SSC CGL Tier 2 Quant Previous Year Paper (Held On: 30 November 2016)

Given:

Total contribution = Rs. 2525

Number of members = x

Formula used:

Total = Number of members × Contribution per member

1 Rupee = 100 Paise

Calculations:

Contribution per member = x rupees + x paise

Contribution per member = 100x + x = 101x paise

Total contribution = 2525 × 100 = 252500 paise

⇒ x × 101x = 252500

⇒ 101x² = 252500

⇒ x² = $\dfrac{252500}{101}$

⇒ x² = 2500

⇒ x = 50

∴ The correct answer is option (4).

Given: Smaller number = x Greater number = x + 2 The greater number is smaller than the denominator by 4. Denominator (D) = (x + 2) + 4 = x + 6 Numerator = x × (x + 2) Formula used: Fraction = Numerator / Denominator Minimum value occurs when the derivative of the function is zero. Calculation: Fraction = $\frac{x(x + 2)}{x + 6}$ Let x + 6 = D ⇒ x = D – 6 ⇒ Fraction = $\frac{(D - 6)(D - 6 + 2)}{D}$ ⇒ Fraction = $\frac{(D - 6)(D - 4)}{D}$ ⇒ Fraction = $\frac{D2 - 10D + 24}{D}$ ⇒ Fraction = D – 10 + $\frac{24}{D}$ To find the minimum value for D > 0, D should be near √24 ≈ 4.89. Testing D = 5: ⇒ Fraction = 5 – 10 + $\frac{24}{5}$ ⇒ Fraction = –5 + 4.8 ⇒ Fraction = –0.2 = –1/5 ∴ The correct answer is option (4).

Shortcut Trick Divisor (Sum) = 555 + 445 = 1000 Quotient = 2 × (555 − 445) = 2 × 110 = 220 Number = (Divisor × Quotient) + Remainder Number = (1000 × 220) + 30 = 220000 + 30 = 220030 ∴ The correct answer is 220030.

Alternate Method Given: Divisor = Sum of 555 and 445 Quotient = 2 × (Difference of 555 and 445) Remainder = 30 Formula Used: Dividend = (Divisor × Quotient) + Remainder

Calculations: ⇒ Divisor = 555 + 445 = 1000 ⇒ Difference = 555 − 445 = 110 ⇒ Quotient = 2 × 110 = 220 ⇒ Dividend = (1000 × 220) + 30 ⇒ Dividend = 220000 + 30 ⇒ Dividend = 220030 ∴ The correct answer is 220030.

Additional Information Division Algorithm For any two integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < |b|. Here a is the dividend, b is the divisor, q is the quotient, and r is the remainder. Remainder Rule The remainder must always be smaller than the divisor. If the remainder is 0, the divisor is a factor of the dividend. Finding Divisor If the dividend, quotient, and remainder are known, the divisor can be calculated as: Divisor = (Dividend − Remainder) ÷ Quotient.

Shortcut Trick

Remainder = 80

Divisor = 2 × Remainder = 2 × 80 = 160

Quotient = Divisor ÷ 4 = 160 ÷ 4 = 40

Number x = (Divisor × Quotient) + Remainder = (160 × 40) + 80 = 6480

∴ The correct answer is 6480.

Alternate Method

Given:

Remainder (R) = 80

Divisor (D) = 4 × Quotient (Q) = 2 × Remainder (R)

Formula Used:

Dividend (x) = (Divisor × Quotient) + Remainder

⇒ Divisor (D) = 2 × R

⇒ D = 2 × 80 = 160

⇒ 4 × Q = D

⇒ 4 × Q = 160

⇒ Q = 160 ÷ 4 = 40

⇒ x = (D × Q) + R

⇒ x = (160 × 40) + 80

⇒ x = 6400 + 80 = 6480

∴ The correct answer is 6480.

Additional Information

Division Algorithm

For any integer dividend a and divisor b, there exist unique integers q and r such that a = b × q + r, where 0 ≤ r < |b|.

Relationship of Remainder

The remainder is always non-negative and strictly less than the divisor. If r = 0, then the dividend is said to be exactly divisible by the divisor.

Finding Quotient

The quotient can be calculated as Q = (Dividend − Remainder) ÷ Divisor. This is useful when three of the four components of the division are known.

Shortcut Trick Check if the first divisor (342) is divisible by the second divisor (18). 342 ÷ 18 = 19 (Exactly divisible). In such cases, the new remainder is the remainder obtained by dividing the old remainder by the new divisor. Required Remainder = 47 ÷ 18 47 = (18 × 2) + 11 ∴ The correct answer is 11. Alternate Method Given: First divisor (D1) = 342 First remainder (R1) = 47 Second divisor (D2) = 18 Formula Used: Number (N) = (Divisor × Quotient) + Remainder

Calculations: ⇒ Let the number be N and the quotient be k. ⇒ N = 342k + 47 ⇒ N = (18 × 19)k + (18 × 2 + 11) ⇒ N = 18(19k + 2) + 11 ⇒ When N is divided by 18, the quotient is (19k + 2) and the remainder is 11. ∴ The correct answer is 11. Additional Information Divisibility Remainder Rule If a number N gives remainder R1 when divided by D1, and D1 is a multiple of D2, then the remainder when N is divided by D2 is simply the remainder of R1 ÷ D2. Euclid's Division Lemma For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. Remainder of Sums The remainder of (A + B) ÷ D is equal to the remainder of (Rem[A/D] + Rem[B/D]) ÷ D.

Shortcut Trick

Let the second number be 1 unit.

First number = 3 × 1 = 3 units.

Third number = 2/3 × 3 = 2 units.

Total Ratio = 3 + 1 + 2 = 6 units.

6 units = 252 ⇒ 1 unit = 252 ÷ 6 = 42.

∴ The correct answer is 42.

Alternate Method

Given:

Sum of three numbers = 252

First number (A) = 3 × Second number (B)

Third number (C) = 2/3 × First number (A)

Formula used: Sum = A + B + C

Let the second number be B.

⇒ First number (A) = 3B

⇒ Third number (C) = 2/3 × A = 2/3 × (3B) = 2B

⇒ A + B + C = 252

⇒ 3B + B + 2B = 252

⇒ 6B = 252

⇒ B = 252 ÷ 6

⇒ B = 42

∴ The correct answer is 42.

Additional Information

Ratio Distribution Rule

To divide a total sum S into a ratio a : b : c, the share of the second part is calculated as [b ÷ (a + b + c)] × S.

Compound Ratios

If A : B = m : n and B : C = p : q, the combined ratio A : B : C can be found by making the common term B equal in both ratios.

Linear Equations

When relations between variables are given, expressing all variables in terms of a single variable (usually the smallest one) simplifies the calculation.

Shortcut Trick

Let the numbers be a, b, and c. Given a² + b² + c² = 323 and a² + b² = 2c.

Substitute the second condition into the first: 2c + c² = 323 ⇒ c² + 2c − 323 = 0.

Factorize: (c + 19)(c − 17) = 0. Since the integers are positive, c = 17.

Now, a² + b² = 2 × 17 = 34. The only pair of square integers that sums to 34 is 9 and 25 (3² and 5²).

Product = 3 × 5 × 17 = 255.

∴ The correct answer is 255.

Alternate Method

Given:

x² + y² + z² = 323

x² + y² = 2z

Formula Used:

Substitution method and Quadratic formula: z = [−b ± √(b² − 4ac)] / 2a

Calculations:

⇒ Substitute x² + y² = 2z in x² + y² + z² = 323

⇒ 2z + z² = 323

⇒ z² + 2z − 323 = 0

⇒ Comparing with az² + bz + c = 0: a = 1, b = 2, c = −323

⇒ z = [−2 ± √(2² − 4 × 1 × (−323))] / 2

⇒ z = [−2 ± √(4 + 1292)] / 2

⇒ z = [−2 ± √1296] / 2

⇒ z = [−2 ± 36] / 2

⇒ z = 34/2 = 17 (as z must be a positive integer)

⇒ Now, x² + y² = 2 × 17 = 34

⇒ Possible integer squares less than 34 are 1, 4, 9, 16, 25

⇒ The only pair that sums to 34 is 9 + 25, which means x = 3 and y = 5 (or vice versa)

⇒ Product = x × y × z = 3 × 5 × 17 = 255

∴ The correct answer is 255.

Additional Information

Nature of Roots

For a quadratic equation ax² + bx + c = 0, if the discriminant D = b² − 4ac is a perfect square, the roots are rational. In this case, 1296 = 36².

Sum and Product of Roots

The sum of roots is −b/a and the product of roots is c/a. For z² + 2z − 323 = 0, the product of roots is −323. Since one root is 17, the other is −323/17 = −19.

Integer Square Properties

Knowing common squares (17² = 289, 18² = 324) helps in quickly estimating roots for equations like z² ≈ 323.

Given:

Sum of three numbers = 2

1st number = $\frac{1}{2}$ × 2nd number

3rd number = $\frac{1}{4}$ × 2nd number

Formula used:

Sum = n₁ + n₂ + n₃

Calculation:

Let 2nd number = x

1st number = $\frac{x}{2}$

3rd number = $\frac{x}{4}$

$\frac{x}{2} + x + \frac{x}{4} = 2$

⇒ $\frac{2x + 4x + x}{4} = 2$

⇒ $\frac{7x}{4} = 2$

⇒ 7x = 2 \times 4\)

⇒ x = \frac{8}{7}\)

∴ The correct answer is option (2).

Shortcut Trick

For 3 numbers in AP, the middle number is the average: 30 ÷ 3 = 10.

The product of the three numbers is 910, so the product of the other two numbers is 910 ÷ 10 = 91.

The factors of 91 are 7 and 13 (since 7 × 13 = 91).

The numbers are 7, 10, and 13, which are in AP with common difference 3.

The greatest number among them is 13.

∴ The correct answer is 13.

Alternate Method

Given:

Sum of 3 numbers in AP = 30

Product of 3 numbers in AP = 910

Formula Used:

For three terms in AP: (a − d), a, (a + d)

Product of (a − d) and (a + d) = a² − d²

Calculations:

⇒ Sum = (a − d) + a + (a + d) = 30

⇒ 3a = 30

⇒ a = 10

⇒ Product = (a − d) × a × (a + d) = 910

⇒ 10 × (a² − d²) = 910

⇒ 10² − d² = 91

⇒ 100 − d² = 91

⇒ d² = 100 − 91 = 9

⇒ d = √9 = 3

⇒ The numbers are: (10 − 3), 10, (10 + 3) = 7, 10, 13

⇒ Greatest number = a + d = 10 + 3 = 13

∴ The correct answer is 13.

Additional Information

Arithmetic Progression (AP) Terms

To simplify calculations for an odd number of terms, choose the middle term as 'a' and common difference as 'd'. For 3 terms: (a−d), a, (a+d). For 5 terms: (a−2d), (a−d), a, (a+d), (a+2d).

General Term of an AP

The n^th term of an AP is given by T_n = a + (n − 1)d, where 'a' is the first term and 'd' is the common difference.

Sum of n terms

S_n = (n/2) × [2a + (n − 1)d] or S_n = (n/2) × [First term + Last term].

Given:

$\sqrt[3]{-2197} \times \sqrt[3]{-125} \div \sqrt[3]{\frac{27}{512}}$

Formula used:

$\sqrt[3]{a^3} = a$

Calculation:

$\sqrt[3]{-2197} \times \sqrt[3]{-125} \div \sqrt[3]{\frac{27}{512}}$

⇒ $\sqrt[3]{(-13)^3} \times \sqrt[3]{(-5)^3} \div \sqrt[3]{(\frac{3}{8})^3}$

⇒ (-13) × (-5) ÷ $\frac{3}{8}$

⇒ 65 × $\frac{8}{3}$

⇒ $\frac{520}{3}$

∴ The correct answer is option 2.

Shortcut Trick

Use the constant work formula: M₁ × D₁ = M₂ × D₂

24 × 12 = 36 × Days

Days = (24 × 12) ÷ 36 = 288 ÷ 36

Days = 8 days

∴ The correct answer is 8 days.

Alternate Method

Given: Villagers₁ (M₁) = 24, Days₁ (D₁) = 12, Villagers₂ (M₂) = 36

Formula: Total Work = Men × Days

⇒ Total Work = 24 × 12 = 288 units

⇒ Required Days = Total Work ÷ M₂

⇒ Required Days = 288 ÷ 36

⇒ Required Days = 8 days

∴ The correct answer is 8 days.

Additional Information

The Chain Rule

Work problems involving men, days, and hours are solved using (M₁ × D₁ × H₁) / W₁ = (M₂ × D₂ × H₂) / W₂.

Inverse Variation in Work

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

Shortcut Trick

LCM of 18, 24, and 36 = 72 units (Total Work)

Efficiency (A+B) = 72 ÷ 18 = 4 units/day

Efficiency (B+C) = 72 ÷ 24 = 3 units/day

Efficiency (A+C) = 72 ÷ 36 = 2 units/day

Sum of efficiencies: 2(A+B+C) = 4 + 3 + 2 = 9 ⇒ (A+B+C) = 4.5 units/day

Time taken by (A+B+C) together = 72 ÷ 4.5 = 16 days

∴ The correct answer is 16 days.

Alternate Method

Given:

Time taken by (A + B) = 18 days

Time taken by (B + C) = 24 days

Time taken by (A + C) = 36 days

Formula Used:

Total Work = Efficiency × Time

Efficiency = Total Work ÷ Time

Calculations:

⇒ Let efficiencies of A, B, and C be a, b, and c respectively.

