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Question 201

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

If a = √7 - √3, and b = √5 - √2, then which of the following is true?

  1. AOption (a)

    a = b

  2. BOption (b)

    a < b

  3. COption (c)

    a > b

  4. DOption (d)

    Cannot be determined

View answer & explanation

Correct answer

Answer: (c)C.

a > b

Why this answer is correct

Shortcut Trick We can use approximate square root values to quickly compare the two given terms. ⇒ a = √7 − √3 ≈ 2.64 − 1.73 ≈ 0.91 ⇒ b = √5 − √2 ≈ 2.23 − 1.41 ≈ 0.82 Comparing the final decimal values, 0.91 > 0.82, which clearly indicates that a > b. ∴ The correct answer is a > b.

Alternate Method Given: a = √7 − √3 b = √5 − √2 Calculation: We can accurately compare a and b by finding their reciprocals and rationalizing the denominators: ⇒ 1/a = 1 / (√7 − √3) ⇒ 1/a = (√7 + √3) / (7 − 3) = (√7 + √3) / 4 ⇒ 1/b = 1 / (√5 − √2) ⇒ 1/b = (√5 + √2) / (5 − 2) = (√5 + √2) / 3 Now, let us compare 1/a and 1/b by equating their denominators (LCM of 3 and 4 is 12): ⇒ 1/a = 3(√7 + √3) / 12 = (√63 + √27) / 12 ⇒ 1/b = 4(√5 + √2) / 12 = (√80 + √32) / 12 Now comparing the numerators of both fractions: ⇒ Since 63 < 80, we have √63 < √80 ⇒ Since 27 < 32, we have √27 < √32 Therefore, (√63 + √27) < (√80 + √32), which directly implies that 1/a < 1/b. Since both a and b are positive numbers, taking the reciprocal reverses the inequality sign: ⇒ a > b ∴ The correct answer is a > b.

Additional Information Rationalization of Surds To remove roots from the denominator, multiply both the numerator and the denominator by the conjugate of the denominator using the formula (x − y)(x + y) = x2 − y2. Reciprocal Inequality Rule For any two positive numbers x and y, if x > y, then taking their reciprocal reverses the inequality resulting in 1/x < 1/y. Comparing Pure Surds When comparing mixed surds like a√b, move the outside coefficient inside by squaring it as √(a2b) to make comparisons entirely based on the inner radicands.

Question 202

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

Evaluate the expression: (14÷12)÷(115+1−910)(\frac{1}{4} \div \frac{1}{2}) \div (\frac{1}{15} + 1 - \frac{9}{10})

  1. AOption (a)

    5

  2. BOption (b)

    6

  3. COption (c)

    3

  4. DOption (d)

    2

View answer & explanation

Correct answer

Answer: (c)C.

3

Why this answer is correct

Shortcut Trick

First bracket: (1/4 ÷ 1/2) = 1/4 × 2 = 1/2

Second bracket: (1/15 + 1 − 9/10) = 2/30 + 30/30 − 27/30 = 5/30 = 1/6

Combine results: 1/2 ÷ 1/6 = 1/2 × 6 = 3

∴ The correct answer is 3.

Alternate Method

Given:

Expression: (1/4 ÷ 1/2) ÷ (1/15 + 1 − 9/10)

Formula Used:

BODMAS Rule: Brackets, Orders, Division, Multiplication, Addition, Subtraction.

Calculations:

⇒ Step 1: Solve the first bracket (1/4 ÷ 1/2)

⇒ 1/4 ÷ 1/2 = 1/4 × 2/1 = 2/4 = 1/2

⇒ Step 2: Solve the second bracket (1/15 + 1 − 9/10)

⇒ Find the Least Common Multiple (LCM) of the denominators 15 and 10, which is 30.

⇒ 1/15 = 2/30

⇒ 1 = 30/30

⇒ 9/10 = 27/30

⇒ (2/30 + 30/30 − 27/30) = (2 + 30 − 27) / 30 = 5/30 = 1/6

⇒ Step 3: Divide the result of the first bracket by the result of the second bracket

⇒ 1/2 ÷ 1/6 = 1/2 × 6/1

⇒ 6 / 2 = 3

∴ The correct answer is 3.

Additional Information

BODMAS Rule

It dictates the order of operations: Brackets first, then Orders (powers/roots), Division, Multiplication, Addition, and finally Subtraction.

