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

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

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

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

∴ The correct answer is ₹459.

Alternate Method

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

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

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

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

⇒ 15x + 19x + 17x = 1377

⇒ 51x = 1377

⇒ x = 1377 ÷ 51 = 27

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

∴ The correct answer is ₹459.

Additional Information

Ratio and Proportion

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

Original Amount Calculation

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

Sum of Ratio Parts

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

Shortcut Trick

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

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

The number with the highest frequency is 7.

∴ The correct answer is 7.

Alternate Method

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

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

⇒ Count of number 7 = 2

⇒ Count of number 8 = 1

⇒ Count of number 9 = 1

⇒ Count of number 10 = 1

⇒ Maximum frequency = 2 (for number 7)

⇒ Mode = 7

∴ The correct answer is 7.

Additional Information

Empirical Relationship

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

Types of Mode

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

No Mode Case

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

Shortcut Trick

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

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

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

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

r² = 3146 ÷ 26 = 121 ⇒ r = 11 cm.

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

∴ The correct answer is 110 cm.

Alternate Method

Given:

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

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

Formula Used:

Surface area of a sphere = 4πr²

Calculations:

⇒ 4πR² + 4πr² = 12584π

⇒ 4π(R² + r²) = 12584π

⇒ R² + r² = 3146

Substitute R = 5r:

⇒ (5r)² + r² = 3146

⇒ 25r² + r² = 3146

⇒ 26r² = 3146

⇒ r² = 121

⇒ r = √121 = 11 cm

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

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

∴ The correct answer is 110 cm.

Additional Information

Volume of a Sphere

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

Hemisphere Properties

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

Ratio of Attributes

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

Shortcut Trick

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

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

Verify the sum: 17 + 19 = 36.

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

∴ The correct answer is 323.

Alternate Method

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

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

⇒ Let the primes be p₁ and p₂.

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

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

⇒ Testing 17 and 19: They are consecutive primes.

⇒ Sum = 17 + 19 = 36 (Satisfied).

⇒ Product = 17 × 19 = 323.

∴ The correct answer is 323.

Additional Information

Prime Numbers

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

Twin Primes

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

Consecutive Primes

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

Product of Primes

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

Shortcut Trick

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

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

Given, 160% = ₹5,040

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

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

∴ The correct answer is ₹1,260.

Alternate Method

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

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

⇒ Amount = P[1 + (R × T₁) ÷ 100]

⇒ 5,040 = P[1 + (20 × 3) ÷ 100]

⇒ 5,040 = P[1 + 0.6] ⇒ 5,040 = 1.6P

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

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

⇒ SI₂ = (3,150 × 20 × 2) ÷ 100

⇒ SI₂ = 3,150 × 0.4 = ₹1,260

∴ The correct answer is ₹1,260.

Additional Information

Simple Interest Formula

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

Relationship between Amount and Principal

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

Constant Interest Property

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

Shortcut Trick

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

Shop I factor: 0.11 × 0.32 = 0.0352

Shop II factor: 0.50 × 0.49 = 0.2450

Shop III factor: 0.75 × 0.89 = 0.6675

Shop IV factor: 0.06 × 0.51 × 0.79 = 0.024174

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

∴ The correct answer is IV.

Alternate Method

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

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

Shop I SP:

⇒ 100 × (1 − 0.89) × (1 − 0.68)

⇒ 100 × 0.11 × 0.32 = ₹3.52

Shop II SP:

⇒ 100 × (1 − 0.50) × (1 − 0.51)

⇒ 100 × 0.50 × 0.49 = ₹24.50

Shop III SP:

⇒ 100 × (1 − 0.25) × (1 − 0.11)

⇒ 100 × 0.75 × 0.89 = ₹66.75

Shop IV SP:

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

⇒ 100 × 0.06 × 0.51 × 0.79 = ₹2.4174

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

∴ The correct answer is IV.

Additional Information

Successive Discount Formula

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

Multiplying Factor Method

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

Marked Price and Selling Price

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

Shortcut Trick

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

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

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

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

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

∴ The correct answer is 152 years.

Alternate Method

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

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

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

⇒ 3 × (12x + 4) = 5 × (7x + 4)

⇒ 36x + 12 = 35x + 20

⇒ 36x − 35x = 20 − 12

⇒ x = 8

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

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

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

∴ The correct answer is 152 years.

Additional Information

Concept of Age Difference

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

Ratio Balancing Method

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

Cross-Multiplication Rule

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

Shortcut Trick

Equalize decimal places: 15.0, 10.0, 0.4.

Remove decimals: 150, 100, 4.

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

LCM = 2² × 3 × 5² = 4 × 3 × 25 = 300.

Re-adjust decimal: 300 ÷ 10 = 30.

∴ The correct answer is 30.

Alternate Method

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

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

Convert the decimal and integers into simple fractions:

⇒ 15 = 15/1

⇒ 10 = 10/1

⇒ 0.4 = 4/10 = 2/5

⇒ Numerators = 15, 10, 2

⇒ Denominators = 1, 1, 5

⇒ LCM(15, 10, 2) = 30

⇒ HCF(1, 1, 5) = 1

⇒ Required LCM = 30 ÷ 1 = 30

∴ The correct answer is 30.

Additional Information

LCM of Fractions

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

HCF of Fractions

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

Property of Two Numbers

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

Shortcut Trick

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

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

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

Remaining work = 12 − 6 = 6 units.

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

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

∴ The correct answer is 5 days.

Alternate Method

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

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

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

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

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

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

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

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

∴ The correct answer is 5 days.

Additional Information

Efficiency and Time Relation

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

Work and Wages

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

Pipes and Cisterns

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

Given:

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

Formula used:

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

Calculation:

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

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

⇒ (5/36) ÷ (11/12) + (1 × 5/33)

⇒ (5/36 × 12/11) + 5/33

⇒ 5/33 + 5/33

⇒ 10/33

∴ The correct answer is option (1).