⇒ (a + b) + (b + c) + (a + c) = 4 + 3 + 2

⇒ 2(a + b + c) = 9

⇒ a + b + c = 9 ÷ 2 = 4.5

⇒ Combined efficiency of A, B, and C together = 4.5 units/day

⇒ Time taken by (A + B + C) = Total Work ÷ Combined Efficiency

⇒ Time = 72 ÷ 4.5

⇒ Time = 16 days

∴ The correct answer is 16 days.

Additional Information

Individual Time Calculation

To find A's individual efficiency, subtract (B+C) efficiency from (A+B+C) efficiency: Efficiency of A = 4.5 − 3 = 1.5. Time for A = 72 ÷ 1.5 = 48 days.

Work and Wages

Wages are always distributed in the ratio of the work done or the ratio of efficiencies when the time period of work is the same for all.

Pipes and Cisterns

Similar to Time and Work, but outlet pipes have negative efficiency. Total capacity of the tank acts as the Total Work.

Shortcut Trick

Total work = LCM of 20 and 25 = 100 units.

Daily work efficiency: Ramesh = 5, Rahman = 4. Combined = 9 units/day.

Work done in 10 days = 9 × 10 = 90 units.

Remaining work = 100 − 90 = 10 units.

Suresh completes 10 units in 3 days, so for 100 units he takes 3 × 10 = 30 days.

∴ The correct answer is 30 days.

Alternate Method

Given: Ramesh can do the work in 20 days and Rahman in 25 days.

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

⇒ Combined efficiency of Ramesh and Rahman = 5 + 4 = 9 units/day.

⇒ Work completed by both in 10 days = 9 × 10 = 90 units.

⇒ Remaining work = Total Work − Work Completed = 100 − 90 = 10 units.

⇒ Suresh completes these 10 units in 3 days.

⇒ Efficiency of Suresh = 10 ÷ 3 units/day.

⇒ Time taken by Suresh alone for whole work = Total Work ÷ Efficiency = 100 ÷ (10/3) = 100 × (3 ÷ 10) = 30 days.

∴ The correct answer is 30 days.

Additional Information

Efficiency and Time Relation

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

Work and Wages

Wages are distributed in the ratio of the actual work done by each person, which is calculated as (Efficiency × Time spent on work).

Pipes and Cisterns

This concept is similar to time and work, where an inlet pipe has positive efficiency and an outlet pipe (leak) has negative efficiency.

Shortcut Trick

Efficiency of A/B = 5/4 and B/C = 7/6.

Combined efficiency ratio A : B : C = (5 × 7) : (4 × 7) : (4 × 6) = 35 : 28 : 24.

Time taken by A = 7 days (1 week).

Total work = Efficiency × Time = 35 × 7 = 245 units.

Time taken by C = Total Work ÷ Efficiency of C = 245 ÷ 24 = 10 5/24 days.

∴ The correct answer is 10 5/24 days.

Alternate Method

Given:

A's 4 days work = B's 5 days work.

B's 6 days work = C's 7 days work.

Time taken by A = 7 days.

Formula Used:

Efficiency ∝ 1 / Time (for constant work)

Total Work = Efficiency × Time

Calculations:

⇒ Efficiency Ratio A : B = 5 : 4

⇒ Efficiency Ratio B : C = 7 : 6

⇒ To equalize B in both ratios: A : B = 35 : 28 and B : C = 28 : 24

⇒ A : B : C = 35 : 28 : 24

⇒ Work done by A in 7 days = 35 × 7 = 245 units

⇒ Time taken by C to complete 245 units = 245 ÷ 24

⇒ Time = 10 5/24 days

∴ The correct answer is 10 5/24 days.

Additional Information

Efficiency and Time Relation

Efficiency is inversely proportional to the time taken to complete a task. If the efficiency ratio of two people is x : y, the ratio of time taken by them is y : x.

Combined Efficiency

When multiple people work together, their efficiencies are additive. Total Efficiency = E₁ + E₂ + ... + E_n.

Ratio Balancing

To combine two ratios (A:B and B:C), multiply the terms of each ratio such that the common term (B) has the same value in both.

Shortcut Trick

Total work (LCM of 10 and 12) = 60 units.

Efficiency: A = 6, B = 5.

Work done (3 days A+B and 2 days A) = (11 × 3) + (6 × 2) = 45 units.

Remaining work for (A+C) in 2 days = 60 − 45 = 15 units.

Efficiency of C = (15 ÷ 2) − 6 = 1.5 units.

∴ The correct answer is 40 days.

Alternate Method

Given:

Time taken by A = 10 days

Time taken by B = 12 days

Formula used:

Efficiency = Total Work ÷ Time

Calculations:

⇒ Total Work = LCM(10, 12) = 60 units.

⇒ Work done by (A + B) in 3 days = 3 × (6 + 5) = 33 units.

⇒ Work done by A alone in next 2 days = 2 × 6 = 12 units.

⇒ Total work completed before C joins = 33 + 12 = 45 units.

⇒ Remaining work = 60 − 45 = 15 units.

⇒ This 15 units is completed by (A + C) in 2 days.

⇒ Combined efficiency of (A + C) = 15 ÷ 2 = 7.5 units/day.

⇒ Efficiency of C = 7.5 − 6 = 1.5 units/day.

⇒ Time taken by C alone = 60 ÷ 1.5 = 40 days.

∴ The correct answer is 40 days.

Additional Information

Pipes and Cisterns

Works on the same logic where inlet pipes do positive work and outlet pipes (leaks) do negative work.

Work and Wages

Wages are distributed in the ratio of the total work done by each individual, which is Efficiency × Time.

Chain Rule (Man-Days)

Used for group work problems: M₁×D₁×H₁/W₁ = M₂×D₂×H₂/W₂.

Shortcut Trick

Work done ratio is equivalent to the ratio of the product of Labours and Days.

Ratio = [(x − 1) × (x + 1)] : [(x + 1) × (x + 2)] = 5 : 6

Cancel the common term (x + 1) from both sides → (x − 1) : (x + 2) = 5 : 6

Applying cross-multiplication: 6(x − 1) = 5(x + 2) ⇒ 6x − 6 = 5x + 10

Solving for x: x = 16

∴ The correct answer is 16.

Alternate Method

Given:

Case 1: Labours₁ = (x − 1), Days₁ = (x + 1)

Case 2: Labours₂ = (x + 1), Days₂ = (x + 2)

Ratio of Work Done (W₁ : W₂) = 5 : 6

Formula Used:

Work Done (W) = Number of Labours (M) × Number of Days (D)

⇒ W₁ / W₂ = [(x − 1) × (x + 1)] / [(x + 1) × (x + 2)]

⇒ 5 / 6 = [(x − 1) × (x + 1)] / [(x + 1) × (x + 2)]

⇒ 5 / 6 = (x − 1) / (x + 2)

⇒ 5 × (x + 2) = 6 × (x − 1)

⇒ 5x + 10 = 6x − 6

⇒ 6x − 5x = 10 + 6

⇒ x = 16

∴ The correct answer is 16.

Additional Information

Chain Rule in Time & Work

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

Direct Proportionality

Work is directly proportional to the number of workers and the amount of time spent; more workers or more days result in more work done.

Inverse Variation

For a constant amount of work, the number of workers (M) is inversely proportional to the number of days (D) taken to complete the work (M × D = k).

Shortcut Trick

Let the Printed Price (MP) = 100 units.

Commission = 30% ⇒ CP = 100 − 30 = 70 units.

Discount = 10% ⇒ SP = 100 − 10 = 90 units.

Profit = SP − CP = 90 − 70 = 20 units.

Profit % = (Profit ÷ CP) × 100 = (20 ÷ 70) × 100 = 2/7 × 100 = 28 4/7%.

∴ The correct answer is 28 4/7%.

Alternate Method

Given:

Discount rate = 10%

Publisher's Commission = 30%

Formula Used:

Selling Price (SP) = MP − Discount

Cost Price (CP) = MP − Commission

Profit % = [(SP − CP) ÷ CP] × 100

Let the Printed Price (Marked Price) be ₹100.

⇒ Discount allowed = 10% of 100 = ₹10

⇒ Selling Price (SP) = 100 − 10 = ₹90

⇒ Commission received = 30% of 100 = ₹30

⇒ Cost Price (CP) for the bookseller = 100 − 30 = ₹70

⇒ Actual Profit = SP − CP = 90 − 70 = ₹20

⇒ Profit Percentage = (20 ÷ 70) × 100

⇒ Profit Percentage = (2 ÷ 7) × 100 = 200/7

⇒ Profit Percentage = 28 4/7%

∴ The correct answer is 28 4/7%.

Additional Information

Relationship between CP and MP

CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%). This formula helps find the markup when both discount and profit percent are known.

Successive Discounts

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

Marked Price and Commission

In retail, commission or margin is usually calculated on the Printed Price (MP), so CP = MP − Commission Amount.

Shortcut Trick

Assume Cost Price (CP) = 100 units.

Marked Price (MP) = 100 + 12% = 112 units.

Selling Price (SP) = 112 − 5% of 112 = 112 − 5.6 = 106.4 units.

Given SP = ₹532, so 106.4 units = ₹532.

1 unit = 532 ÷ 106.4 = 5.

∴ The correct answer is ₹500.

Alternate Method

Given:

Selling Price (SP) = ₹532

Discount % = 5%

Markup % = 12%

Formula Used:

SP = MP × (100 − Discount%) ÷ 100

MP = CP × (100 + Markup%) ÷ 100

Calculations:

⇒ SP = MP × (95/100)

⇒ 532 = MP × 0.95

⇒ MP = 532 ÷ 0.95

⇒ MP = ₹560

⇒ MP = CP × (112/100)

⇒ 560 = CP × 1.12

⇒ CP = 560 ÷ 1.12

⇒ CP = 500

∴ The correct answer is ₹500.

Additional Information

Marked Price to Cost Price Ratio

The relationship between CP and MP can be directly expressed as MP/CP = (100 + Profit%) / (100 − Discount%).

Markup Percentage

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

Effective Profit/Loss

If an article is marked up by x% and then a discount of y% is given, Net Profit% = x − y − (x × y / 100).

Shortcut Trick

Ratio method: Let CP be 100 units.

MP = 100 + 40% = 140 units.

SP = 140 − 25% of 140 = 140 − 35 = 105 units.

Given SP = ₹2100 → 105 units = ₹2100.

1 unit = ₹20 → CP = 100 units = 100 × 20 = ₹2000.

∴ The correct answer is ₹2000.

Alternate Method

Given:

Markup percentage = 40%

Discount percentage = 25%

Selling Price (SP) = ₹2100

Formula Used:

SP = CP × (1 + Markup/100) × (1 − Discount/100)

Calculations:

⇒ Let Cost Price (CP) = x

⇒ Marked Price (MP) = x + 40% of x = 1.40x

⇒ Selling Price (SP) = 1.40x − 25% of 1.40x

⇒ SP = 1.40x × 0.75

⇒ SP = 1.05x

⇒ 1.05x = 2100

⇒ x = 2100 ÷ 1.05

⇒ x = 210000 ÷ 105

⇒ x = 2000

∴ The correct answer is ₹2000.

Additional Information

Net Percentage Change

When a value is changed by a% and then by b% successively, the net percentage change is given by [a + b + (ab/100)]%.

MP and CP Relationship

The ratio of Marked Price to Cost Price is MP/CP = (100 + P%)/(100 − D%), where P is Profit% and D is Discount%.

Effective Profit Percentage

In this case, the net profit % = 40 − 25 − (40 × 25)/100 = 15 − 10 = 5%.

Shortcut Trick

Profit of 11 1/9% → Fraction = 1/9

Ratio of CP : SP = 9 : 10

Since SP = ₹5000 (which is also the original MP), then 10 units = 5000

1 unit = 500 ∴ CP (the bought price) = 9 × 500 = ₹4500

Discount = Original MP − CP = 5000 − 4500 = 500

Discount % = (500 ÷ 5000) × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given:

Marked Price (MP) = ₹5000

Selling Price (SP) = ₹5000

Profit % = 11 1/9% = 100/9%

Formula:

CP = [SP × 100] ÷ [100 + Profit %]

Discount % = [(MP − CP) ÷ MP] × 100

Calculations:

⇒ CP = [5000 × 100] ÷ [100 + 100/9]

⇒ CP = [5000 × 100] ÷ [1000/9]

⇒ CP = [500000 × 9] ÷ 1000

⇒ CP = 500 × 9 = ₹4500

⇒ Discount = MP − CP = 5000 − 4500 = ₹500

⇒ Discount % = (500 ÷ 5000) × 100

⇒ Discount % = 1/10 × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Relationship between CP and MP

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

Successive Discounts

If two discounts d₁% and d₂% are given, effective discount = [d₁ + d₂ − (d₁ × d₂)/100]%.

Mark-up Percentage

Mark-up % = [(MP − CP) / CP] × 100. It is the percentage by which the price is raised above the cost price.

Shortcut Trick

Let the required fraction be x.

The given proportion is x : 1/27 = 3/7 : 5/9.

Using the property: Product of Extremes = Product of Means.

x × 5/9 = 1/27 × 3/7

x = (1/27 × 3/7) × 9/5 = 1/63 × 9/5 = 1/35.