Fraction Division

To divide by a fraction, multiply by its reciprocal: (a/b) ÷ (c/d) = (a/b) × (d/c).

Fraction Addition & Subtraction

Always find the LCM of the denominators to make them identical before adding or subtracting the numerators directly.

Question 203

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

The number of red, blue, and green marbles in a bag is in the ratio 3 : 4 : 6. If 20 red marbles, 15 blue marbles, and an unknown number of green marbles are added to the bag, the ratio of red, blue, and green marbles becomes 4 : 5 : 7. Determine the number of green marbles added.

  1. AOption (a)

    3

  2. BOption (b)

    5

  3. COption (c)

    7

  4. DOption (d)

    9

View answer & explanation

Correct answer

Answer: (b)B.

5

Why this answer is correct

Shortcut Trick

Initial R : B = 3 : 4 and final R : B = 4 : 5. Quantities added are +20 and +15.

Using cross-multiplication: 1 unit = (20 × 5 − 15 × 4) / (4 × 4 − 3 × 5) = (100 − 60) / 1 = 40.

Initial green marbles = 6 units = 6 × 40 = 240. New red marbles = (3 × 40) + 20 = 140.

New red (140) equals 4 parts in final ratio. Thus, 1 final part = 140 ÷ 4 = 35.

New green = 7 final parts = 7 × 35 = 245. Green marbles added = 245 − 240 = 5.

∴ The correct answer is 5.

Alternate Method

Given:

Initial ratio of red, blue, and green marbles = 3 : 4 : 6

Red marbles added = 20, Blue marbles added = 15

Final ratio of red, blue, and green marbles = 4 : 5 : 7

Ratio Change of MarblesInitial Ratio3y (R)4y (B)6y (G)Added+ 20+ 15+ xFinal Ratio4 parts5 parts7 parts(3y + 20) : (4y + 15) = 4 : 5

Calculations:

⇒ Let the initial number of red, blue, and green marbles be 3y, 4y, and 6y respectively.

⇒ After adding the marbles, new number of red marbles = 3y + 20

⇒ New number of blue marbles = 4y + 15

⇒ The new ratio of red to blue marbles is 4 : 5.

⇒ (3y + 20) ÷ (4y + 15) = 4 ÷ 5

⇒ 5 × (3y + 20) = 4 × (4y + 15)

⇒ 15y + 100 = 16y + 60

⇒ 16y − 15y = 100 − 60

⇒ y = 40

⇒ Initial number of green marbles = 6 × 40 = 240

⇒ New number of red marbles = (3 × 40) + 20 = 140

⇒ The final ratio of red to green marbles is 4 : 7.

⇒ Let the new number of green marbles be G.

⇒ 140 ÷ G = 4 ÷ 7

⇒ 4 × G = 140 × 7

⇒ G = 980 ÷ 4 = 245

⇒ Number of green marbles added = New green marbles − Initial green marbles

⇒ Green marbles added = 245 − 240 = 5

∴ The correct answer is 5.

Additional Information

Ratio Comparison Method

When individual components of a ratio are increased by specific quantities, establishing an equation with any two known components allows determining the exact multiplier of the ratio.

Cross-Multiplication Trick

For an initial ratio a:b becoming c:d after adding quantities p and q respectively, 1 ratio unit can be calculated directly as (pd − qc) ÷ (ad − bc).

Proportional Value Assignment

Once a new concrete value corresponding to a final ratio part is found (like 140 equivalent to 4 parts), the exact quantities of all other parts in the final state can be determined by direct proportion.

Question 204

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A invests ₹25,000 for 12 months, B invests ₹50,000 for 6 months. What is A's share of ₹30,000 profit?

  1. AOption (a)

    ₹10,000

  2. BOption (b)

    ₹12,000

  3. COption (c)

    ₹15,000

  4. DOption (d)

    ₹18,000

View answer & explanation

Correct answer

Answer: (c)C.

₹15,000

Why this answer is correct

Shortcut Trick

Profit ratio = (Capital of A × Time of A) : (Capital of B × Time of B)

Profit ratio = (25000 × 12) : (50000 × 6) = 300000 : 300000 = 1 : 1

Total profit ratio parts = 1 + 1 = 2 parts.

2 parts = ₹30,000 ⇒ 1 part = ₹15,000.

A's share is 1 part = ₹15,000.

