ₹1200.

Principal (P) = 4000 − 1200 = ₹2800.

Rate (R) = (600 ÷ 2800) × 100 = 150 ÷ 7 = 21³/₇%.

∴ The correct answer is 21³/₇%.

Alternate Method

Given:

Amount after 2 years (A₁) = ₹4000

Amount after 4 years 6 months (A₂) = ₹5500

Time period 1 (T₁) = 2 years

Time period 2 (T₂) = 4.5 years

Formula Used:

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

Amount (A) = Principal (P) + Simple Interest (SI)

Calculations:

⇒ SI for (4.5 − 2) years = 5500 − 4000

⇒ SI for 2.5 years = ₹1500

⇒ SI for 1 year = 1500 ÷ 2.5 = ₹600

⇒ SI for 2 years = 600 × 2 = ₹1200

⇒ Principal (P) = A₁ − SI for 2 years

⇒ P = 4000 − 1200 = ₹2800

⇒ Rate (R) = (SI for 1 year × 100) ÷ P

⇒ R = (600 × 100) ÷ 2800

⇒ R = 600 ÷ 28 = 150 ÷ 7

⇒ R = 21³/₇%

∴ The correct answer is 21³/₇%.

Additional Information

Simple Interest Property

In Simple Interest, the interest earned remains constant for every year as it is always calculated on the original Principal amount.

Rate of Interest Formula

Rate % = (SI × 100) ÷ (P × T). It represents the interest earned on ₹100 for a period of one year.

Principal from Amount

Principal (P) = Amount (A) − Simple Interest (SI). For multiple periods, P = A₁ − SI₁ = A₂ − SI₂.

Shortcut Trick

Volume of iron = π × (R² − r²) × h

Given: R = 4 cm (external), r = 3 cm (internal), h = 20 cm

Volume = (22/7) × (4² − 3²) × 20

Volume = (22/7) × 7 × 20 = 440 cm³

∴ The correct answer is 440 cm³.

Alternate Method

Given:

Length of the tube (h) = 20 cm

External diameter = 8 cm ⇒ External radius (R) = 8/2 = 4 cm

Internal diameter = 6 cm ⇒ Internal radius (r) = 6/2 = 3 cm

Formula Used:

Volume of hollow cylinder = πh(R² − r²)

Calculations:

⇒ Volume of iron = (22/7) × 20 × (4² − 3²)

⇒ Volume of iron = (22/7) × 20 × (16 − 9)

⇒ Volume of iron = (22/7) × 20 × 7

⇒ Volume of iron = 22 × 20

⇒ Volume of iron = 440 cm³

∴ The correct answer is 440 cm³.

Additional Information

Curved Surface Area (CSA)

The total CSA of a hollow cylinder is the sum of the external and internal curved surfaces: CSA = 2πRh + 2πrh = 2πh(R + r).

Total Surface Area (TSA)

The TSA includes both the CSAs and the area of the two circular rings at the ends: TSA = 2π(R + r)(h + R − r).

Volume of Solid Cylinder

If the cylinder was solid without a hole, the volume would simply be V = πr²h.

Shortcut Trick

For a rectangular box, Volume (V) = √(Area₁ × Area₂ × Area₃)

Given Areas: A₁ = 12, A₂ = 15, A₃ = 20

V = √(12 × 15 × 20)

V = √3600 = 60 cm³

∴ The correct answer is 60 cm³.

Alternate Method

Given:

Area of face 1 (l × b) = 12 cm²

Area of face 2 (b × h) = 15 cm²

Area of face 3 (h × l) = 20 cm²

Formula Used:

Volume of a cuboid (V) = l × b × h

Calculations:

⇒ Let the three sides be l, b, and h.

⇒ (l × b) × (b × h) × (h × l) = 12 × 15 × 20

⇒ l² × b² × h² = 3600

⇒ (l × b × h)² = 3600

⇒ l × b × h = √3600

⇒ Volume = 60 cm³

∴ The correct answer is 60 cm³.

