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Total frequency N = 52, so N/2 = 26.

Cumulative frequency (CF) for class 12-18 is 22, and for 18-24 is 35.

The median class is 18−24 where L = 18, CF = 22, f = 13, and h = 6.

Median = 18 + ((26 − 22) ÷ 13) × 6 = 18 + (4 × 6) ÷ 13 = 18 + 1.846...

∴ The correct answer is 19.85.

Alternate Method

Given Data:

Formula Used:

Median = L + [((N ÷ 2) − CF) ÷ f] × h

Where:

L = Lower limit of median class = 18

N = Total frequency = 52

CF = Cumulative frequency of the class preceding the median class = 22

f = Frequency of the median class = 13

h = Class width = 6

Calculations:

⇒ N ÷ 2 = 52 ÷ 2 = 26

⇒ Median = 18 + [(26 − 22) ÷ 13] × 6

⇒ Median = 18 + [4 ÷ 13] × 6

⇒ Median = 18 + (24 ÷ 13)

⇒ Median = 18 + 1.84615...

⇒ Median ≈ 19.846

∴ The correct answer is 19.85.

Additional Information

Empirical Formula

Relationship between mean, median, and mode: Mode = 3 × Median − 2 × Mean.

Median Class Identification

The median class is the class interval whose cumulative frequency is just greater than or equal to N/2.

Mean of Grouped Data

The mean (μ) is calculated using Σf_ix_i ÷ Σf_i, where x_i is the class mark.

Mode for Grouped Data

Mode = L + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h.

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Compare the net discount percentage for each scheme to find the least beneficial (lowest discount):

Scheme A: 18 + 39 − (18 × 39)/100 = 57 − 7.02 = 49.98%

Scheme B: Net multiplier = 0.71 × 0.55 × 0.96 ≈ 0.375 ⇒ Discount ≈ 62.5%

Scheme C: For 8 toys, the discount is a flat 29%.

Scheme D: 7 free on 8 bought ⇒ 7/15 × 100 ≈ 46.67%

∴ The correct answer is Scheme C.

Alternate Method

Given: Number of toys to buy = 8. All toys have the same Marked Price (MP).

Formula Used:

1. Successive discounts x% and y% = [x + y − (xy/100)]%
2. Buy X, Get Y Free discount % = [Y / (X + Y)] × 100

Calculation:

⇒ Scheme A: Net Discount = 18 + 39 − (18 × 39 / 100) = 57 − 7.02 = 49.98%

⇒ Scheme B: Successive 29%, 45%, 4%. Price multiplier = 0.71 × 0.55 × 0.96 = 0.37488. Discount = (1 − 0.37488) × 100 = 62.512%

⇒ Scheme C: Customer buys exactly 8 toys. Discount offered on the first 8 toys = 29%.

⇒ Scheme D: Buy 8, Get 7 Free. Total toys = 15. Discount = (7 / 15) × 100 = 46.67%

⇒ Comparing all: 29% (C) < 46.67% (D) < 49.98% (A) < 62.51% (B).

⇒ The least beneficial scheme for the customer is the one with the minimum discount.

∴ The correct answer is Scheme C.

Additional Information

Successive Discounts

The effective discount for two successive discounts a% and b% is calculated as [a + b − (ab/100)]%. For three or more, use the multiplier method: 100 − [100 × (100−d₁)/100 × (100−d₂)/100 × ...].

Buy X Get Y Free

In this retail scheme, the customer pays for X items but receives X + Y items in total. The discount percentage is (Free Items / Total Items) × 100 = [Y / (X + Y)] × 100.

Customer vs. Shopkeeper Benefit

A customer always seeks the maximum discount (most beneficial), while the shopkeeper seeks the minimum discount to maximize profit.

Shortcut Trick

Tap 2 fills (18 hr) and Tap 3 empties (18 hr) at the same rate.

Their net contribution to the cistern is zero (1/18 − 1/18 = 0).

Only Tap 1 effectively fills the cistern at a rate of 4 hr per full tank.

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

∴ The correct answer is 1 hr.

Alternate Method

Given:

Time taken by Tap 1 (A) = 4 hr

Time taken by Tap 2 (B) = 18 hr

Time taken by Tap 3 (C) = 18 hr (emptying)

Formula Used:

Efficiency = Total Work ÷ Time

Time Taken = Work to be done ÷ Combined Efficiency

Calculations:

⇒ Total work (LCM of 4, 18, 18) = 36 units

⇒ Work to be completed (one-fourth of cistern) = 36 ÷ 4 = 9 units

⇒ Combined efficiency of all three taps = 9 + 2 − 2 = 9 units/hr

⇒ Time required = 9 ÷ 9 = 1 hr

∴ The correct answer is 1 hr.

