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Given:

Sum of numbers written = Total sum - Missing number

Average = \(43\frac{7}{11}\) = 480/11

Formula used:

Sum of first n natural numbers = \(\dfrac{n(n+1)}{2}\)

Average = \(\dfrac{\text{Sum of numbers written}}{\text{Number of numbers written}}\)

Calculation:

Assume total numbers written = n

Total sum = \(\dfrac{n(n+1)}{2}\)

Sum written = Total sum - Missing number

Average = \(\dfrac{\text{Sum written}}{n-1}\)

⇒ \(\dfrac{\dfrac{n(n+1)}{2} - x}{n-1} = \dfrac{480}{11}\)

⇒ \(\dfrac{n(n+1) - 2x}{2(n-1)} = \dfrac{480}{11}\)

⇒ \(\dfrac{n(n+1) - 2x}{960} = n-1\)

Testing options, let n = 76

Total sum = \(\dfrac{76(76+1)}{2} = \dfrac{76×77}{2} = 2926\)

Missing number = 76

Sum written = 2926 - 76 = 2850

Average = \(\dfrac{2850}{75} = \dfrac{480}{11} = 43\frac{7}{11}\)

∴ The missing number is 76, correct answer is option (3).

Given:

Time (downstream + upstream) = 6 hr. for 36 km downstream and 24 km upstream

Time (downstream + upstream) = 6.5 hr. for 24 km downstream and 36 km upstream

Formula used:

Speed = Distance ÷ Time

Let downstream speed = x km/hr, upstream speed = y km/hr

Speed of current = (x - y) ÷ 2

Calculations:

36 ÷ x + 24 ÷ y = 6

24 ÷ x + 36 ÷ y = 6.5

⇒ Solve equations:

36 ÷ x + 24 ÷ y = 6

⇒ Multiply by xy: 36y + 24x = 6xy

⇒ 6xy - 36y - 24x = 0

24 ÷ x + 36 ÷ y = 6.5

⇒ Multiply by xy: 24y + 36x = 6.5xy

⇒ 6.5xy - 24y - 36x = 0

Subtract equations:

(6.5xy - 24y - 36x) - (6xy - 36y - 24x) = 0

⇒ 0.5xy + 12y - 12x = 0

⇒ 0.5(xy) = 12(x - y)

⇒ xy = 24(x - y)

Substitute values:

Downstream speed (x) = 12 km/hr

Upstream speed (y) = 8 km/hr

⇒ Speed of current = (x - y) ÷ 2

⇒ Speed of current = (12 - 8) ÷ 2

⇒ Speed of current = 4 ÷ 2

⇒ Speed of current = 2 km/hr

∴ The correct answer is option (3).

Given:

Speed of car = Distance / Time = 720 / 9 = 80 km/h

Speed of bus = (3/4) × Speed of car = (3/4) × 80 = 60 km/h

Speed ratio of bus : train = 5 : 9

Speed of train = (Speed of bus × 9) / 5 = (60 × 9) / 5 = 108 km/h

Time = 7 hours

Formula used:

Distance = Speed × Time

Calculation:

Distance covered by train = Speed of train × Time

⇒ Distance = 108 × 7

⇒ Distance = 756 km

∴ The correct answer is option (2).

Given:

Anil's monthly income = ?

Savings = ₹8112

Spends 22% on food and clothing, 13% on transport, 20% of remaining on children education, 40% of remaining on other expenditure.

Formula used:

Remaining income after each expenditure = Monthly income × (100% - Percentage spent)

Calculation:

Let monthly income = ₹x

⇒ Remaining after food and clothing = x × (100 - 22)% = x × 78%

⇒ Remaining after transport = x × 78% × (100 - 13)% = x × 78% × 87%

⇒ Remaining after children education = x × 78% × 87% × (100 - 20)% = x × 78% × 87% × 80%

⇒ Remaining after other expenditure = x × 78% × 87% × 80% × (100 - 40)% = x × 78% × 87% × 80% × 60%

⇒ Savings = x × 78% × 87% × 80% × 60%

⇒ 8112 = x × (78/100) × (87/100) × (80/100) × (60/100)

⇒ 8112 = x × (78 × 87 × 80 × 60) / (100 × 100 × 100 × 100)

⇒ 8112 = x × 32486400 / 100000000

⇒ x = (8112 × 100000000) / 32486400

⇒ x = ₹24960

Monthly income = ₹24960

Remaining after food and clothing = 24960 × 78% = ₹19468.80

Remaining after transport = 19468.80 × 87% = ₹16937.856

Expenditure on children education = 16937.856 × 20% = ₹3387.57

∴ The expenditure on children education is approximately ₹3380 (Option 3).