Additional Information

Total Surface Area of Cuboid

The total surface area is given by the formula: 2(lb + bh + hl), where l, b, and h are the dimensions of the rectangular box.

Diagonal of a Cuboid

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

Special Case: Cube

If l = b = h = a, then the volume is a³ and the total surface area is 6a².

Shortcut Trick

Perimeter = Speed × Time = (12000 m ÷ 60 min) × 8 min = 1600 m.

L + B = 1600 ÷ 2 = 800 m.

Ratio 3 : 2 ⇒ Total 5 parts = 800 m ⇒ 1 part = 160 m.

L = 3 × 160 = 480 m and B = 2 × 160 = 320 m.

Area = 480 × 320 = 153600 m².

∴ The correct answer is 153600 m².

Alternate Method

Given:

Ratio of Length (L) : Breadth (B) = 3 : 2

Cycling Speed = 12 km/h

Time taken for 1 round = 8 minutes

Formula Used:

Perimeter of Rectangle = 2 × (Length + Breadth)

Area of Rectangle = Length × Breadth

Distance = Speed × Time

Calculations:

⇒ Speed in m/min = (12 × 1000) ÷ 60 = 200 m/min

⇒ Distance (Perimeter) = 200 × 8 = 1600 m

⇒ Let Length = 3x and Breadth = 2x

⇒ 2(3x + 2x) = 1600

⇒ 10x = 1600

⇒ x = 160

⇒ Length = 3 × 160 = 480 m

⇒ Breadth = 2 × 160 = 320 m

⇒ Area = 480 × 320

⇒ Area = 153600 m²

∴ The correct answer is 153600 m².

Additional Information

Perimeter and Area of Rectangle

The perimeter is the total boundary length 2(L + B) and area is the region enclosed L × B.

Unit Conversion

To convert km/h to m/s, multiply by 5/18. To convert km/h to m/min, multiply by 1000/60 or 50/3.

Diagonal of Rectangle

The length of the diagonal d is calculated using Pythagoras theorem: d = √(L² + B²).

Shortcut Trick

Curved Surface Area (CSA) of a cylinder depends on the product of radius (r) and height (h).

Successive percentage change formula: Net Change % = a + b + (a × b)/100

Given: r decreases by 25% (a = −25) and h increases by 25% (b = +25).

Net Change = −25 + 25 + (−25 × 25)/100 = 0 − 6.25 = −6.25%.

The negative sign indicates a decrease.

∴ The correct answer is decreased by 6.25%.

Alternate Method

Given:

Change in radius (r) = −25%

Change in height (h) = +25%

Formula Used:

Curved Surface Area (CSA) of cylinder = 2 × π × r × h

Calculations:

⇒ Let original radius = r and original height = h.

⇒ Original CSA = 2 × π × r × h

⇒ New radius (r₁) = r − 25% of r = 0.75r

⇒ New height (h₁) = h + 25% of h = 1.25h

⇒ New CSA = 2 × π × r₁ × h₁

⇒ New CSA = 2 × π × (0.75r) × (1.25h)

⇒ New CSA = (0.75 × 1.25) × (2 × π × r × h)

⇒ New CSA = 0.9375 × (Original CSA)

⇒ Percentage Change = [(New CSA − Original CSA) / Original CSA] × 100

⇒ Percentage Change = (0.9375 − 1) × 100 = −0.0625 × 100 = −6.25%

⇒ The negative value signifies a decrease.

∴ The correct answer is decreased by 6.25%.

Additional Information

Volume Change in Cylinder

The volume of a cylinder is V = πr²h. If the radius changes by x% and height by y%, the net change in volume is calculated by applying the successive percentage formula twice for r and once for h.

Total Surface Area (TSA)

For a closed cylinder, TSA = 2πr(h + r). Unlike CSA, the percentage change in TSA depends on the absolute values of r and h, not just their percentage changes.

Successive Percentage Change

When a quantity depends on the product of two variables (A = x × y), and x changes by a% and y changes by b%, the net percentage change in A is always [a + b + (ab/100)]%.