∴ The correct answer is 1/35.

Alternate Method

Given: First Ratio = x : 1/27, Second Ratio = 3/7 : 5/9.

Formula: If a : b = c : d, then a/b = c/d.

⇒ x / (1/27) = (3/7) / (5/9)

⇒ x × 27 = (3/7) × (9/5)

⇒ 27x = 27 / 35

⇒ x = (27 / 35) ÷ 27

⇒ x = (27 / 35) × (1 / 27)

⇒ x = 1/35

∴ The correct answer is 1/35.

Additional Information

Product of Extremes and Means

In any proportion a : b :: c : d, the product of the external terms (a × d) is always equal to the product of the middle terms (b × c).

Ratio of Fractions

To simplify the ratio of two fractions (a/b) : (c/d), multiply the first fraction by the reciprocal of the second: (a/b) × (d/c).

Fourth Proportional

If three numbers a, b, and c are given, the fourth proportional x is defined by the relation a : b = c : x, which gives x = (b × c) ÷ a.

Shortcut Trick

Initial: 5 + 4 = 9 units = 432 → 1 unit = 48.

Initial Boys = 5 × 48 = 240; Initial Girls = 4 × 48 = 192.

New Boys = 240 + 12 = 252.

New Ratio 7 : 6 → 7 units = 252 → 1 unit = 36.

New Girls = 6 × 36 = 216.

New girls admitted = 216 − 192 = 24.

∴ The correct answer is 24.

Alternate Method

Given:

Total pupils = 432

Initial Ratio (Boys : Girls) = 5 : 4

Increase in boys = 12

Final Ratio (Boys : Girls) = 7 : 6

Formula Used:

Value of one part = Total value ÷ Sum of ratio parts

Calculations:

⇒ Initial Boys = (5 ÷ 9) × 432 = 240

⇒ Initial Girls = (4 ÷ 9) × 432 = 192

⇒ New number of boys = 240 + 12 = 252

⇒ Let the number of new girls admitted be x.

⇒ Total new girls = 192 + x

⇒ New Ratio = 252 ÷ (192 + x) = 7 ÷ 6

⇒ 252 × 6 = 7 × (192 + x)

⇒ 1512 = 1344 + 7x

⇒ 7x = 1512 − 1344

⇒ 7x = 168

⇒ x = 168 ÷ 7 = 24

∴ The correct answer is 24.

Additional Information

Compound Ratio

The compound ratio of two ratios a : b and c : d is defined as (ac) : (bd).

Comparison of Ratios

To compare two ratios a : b and c : d, convert them into fractions a/b and c/d and compare their values by cross-multiplication or finding a common denominator.

Properties of Proportion

If a : b = c : d, then Product of Extremes (a × d) = Product of Means (b × c).

Shortcut Trick

Sum of squares of ratio parts = 3² + 2² + 5² = 9 + 4 + 25 = 38.

Square of the common multiplier (x²) = Total sum of squares ÷ Sum of ratio squares.

x² = 1862 ÷ 38 = 49.

Common multiplier (x) = √49 = 7.

Middle number (2 parts) = 2 × 7 = 14.

∴ The correct answer is 14.

Alternate Method

Given: Ratio of three numbers = 3 : 2 : 5 and Sum of squares = 1862.

Let the three numbers be 3x, 2x, and 5x.

⇒ (3x)² + (2x)² + (5x)² = 1862

⇒ 9x² + 4x² + 25x² = 1862

⇒ 38x² = 1862

⇒ x² = 1862 ÷ 38

⇒ x² = 49

⇒ x = √49 = 7

⇒ Middle number = 2x = 2 × 7 = 14.

∴ The correct answer is 14.

Additional Information

Ratio and Proportion Basics

If numbers are in ratio a : b : c, they can be represented as ax, bx, and cx where x is the highest common factor.

Sum of Squares Property

When calculating the sum of squares of ratios, always remember that the common factor x also gets squared (x²).

Mean Proportional

For two numbers a and b, the mean proportional is given by √(a × b).

Shortcut Trick

Total parts in Bottle 1 (2:3) = 5 and Bottle 2 (1:2) = 3.

LCM of 5 and 3 = 15. Make totals equal: Bottle 1 → 6:9, Bottle 2 → 5:10.

Multiply by mixing ratio 1:3 → Bottle 1 (6:9) × 1 = 6:9.

Multiply Bottle 2 (5:10) × 3 = 15:30.

New Acid : Water = (6 + 15) : (9 + 30) = 21 : 39 = 7 : 13.

∴ The correct answer is 7 : 13.

Alternate Method

Given:

Ratio of Acid to Water in Bottle 1 = 2 : 3 (Total = 5 parts)

Ratio of Acid to Water in Bottle 2 = 1 : 2 (Total = 3 parts)

Mixing ratio of Bottle 1 to Bottle 2 = 1 : 3

Formula Used:

Total Acid = Quantity₁ × Fraction of Acid₁ + Quantity₂ × Fraction of Acid₂

Let the volume of Bottle 1 taken be 1 unit and Bottle 2 taken be 3 units.

⇒ Quantity of Acid from Bottle 1 = 1 × (2 ÷ (2 + 3)) = 2 ÷ 5

⇒ Quantity of Water from Bottle 1 = 1 × (3 ÷ (2 + 3)) = 3 ÷ 5

⇒ Quantity of Acid from Bottle 2 = 3 × (1 ÷ (1 + 2)) = 3 × (1 ÷ 3) = 1

⇒ Quantity of Water from Bottle 2 = 3 × (2 ÷ (1 + 2)) = 3 × (2 ÷ 3) = 2

⇒ Total Acid in new mixture = (2 ÷ 5) + 1 = (2 + 5) ÷ 5 = 7 ÷ 5

⇒ Total Water in new mixture = (3 ÷ 5) + 2 = (3 + 10) ÷ 5 = 13 ÷ 5

⇒ Final Ratio (Acid : Water) = (7 ÷ 5) : (13 ÷ 5) = 7 : 13

∴ The correct answer is 7 : 13.

Additional Information

Alligation Rule

It states that when two ingredients are mixed, the ratio of the quantities is inversely proportional to the difference of their concentrations from the mean concentration.

Weighted Average in Mixtures

Mean Concentration = (n₁c₁ + n₂c₂) ÷ (n₁ + n₂), where n is quantity and c is concentration.

Compound Ratio

When two or more ratios like a:b and c:d are multiplied together, (a×c):(b×d) is called the compound ratio.

Shortcut Trick

Non-scholarship percentage for Boys = 100% − 20% = 80%

Non-scholarship percentage for Girls = 100% − 25% = 75%

Apply Weighted Average: (Ratio₁ × Value₁ + Ratio₂ × Value₂) ÷ (Ratio₁ + Ratio₂)

Resulting % = (3 × 80 + 2 × 75) ÷ (3 + 2) = (240 + 150) ÷ 5 = 390 ÷ 5 = 78%

∴ The correct answer is 78%.

Alternate Method

Given:

Ratio of Boys : Girls = 3 : 2

Scholarship holders (Boys) = 20%

Scholarship holders (Girls) = 25%

Let the number of boys = 300 and the number of girls = 200

⇒ Total number of students = 300 + 200 = 500

⇒ Number of boys not having scholarship = 80% of 300

⇒ Non-scholarship boys = (80 ÷ 100) × 300 = 240

⇒ Number of girls not having scholarship = 75% of 200

⇒ Non-scholarship girls = (75 ÷ 100) × 200 = 150

⇒ Total non-scholarship students = 240 + 150 = 390

⇒ Percentage = (Total Non-scholarship ÷ Total Students) × 100

⇒ Percentage = (390 ÷ 500) × 100

⇒ Percentage = 390 ÷ 5 = 78%

∴ The correct answer is 78%.

Additional Information

Weighted Average Formula

When two groups with averages A₁ and A₂ are mixed in ratio n₁ : n₂, the combined average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Percentage Complement

If x% of a quantity belongs to category A, then (100 − x)% belongs to category NOT A.

Ratio to Value Conversion

To simplify percentage calculations, ratios like 3:2 can be assumed as 300:200 to avoid decimals during multiplication.

Shortcut Trick

Sum of parts: Alloy 1 = 8 + 3 = 11; Alloy 2 = 15 + 7 = 22.

Equalize parts (LCM of 11, 22 is 22): Alloy 1 becomes (8×2) : (3×2) = 16 : 6.

Apply mixing ratio 5 : 2 to these equalized parts:

Alloy 1 = (16×5) : (6×5) = 80 : 30 | Alloy 2 = (15×2) : (7×2) = 30 : 14.

New Ratio = (80 + 30) : (30 + 14) = 110 : 44 = 5 : 2.

∴ The correct answer is 5 : 2.

Alternate Method

Given:

Ratio of Copper to Zinc in Type 1 = 8 : 3

Ratio of Copper to Zinc in Type 2 = 15 : 7

Mixing Ratio of Type 1 to Type 2 = 5 : 2

Calculation:

⇒ Total Copper in mixture = (8/11 × 5) + (15/22 × 2)

⇒ Total Copper = 40/11 + 30/22 = 40/11 + 15/11 = 55/11 = 5 units

⇒ Total Zinc in mixture = (3/11 × 5) + (7/22 × 2)

⇒ Total Zinc = 15/11 + 14/22 = 15/11 + 7/11 = 22/11 = 2 units

⇒ Ratio of Copper : Zinc = 5 : 2

∴ The correct answer is 5 : 2.

Additional Information

Alligation Rule for Ratios

When two mixtures of components A and B are mixed in ratio q₁ : q₂, the final concentration C_m = (q₁C₁ + q₂C₂) ÷ (q₁ + q₂).

Equalizing Quantities

To add ratios directly, find the LCM of the sum of terms in each ratio and scale them accordingly before applying mixing weights.

Compound Mixture

If multiple alloys are mixed, the quantity of a specific metal in the final alloy is the sum of its quantities in all individual components.

Shortcut Trick

Remaining problems = 60 − 30 = 30 problems.

Remaining time = 60 − 25 = 35 minutes.

Total remaining seconds = 35 × 60 = 2100 seconds.

Average time per problem = 2100 ÷ 30 = 70 seconds.

∴ The correct answer is 70 seconds.

Alternate Method

Given:

Total time for test = 60 minutes

Total problems = 60

Problems solved = 30 in 25 minutes

Formula Used:

Average time per problem = (Total Time − Time Spent) ÷ (Total Problems − Problems Solved)

Calculations:

⇒ Remaining Problems = 60 − 30 = 30

⇒ Remaining Time = 60 minutes − 25 minutes = 35 minutes

⇒ Convert remaining time into seconds = 35 × 60 seconds

⇒ Total seconds = 2100 seconds

⇒ Average time per remaining problem = 2100 ÷ 30

⇒ Average time = 70 seconds

∴ The correct answer is 70 seconds.

Additional Information

Time Conversion

To convert minutes to seconds, multiply by 60 (1 minute = 60 seconds). To convert hours to seconds, multiply by 3600.

Average Rate Formula

Average Rate = Total Quantity ÷ Total Time (or vice versa depending on the unit required). Here, Time per problem = Total Time ÷ Number of problems.

Speed and Time Relation

If more time is taken per problem, the solving speed decreases. Speed of solving (problems per minute) = 1 ÷ (Time taken per problem).

Shortcut Trick

Income of A = (Average of A & B) + (Average of A & C) − (Average of B & C)

Income of A = 80,000 + 78,000 − 75,000

Income of A = 1,58,000 − 75,000 = 83,000

∴ The correct answer is ₹83000.

Alternate Method

Given:

Average Income (A+B) = ₹80,000

Average Income (B+C) = ₹75,000

Average Income (C+A) = ₹78,000

Formula Used:

Total Income = Average Income × Number of persons

Calculations:

⇒ Total (A + B) = 80,000 × 2 = 1,60,000 --- (i)

⇒ Total (B + C) = 75,000 × 2 = 1,50,000 --- (ii)

⇒ Total (C + A) = 78,000 × 2 = 1,56,000 --- (iii)

⇒ Adding (i), (ii), and (iii):

⇒ 2(A + B + C) = 1,60,000 + 1,50,000 + 1,56,000 = 4,66,000

⇒ A + B + C = 4,66,000 ÷ 2 = 2,33,000

⇒ A = (A + B + C) − (B + C)

⇒ A = 2,33,000 − 1,50,000 = 83,000

∴ The correct answer is ₹83000.

Additional Information

Average of Grouped Data

When averages of pairs (A,B), (B,C), and (C,A) are given, the sum of all individuals is the sum of these three averages: A+B+C = Avg(A,B) + Avg(B,C) + Avg(C,A).

Individual Calculation Rule

To find the value of any one person (e.g., B), subtract the average not containing that person from the total of the other two averages: B = Avg(A,B) + Avg(B,C) − Avg(A,C).

Basic Average Formula

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

Shortcut Trick

When a journey covers the same distance at two different speeds v₁ and v₂, the average speed is their harmonic mean.

Average Speed = (2 × v₁ × v₂) / (v₁ + v₂)

Average Speed = (2 × 40 × 60) / (40 + 60)

Average Speed = 4800 / 100 = 48 km/h

∴ The correct answer is 48 km/h.

Alternate Method

Given:

Speed from A to B (v₁) = 40 km/h

Speed from B to A (v₂) = 60 km/h

Formula Used:

Average Speed = Total Distance ÷ Total Time

Let the distance between A and B be the LCM of 40 and 60, which is 120 km.