Additional Information

Negative Work in Cisterns

An outlet pipe or a leak is considered to do negative work because it removes water from the cistern. Its efficiency is subtracted from the inlet pipes.

Combined Efficiency

When multiple pipes operate together, their individual efficiencies (work done per unit time) are added algebraically to find the net filling or emptying rate.

Time and Efficiency Relation

Time is inversely proportional to efficiency. If a tap fills a tank in 't' hours, its 1-hour efficiency is 1/t of the total capacity.

Shortcut Trick

LCM of ratio parts 5, 14, and 9 = 5 × 14 × 9 = 630 units.

Given LCM = 7560.

1 ratio unit (which represents the HCF) = 7560 ÷ 630 = 12.

∴ The correct answer is 12.

Alternate Method

Given:

Ratio of three numbers = 5 : 14 : 9

LCM of the numbers = 7560

Formula Used:

For numbers in ratio a : b : c, the LCM of the numbers is given by: x × LCM(a, b, c), where x is the HCF.

Calculations:

⇒ Let the numbers be 5x, 14x, and 9x, where x is the HCF.

⇒ LCM(5x, 14x, 9x) = x × LCM(5, 14, 9)

⇒ Since 5, 14, and 9 are pairwise co-prime, their LCM = 5 × 14 × 9 = 630.

⇒ Given LCM = 7560

⇒ 630 × x = 7560

⇒ x = 7560 ÷ 630

⇒ x = 12

∴ The correct answer is 12.

Additional Information

HCF and LCM Relation

For any two numbers a and b, the product of the numbers is equal to the product of their HCF and LCM (a × b = HCF × LCM).

Co-prime Numbers

Numbers are said to be co-prime if their HCF is 1. If ratio parts are co-prime, their LCM is simply their product.

LCM and HCF of Fractions

LCM of fractions = LCM of numerators ÷ HCF of denominators; HCF of fractions = HCF of numerators ÷ LCM of denominators.

Shortcut Trick

Net Compound Interest (CI) rate for 2 years at 10% = 10 + 10 + (10 × 10)/100 = 21%.

Net Simple Interest (SI) rate for 2 years at 10% = 10 × 2 = 20%.

Ratio of CI to SI = 21 : 20.

Given 21 units = ₹178.5 ⇒ 1 unit = 178.5 ÷ 21 = 8.5.

Value of 20 units (SI) = 20 × 8.5 = ₹170.

∴ The correct answer is ₹170.

Alternate Method

Given: Rate (R) = 10%, Time (T) = 2 years, Compound Interest (CI) = ₹178.5.

Formula Used: CI = P[(1 + R/100)^T − 1] and SI = (P × R × T) ÷ 100

⇒ CI = P[(1 + 10/100)² − 1]

⇒ 178.5 = P[(1.1)² − 1]

⇒ 178.5 = P[1.21 − 1]

⇒ 178.5 = 0.21P

⇒ P = 178.5 ÷ 0.21 = ₹850

⇒ SI = (850 × 10 × 2) ÷ 100

⇒ SI = 850 × 0.2 = ₹170

∴ The correct answer is ₹170.

Additional Information

Difference between CI and SI for 2 years

The difference (D) is given by the formula D = P × (R/100)².

CI and SI Ratio for 2 years

The ratio of Compound Interest to Simple Interest for 2 years is CI / SI = (200 + R) / 200.

Effective Rate of Interest

For CI, the effective rate for two successive years x% and y% is [x + y + (xy/100)]%.

Let the initial price be P.

Price at end of 1^st year (P1) = P × 1.20

Price at end of 2^nd year (P2) = P1 × 0.80 = 0.96 P1

Price at end of 3^rd year (P3) = P2 × 1.20 = 1.152 P1

Difference |P3 - P2| = |1.152 P1 - 0.96 P1| = 0.192 P1

0.192 P1 = 1920

P1 = 10,000

40% of P1 = 0.40 × 10,000 = ₹4,800.

18 - [12 - 4] + 7 = 18 - 8 + 7 = 10 + 7 = 17.

Cost Price (CP) = ₹2,850

Number of lamps sold = 7 - 2 = 5

Selling Price (SP) = 5 × 360 = ₹1,800

Loss = CP - SP = 2,850 - 1,800 = ₹1,050.

Using the formula S1/S2 = √(T2/T1):

60/S2 = √(8/6) = √(4/3) = 2/√3

2 × S2 = 60√3

S2 = 30√3 km/h.

If p = 2, p²-1 = 3 and p²+1 = 5. GCD(3, 5) = 1.

If p is any odd prime, p² is odd, so p²-1 and p²+1 are consecutive even numbers. The GCD of any two consecutive even numbers is always 2.