Shortcut Trick

Volume of material removed = Volume of Cylinder − Volume of Cone (for the same height h and radius r).

Trick formula: Volume removed = (2/3) × π × r² × h

Given: r = 1.2 ÷ 2 = 0.6 cm, h = 1.4 cm.

Calculation: (2/3) × (22/7) × 0.6 × 0.6 × 1.4 = 2 × 22 × 0.12 × 0.2 = 1.056 cm³.

∴ The correct answer is 1.056 cm³.

Alternate Method

Given data:

Diameter of the pencil = 1.2 cm

Radius (r) = 1.2 ÷ 2 = 0.6 cm

Height of the conical part (h) = 1.4 cm

Formula used: Volume of cylinder = πr²h and Volume of cone = (1/3)πr²h

Calculation steps:

⇒ Volume of the cylindrical portion before sharpening = πr²h

⇒ Volume of the conical tip after sharpening = (1/3)πr²h

⇒ Volume of material removed = πr²h − (1/3)πr²h

⇒ Volume of material removed = (2/3) × π × r² × h

⇒ Volume = (2/3) × (22/7) × (0.6)² × 1.4

⇒ Volume = (2/3) × (22/7) × 0.36 × 1.4

⇒ Volume = (2/3) × 22 × 0.36 × 0.2

⇒ Volume = 2 × 22 × 0.12 × 0.2

⇒ Volume = 44 × 0.024 = 1.056 cm³

∴ The correct answer is 1.056 cm³.

Additional Information

Volume of Cylinder

The volume of a cylinder is calculated as the area of the base times height: V = πr²h.

Volume of Cone

The volume of a cone is exactly one-third of the volume of a cylinder with the same radius and height: V = (1/3)πr²h.

Ratio of Volumes

For a cylinder and a cone of the same dimensions, the ratio of their volumes is 3 : 1. The volume removed when carving a cone from a cylinder is 2/3 of the cylinder's volume.

Curved Surface Area of Cone

The curved surface area (CSA) of a cone is given by πrl, where l = √(r² + h²) is the slant height.

Shortcut Trick

For two perpendicular crossroads in the middle of a rectangle, Area of Lawn = (Length − Width of road) × (Breadth − Width of road).

Let width of road be x. Then, (60 − x) × (40 − x) = 2109.

Testing Option (1): If x = 3 m → (60 − 3) × (40 − 3) = 57 × 37 = 2109 m².

The condition is satisfied for x = 3 m.

∴ The correct answer is 3 m.

Alternate Method

Given: Length (L) = 60 m, Breadth (B) = 40 m, Area of lawn = 2109 m².

Formula for Area of crossroads: x(L + B − x), where x is the width of the road.

⇒ Total Area of park = 60 × 40 = 2400 m²

⇒ Area of crossroads = Total Area − Area of Lawn = 2400 − 2109 = 291 m²

⇒ x(60 + 40 − x) = 291

⇒ x(100 − x) = 291

⇒ 100x − x² = 291

⇒ x² − 100x + 291 = 0

⇒ x² − 97x − 3x + 291 = 0

⇒ x(x − 97) − 3(x − 97) = 0

⇒ (x − 97)(x − 3) = 0

⇒ x = 3 or x = 97. Since x cannot be greater than the park breadth (40 m), x = 3 m.

∴ The correct answer is 3 m.

Additional Information

Area of Path around a Rectangular Field (Inside)

If a path of width x is built inside a rectangle of L × B, Area of path = 2x(L + B − 2x).

Area of Path around a Rectangular Field (Outside)

If a path of width x is built outside a rectangle of L × B, Area of path = 2x(L + B + 2x).

Crossroads Calculation Shortcut

Area of two perpendicular crossroads = (L × x) + (B × x) − x². The subtraction of x² is necessary because the central square intersection is counted twice.

Shortcut Trick

Radius of each circle (r) = Side of square ÷ 2 = 140 ÷ 2 = 70 cm.