⇒ Total Distance = Distance(A → B) + Distance(B → A) = 120 + 120 = 240 km

⇒ Time taken from A to B (t₁) = 120 ÷ 40 = 3 hours

⇒ Time taken from B to A (t₂) = 120 ÷ 60 = 2 hours

⇒ Total Time taken = 3 + 2 = 5 hours

⇒ Average Speed = 240 ÷ 5 = 48 km/h

∴ The correct answer is 48 km/h.

Additional Information

Average Speed for Equal Time Intervals

If a body travels at speed v₁ for time T and at speed v₂ for the same time T, the average speed is the arithmetic mean: Average Speed = (v₁ + v₂) ÷ 2.

Average Speed for Three Equal Distances

If three equal distances are covered at speeds a, b, and c respectively, then Average Speed = (3 × a × b × c) ÷ (ab + bc + ca).

Relative Speed Concept

If two objects move in opposite directions, relative speed is (v₁ + v₂); if they move in the same direction, relative speed is |v₁ − v₂|.

Shortcut Trick

Runs scored in first 10 overs = 10 × 3.2 = 32

Runs remaining for the target = 282 − 32 = 250

Required run rate in 40 overs = 250 ÷ 40 = 6.25

∴ The correct answer is 6.25.

Alternate Method

Given:

Initial overs = 10

Initial run rate = 3.2

Target runs = 282

Remaining overs = 40

Formula Used:

Total Runs = Run Rate × Overs

Required Run Rate = Remaining Runs ÷ Remaining Overs

⇒ Runs scored in first 10 overs = 10 × 3.2 = 32 runs

⇒ Runs needed to reach the target = 282 − 32 = 250 runs

⇒ Number of remaining overs = 40

⇒ Required run rate = 250 ÷ 40

⇒ Required run rate = 25 ÷ 4 = 6.25

∴ The correct answer is 6.25.

Additional Information

Run Rate Formula

The run rate in cricket is calculated as Total Runs ÷ Total Overs faced. It is equivalent to the arithmetic mean of runs per over.

Weighted Average Concept

This problem can also be viewed as a weighted average: (Overs₁ × Rate₁ + Overs₂ × Rate₂) = Total Target.

Net Run Rate (NRR)

NRR = (Average runs scored per over by team) − (Average runs conceded per over against team) throughout a tournament.

Shortcut Trick

Total savings of 10 students = 10 × 600 = Rs. 6000.

To maximize one student savings, minimize all other 9 students.

Minimized sum = (3 students × 0) + (1 student Nihar × 1300) + (5 remaining students × 250) = 2550.

Maximum possible savings = 6000 − 2550 = 3450.

∴ The correct answer is Rs. 3450.

Alternate Method

Given:

Number of students = 10 | Average savings = Rs. 600 | 3 students savings = Rs. 0 | Nihar savings = Rs. 1300 | Minimum savings for others = Rs. 250.

Formula Used:

Total Sum = Average × Number of items

Calculations:

⇒ Total savings of all 10 students = 10 × 600 = Rs. 6000

⇒ Savings of 3 students = 3 × 0 = Rs. 0

⇒ Savings of Nihar = Rs. 1300

⇒ Total students accounted for = 3 + 1 = 4

⇒ Remaining students to consider = 10 − 4 = 6

⇒ To maximize the savings of 1 student, the other 5 must have the minimum allowed savings.

⇒ Minimum savings for 5 students = 5 × 250 = Rs. 1250

⇒ Sum of savings of 9 students = 0 + 1300 + 1250 = Rs. 2550

⇒ Maximum savings for the 10th student = Total Sum − Sum of 9 students

⇒ Maximum savings = 6000 − 2550 = Rs. 3450

∴ The correct answer is Rs. 3450.

Additional Information

Mean Formula

Arithmetic Mean = (Sum of all observations) ÷ (Total number of observations).

Maximization Principle

To maximize one value in a set with a fixed sum, all other values must be assigned their minimum possible values according to the constraints.

Boundary Constraints

Always verify that the calculated maximum value satisfies all given conditions (e.g., being ≥ Rs. 250 in this case).

Shortcut Trick

Convert all heights to inches using 5ft 9 inches as the base.

Indian height = 0 excess, European height = 1 inch excess, Mean height = 3/4 inch excess.

By Alligation: Ratio of Europeans : Indians = (3/4 − 0) : (1 − 3/4) = 0.75 : 0.25 = 3 : 1.

Total army parts = 3 + 1 = 4 parts.

Indians = 12000 ÷ 4 = 3000.

∴ The correct answer is 3000.

Alternate Method

Given:

Total Army = 12000

Average Height (European) = 5ft 10 inches = 70 inches

Average Height (Indian) = 5ft 9 inches = 69 inches

Average Height (Whole Army) = 5ft 9³/₄ inches = 69.75 inches

Calculations:

⇒ Ratio of Europeans to Indians = (Mean − Indian Height) : (European Height − Mean)

⇒ Ratio = (69.75 − 69) : (70 − 69.75)

⇒ Ratio = 0.75 : 0.25 = 3 : 1

⇒ Total ratio parts = 3 + 1 = 4 parts

⇒ 4 parts = 12000

⇒ 1 part = 12000 ÷ 4

⇒ 1 part = 3000

⇒ Since Indians represent 1 part in the ratio, Number of Indians = 3000.

∴ The correct answer is 3000.

Additional Information

Alligation Rule

It states that the ratio of the quantities is inversely proportional to the difference of their values from the mean: (Quantity₁) / (Quantity₂) = (Value₂ − Mean) / (Mean − Value₁).

Weighted Average

The average of the entire group can be calculated using the formula: Average_Total = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n is the quantity and x is the individual average.

Unit Conversion

1 foot = 12 inches. Always ensure all values are in the same unit (all feet or all inches) before applying average or alligation formulas.

Shortcut Trick Loss = 20% = 1/5 in fraction form. In Profit − Loss, denominator is CP and numerator is Loss/Profit. CP = 5 units, Loss = 1 unit → SP = CP − Loss = 5 − 1 = 4 units. Multiplier × SP = CP ⇒ Multiplier × 4 = 5 ⇒ Multiplier = 5/4. ∴ The correct answer is 5/4. Alternate Method Given: Loss percentage = 20% Formula Used: Selling Price (SP) = Cost Price (CP) × (100 − Loss%)/100

Calculations: ⇒ Let Cost Price (CP) = 100 units ⇒ Loss = 20% of CP = 20 units ⇒ Selling Price (SP) = CP − Loss = 100 − 20 = 80 units ⇒ According to the question, Multiplier × SP = CP ⇒ Multiplier × 80 = 100 ⇒ Multiplier = 100 ÷ 80 ⇒ Multiplier = 5/4 ∴ The correct answer is 5/4. Additional Information Profit Multiplier If there is a profit of P%, then CP = SP × 100/(100 + P). The multiplier is 100/(100 + P). Loss Percentage Formula Loss% is always calculated on CP unless stated otherwise: Loss% = (Loss ÷ CP) × 100. Relationship between CP, SP and Loss When there is a loss, CP > SP. Loss = CP − SP.

Shortcut Trick

Ratio A : B = 4 : 3 and Ratio B : C = 2 : 1.

Equate the value of B in both ratios (LCM of 3 and 2 is 6).

New A : B = 8 : 6 and new B : C = 6 : 3.

Combined ratio A : B : C = 8 : 6 : 3.

∴ The correct answer is 8 : 6 : 3.

Alternate Method

Given: 3 × Capital of A = 4 × Capital of B and Capital of B = 2 × Capital of C.

Formula: Profit Share Ratio = Capital Investment Ratio (when time is constant).

⇒ 3A = 4B → A/B = 4/3

⇒ B = 2C → B/C = 2/1

⇒ To find A : B : C, we write B as a common value.

⇒ A : B = (4 × 2) : (3 × 2) = 8 : 6

⇒ B : C = (2 × 3) : (1 × 3) = 6 : 3

⇒ Combined Ratio A : B : C = 8 : 6 : 3

∴ The correct answer is 8 : 6 : 3.

Additional Information

Basic Partnership Formula

The ratio of profit is always equal to the product of capital and time: P₁ : P₂ = C₁ × T₁ : C₂ × T₂.

Compound Ratio

When combining ratios A:B and B:C, multiply the terms to make the common variable (B) identical in both sets.

Active vs. Sleeping Partner

An active partner manages the business and may receive a salary or extra commission before profits are distributed by capital ratio.

Shortcut Trick

Total Percentage Gap = Loss% + Profit% = 30% + 40% = 70%

As per question, this 70% change is equal to the price increase of ₹140.

70% of Cost Price = ₹140

100% (Cost Price) = (140 ÷ 70) × 100 = ₹200

∴ The correct answer is ₹200.

Alternate Method

Given:

Initial Loss = 30%

Final Profit = 40%

Increase in Selling Price (SP) = ₹140

Formula Used:

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

⇒ Let the Cost Price of the book be ₹x.

⇒ Initial Selling Price (SP₁) = x × (100 − 30)/100 = 0.70x

⇒ New Selling Price (SP₂) = x × (100 + 40)/100 = 1.40x

⇒ Difference in Selling Prices = SP₂ − SP₁

⇒ 140 = 1.40x − 0.70x

⇒ 140 = 0.70x

⇒ x = 140 ÷ 0.70

⇒ x = 200

∴ The correct answer is ₹200.

Additional Information

Relation between CP and MP

CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%). This helps find the marked price when profit and discount are known.

Successive Discount Formula

For two successive discounts of a% and b%, the equivalent single discount is (a + b − (a × b) ÷ 100)%.

Dishonest Dealer Concept

Profit% = (Error ÷ (True Value − Error)) × 100. This is used when a dealer uses less weight than actual.

The correct answer is Rs. 25.50 per dozen

Key Points

• Total eggs purchased = 510 eggs. Since 1 dozen = 12 eggs, the total quantity in dozens is 510 ÷ 12 = 42.5 dozens.
• The Cost Price (CP) of the eggs is Rs. 20 per dozen. Therefore, the Total Cost Price = 42.5 × 20 = Rs. 850.
• Broken eggs = 30 eggs. The remaining eggs available for sale = 510 − 30 = 480 eggs.
• Remaining dozens to be sold = 480 ÷ 12 = 40 dozens.
• The shopkeeper desires a gain of 20%. To find the required Total Selling Price (SP), we use the formula: Total SP = CP + (Profit% of CP).
  

Total SP = 850 + (20 ÷ 100 × 850) = 850 + 170 = Rs. 1020.

• To find the Selling Price per dozen for the remaining eggs, divide the Total SP by the number of remaining dozens:
  

SP per dozen = 1020 ÷ 40 = Rs. 25.50.

• Comparing this with the options, Option 2 is the correct rate the shopkeeper must set to achieve the target profit.

Additional Information

• Profit and Loss Formula: The Selling Price can also be directly calculated as SP = CP × (100 + Gain%) ÷ 100. For this problem: SP = 850 × (120 ÷ 100) = 850 × 1.2 = Rs. 1020.
• Breakage Impact: In business math, when items are lost or broken, the cost of those items is distributed over the remaining saleable inventory, which increases the necessary break-even price per unit.
• Unit Conversion: Always ensure the units (individual items vs. dozens) are consistent throughout the calculation to avoid errors.

Shortcut Trick

Net change formula: a + b + (a × b / 100)

Here, a = −12% (loss) and b = +12.5% (profit)

Net change = −12 + 12.5 + (−12 × 12.5 / 100)

Net change = 0.5 − 1.5 = −1%

Negative sign indicates a loss.

∴ The correct answer is 1% loss.

Alternate Method

Given:

Loss of A = 12%

Profit of B = 12.5% (or 1/8 in fraction)

Hypothetical SP of A = CP of C

Let the Cost Price (CP) for A be ₹100.

⇒ A sells to B at 12% loss.

⇒ Selling Price (SP) for A = CP for B = 100 − 12% of 100 = ₹88.

⇒ B sells it to C at 12.5% profit.

⇒ SP for B = CP for C = 88 + (12.5/100) × 88

⇒ CP for C = 88 + (1/8) × 88 = 88 + 11 = ₹99.

⇒ If A sells to B at CP of C, then new SP for A = ₹99.

⇒ Profit/Loss for A = New SP − Original CP = 99 − 100 = −1.

⇒ Loss Percentage = (1 / 100) × 100 = 1%.

∴ The correct answer is 1% loss.

Additional Information

Successive Percentage Change

When a value changes by x% and then by y%, the net percentage change is given by [x + y + (xy/100)]%. Use positive for profit and negative for loss.

Relation between CP and SP

SP = CP × (100 + Profit%)/100 or SP = CP × (100 − Loss%)/100. For fractional calculations, SP = CP × (1 + Profit Fraction).

Marked Price and Discount

SP = MP × (100 − Discount%)/100. Discount is always calculated on the Marked Price (MP).

Shortcut Trick

Overall profit % = (Total Profit ÷ Total CP) × 100 = (30 ÷ 900) × 100 = 10/3%.

Using Alligation: (Type-I Profit: 15%) vs (Type-II Loss: −10%) with Mean (10/3%).

CP Ratio (3 Type-I : 6 Type-II) = (10/3 − (−10)) : (15 − 10/3) = 40/3 : 35/3 = 8 : 7.

Total 15 units = ₹900 ⇒ 1 unit = ₹60.

