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Using the fractional reduction rule for constant expenditure:

Price increase = 40% = 2/5

If price increases by a fraction a/b, consumption must decrease by a/(a+b) to keep expenditure constant.

Required consumption reduction = 2 / (2 + 5) = 2/7

Percentage reduction = (2 ÷ 7) × 100 = 28.57% ≈ 28.5%

∴ The correct answer is 28.5%.

Alternate Method

Given:

Percentage increase in the price of sugar = 40%

Concept Used:

When Expenditure remains constant, Price and Consumption are inversely proportional (Expenditure = Price × Consumption).

Calculations:

⇒ Let the initial price of sugar be ₹100 per unit and initial consumption be 100 units.

⇒ Initial Expenditure = 100 × 100 = ₹10000

⇒ New price of sugar after 40% increase = ₹140 per unit

⇒ To keep the expenditure constant at ₹10000, New Consumption = 10000 ÷ 140 = 500 / 7 units

⇒ Reduction in consumption = Original Consumption − New Consumption

⇒ Reduction in consumption = 100 − (500 / 7) = (700 − 500) / 7 = 200 / 7 units

⇒ Percentage reduction = [(Reduction in consumption) ÷ (Original Consumption)] × 100

⇒ Percentage reduction = [(200 / 7) ÷ 100] × 100

⇒ Percentage reduction = 200 / 7 % = 28.57%

⇒ Looking at the options, it is approximately 28.5%.

∴ The correct answer is 28.5%.

Additional Information

Price & Consumption Rule (Increase)

If the price of a commodity increases by R%, the required percentage reduction in consumption so as not to increase the overall expenditure is given by [R ÷ (100 + R)] × 100%.

Price & Consumption Rule (Decrease)

If the price of a commodity decreases by R%, the permissible percentage increase in consumption to maintain the exact same expenditure is given by [R ÷ (100 − R)] × 100%.

Constant Product Rule

For any equation of the form A × B = Constant, if variable A increases by a fraction of x/y, then variable B must decrease by the fraction x/(x+y) to keep their product constant.

FOR COMPOUND INTEREST

We know the formula for compound interest-         

\(⇒ {\bf{CI}} = \left[ {{\bf{P}}\left{ {{{\left( {1 + \frac{{\bf{R}}}{{100}}} \right)}^T} - 1} \right}} \right]\)

CI = Compound interest

P = Principal

R = Rate of interest

T = Time period

We have

\(\begin{array}{l} {\rm{CI}} = \left[ {{\rm{P}}\left{ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right}} \right]\ {\rm{CI}} = \left[ {{\rm{P}}\left{ {1.1025 - 1} \right}} \right]\ {\rm{CI}} = \left[ {0.1025 × {\rm{P}}} \right] \end{array}\)

FOR SIMPLE INTEREST

We know that, formula for simple interest:

\(S.I. = \frac{{P × R × T}}{{100}}\)

S.I. = Simple Interest

P = principal

R = Rate of interest

T = Time

\(\begin{array}{l} S.I. = \frac{{P × 5 × 2}}{{100}}\ S.I. = 0.1 × P \end{array}\)

We have CI – SI = Rs 6

⇒ 0.1025P – 0.1P = Rs 6

⇒ 0.0025P = Rs 6

\(⇒ P = \frac{{6}}{{0.0025}}\)

⇒ P = Rs 2400

 

 

In the case of compound interest, the rate of interest increased successively. 

And in the case of the simple interest rate of interest is the same for every year.

The effective rate of interest of two years in the case of compound interest.

{5 + 5 + (5×5)/100} = 10.25%

Effective rate of  interest for two years in case of simple interest

5 + 5 = 10%

According to the question:

Difference of Compound interest and simple interest = Rs.6

As we calculated:

10.25% - 10% = 0.25% 

⇒ 0.25% = Rs.6

⇒ 100% = 2400

∴ the principal is Rs.2400

Formula used:

Difference between C.I and S.I for 2 years = PR²/100²

Where P = Principal, R = Rate of interest

Calculation:

 Let principal be P

So, from question,

(P × 5²)/100² = 6

⇒ P = 6 × 4 × 100 = 2400

∴ Principal is Rs. 2400

The number of students selected in exam from school B, C and D together = (18% + 22% + 17%) of 3200

⇒ 57% of 3200

⇒ 1824

The number of students selected in exam from school A, E and F together = (19% + 13% + 11%) of 3200

⇒ 43% of 3200

⇒ 1376

∴ required difference = (1824 – 1376) = 448

Given:

Principal of Ram = Rs. 8000

Rate of interest (For Ram) = 12%

Principal of Sham = Rs. 9100 

Rate of interest (For Sham) = 10%

Amount of Ram and Sham are equal

Formula used:

SI \(= \frac{{{\rm{P;}} \times {\rm{;R;}} \times {\rm{;T}}}}{{100}}\)

Amount = P + S.I.

Calculation:

Let the required time be y years.