Area of space enclosed = Area of Square − Area of 4 Quadrants

Area of 4 Quadrants = Area of 1 complete circle = πr²

Area enclosed = (140 × 140) − (22/7 × 70 × 70) = 19600 − 15400 = 4200 cm².

∴ The correct answer is 4200 cm².

Alternate Method

Given:

Side of the square (a) = 140 cm

Radius of each circle (r) = a ÷ 2 = 70 cm

Value of π = 22/7

Formula Used:

Area of square = Side²

Area of circle = π × r²

Area of four quadrants = 4 × (1/4 × πr²) = πr²

Calculations:

⇒ Area of the square = 140 × 140 = 19600 cm²

⇒ Area of the four circle sectors = (22/7) × 70 × 70

⇒ Area of the four circle sectors = 22 × 10 × 70 = 15400 cm²

⇒ Enclosed space = Area of square − Area of four sectors

⇒ Enclosed space = 19600 − 15400 = 4200 cm²

∴ The correct answer is 4200 cm².

Additional Information

Area of a Circle Sector

The area of a sector with central angle θ is given by (θ/360) × πr². For a quadrant, θ = 90°.

Property of Tangent Circles in a Square

When four equal circles are placed at the corners of a square and touch each other, the radius of each circle is exactly half the side of the square.

General Area Formula for Enclosed Space

The area of the space enclosed between four such circles is always equal to (1 − π/4) × Side² or approx 3/14 × Side².

Shortcut Trick

Radius ($r$) = Perimeter ÷ 2π = 8.8 ÷ (2 × 22/7) = 1.4 m.

Volume of Cylinder = (Curved Surface Area × Radius) ÷ 2.

Concrete required (Volume) = (17.6 × 1.4) ÷ 2 = 12.32 m³.

∴ The correct answer is 12.32 m³.

Alternate Method

Given:

Perimeter of base = 8.8 m

Curved Surface Area (CSA) = 17.6 m²

π = 22/7

Formula Used:

Perimeter = 2π$r$

CSA = 2π$rh$

Volume = π$r$²$h$

Calculations:

⇒ 2 × (22/7) × $r$ = 8.8

⇒ (44/7) × $r$ = 8.8

⇒ $r$ = (8.8 × 7) ÷ 44 = 0.2 × 7 = 1.4 m

⇒ CSA = 2π$rh$ = 17.6

⇒ 8.8 × $h$ = 17.6 (since 2π$r$ = 8.8)

⇒ $h$ = 17.6 ÷ 8.8 = 2 m

⇒ Volume of concrete = π$r$²$h$

⇒ Volume = (22/7) × 1.4 × 1.4 × 2

⇒ Volume = 22 × 0.2 × 1.4 × 2

⇒ Volume = 4.4 × 2.8 = 12.32 m³

∴ The correct answer is 12.32 m³.

Additional Information

Total Surface Area of Cylinder

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

Relationship between Volume and CSA

For any cylinder, the ratio of Volume to Curved Surface Area is always $r$/2.

Hollow Cylinder Volume

The volume of a hollow pipe or cylinder is given by π$h$($R$² − $r$²), where $R$ is outer and $r$ is inner radius.

The correct answer is 54

Key Points

• To find the number of bottles required, we need to divide the total volume of the liquid in the bowl by the volume of one cylindrical bottle.
• The formula for the Volume of a Hemisphere is (2/3) π R³, where R is the internal radius.
• Given: Internal radius of the bowl (R) = 9 cm.
• Volume of the liquid in the bowl = (2/3) π (9)³ = (2/3) π (729) = 2 π (243) = 486π cm³.
• The formula for the Volume of a Cylinder is π r² h, where r is the radius and h is the height.
• Given: Diameter of the bottle = 3 cm, so the radius (r) = 3/2 = 1.5 cm. The height (h) = 4 cm.
• Volume of one cylindrical bottle = π (1.5)² (4) = π (2.25) (4) = 9π cm³.
• Number of bottles = (Volume of the bowl) ÷ (Volume of one bottle) = 486π ÷ 9π = 54.
• Thus, 54 bottles are necessary to empty the bowl.