CP of 1 Type-I cake = (8 × 60) ÷ 3 = ₹160; CP of 1 Type-II cake = (7 × 60) ÷ 6 = ₹70.

∴ The correct answer is ₹160, ₹70.

Alternate Method

Given:

Total cakes = 3 (Type-I) + 6 (Type-II)

Total CP = ₹900

Type-I Profit = 15%, Type-II Loss = 10%

Overall Profit = ₹30

Formula Used:

Profit = CP × Profit% | Loss = CP × Loss%

Let CP of one Type-I cake be x and one Type-II cake be y.

⇒ 3x + 6y = 900

⇒ x + 2y = 300 → (Eq. 1)

⇒ Total Profit/Loss: 0.15(3x) − 0.10(6y) = 30

⇒ 0.45x − 0.60y = 30

⇒ 45x − 60y = 3000

⇒ 3x − 4y = 200 → (Eq. 2)

Multiplying Eq. 1 by 2: 2x + 4y = 600

Adding with Eq. 2: (2x + 4y) + (3x − 4y) = 600 + 200

⇒ 5x = 800 ⇒ x = ₹160

Substitute x in Eq. 1: 160 + 2y = 300

⇒ 2y = 140 ⇒ y = ₹70

∴ The correct answer is ₹160, ₹70.

Additional Information

Alligation Rule in Profit/Loss

Ratio of Cost Prices = (Mean % − Profit%₂) : (Profit%₁ − Mean %). Note: Treat loss as negative profit.

Net Profit Percentage

Overall Profit % = (Net Profit ÷ Total Cost Price) × 100.

Linear Equations in Two Variables

Useful when items are bought in different quantities; solve using substitution or elimination.

Shortcut Trick

20% increase is equivalent to the fraction 1/5.

To reverse an increase of fraction a/b, we must decrease by a/(a+b).

Here, a = 1 and b = 5.

Required reduction = 1 / (1 + 5) = 1/6.

Percentage reduction = (1/6) × 100 = 16 2/3%.

∴ The correct answer is 16 2/3%.

Alternate Method

Given:

Increase in number = 20%

Formula Used:

Percentage Reduction = [r / (100 + r)] × 100, where r is the percentage increase.

Calculations:

⇒ Let the original number be 100.

⇒ Increased number = 100 + (20% of 100) = 100 + 20 = 120.

⇒ To return to the original number (100), the reduction needed = 120 − 100 = 20.

⇒ Percentage reduction needed on the increased number = (Reduction / Increased Number) × 100

⇒ Percentage reduction = (20 / 120) × 100

⇒ Percentage reduction = (1 / 6) × 100

⇒ Percentage reduction = 50 / 3 = 16 2/3%

∴ The correct answer is 16 2/3%.

Additional Information

Percentage Increase and Decrease Relationship

If A is r% more than B, then B is [r / (100 + r)] × 100% less than A. If A is r% less than B, then B is [r / (100 − r)] × 100% more than A.

Successive Percentage Changes

If a value is changed by x% and then by y%, the net change is given by the formula: [x + y + (xy / 100)]%.

Fraction to Percentage Conversion

Common fractions used in exams: 1/2 = 50%, 1/3 = 33 1/3%, 1/4 = 25%, 1/5 = 20%, 1/6 = 16 2/3%, 1/7 = 14 2/7%, 1/8 = 12 1/2%.

Shortcut Trick

Let the original number of goats be 100%.

After 12% flood loss, remainder = 88%.

Loss from diseases = 5% of 88% = 4.4%.

Final remaining goats = 88% − 4.4% = 83.6%.

Given 83.6% = 8360 → 1% = 100 → 100% = 10000.

∴ The correct answer is 10000 goats.

Alternate Method

Given:

First loss = 12% of total

Second loss = 5% of remainder

Final number of goats = 8360

Formula Used:

Final Value = Initial Value × (1 − r₁/100) × (1 − r₂/100)

Let the original number of goats be x.

⇒ x × (1 − 12/100) × (1 − 5/100) = 8360

⇒ x × (88/100) × (95/100) = 8360

⇒ x × (22/25) × (19/20) = 8360

⇒ x × (418/500) = 8360

⇒ x = (8360 × 500) ÷ 418

⇒ x = 20 × 500

⇒ x = 10000

∴ The correct answer is 10000 goats.

Additional Information

Successive Percentage Change

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

Percentage Reduction Formula

New Value = Old Value × [ (100 − loss%)/100 ]. For successive losses, multiply the factors repeatedly.

Effective Discount / Loss

Two successive losses of 12% and 5% are not equal to a single loss of 17%, but actually 16.4% because the second loss is calculated on the already reduced value.

Shortcut Trick

Weight ratio of the two papers = 900 : 700 = 9 : 7

Apply weighted average: (Percentage₁ × Weight₁ + Percentage₂ × Weight₂) ÷ (Weight₁ + Weight₂)

Combined Percentage = (72 × 9 + 80 × 7) ÷ (9 + 7)

Combined Percentage = (648 + 560) ÷ 16 = 1208 ÷ 16 = 75.5%

∴ The correct answer is 75.5%.

Alternate Method

Given:

Maximum marks Paper 1 = 900, Scored = 72%

Maximum marks Paper 2 = 700, Scored = 80%

Formula Used:

Combined Percentage = (Total Marks Scored ÷ Total Maximum Marks) × 100

Calculations:

⇒ Marks scored in Paper 1 = 72% of 900 = (72 ÷ 100) × 900 = 648

⇒ Marks scored in Paper 2 = 80% of 700 = (80 ÷ 100) × 700 = 560

⇒ Total Marks Scored = 648 + 560 = 1208

⇒ Total Maximum Marks = 900 + 700 = 1600

⇒ Combined Percentage = (1208 ÷ 1600) × 100

⇒ Combined Percentage = 1208 ÷ 16 = 75.5%

∴ The correct answer is 75.5%.

Additional Information

Weighted Average

When values have different weights (max marks), the average is calculated as Σ(w_i × x_i) ÷ Σw_i, where w is weight and x is value.

Percentage Calculation

To find percentage: (Value ÷ Total) × 100. Always ensure the base (denominator) represents the correct total for the context.

Successive Percentage

If a value changes by a% then b%, the net change is (a + b + ab/100)%. This is different from combined percentage of two different entities.

Shortcut Trick

Each 10% reduction means the strength becomes 9/10 of the previous state.

Successive remaining factor = (9/10) × (9/10) × (9/10) = 729/1000

Given final strength = 7,29,000

Equivalent units: 729 units = 7,29,000

Original strength (1000 units) = 1000 × 1000 = 10,00,000

∴ The correct answer is 1000000.

Alternate Method

Given:

Final active men = 7,29,000

Successive reduction rates = 10%, 10%, and 10%

Formula Used:

Final Value = Initial Value × (1 − r₁/100) × (1 − r₂/100) × (1 − r₃/100)

Calculations:

⇒ Let original strength be x.

⇒ x × (90/100) × (90/100) × (90/100) = 7,29,000

⇒ x × (0.9 × 0.9 × 0.9) = 7,29,000

⇒ x × 0.729 = 7,29,000

⇒ x = 7,29,000 ÷ 0.729

⇒ x = 10,00,000

∴ The correct answer is 1000000.

Additional Information

Successive Percentage Change

For two successive changes of a% and b%, the effective change is (a + b + ab/100)%. For three changes, the formula is applied twice.

Percentage Multiplier Method

If a value decreases by r%, the new value is obtained by multiplying the original by (100 − r)/100.

Population Depreciation Rule

The value after n years/steps with a constant reduction rate r is P × (1 − r/100)ⁿ.

Shortcut Trick

Total Distance = 150 + (60 × 2) = 270 km

Total Time = 3 + 2 = 5 hours

Average Speed = Total Distance ÷ Total Time = 270 ÷ 5 = 54 km/hr

∴ The correct answer is 54 km/hr.

Alternate Method

Given:

Distance of 1^st part (D₁) = 150 km

Time of 1^st part (T₁) = 3 hours

Time of 2^nd part (T₂) = 2 hours

Speed of 2^nd part (S₂) = 60 km/hr

Formula Used:

Average Speed = Total Distance ÷ Total Time

Distance = Speed × Time

Calculations:

⇒ Distance covered in the second part (D₂) = S₂ × T₂

⇒ D₂ = 60 × 2 = 120 km

⇒ Total Distance = D₁ + D₂ = 150 + 120 = 270 km

⇒ Total Time = T₁ + T₂ = 3 + 2 = 5 hours

⇒ Average Speed = 270 ÷ 5

⇒ Average Speed = 54 km/hr

∴ The correct answer is 54 km/hr.

Additional Information

Average Speed (Equal Distances)

If a journey is divided into two equal distances traveled at speeds x and y, Average Speed = (2 × x × y) ÷ (x + y).

Average Speed (Equal Time Intervals)

If an object travels for equal time intervals at speeds x and y, Average Speed = (x + y) ÷ 2.

Relative Speed

When two bodies move in the same direction, relative speed = |s₁ − s₂|. In opposite directions, relative speed = s₁ + s₂.

Shortcut Trick

Speed Ratio (Walking : Cycling) = 5 : 24

Since distance is constant, Time Ratio = Inverse of Speed Ratio = 24 : 5

Walking time = 3 hours 36 minutes = (3 × 60) + 36 = 216 minutes

24 units → 216 minutes

1 unit → 216 ÷ 24 = 9 minutes

Cycling time (5 units) → 5 × 9 = 45 minutes

∴ The correct answer is 45 minutes.

Alternate Method

Given:

Speed while walking (S₁) = 5 km/hr

Time taken while walking (T₁) = 3 hours 36 minutes

Speed while cycling (S₂) = 24 km/hr

Formula Used:

Distance = Speed × Time

Time = Distance ÷ Speed

Calculations:

⇒ Convert T₁ into hours: 3 hours + (36 ÷ 60) hours = 3 + 0.6 = 3.6 hours

⇒ Total Distance = S₁ × T₁

⇒ Distance = 5 × 3.6 = 18 km

⇒ New Time (T₂) = Distance ÷ S₂

⇒ T₂ = 18 ÷ 24 = 0.75 hours

⇒ Convert T₂ into minutes: 0.75 × 60 = 45 minutes

∴ The correct answer is 45 minutes.

Additional Information

Speed-Time Relationship

When distance is constant, speed is inversely proportional to time (S ∝ 1/T). This means S₁/S₂ = T₂/T₁.

Unit Conversion

To convert hours to minutes, multiply by 60. To convert km/hr to m/s, multiply the speed by 5/18.

Average Speed

If an object covers two equal distances at speeds x and y, the average speed is (2 × x × y) ÷ (x + y).

Shortcut Trick

Using the distance formula when speed difference and time difference are given:

Distance = [S × (S − 300) ÷ 300] × Time Difference

⇒ 1200 = [S × (S − 300) ÷ 300] × 2

⇒ S × (S − 300) = 600 × 300 ⇒ Original Speed (S) = 600 km/hr

Scheduled duration = Distance ÷ Speed = 1200 ÷ 600 = 2 hours.

∴ The correct answer is 2 hours.

Alternate Method

Given:

Total distance to the destination = 1200 km

Reduction in flight speed = 300 km/hr

Increase in travel time = 2 hours

Distance = 1200 km

Original Speed = v, Time = t

New Speed = v − 300, Time = t + 2

Formula Used:

Time = Distance ÷ Speed

Speed = Distance ÷ Time

Calculations:

Let the originally scheduled time for the flight be t hours.

Original scheduled speed = 1200 ÷ t

New travel time = (t + 2) hours

New flight speed = 1200 ÷ (t + 2)

According to the given condition, the new speed is 300 km/hr less than the original speed.

⇒ (1200 ÷ t) − [1200 ÷ (t + 2)] = 300

Dividing the entire equation by 300 to simplify:

⇒ (4 ÷ t) − [4 ÷ (t + 2)] = 1

⇒ [4(t + 2) − 4t] ÷ [t(t + 2)] = 1

⇒ (4t + 8 − 4t) ÷ (t² + 2t) = 1

⇒ 8 ÷ (t² + 2t) = 1

⇒ t² + 2t = 8

⇒ t² + 2t − 8 = 0

⇒ t² + 4t − 2t − 8 = 0

⇒ t(t + 4) − 2(t + 4) = 0

⇒ (t − 2)(t + 4) = 0

Since travel time cannot be negative, we reject t = −4.

⇒ t = 2 hours.

∴ The correct answer is 2 hours.

Additional Information

Distance Formula with Time Difference

When the time difference (Δt) and respective speeds (S₁, S₂) for a journey are known, the distance can be calculated directly as: Distance = [(S₁ × S₂) ÷ |S₁ − S₂|] × Δt.

Relative Speed

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

Average Speed

For a journey divided into two equal distances travelled at different speeds S₁ and S₂, the overall Average Speed is given by the harmonic mean: (2 × S₁ × S₂) ÷ (S₁ + S₂).

Shortcut Trick

Gap between B and C when A finishes = 18 − 12 = 6 m.

Gap between B and C when B finishes = 8 m.

B covers the remaining 12 m to finish and the gap increases by (8 − 6) = 2 m.

Since the gap increases linearly, a 2 m gap is created in 12 m of B's travel.

To create the total gap of 8 m from the start, B travels = (12 ÷ 2) × 8 = 48 m.

∴ The correct answer is 48 Metres.

