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The difference between a single discount of (a + b)% and two successive discounts of a% and b% is always (ab ÷ 100)% of the Marked Price.

Here, a = 25 and b = 5. Note that 25 + 5 = 30, which matches the single discount.

Difference % = (25 × 5) ÷ 100 = 1.25%.

Difference in value = 1.25% of ₹3,840 = (1.25 ÷ 100) × 3,840 = 3,840 ÷ 80 = ₹48.

∴ The correct answer is ₹48.

Alternate Method

Given: Shopping amount (MP) = ₹3,840, Case 1 Discount = 30%, Case 2 Successive Discounts = 25% and 5%.
Formulas: Net Discount = a + b − (ab ÷ 100) | Value = (Discount% ÷ 100) × MP.

⇒ Case 1 Discount: 30% of 3,840 = (30 ÷ 100) × 3,840 = ₹1,152.
⇒ Case 2 Net Discount%: 25 + 5 − (25 × 5 ÷ 100) = 30 − 1.25 = 28.75%.
⇒ Case 2 Discount Value: (28.75 ÷ 100) × 3,840 = ₹1,104.
⇒ Difference: 1,152 − 1,104 = ₹48.

∴ The correct answer is ₹48.

Additional Information

Successive Discounts

When two discounts d₁ and d₂ are given, the effective discount is always less than the sum of the individual discounts because the second discount is applied to the reduced price.

Difference between Single and Successive

If D = a + b, the difference between a single discount of D% and successive discounts of a% and b% is simply ab/100 % of the Marked Price.

Given:

The successive discounts are 15% and 5%.

Formula used:

Successive discount = X + Y - (XY)/100

Where,

X = First discount

Y = Second discount

Calculation:

Successive discount = X + Y - (XY)/100

X = 15%, Y = 15%

⇒ 15 + 5 - 75/100

⇒ 19.25%

∴ The single equivalent discount is 19.25%

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When an original owner repurchases an item, their net financial gain is exactly equal to the second person's cash loss.

Montoo's Cost Price = 150000 × (100 + 5) ÷ 100 = Rs. 157500.

Montoo's Loss = 2% of 157500 = Rs. 3150. This loss amount represents Raju's exact net profit.

∴ The correct answer is Rs. 3150.

Alternate Method

Given:

Original CP of the car for Raju = Rs. 150000

Raju's profit percentage = 5%

Montoo's loss percentage = 2%

Calculations:

⇒ Selling Price (SP) of Raju = CP × (100 + Profit%) ÷ 100

⇒ SP of Raju = 150000 × (100 + 5) ÷ 100 = 150000 × 105 ÷ 100 = Rs. 157500

⇒ This SP becomes the Cost Price for Montoo.

⇒ SP of Montoo = 157500 × (100 − 2) ÷ 100 = 157500 × 98 ÷ 100 = Rs. 154350

⇒ Raju initially made a profit of: 157500 − 150000 = Rs. 7500

⇒ To repurchase the car, Raju paid an extra amount over the original CP: 154350 − 150000 = Rs. 4350

⇒ Raju's net financial gain = Initial Profit − Extra amount paid

⇒ Net gain = 7500 − 4350 = Rs. 3150

∴ The correct answer is Rs. 3150.

Additional Information

Closed Loop Repurchase

If person A sells an item to person B and later buys it back, A's net physical assets are restored. Their net financial gain exactly equals B's net cash loss.

Successive Transactions

When an item changes hands sequentially, the final price is determined by chaining the percentage changes: Final Price = Initial Price × (1 ± P₁/100) × (1 ± P₂/100).

Profit and Loss Basics

Profit occurs when Selling Price > Cost Price (Profit = SP − CP). Loss occurs when Cost Price > Selling Price (Loss = CP − SP). Both are fundamentally calculated based on the Cost Price.

Given:

Total students in the college = 250

The average weight of boys = 50

The average weight of girls = 45

Average of total students = 48

Concept used:

Average = Total sum of observations/Total number of observations

Calculation:

We can use allegation method to solve this problem as -

The ratio of boys to girls = 3 ∶ 2

So, the number of boys = (3/5) × 250 = 150

∴ The number of boys in the college is 150.

Given:

The average weight of the first 14 persons out of 15 persons is 75 kg.

The weight of the 15^th person is 28 kg more than the average weight of all the 15 persons.

Formula used:

Average = Sum of observation/Number of observation

Calculation:

Let the average weight of all the 15 persons be x kg,

As given, the weight of 15^th person will be (x + 28) kg

⇒ Average weight of 15 persons = (The sum of the weight of all the 15 persons)/15

⇒ x = {(14 × 75) + (x + 28)}/15

⇒ 15x = 1050 + x + 28

⇒ 14x = 1078

⇒ x = 77

∴ The weight of 15^th person = 77 + 28 is 105 kg.

Given:

The average of 9 numbers = 16

Average of last two numbers = 18

Formula used:

Average of n numbers = (Sum of n numbers) ÷ n

Calculation:

According to the question,

The average of 9 numbers = 16

The sum of 9 numbers = 16 × 9 = 144

The average of the last two numbers = 18

The sum of the last two numbers = 18 × 2 = 36

The sum of first 7 numbers = 144 - 36 = 108

The average of the first 7 numbers = \(\frac{108}{7}\)

Therefore, '\(\frac{108}{7}\)' is the required answer.

Given:

Average of 5 numbers (A1) = 12

Average of 3 of these numbers (A2) = 10

Formulae: Sum = Average × Number of items Average = Sum / Number of items

Calculation:

Sum1 = A1 × 5 = 12 × 5 = 60

Sum2 = A2 × 3 = 10 × 3 = 30

Sum3 = Sum1 - Sum2 = 60 - 30 = 30

A3 = Sum3 / 2 = 30 / 2 = 15

So, the average of the other two numbers is 15.

Follow BODMAS rule to solve this question, as per the order given

.

⇒ 7 ÷ 2 – [3 of 7 ÷ 4 ÷ {(2 ÷ 5) × (25 ÷ 8) ÷ (5 ÷ 2)}]

⇒ 7 ÷ 2 – [21 ÷ 4 ÷ {2/5 × 25/8 ÷ 5/2}]

⇒ 7 ÷ 2 – [21 ÷ 4 ÷ {2/5 × 5/4}]

⇒ 7 ÷ 2 – [21 ÷ 4 ÷ 1/2]

⇒ 7 ÷ 2 – [21 × 1/4 × 2]

⇒ 7 × 1/2 - 21/2

⇒ 7/2 – 21/2 = -14/2 = -7

Concept used:

CP/MP = (100 - Discount%) / (100 + Profit%)

Calculation:

According to the given concept,

CP/MP = (100 - 18)/(100 + 30) = 82/130 = 41/65

So, CP = 41 unit and MP = 65 unit

Now, he sells his goods at 12% discount, SP = 65 × 88/100 = 57.2

So profit% = (57.2 - 41)/41 × 100 = 39.51%

∴ The correct answer is 39.51%

Given:

250 - [90 - (120 ÷ 10 - (30 - 60 ÷ 3) ÷ 10)]

Concept used:

Calculation:

250 - [90 - (120 ÷ 10 - (30 - 60 ÷ 3) ÷ 10)]

⇒  250 - [90 - (12 - (30 - 20) ÷ 10)]

⇒  250 - [90 - 11]

⇒ 250 - 79

⇒ 171

∴ Option 2 is the correct answer.