Additional Information

• A Hemisphere is a three-dimensional geometric shape that is half of a sphere with one flat side. Its Curved Surface Area (CSA) is 2πr² and its Total Surface Area (TSA) is 3πr².
• A Cylinder is a solid with two parallel circular bases. The Curved Surface Area of a cylinder is 2πrh and the Total Surface Area is 2πr(r + h).
• When transferring liquid from one container to another, the total volume remains constant, which is the fundamental principle used to solve this problem.
• Ensure the units for radius and height are the same (in this case, cm) before performing calculations.

Shortcut Trick

Rise in level = (Area of pipe × Speed of water × Time) ÷ (Base Area of tank)

Volume flow = 40/10000 m² × 10000 m/hr × 0.5 hr = 20 m³

Rise (h) = 20 ÷ (80 × 40) = 20/3200 = 1/160 m

Rise in cm = (1/160) × 100 = 10/16 = 5/8 cm

∴ The correct answer is 5/8 cm.

Alternate Method

Given:

Dimensions of tank = 80 m × 40 m

Area of pipe opening = 40 cm²

Speed of water = 10 km/hr

Time = 30 minutes = 0.5 hours

Formula Used:

Volume of water = Cross-sectional area × Velocity × Time

Rise in level = Volume ÷ Base Area of Tank

Calculations:

⇒ Area of pipe = 40 cm² = 40 ÷ (100 × 100) m² = 0.004 m²

⇒ Speed of water = 10 km/hr = 10,000 m/hr

⇒ Volume of water in 0.5 hours = 0.004 × 10000 × 0.5

⇒ Volume = 40 × 0.5 = 20 m³

⇒ Let the rise in water level be h meters.

⇒ Volume = Base Area × h

⇒ 20 = (80 × 40) × h

⇒ 20 = 3200 × h

⇒ h = 20 ÷ 3200 = 1/160 m

⇒ Convert h to cm: (1/160) × 100 = 100/160 = 5/8 cm

∴ The correct answer is 5/8 cm.

Additional Information

Volume Flow Rate

The volume of liquid flowing through a pipe per unit time is given by Q = A × v, where A is the cross-sectional area and v is the velocity.

Unit Conversions

1 m² = 10,000 cm² and 1 km = 1,000 m. Always ensure all dimensions are in the same unit before calculation.

Rise in Liquid Level

The rise in level (h) of a liquid in a container with a uniform base area is h = Volume ÷ Base Area.

Shortcut Trick

Area of a square inscribed in a circle of radius r = 2 × r².

Area of a regular hexagon inscribed in a circle of radius r = (3√3 / 2) × r².

Required Ratio = 2r² : (3√3 / 2)r²

Multiply both sides by 2 ⇒ 4 : 3√3.

∴ The correct answer is 4 : 3√3.

Alternate Method

Given:

Radius of the circle = r cm

Square and regular hexagon are inscribed in this circle.

Case 1: Inscribed Square

For a square inscribed in a circle, the diagonal (d) of the square is equal to the diameter of the circle.

⇒ Diagonal of square (d) = 2 × r

⇒ Area of square = d² ÷ 2

⇒ Area of square = (2r)² ÷ 2 = 4r² ÷ 2 = 2r²

Case 2: Inscribed Regular Hexagon

For a regular hexagon inscribed in a circle, the side length (a) of the hexagon is equal to the radius of the circle.

⇒ Side of hexagon (a) = r

⇒ Area of regular hexagon = 3√3/2 × a²

⇒ Area of regular hexagon = 3√3/2 × r²

Ratio Calculation:

⇒ Ratio = Area of square : Area of hexagon

⇒ Ratio = 2r² : (3√3 / 2)r²

⇒ Ratio = 2 : 3√3 / 2

⇒ Ratio = 4 : 3√3

∴ The correct answer is 4 : 3√3.

Additional Information

Area of Circle

The area of the circumscribed circle is πr². The area of any inscribed regular polygon is always less than the area of the circle.