Alternate Method

Given:

When A finishes (distance L), B is at (L − 12) m and C is at (L − 18) m.

When B finishes (distance L), C is at (L − 8) m.

Formula Used:

Ratio of speeds (V) = Ratio of distances (D) covered in the same time: V₁/V₂ = D₁/D₂

Calculations:

From the first condition (A finishes):

⇒ Ratio of speeds of B and C = V_B/V_C = (L − 12) / (L − 18) --- (i)

From the second condition (B finishes):

⇒ Ratio of speeds of B and C = V_B/V_C = L / (L − 8) --- (ii)

Equating (i) and (ii):

⇒ (L − 12) / (L − 18) = L / (L − 8)

⇒ (L − 12)(L − 8) = L(L − 18)

⇒ L² − 8L − 12L + 96 = L² − 18L

⇒ L² − 20L + 96 = L² − 18L

⇒ 20L − 18L = 96

⇒ 2L = 96

⇒ L = 48 m

∴ The correct answer is 48 Metres.

Additional Information

Direct Proportionality of Speed and Distance

When time is constant, the distance covered is directly proportional to the speed (D ∝ S). This allows using ratios of distances covered as ratios of speeds.

Head Start Concept

If A gives B a head start of x meters in a race of L meters, B only needs to cover (L − x) meters to reach the finish line.

Dead Heat

A dead heat occurs when two or more participants finish the race at the exact same time, meaning they have covered their respective required distances in the same duration.

Shortcut Trick

For 10% rate and 4 years, the effective compound interest rate is 46.41%.

CI = Principal × Effective Rate

CI = 4000 × 46.41% = 40 × 46.41 = 1856.40

∴ The correct answer is Rs. 1856.40.

Alternate Method

Given:

Principal (P) = Rs. 4000

Rate (R) = 10% per annum

Time (n) = 4 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Compound Interest (CI) = A − P

Calculations:

⇒ A = 4000 × (1 + 10/100)⁴

⇒ A = 4000 × (1.1)⁴

⇒ A = 4000 × (1.4641)

⇒ A = 5856.40

⇒ CI = 5856.40 − 4000

⇒ CI = 1856.40

∴ The correct answer is Rs. 1856.40.

Additional Information

Effective Rate for 10% Rate

For 2 years: 21%, For 3 years: 33.1%, For 4 years: 46.41%.

Difference between CI and SI (2 years)

Difference = P × (R/100)².

Multiplying Factor Method (4 years)

Ratio for 4 years CI is 4 : 6 : 4 : 1. CI = 4×A + 6×B + 4×C + 1×D, where A is interest on P, B on A, C on B, etc.

Shortcut Trick

Overall interest rate = (Total Interest ÷ Total Principal) × 100

Overall Rate = (352 ÷ 4000) × 100 = 8.8%

Using Alligation: (Rate 2 − Average Rate) : (Average Rate − Rate 1)

Ratio of sums = (10 − 8.8) : (8.8 − 8) = 1.2 : 0.8 = 3 : 2

Sum at 8% = (3 ÷ 5) × 4000 = ₹2400

∴ The correct answer is ₹2400.

Alternate Method

Given:

Total Sum (P) = ₹4000

Rate 1 (R₁) = 8% | Rate 2 (R₂) = 10%

Total Annual Interest = ₹352

Formula Used:

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

Alligation Method: Ratio of Parts = (R₂ − R_avg) : (R_avg − R₁)

Calculations:

⇒ Let Part 1 be x and Part 2 be (4000 − x)

⇒ Total Interest = (x × 8 × 1)/100 + ((4000 − x) × 10 × 1)/100 = 352

⇒ 8x + 40000 − 10x = 35200

⇒ −2x = 35200 − 40000

⇒ −2x = −4800

⇒ x = 4800 ÷ 2

⇒ x = ₹2400

∴ The correct answer is ₹2400.

Additional Information

Simple Interest (SI)

The interest calculated only on the principal amount for the entire period. Formula: SI = (P × R × T) ÷ 100.

Alligation in Interest

When a sum is divided into two parts with different rates, the ratio of the two sums is inversely proportional to the deviation of their rates from the average rate.

Effective Annual Rate

If interest is compounded or multiple rates apply, the effective rate is the single rate that produces the same total interest over the year.

Shortcut Trick

Effective rate of SI for 10% for 3 years = 30%

Effective rate of CI for 10% for 3 years = 33.1%

Difference in effective rates = 33.1% − 30% = 3.1%

Given: 3.1% of Sum = Rs. 186

Sum = (186 ÷ 3.1) × 100 = Rs. 6000

∴ The correct answer is Rs. 6000.

Alternate Method

Given: Difference (D) = Rs. 186, Rate (R) = 10%, Time (T) = 3 years

Formula: Difference = P(R/100)² × (3 + R/100)

⇒ 186 = P × (10 ÷ 100)² × (3 + 10 ÷ 100)

⇒ 186 = P × (1 ÷ 100) × (31 ÷ 10)

⇒ 186 = P × (31 ÷ 1000)

⇒ P = (186 × 1000) ÷ 31

⇒ P = 6 × 1000

⇒ P = 6000

∴ The correct answer is Rs. 6000.

Additional Information

Difference for 2 years

The difference between Compound Interest and Simple Interest for 2 years is calculated as P(R/100)².

Effective Interest Rate

Effective rate for 3 years at R% can be calculated using the successive percentage formula: R + R + (R×R)/100 followed by another iteration with R.

Amount Formula

Compound Interest Amount A = P(1 + R/100)^T. Simple Interest I = (P × R × T)/100.

Shortcut Trick

Effective rate (Annual) for 2 years at 20% = 20 + 20 + (20 × 20) ÷ 100 = 44%

Effective rate (Half-yearly) for 4 periods at 10% = 46.41%

Difference in interest rates = 46.41% − 44% = 2.41%

2.41% of Sum = ₹ 723

Sum = (723 ÷ 2.41) × 100 = ₹ 30,000

∴ The correct answer is ₹30,000.

Alternate Method

Given:

Rate (R) = 20% per annum

Time (T) = 2 years

Interest difference = ₹ 723

Formula Used:

Compound Interest (CI) = P[(1 + R/100)ⁿ − 1]

Calculations:

For Annual compounding:

⇒ CI_annual = P[(1 + 20/100)² − 1]

⇒ CI_annual = P[(1.2)² − 1] = P[1.44 − 1] = 0.44P

For Half-yearly compounding:

⇒ New Rate = 20 ÷ 2 = 10%

⇒ New Time = 2 × 2 = 4 periods

⇒ CI_half-yearly = P[(1 + 10/100)⁴ − 1]

⇒ CI_half-yearly = P[(1.1)⁴ − 1] = P[1.4641 − 1] = 0.4641P

Difference in interest:

⇒ 0.4641P − 0.44P = 723

⇒ 0.0241P = 723

⇒ P = 723 ÷ 0.0241

⇒ P = 30000

∴ The correct answer is ₹30,000.

Additional Information

Effective Interest Rate

When interest is compounded n times a year, the effective annual rate is (1 + r/n)ⁿ − 1.

Successive Percentage Formula

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

CI vs SI for 2 Years

The difference between CI and SI for 2 years is P × (R/100)².

Shortcut Trick

The area of an equilateral triangle can be directly calculated from its height using the formula: Area = h² ÷ √3

Given height, h = 18 cm

⇒ Area = 18² ÷ √3

⇒ Area = 324 ÷ √3 = 108√3 sq. cm.

∴ The correct answer is 108√3 sq. cm.

Alternate Method

Given:

Height of the equilateral triangle (h) = 18 cm

Formula Used:

Height of equilateral triangle = (√3 ÷ 2) × a (where a is the side of the triangle)

Area of equilateral triangle = (√3 ÷ 4) × a²

Calculations:

⇒ h = (√3 ÷ 2) × a

⇒ 18 = (√3 ÷ 2) × a

⇒ a = (18 × 2) ÷ √3

⇒ a = 36 ÷ √3

⇒ a = 12√3 cm

Now, substitute the value of side 'a' into the area formula:

⇒ Area = (√3 ÷ 4) × a²

⇒ Area = (√3 ÷ 4) × (12√3)²

⇒ Area = (√3 ÷ 4) × (144 × 3)

⇒ Area = (√3 ÷ 4) × 432

⇒ Area = 108√3 sq. cm.

∴ The correct answer is 108√3 sq. cm.

Additional Information

Properties of an Equilateral Triangle

All three sides are equal, and all internal angles are exactly 60°. The height also acts as the median, angle bisector, and perpendicular bisector of the base.

Inradius of Equilateral Triangle

The radius of the inscribed circle (inradius) is directly proportional to the height and is given by r = h ÷ 3, or r = a ÷ (2√3).

Circumradius of Equilateral Triangle

The radius of the circumscribed circle (circumradius) is given by R = 2h ÷ 3, or R = a ÷ √3. The ratio of circumradius to inradius is always 2:1.

Shortcut Trick

Total Surface Area (TSA) = 2πr(r + h)

880 = 2 × (22/7) × r × 20

r = (880 × 7) ÷ (44 × 20) = 7 cm

Given r + h = 20 ⇒ h = 20 − 7 = 13 cm

Volume = πr²h = (22/7) × 7 × 7 × 13 = 154 × 13 = 2002 cm³

∴ The correct answer is 2002 cm³.

Alternate Method

Given:

Sum of radius (r) and height (h) = 20 cm

Total Surface Area (TSA) = 880 cm²

Formula Used:

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

Volume of Cylinder = πr²h

⇒ 2 × (22/7) × r × (r + h) = 880

⇒ 2 × (22/7) × r × 20 = 880

⇒ (880/7) × r = 880

⇒ r = (880 × 7) ÷ 880

⇒ r = 7 cm

⇒ h = 20 − r = 20 − 7 = 13 cm

⇒ Volume = πr²h = (22/7) × 7 × 7 × 13

⇒ Volume = 22 × 7 × 13 = 154 × 13

⇒ Volume = 2002 cm³

∴ The correct answer is 2002 cm³.

Additional Information

Curved Surface Area (CSA)

The lateral surface area of a cylinder excludes the two circular bases and is given by 2πrh.

Total Surface Area (TSA)

The sum of curved surface area and the area of the two circular bases: TSA = 2πrh + 2πr² = 2πr(h + r).

Volume of Cylinder

The amount of space occupied by the cylinder is given by Base Area × Height = πr²h.

Hollow Cylinder Volume

Volume of a hollow cylinder with external radius R and internal radius r is given by πh(R² − r²).

Shortcut Trick

Total Surface Area (TSA) ratio Sphere : Hemisphere = 4πR² : 3πr² = 1 : 1.

Radius ratio: (R/r)² = 3/4 ⇒ R/r = √3/2.

Volume ratio: V_sphere : V_hemisphere = (4/3 πR³) : (2/3 πr³) = 2R³ : r³.

Substituting R/r: 2 × (√3)³ : (2)³ = 2 × 3√3 : 8 = 3√3 : 4.

∴ The correct answer is 3√3 : 4.

Alternate Method

Given:

TSA of Sphere = TSA of Solid Hemisphere

Formula used:

TSA of Sphere = 4πR² | TSA of Solid Hemisphere = 3πr²

Volume of Sphere = (4/3)πR³ | Volume of Hemisphere = (2/3)πr³

⇒ 4πR² = 3πr²

⇒ R²/r² = 3/4

⇒ R/r = √3/2

⇒ Ratio of Volumes = (4/3 πR³) / (2/3 πr³)

⇒ Ratio = (4/3 ÷ 2/3) × (R/r)³

⇒ Ratio = 2 × (√3/2)³

⇒ Ratio = 2 × (3√3 / 8)

⇒ Ratio = 3√3 / 4

∴ The correct answer is 3√3 : 4.

Additional Information

Curved Surface Area of Hemisphere

The curved surface area of a hemisphere is exactly half of the sphere surface area, CSA = 2πr².

Volume of a Sphere

The volume represents the space occupied by the sphere, calculated as V = (4/3)πr³.

Properties of Solid Hemisphere

A solid hemisphere has two surfaces: one curved (2πr²) and one flat circular base (πr²), making TSA = 3πr².

Shortcut Trick

Simplify ratio 1/2 : 1/3 : 1/4 by multiplying with LCM(2, 3, 4) = 12.

New ratio = 6 : 4 : 3.

Total units = 6 + 4 + 3 = 13 units.

Given: 13 units = 104 cm ⇒ 1 unit = 8 cm.

Longest side = 6 units = 6 × 8 = 48 cm.

∴ The correct answer is 48 cm.

Alternate Method

Given: Ratio of sides = 1/2 : 1/3 : 1/4 and Perimeter = 104 cm.

Formula Used: Perimeter of a triangle = Sum of all three sides.

⇒ Simplifying the ratio of sides = (1/2 × 12) : (1/3 × 12) : (1/4 × 12)

⇒ Simplified ratio = 6 : 4 : 3

⇒ Let the sides of the triangle be 6x, 4x, and 3x.

⇒ Perimeter = 6x + 4x + 3x = 104

⇒ 13x = 104

⇒ x = 104 ÷ 13

⇒ x = 8

⇒ Length of longest side = 6x = 6 × 8 = 48 cm

∴ The correct answer is 48 cm.

Additional Information

Ratio Simplification

To convert a fractional ratio into an integer ratio, multiply each term by the Least Common Multiple (LCM) of the denominators.