Amount of Ram = Amount of Sham

8000 + [8000 × (12/100) × y] = 9100 + [9100 × (10/100) × y]

⇒ 8000 + 960 y = 9100 + 910 y

⇒ 960 y - 910 y = 9100 - 8000

⇒ 50 y = 1100

⇒ y = 22 years

∴ The time taken to debt become equal is 22 years

Given:

Average age of 20 boys = 15

Calculation:

Let the average age of girls be x years

According to the question,

20 × 15 + 10x = 420

⇒ 300 + 10x = 420

⇒ 10x = 120

⇒ x = 12

⇒ Average = 12 years

∴ The average age of the girls will be 12 years.

Shortcut TrickAverage decreased by 1 year

Total decreased = (20 + 10) × 1 [20 boys and 10 girls]

⇒ 30

Average = Previous average – (Total change/No. of new students)

⇒ 15 – 30/10

⇒ 15 – 3

⇒ 12

∴ The average age of the girls will be 12 years.

Calculation:

Total expenditure = 48 cr

Expenditure of company Q & R together = (22 + 18)% = 40% of Total expenditure

⇒ (\(\frac{40}{{100}}\) × 48) = 19.2

Expenditure of company P = (24% of 48) = (\(\frac{24}{{100}}\) × 48) = 11.52

Percentage of expenditure of Q & R greater in terms of P

= [(19.2 - 11.52)/11.52] × 100

⇒ (\(\frac{7.68}{{11.52}}\) × 100) = (\(\frac{768}{{11.52}}\)) = 66.67%

∴ The required percentage is 66.67%

Given: 

The boat can cover 15 km upstream and 18 km downstream distance in total 4 hours

Also, it can cover 18 km upstream and 15 km downstream distance in total 4 hours 15 minutes(17/4 hours)

Formula used:

Speed = distance/time

Upstream speed(u) = b - w

Downstream speed(d) = b + w

b = speed of the boat in still water

w = speed of the stream

Calculation:

According to the question:

(15/u) + (18/d) = 4

⇒ 15d + 18u = 4ud     ----eq.3

⇒ 255d + 306u = 68ud    ----eq.1

(18/u) + (15/d) = 17/4

⇒ 18d + 15u = 17/4(ud)

⇒ 72d + 60u = 17ud

⇒ 288d + 240u = 68ud     ----eq.2

By comparing eq.1 and eq.2:

288d + 240u = 255d + 306u

⇒ d/u = 2/1

Let d = 2x and u = x

Put in eq.3:

30x + 18x = 8x2

⇒ x = 6

⇒ u = b - w = 6     ----eq.4

⇒ d = b + w = 12     ----eq.5

From eq. 4 and eq.5, b = 9

∴ Speed of the boat in still water = 9 km/h

Shortcut Trick

Profit sharing ratio = (Investment × Time) ratio

Ravi : Raju = (50,000 × 12) : (1,00,000 × 6)

Ravi : Raju = 6,00,000 : 6,00,000 = 1 : 1

Since the ratio is 1 : 1, the profit is divided equally.

Raju profit share = ₹63,000 ÷ 2 = ₹31,500

∴ The correct answer is ₹31,500.

Alternate Method

Given:

Investment of Ravi = ₹50,000 | Time period = 12 months

Investment of Raju = ₹1,00,000 | Time period = 6 months (Joined after 6 months)

Total Annual Profit = ₹63,000

Formula Used:

Profit Ratio = (Capital₁ × Time₁) : (Capital₂ × Time₂)

⇒ Ravi profit equivalent = 50,000 × 12 = 6,00,000

⇒ Raju profit equivalent = 1,00,000 × 6 = 6,00,000

⇒ Profit Ratio (Ravi : Raju) = 6,00,000 : 6,00,000

⇒ Ratio = 1 : 1

⇒ Total Ratio Parts = 1 + 1 = 2 units

⇒ 2 units = ₹63,000

⇒ 1 unit = ₹63,000 ÷ 2 = ₹31,500

⇒ Raju share (1 unit) = ₹31,500

∴ The correct answer is ₹31,500.

Additional Information

Partnership Profit Rule

In any business partnership, the Profit (P) is directly proportional to the product of Capital (C) and Time (T). Formula: P₁ : P₂ = C₁T₁ : C₂T₂.

Types of Partners

An Active Partner manages the business and may take a salary before profit distribution, while a Sleeping Partner only provides capital.

Compound Partnership

When partners invest different amounts for different durations of time, it is called a compound partnership, where effective capital is calculated per unit of time.

Given:

On selling 32 articles gain is equal to selling price 12 articles.

Formula Used:

Gain = SP – CP

Gain percent = Gain/CP × 100

Calculation:

According to the question,

⇒ Selling price of 32 articles = Cost price of 32 articles + Selling price of 12 articles

⇒ 32SP = 32CP + 12SP

⇒ 20SP = 32CP

⇒ 5SP = 8CP

SP = 8 and CP = 5 

Gain = SP – CP

Gain = 8 – 5 = 3

Percent gain = Gain/CP × 100

⇒ Percent gain = 3/5 × 100

⇒ Percent gain = 60%

∴ His percent gain is 60%.

The pattern followed here is:

AMS : UOC 

PGL : NIR

similarly,

BMH: JOD

Hence, the correct answer is "Option (3)".