Inscribed Equilateral Triangle

For an equilateral triangle inscribed in a circle of radius r, the side length is r√3 and the area is (3√3 / 4)r².

General Formula for Inscribed Polygon

Area of a regular n-sided polygon inscribed in a circle of radius r = (n/2) × r² × sin(360°/n).

Shortcut Trick

Ratio of CSA to TSA is 2 : 3 ⇒ CSA / (CSA + 2 × Base Area) = 2/3.

Solving the ratio: 3 × CSA = 2 × CSA + 4 × Base Area ⇒ CSA = 4 × Base Area.

TSA = CSA + 2 × Base Area = 4 × Base Area + 2 × Base Area = 6 × Base Area.

6 × (πr²) = 231 ⇒ πr² = 38.5 sq. cm and r = 3.5 cm.

Since CSA = 2πrh = 4πr², we get h = 2r = 7 cm. Volume = πr²h = 38.5 × 7 = 269.5 cu.cm.

∴ The correct answer is 269.5 cu.cm.

Alternate Method

Given:

Total Surface Area (TSA) = 231 cm²

Curved Surface Area (CSA) = 2/3 × TSA

Formula Used:

TSA = 2πr(r + h) = 2πrh + 2πr²

CSA = 2πrh

Volume = πr²h

Calculations:

⇒ CSA = (2/3) × 231 = 154 cm²

⇒ Total Surface Area = Curved Surface Area + 2 × Base Area

⇒ 231 = 154 + 2 × (πr²)

⇒ 2 × πr² = 231 − 154 = 77

⇒ πr² = 77 / 2 = 38.5 cm²

⇒ (22/7) × r² = 38.5 ⇒ r² = (38.5 × 7) / 22 = 12.25

⇒ r = √12.25 = 3.5 cm

⇒ CSA = 2πrh = 154

⇒ 2 × (22/7) × 3.5 × h = 154

⇒ 22 × h = 154 ⇒ h = 7 cm

⇒ Volume = πr²h = 38.5 × 7 = 269.5 cu.cm

∴ The correct answer is 269.5 cu.cm.

Additional Information

Surface Areas of Cylinder

Curved Surface Area = 2πrh and Total Surface Area = 2πr(r + h). The difference between TSA and CSA is always equal to twice the area of the circular base (2πr²).

Volume of Cylinder

The volume of a solid cylinder is the product of the area of its base and its height, V = πr²h.

Related Shapes

Volume of a cone with same radius and height = (1/3)πr²h. Volume of a sphere = (4/3)πr³.

Shortcut Trick

Base radius R = √16 = 4 cm.

Upper radius r = Diameter ÷ 2 = 4 ÷ 2 = 2 cm.

Lateral Surface Area (LSA) = π × (R + r) × l

LSA = π × (4 + 2) × 6 = 36π cm².

∴ The correct answer is 36π cm².

Alternate Method

Given:

Area of base = 16π cm²

Diameter of upper surface = 4 cm

Slant height (l) = 6 cm

Formula Used:

Lateral Surface Area (LSA) of frustum = π(R + r)l

Area of circular base = πR²

Calculations:

⇒ Area of base = πR² = 16π

⇒ R² = 16 ⇒ R = 4 cm

⇒ Upper radius (r) = Diameter ÷ 2 = 4 ÷ 2 = 2 cm

⇒ LSA = π(4 + 2) × 6

⇒ LSA = π × 6 × 6

⇒ LSA = 36π cm²

∴ The correct answer is 36π cm².

Additional Information

Total Surface Area of Frustum

TSA = LSA + Area of Top + Area of Base = π(R + r)l + πR² + πr².

Volume of Frustum

V = (1/3) × π × h × (R² + r² + Rr), where h is the vertical height.

Slant Height Relation

The relation between slant height (l), vertical height (h), and radii (R, r) is l² = h² + (R − r)².