Triangle Inequality Theorem

In any triangle, the sum of the lengths of any two sides must be strictly greater than the length of the third side (a + b > c).

Semi-perimeter (s)

The semi-perimeter is half of the total perimeter: s = (a + b + c) ÷ 2. It is used in Heron’s formula for calculating area: Area = √(s(s−a)(s−b)(s−c)).

Shortcut Trick

Total Painting Area = Area of 4 walls + Area of ceiling = 2h(l + b) + lb

Total Area = 2 × 10 × (25 + 12) + (25 × 12) = 740 + 300 = 1040 m²

Efficiency of A = 200 ÷ 5 = 40 m²/day

Efficiency of B = 250 ÷ 2 = 125 m²/day

Combined Efficiency = 40 + 125 = 165 m²/day

Time Taken = 1040 ÷ 165 = 208/33 = 6 10/33 days

∴ The correct answer is 6 10/33 days.

Alternate Method

Given:

Length (l) = 25 m, Breadth (b) = 12 m, Height (h) = 10 m

Painter A: 200 m² in 5 days

Painter B: 250 m² in 2 days

Formula Used:

Area to be painted = 2h(l + b) + lb

Time = Total Work ÷ Total Efficiency

Calculations:

⇒ Area of 4 walls = 2 × 10 × (25 + 12) = 20 × 37 = 740 m²

⇒ Area of ceiling = 25 × 12 = 300 m²

⇒ Total Work (Area) = 740 + 300 = 1040 m²

⇒ Days for A to finish 1040 m² = 1040 ÷ (200 ÷ 5) = 1040 ÷ 40 = 26 days

⇒ Days for B to finish 1040 m² = 1040 ÷ (250 ÷ 2) = 1040 ÷ 125 = 8.32 days

⇒ Combined efficiency of A and B = 40 + 125 = 165 m²/day

⇒ Time taken together = Total Work ÷ Combined Efficiency

⇒ Time = 1040 ÷ 165 = 208/33 = 6 10/33 days

∴ The correct answer is 6 10/33 days.

Additional Information

Surface Area of a Room

Lateral Surface Area (4 walls) = 2h(l + b). Total Surface Area (including floor and ceiling) = 2(lb + bh + hl).

Work-Rate Relationship

Work = Efficiency × Time. When working together, individual efficiencies (rates) are added to find the combined rate.

Alternate Day Work

If A and B work on alternate days, find the work done in a 2-day cycle and divide total work by the cycle efficiency.

Shortcut Trick

Volume of a Prism = Base Area × Height

Area of Trapezium Base = 1/2 × (Sum of parallel sides) × Perpendicular distance

Area of Base = 1/2 × (25 + 11) × 16 = 18 × 16 = 288 cm²

Volume = 288 × 10 = 2880 cm³

∴ The correct answer is 2880 cu.cm.

Alternate Method

Given:

Parallel sides of the trapezium base (a) = 25 cm and (b) = 11 cm

Distance between parallel sides (h_base) = 16 cm

Height of the prism (H) = 10 cm

Formula Used:

Volume of Right Prism = [1/2 × (a + b) × h_base] × H

Calculations:

⇒ Area of Trapezoidal base = 1/2 × (25 + 11) × 16

⇒ Area of Trapezoidal base = 1/2 × 36 × 16

⇒ Area of Trapezoidal base = 18 × 16 = 288 cm²

⇒ Volume of Prism = Area of base × Height of prism

⇒ Volume = 288 × 10

⇒ Volume = 2880 cm³

∴ The correct answer is 2880 cu.cm.

Additional Information

Surface Area of a Prism

Lateral Surface Area (LSA) = Perimeter of base × height. Total Surface Area (TSA) = LSA + 2 × Base Area.

Volume of a Pyramid

The volume of a pyramid with the same base and height as a prism is exactly 1/3 of the volume of the prism (V = 1/3 × Base Area × Height).

Regular Polygon Base

If the base is a regular polygon with side 'a' and 'n' sides, the Area = (n × a²) / [4 × tan(180°/n)].

Shortcut Trick

Volume of material remains constant during melting: Volume of Hollow Cylinder = Volume of Solid Cylinder.

π × (R² − r²) × h₁ = π × R_s² × h₂

(R + r)(R − r) × h₁ = R_s² × (h₁ ÷ 2)

(6.75 + 5.25)(6.75 − 5.25) × 15 = R_s² × 7.5

12 × 1.5 × 2 = R_s² ⇒ 36 = R_s²

R_s = √36 = 6 cm.

∴ The correct answer is 6 cm.

Alternate Method

Given:

External radius of hollow cylinder (R) = 6.75 cm

Internal radius of hollow cylinder (r) = 5.25 cm

Height of hollow cylinder (h₁) = 15 cm

Height of solid cylinder (h₂) = 15 ÷ 2 = 7.5 cm

Formula:

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

Volume of solid cylinder = πR²h

According to the problem, Volume of Hollow Cylinder = Volume of Solid Cylinder

⇒ π × (R² − r²) × h₁ = π × R_solid² × h₂

⇒ (6.75² − 5.25²) × 15 = R_solid² × 7.5

⇒ (6.75 + 5.25)(6.75 − 5.25) × 15 = R_solid² × 7.5

⇒ 12 × 1.5 × 15 = R_solid² × 7.5

⇒ 18 × 15 = R_solid² × 7.5

⇒ R_solid² = (18 × 15) ÷ 7.5

⇒ R_solid² = 18 × 2

⇒ R_solid² = 36

⇒ R_solid = √36 = 6 cm.

∴ The correct answer is 6 cm.

Additional Information

Volume Invariance

When a 3D object is melted and recast into another shape, its total volume remains constant, provided there is no wastage during the process.

Surface Area of Hollow Cylinder

The Total Surface Area (TSA) of a hollow cylinder is 2πRh + 2πrh + 2π(R² − r²), which includes inner, outer, and base areas.

Cylinder Volume Formula

Volume of any cylinder is Area of Base × Height. For a solid cylinder, it is πr²h.

Shortcut Trick

Perimeter = Total Cost ÷ Rate = 6000 ÷ 7.50 = 800 m.

Ratio sum parts for perimeter = 2 × (5 + 3) = 16 units.

16 units = 800 m ⇒ 1 unit = 50 m.

Difference (5 − 3) = 2 units = 2 × 50 = 100 m.

∴ The correct answer is 100 metres.

Alternate Method

Given: Ratio of Length (L) to Breadth (B) = 5 : 3, Cost of fencing = ₹6000, Rate = ₹7.50/m.

Formula Used: Perimeter of rectangle = 2(L + B); Total Cost = Perimeter × Rate.

⇒ Let the Length be 5x and Breadth be 3x.

⇒ Perimeter = Total Cost ÷ Rate per metre

⇒ Perimeter = 6000 ÷ 7.50 = 800 metres.

⇒ According to formula: 2(5x + 3x) = 800

⇒ 2(8x) = 800

⇒ 16x = 800

⇒ x = 800 ÷ 16 = 50.

⇒ Difference between Length and Breadth = 5x − 3x = 2x

⇒ Difference = 2 × 50 = 100 metres.

∴ The correct answer is 100 metres.

Additional Information

Perimeter of Rectangle

The total distance around the rectangle is given by P = 2 × (Length + Breadth).

Area of Rectangle

The surface area enclosed by the rectangle is given by A = Length × Breadth.

Diagonal of Rectangle

The length of the diagonal connecting opposite vertices is d = √(Length² + Breadth²).

Shortcut Trick

Let the side of the square be s. Then the diameter of the circle is also s.

Area of Square : Area of Circle = s² : π × (s/2)²

Ratio = s² : (π × s²) / 4 = 1 : π/4 = 4 : π

Ratio = 4 : (22/7) = 28 : 22 = 14 : 11.

∴ The correct answer is 14 : 11.

Alternate Method

Given: Side of square (s) = Diameter of circle (d)

Formula: Area of Square = s² | Area of Circle = πr² (where r = d/2)

Let the side of the square (s) = 14 units

⇒ Diameter of the circle (d) = 14 units

⇒ Radius of the circle (r) = d ÷ 2 = 14 ÷ 2 = 7 units

⇒ Area of the square = s² = 14 × 14 = 196 sq. units

⇒ Area of the circle = πr² = (22/7) × 7 × 7 = 22 × 7 = 154 sq. units

⇒ Required Ratio = 196 : 154

⇒ Dividing both by 14: 14 : 11

∴ The correct answer is 14 : 11.

Additional Information

Circle Inscribed in a Square

When a circle is inscribed in a square, the diameter of the circle is equal to the side of the square (d = s). The ratio of areas is always 4 : π.

Square Inscribed in a Circle

When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle (s√2 = d). The ratio of area of circle to area of square is π : 2.

Area and Perimeter Formulas

Square Area = s², Perimeter = 4s. Circle Area = πr², Circumference = 2πr.

Shortcut Trick For a fixed perimeter (constant length of wire), the area of a closed geometric shape increases as the number of sides increases. Comparing the given shapes: Equilateral Triangle (3 sides) < Square (4 sides) < Circle (limiting case with infinite sides). Area Hierarchy: Area of Triangle < Area of Square < Area of Circle. Since the length of the wire is constant at 132 cm, the Circle will always yield the largest area. ∴ The correct answer is Circle. Alternate Method Given: Total length of wire (Perimeter) = 132 cm Formula Used: Area of Equilateral Triangle = (√3 / 4) × (side)2 Area of Square = (side)2 Area of Circle = πr2

Calculations:

1. For Equilateral Triangle: ⇒ Side = 132 ÷ 3 = 44 cm ⇒ Area = (√3 / 4) × 44 × 44 = 484√3 ≈ 484 × 1.732 ≈ 838.31 cm2
2. For Square: ⇒ Side = 132 ÷ 4 = 33 cm ⇒ Area = 33 × 33 = 1089 cm2
3. For Circle: ⇒ Circumference = 2 × (22 / 7) × r = 132 ⇒ r = (132 × 7) / (2 × 22) = 21 cm ⇒ Area = (22 / 7) × 21 × 21 = 22 × 3 × 21 = 1386 cm2 Comparing the areas: 1386 cm2 (Circle) > 1089 cm2 (Square) > 838.31 cm2 (Triangle). ∴ The correct answer is Circle. Additional Information Isoperimetric Property Among all closed curves of a given length, the circle encloses the maximum area. This is a fundamental principle in geometry. Area of Regular Polygon For a regular polygon with n sides and perimeter P, Area = P2 / [4 × n × tan(180°/n)]. As n increases, the area increases and approaches πr2. Square vs Rectangles For a constant perimeter, a square has a larger area than any other rectangle. Similarly, for a constant area, the square has the smallest perimeter.

Shortcut Trick

If a solid is cut parallel to its base, the ratio of the volumes is the cube of the ratio of their corresponding heights.

Since the plane passes through the midpoint, the ratio of the height of the small cone to the original cone is 1 : 2.

Volume ratio (Small Cone : Original Cone) = (1)³ : (2)³ = 1 : 8.

Volume of the upper part = 1 unit; Volume of the lower part (frustum) = 8 − 1 = 7 units.

The ratio of the volume of the 2 parts is 1 : 7.

∴ The correct answer is 1 : 7.

Alternate Method

Given:

A cone divided at the midpoint of its axis (height) by a plane parallel to the base.

Let the total height of the cone be H and the base radius be R.

Formula Used:

Volume of a cone = (1/3) × π × r² × h

Volume of a Frustum = (1/3) × π × h × (R² + r² + R × r)

Calculations:

⇒ Let height of small cone (h₁) = H/2

⇒ By similar triangles, radius of small cone (r) = R/2

⇒ Volume of small cone (V₁) = (1/3) × π × (R/2)² × (H/2)

⇒ V₁ = (1/3) × π × (R²/4) × (H/2) = (1/24) × π × R² × H

⇒ Volume of original cone (V_total) = (1/3) × π × R² × H

⇒ Volume of lower part (V₂) = V_total − V₁

⇒ V₂ = (1/3)πR²H − (1/24)πR²H = (7/24) × π × R² × H

⇒ Ratio = V₁ : V₂ = [(1/24)πR²H] : [(7/24)πR²H]

⇒ Ratio = 1 : 7

∴ The correct answer is 1 : 7.

Additional Information

Volume Similarity Property

For any similar 3D figures, the ratio of their volumes is the cube of the ratio of their linear dimensions: V₁/V₂ = (h₁/h₂)³ = (r₁/r₂)³.

Surface Area Ratio

The ratio of the surface areas (or cross-sectional areas) of similar solids is the square of the ratio of their linear dimensions: A₁/A₂ = (h₁/h₂)².

Frustum of a Cone

The volume of a frustum of height 'h' and radii 'R' and 'r' is given by V = (πh/3)(R² + r² + Rr).

Shortcut Trick

Ratio of sides n₁ : n₂ = 1 : 2 ⇒ Let sides be n and 2n.

Shortcut for ratio of interior angles of n and 2n sides: (n − 2) / (n − 1) = Ratio.

Here, (n − 2) / (n − 1) = 3 / 4.

Solving for n: 4n − 8 = 3n − 3 ⇒ n = 5.

Number of sides = 5 and 2 × 5 = 10.

∴ The correct answer is 5, 10.

Alternate Method

Given:

Ratio of number of sides = 1 : 2

Ratio of interior angles = 3 : 4

Formula Used:

Measure of each interior angle of a regular polygon with n sides = [(n − 2) × 180] ÷ n

Calculations:

⇒ Let the number of sides of the two polygons be n and 2n respectively.