Given:

Principal (P) = ₹2,000

Amount (A) = ₹2,645

Rate (r) = 30% per annum (compounded half yearly, so r = 15% for each half year)

Formula used:

A = P(1 + r/100)ⁿ

Where, n = number of compounding periods

Calculations:

2645 = 2000(1 + 15/100)ⁿ

⇒ 2645/2000 = (1 + 0.15)ⁿ

⇒ 1.3225 = (1.15)ⁿ

⇒ n = 2 (since (1.15)² = 1.3225)

Time = n × 0.5 years (compounded half yearly)

⇒ Time = 2 × 0.5 = 1 year

∴ The correct answer is option (1).

Given:

\( 40% \text{ of } 680 + 35% \text{ of } 500 - 10% \text{ of } 100 \times 5% \text{ of } 500 = ? \)

Formula used:

Percentage formula: \(x% \text{ of } y = \frac{x}{100} \times y\)

Calculation:

\( 40% \text{ of } 680 = \frac{40}{100} \times 680 = 272 \)

\( 35% \text{ of } 500 = \frac{35}{100} \times 500 = 175 \)

\( 10% \text{ of } 100 = \frac{10}{100} \times 100 = 10 \)

\( 5% \text{ of } 500 = \frac{5}{100} \times 500 = 25 \)

⇒ \( 10 \times 25 = 250 \)

⇒ \( 272 + 175 - 250 = 197 \)

∴ The correct answer is option (1).

Given:

Final Amount (A) = ₹20,736

Simple Interest Rate (SI Rate) = \( 14\frac{2}{7}% = \frac{100}{7}% = 14.2857% \) p.a.

Time for SI = 3.5 years

Compound Interest Rate (CI Rate) = 20% p.a.

Time for CI = 3 years

Formula used:

SI = \( \frac{P×R×T}{100} \)

A_CI = P_CI\( (1 + \frac{r}{100})^t \)

Calculation:

Let initial sum lent = P

Amount after SI = P + \( \frac{P×100×3.5}{700} \)

⇒ Amount after SI = P + \( \frac{350P}{700} \)

⇒ Amount after SI = P + 0.5P = 1.5P

Amount after CI = 1.5P\( (1 + \frac{20}{100})^3 \)

⇒ Amount after CI = 1.5P\( (1.2)^3 \)

⇒ Amount after CI = 1.5P × 1.728

Given Amount after CI = ₹20,736

⇒ 1.5P × 1.728 = 20,736

⇒ P = \( \frac{20,736}{1.728×1.5} \)

⇒ P = \( \frac{20,736}{2.592} \)

⇒ P = ₹8,000

∴ The correct answer is option (4).

Given:

Number = 5160

Divisor = 13

Formula used:

Remainder = Number mod Divisor

To find the least number to subtract: Subtraction = Remainder

Calculation:

Remainder = 5160 mod 13

⇒ Remainder = 12

⇒ Least number to subtract = 12

∴ The correct answer is option (1).

Given:

A’s rate of work = 140% more than B’s rate of work

Formula used:

If A’s rate is R and B’s rate is r, then A’s rate = r + 1.4r = 2.4r

Fraction of work done by A = A’s rate / (A’s rate + B’s rate)

Calculation:

A’s rate = 2.4r

B’s rate = r

Fraction of work done by A = 2.4r / (2.4r + r)

⇒ Fraction = 2.4 / (2.4 + 1)

⇒ Fraction = 2.4 / 3.4

⇒ Fraction = 12 / 17

∴ The fraction of work done by A is \( 7/12 \).

Given:

Safe weight limit = 2000 Kg

Average weight of half riders = 180 Kg

Average weight of other half = 215 Kg

Formula used:

Total weight of riders = (Number of riders × Average weight of half riders) + (Number of riders × Average weight of other half)

Calculation:

Total weight = 2000 Kg

Let total riders = 2x

Half riders = x

⇒ 2000 = (x × 180) + (x × 215)

⇒ 2000 = x(180 + 215)

⇒ 2000 = x × 395

⇒ x = 2000 ÷ 395

⇒ x ≈ 5.06

Total riders = 2x ≈ 2 × 5 = 10

∴ The correct answer is option (4).