Shortcut Trick

Ratio of radii R₁ : R₂ = 2 : 1 (since Diameter ratio is 2 : 1)

Given: Surface Area S₁ = Volume V₂

⇒ 4 × π × R₁² = (4/3) × π × R₂³

Substitute R₂ = R₁/2 ⇒ 4R₁² = (4/3) × (R₁/2)³

⇒ 4R₁² = R₁³/6 ⇒ R₁ = 24

∴ The correct answer is 24 units.

Alternate Method

Given:

Diameter of Sphere 1 (D) = 2 × Diameter of Sphere 2 (d)

Radius of Sphere 1 (R) = 2 × Radius of Sphere 2 (r) ⇒ r = R/2

Surface Area of Sphere 1 (S₁) = Volume of Sphere 2 (V₂)

Formula Used:

Surface Area of a Sphere = 4 × π × radius²

Volume of a Sphere = (4/3) × π × radius³

Calculations:

⇒ S₁ = V₂

⇒ 4 × π × R² = (4/3) × π × r³

⇒ 4R² = (4/3) × (R/2)³

⇒ 4R² = (4/3) × (R³ / 8)

⇒ 4R² = (4 × R³) / (3 × 8)

⇒ 4R² = R³ / 6

⇒ 4 × 6 = R³ / R²

⇒ R = 24

∴ The correct answer is 24 units.

Additional Information

Properties of Sphere

A sphere is a perfectly round geometrical object in three-dimensional space. Its Surface Area is 4πr² and its Volume is (4/3)πr³.

Hemisphere Formulas

For a hemisphere of radius r: Curved Surface Area (CSA) = 2πr², Total Surface Area (TSA) = 3πr², and Volume = (2/3)πr³.

Ratio of Areas and Volumes

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

Shortcut Trick

Number of cones (n) = Volume of Cylinder ÷ Volume of (Cone + Hemisphere)

n = [π × R² × H] ÷ [&frac{1}{3} π r² (h + 2r)]

Since R = 3r (i.e., 10.5 = 3 × 3.5) and H = 38, h = 12:

n = [ (3r)² × 38 ] ÷ [ &frac{1}{3} r² × (12 + 7) ] = [ 9r² × 38 ] ÷ [ &frac{1}{3} r² × 19 ]

n = [ 9 × 38 × 3 ] ÷ 19 = 9 × 2 × 3 = 54

∴ The correct answer is 54.

Alternate Method

Given: Cylinder diameter = 21 cm ⇒ Radius (R) = 10.5 cm, Height (H) = 38 cm.

Given: Cone diameter = 7 cm ⇒ Radius (r) = 3.5 cm, Height (h) = 12 cm.

Formula: Volume of Cylinder = πR²H | Volume of Cone = &frac{1}{3}πr²h | Volume of Hemisphere = &frac{2}{3}πr³

⇒ Volume of Cylinder = π × (10.5)² × 38

⇒ Volume of 1 Ice-cream (Cone + Hemisphere) = &frac{1}{3}πr²h + &frac{2}{3}πr³

⇒ Volume = &frac{1}{3} × π × (3.5)² × (h + 2r) = &frac{1}{3} × π × (3.5)² × (12 + 2 × 3.5)

⇒ Volume = &frac{1}{3} × π × (3.5)² × 19

⇒ Number of cones (n) = Volume of Cylinder ÷ Volume of 1 Ice-cream

⇒ n = [π × (10.5)² × 38] ÷ [(&frac{1}{3}) × π × (3.5)² × 19]

⇒ n = [ (10.5/3.5)² × 38 ] ÷ (19/3) = [ 3² × 38 ] ÷ (19/3)

⇒ n = [ 9 × 38 × 3 ] ÷ 19 = 54

∴ The correct answer is 54.

Additional Information

Volume of Solids

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

Composite Figures

For objects made of multiple shapes, the total volume is the sum of individual volumes: V_total = V_cone + V_hemisphere = &frac{1}{3}πr²h + &frac{2}{3}πr³.

Sphere and Hemisphere

Volume of a sphere is &frac{4}{3}πr³, hence the volume of a hemisphere is exactly half of it, which is &frac{2}{3}πr³.