⇒ Ratio of interior angles = [{(n − 2) × 180} ÷ n] ÷ [{(2n − 2) × 180} ÷ 2n] = 3 ÷ 4

⇒ [(n − 2) ÷ n] × [2n ÷ (2n − 2)] = 3 ÷ 4

⇒ [2(n − 2)] ÷ [2(n − 1)] = 3 ÷ 4

⇒ (n − 2) ÷ (n − 1) = 3 ÷ 4

⇒ 4 × (n − 2) = 3 × (n − 1)

⇒ 4n − 8 = 3n − 3

⇒ 4n − 3n = 8 − 3

⇒ n = 5

⇒ Number of sides of first polygon = 5

⇒ Number of sides of second polygon = 2 × 5 = 10

∴ The correct answer is 5, 10.

Additional Information

Sum of Interior Angles

The sum of all interior angles of a polygon with n sides is given by (n − 2) × 180°.

Exterior Angle Property

For any regular polygon, each exterior angle = 360° ÷ n. The sum of interior and exterior angle is always 180°.

Sum of Exterior Angles

The sum of the exterior angles of any convex polygon, taking one at each vertex, is always 360°.

Shortcut Trick

Let the third side of the isosceles triangle be x. The equal sides are 2x each.

Perimeter (P) = 2x + 2x + x = 5x. Area of Isosceles = (x/4)√(4(2x)² − x²) = (x²√15)/4.

Equilateral triangle has same perimeter P = 5x, so side s = 5x/3.

Area of Equilateral = (√3/4) × (5x/3)² = 25√3x²/36.

Ratio = [(x²√15)/4] : [25√3x²/36] = 9√15 : 25√3 = 9√5 : 25 = 36√5 : 100.

∴ The correct answer is 36√5 : 100.

Alternate Method

Given: Equal sides of isosceles triangle = 2 × Base. Perimeter of isosceles triangle = Perimeter of equilateral triangle.

Let the base of the isosceles triangle be b.

⇒ Equal sides (a) = 2b.

⇒ Perimeter of isosceles triangle = a + a + b = 2b + 2b + b = 5b.

Area of Isosceles triangle = (b/4) × √(4a² − b²)

⇒ Area_iso = (b/4) × √(4(2b)² − b²) = (b/4) × √(16b² − b²) = (b/4) × √15b² = (b²√15)/4.

Side of equilateral triangle (s) = Perimeter/3 = 5b/3.

Area of Equilateral triangle = (√3/4) × s²

⇒ Area_equi = (√3/4) × (5b/3)² = (√3/4) × (25b²/9) = (25√3b²)/36.

Ratio of areas = [(b²√15)/4] : [(25√3b²)/36]

⇒ Ratio = (√15/4) : (25√3/36)

⇒ Ratio = (9√15) : (25√3)

⇒ Ratio = (9 × √5 × √3) : (25 × √3)

⇒ Ratio = 9√5 : 25

To match the options, multiply both sides by 4:

⇒ Ratio = (9√5 × 4) : (25 × 4) = 36√5 : 100.

∴ The correct answer is 36√5 : 100.

Additional Information

Area of Isosceles Triangle

For an isosceles triangle with equal sides 'a' and base 'b', the altitude h = √(a² − (b/2)²) and Area = (b/4)√(4a² − b²).

Area of Equilateral Triangle

For an equilateral triangle with side 's', all sides are equal and all angles are 60°. Area = (√3/4)s² and height h = (√3/2)s.

Heron's Formula

Area = √[s(s − a)(s − b)(s − c)], where s = (a + b + c)/2 is the semi-perimeter and a, b, c are the lengths of the sides.

Shortcut Trick

Volume of water rise in cylinder = Total volume of two immersed spheres.

π × R² × h = (4/3) π × (r₁³ + r₂³)

Let radii be r and 2r. Cylinder Radius R = 18/2 = 9 cm, Height rise h = 4 cm.

9² × 4 = (4/3) × (r³ + (2r)³) ⇒ 81 × 4 = (4/3) × 9r³

324 = 12r³ ⇒ r³ = 27 ⇒ r = 3 cm.

∴ The correct answer is 3 cm and 6 cm.

Alternate Method

Given:

Diameter of cylinder = 18 cm ⇒ Radius (R) = 9 cm

Rise in water height (h) = 4 cm

Let radius of smaller ball = r cm

Radius of larger ball = 2r cm

Formula Used:

Volume of cylinder = πR²h

Volume of sphere = (4/3)πr³

Calculations:

⇒ Volume of water displaced = Volume of ball 1 + Volume of ball 2

⇒ π × (9)² × 4 = (4/3) × π × (r)³ + (4/3) × π × (2r)³

⇒ 81 × 4 = (4/3) × [r³ + 8r³]

⇒ 324 = (4/3) × 9r³

⇒ 324 = 12r³

⇒ r³ = 324 ÷ 12

⇒ r³ = 27

⇒ r = 3 cm

⇒ Larger radius = 2r = 2 × 3 = 6 cm

∴ The correct answer is 3 cm and 6 cm.

Additional Information

Archimedes Principle

An object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. In terms of volume, the volume of displaced liquid equals the volume of the submerged portion of the object.

Volume of Hollow Cylinder

The volume of a hollow cylinder with external radius R and internal radius r is given by V = πh(R² − r²).

Surface Area of Sphere

The surface area of a sphere of radius r is 4πr², while its volume is (4/3)πr³.

Shortcut Trick

Total Lateral Surface Area = Total Surface Area − Base Area

Lateral Surface Area = 340 − 100 = 240 cm²

Number of lateral faces = Total Lateral Area ÷ Area of each face

Number of lateral faces = 240 ÷ 30 = 8

∴ The correct answer is 8.

Alternate Method

Given:

Total Surface Area (TSA) = 340 cm²

Area of base = 100 cm²

Area of one lateral face = 30 cm²

Formula Used:

Total Surface Area of Pyramid = Area of Base + (Number of faces × Area of each lateral face)

Calculations:

⇒ Total Lateral Surface Area (LSA) = TSA − Base Area

⇒ LSA = 340 − 100

⇒ LSA = 240 cm²

⇒ Let n be the number of lateral faces.

⇒ n × Area of each lateral face = LSA

⇒ n × 30 = 240

⇒ n = 240 ÷ 30

⇒ n = 8

∴ The correct answer is 8.

Additional Information

Total Surface Area of a Pyramid

For any pyramid, the Total Surface Area (TSA) is the sum of the base area and the lateral surface area (sum of all triangular face areas).

Lateral Surface Area of Regular Pyramid

LSA = ½ × Perimeter of base × Slant Height (l). This applies when all lateral faces are congruent triangles.

Volume of a Pyramid

Volume = 1/3 × Base Area × Vertical Height (h). It is always one-third the volume of a prism with the same base and height.

Shortcut Trick

The expression P(P² + 3P + 3) can be written as P³ + 3P² + 3P.

Using the algebraic identity (a + b)³ = a³ + 3a²b + 3ab² + b³, we see that:

(P + 1)³ − 1 = P³ + 3P² + 3P + 1 − 1 = P³ + 3P² + 3P.

Substitute P = 99: (99 + 1)³ − 1 = 100³ − 1 = 1,000,000 − 1 = 999,999.

∴ The correct answer is 999999.

Alternate Method

Given: P = 99

Formula Used: (a + b)³ = a³ + 3a²b + 3ab² + b³

Calculations:

Required value = P(P² + 3P + 3)

⇒ P³ + 3P² + 3P

⇒ (P³ + 3P² + 3P + 1) − 1

⇒ (P + 1)³ − 1

⇒ (99 + 1)³ − 1

⇒ (100)³ − 1

⇒ 1000000 − 1

⇒ 999999

∴ The correct answer is 999999.

Additional Information

Cube of a Sum Identity

(a + b)³ = a³ + 3a²b + 3ab² + b³. In this problem, putting b = 1 gives a³ + 3a² + 3a + 1.

Cube of a Difference Identity

(a − b)³ = a³ − 3a²b + 3ab² − b³. For b = 1, it results in a³ − 3a² + 3a − 1.

Sum and Difference of Cubes

a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²).

Shortcut Trick The equation x + 1/x = c + 1/c is symmetric with respect to x and c. By inspection, if we substitute x = c, both sides become identical (c + 1/c = c + 1/c). In reciprocal equations of the form f(x) + 1/f(x) = k, if x is a root, then 1/x is also a root. By substituting x = 1/c, the equation becomes 1/c + c = c + 1/c, which is also true. Thus, the two values satisfying the equation are c and its reciprocal 1/c. ∴ The correct answer is c, 1/c. Alternate Method Given: x + 1/x = c + 1/c Formula Used: Quadratic rearrangement and factorisation. ⇒ x − c + 1/x − 1/c = 0 ⇒ (x − c) + (c − x) / (cx) = 0 ⇒ (x − c) − (x − c) / (cx) = 0 ⇒ (x − c) [1 − 1/(cx)] = 0 ⇒ (x − c) [(cx − 1) / (cx)] = 0 ⇒ x − c = 0 or cx − 1 = 0 ⇒ x = c or x = 1/c ∴ The correct answer is c, 1/c. Additional Information Reciprocal Equations Equations where the reciprocal of any root is also a root of the equation. These often take the form x + 1/x = k. Product of Roots in Reciprocal Form The equation x + 1/x = a can be written as x2 − ax + 1 = 0. According to Vieta's formulas, the product of roots (c/a) is 1/1 = 1. Nature of Roots For a quadratic equation ax2 + bx + c = 0, the roots are real and distinct if the discriminant D = b2 − 4ac > 0.

Shortcut Trick

Test integer pairs whose sum is 9: (1,8), (2,7), (3,6), (4,5).

Check squares: 4² + 5² = 16 + 25 = 41. (Condition satisfied)

Sum of cubes: 4³ + 5³ = 64 + 125 = 189.

∴ The correct answer is 189.

Alternate Method

Given: Sum of numbers (x + y) = 9

Sum of squares (x² + y²) = 41

Formula used: (x + y)² = x² + y² + 2xy

Formula used: x³ + y³ = (x + y)(x² + y² − xy)

⇒ (x + y)² = x² + y² + 2xy

⇒ 9² = 41 + 2xy

⇒ 81 = 41 + 2xy

⇒ 2xy = 81 − 41 = 40

⇒ xy = 20

⇒ x³ + y³ = (x + y)(x² + y² − xy)

⇒ x³ + y³ = 9 × (41 − 20)

⇒ x³ + y³ = 9 × 21 = 189

∴ The correct answer is 189.

Additional Information

Algebraic Identity for Cubes

The sum of cubes can also be written as: x³ + y³ = (x + y)³ − 3xy(x + y).

Difference of Cubes

The identity for the difference of two cubes is: x³ − y³ = (x − y)(x² + xy + y²).

Product of Numbers

Given the sum and sum of squares, the product of two numbers is always xy = [(x + y)² − (x² + y²)] ÷ 2.

Shortcut Trick

Identify the expression as the sum of two squares: x⁴ + 64 = (x²)² + 8².

Apply the Sophie Germain Identity: a⁴ + 4b⁴ = (a² + 2b² + 2ab)(a² + 2b² − 2ab).

Here, a = x and 4b⁴ = 64 ⇒ b⁴ = 16 ⇒ b = 2.

Substituting values: (x² + 2 × 2² + 2 × x × 2)(x² + 2 × 2² − 2 × x × 2).

Simplifying the terms: (x² + 8 + 4x)(x² + 8 − 4x) → (x² + 4x + 8)(x² − 4x + 8).

∴ The correct answer is (x² + 4x + 8)(x² − 4x + 8).

Alternate Method

Given: x⁴ + 64

Formula Used: a² + b² = (a + b)² − 2ab and a² − b² = (a + b)(a − b)

⇒ x⁴ + 64 = (x²)² + (8)² ⇒ (x² + 8)² − 2 × x² × 8 ⇒ (x² + 8)² − 16x² ⇒ (x² + 8)² − (4x)² ⇒ Using the difference of squares identity where a = (x² + 8) and b = 4x ⇒ (x² + 8 + 4x)(x² + 8 − 4x) ⇒ (x² + 4x + 8)(x² − 4x + 8)

∴ The correct answer is (x² + 4x + 8)(x² − 4x + 8).

Additional Information

Sophie Germain Identity

The identity states that a⁴ + 4b⁴ = (a² + 2b² + 2ab)(a² + 2b² − 2ab). This is useful for factorizing sums of even powers.

Difference of Two Squares

For any two terms, a² − b² = (a − b)(a + b). It is a fundamental tool in polynomial factorization.

Perfect Square Trinomial

Expanding (a + b)² gives a² + 2ab + b². In this problem, we added and subtracted the middle term 16x² to create this form.

Shortcut Trick

For algebraic identities given with a condition like a + b = 1, use the value substitution method.

Let a = 1 and b = 0, which satisfies the condition 1 + 0 = 1.

Substitute these values into the expression: 1⁴ + 0⁴ − 1³ − 0³ − 2(1² × 0²) + (1 × 0)

⇒ 1 + 0 − 1 − 0 − 0 + 0 = 0.

∴ The correct answer is 0.

Alternate Method

Given: a + b = 1

Expression: a⁴ + b⁴ − a³ − b³ − 2a²b² + ab