∴ The correct answer is ₹15,000.

Alternate Method

Given:

A's Investment = ₹25,000, Time = 12 months

B's Investment = ₹50,000, Time = 6 months

Total Profit = ₹30,000

Formula Used:

Profit Ratio = (C1 × T1) : (C2 × T2)

Partner ACap: ₹25,000Time: 12 monthsWeight: 3,00,000Partner BCap: ₹50,000Time: 6 monthsWeight: 3,00,000Profit Ratio (A : B) = 1 : 1

Calculations:

⇒ Profit ratio of A and B = (25000 × 12) : (50000 × 6)

⇒ Profit ratio = 300000 : 300000 = 1 : 1

⇒ Sum of ratio parts = 1 + 1 = 2

⇒ A's share = (1 ÷ 2) × Total Profit

⇒ A's share = (1 ÷ 2) × 30000

⇒ A's share = 15000

∴ The correct answer is ₹15,000.

Additional Information

Simple Partnership

When all partners invest their capital for the same duration of time, the profit is distributed strictly in the ratio of their original investments.

Compound Partnership

When partners invest for different periods, the equivalent capital for each is calculated as Investment × Time. Profit is distributed in the ratio of these final products.

Working Partner vs Sleeping Partner

A working partner actively manages the business and is typically paid a fixed salary or a certain profit percentage before the rest of the profit is divided among all active and sleeping partners.

Question 205

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A and B start a business with capitals in the ratio 5:3. After 6 months, A withdraws ₹10,000 and B doubles his investment. If the initial investment of A was ₹50,000, and the total profit after 1 year is ₹84,000, find the profit share of A.

  1. AOption (a)

    ₹42,000

  2. BOption (b)

    ₹50,000

  3. COption (c)

    ₹60,000

  4. DOption (d)

    ₹45,000

View answer & explanation

Correct answer

Answer: (a)A.

₹42,000

Why this answer is correct

Shortcut Trick

Initial capital ratio = 5 : 3. Since A's capital = ₹50,000, B's = ₹30,000.

Equivalent capital of A = (50k × 6) + (40k × 6) = 540k

Equivalent capital of B = (30k × 6) + (60k × 6) = 540k

Profit ratio = 540k : 540k = 1 : 1

Profit share of A = (1 ÷ 2) × ₹84,000 = ₹42,000

∴ The correct answer is ₹42,000.

Alternate Method

Given:

Initial capital ratio of A : B = 5 : 3

Initial capital of A = ₹50,000

Total profit after 1 year = ₹84,000

Formula Used:

Profit Ratio = (CapitalA × TimeA) : (CapitalB × TimeB)

Partner A's Timeline₹50,000 × 6m₹40,000 × 6mPartner B's Timeline₹30,000 × 6m₹60,000 × 6mTotal Equivalent Capitals: A = ₹5,40,000 | B = ₹5,40,000

Calculations:

⇒ Initial capital of A = 5 units = ₹50,000

⇒ 1 unit = ₹50,000 ÷ 5 = ₹10,000

⇒ Initial capital of B = 3 units = 3 × ₹10,000 = ₹30,000

⇒ After 6 months, A withdraws ₹10,000, so new capital = ₹50,000 − ₹10,000 = ₹40,000

⇒ After 6 months, B doubles his investment, so new capital = ₹30,000 × 2 = ₹60,000

⇒ Equivalent capital of A = (₹50,000 × 6) + (₹40,000 × 6) = ₹3,00,000 + ₹2,40,000 = ₹5,40,000

⇒ Equivalent capital of B = (₹30,000 × 6) + (₹60,000 × 6) = ₹1,80,000 + ₹3,60,000 = ₹5,40,000

⇒ Ratio of profit share = 5,40,000 : 5,40,000 = 1 : 1

⇒ Sum of profit ratios = 1 + 1 = 2 units

⇒ A's profit share = (1 ÷ 2) × ₹84,000 = ₹42,000

∴ The correct answer is ₹42,000.

Additional Information

Compound Partnership

When capitals are invested for unequal time periods, profits are shared in the ratio of the product of capital and time, known as equivalent capital.

Simple Partnership

When partners invest their capitals for the exact same period, the profit is purely distributed in the respective ratio of their original investments.

Working vs Sleeping Partner

A working active partner manages the business and usually receives a salary or fixed percentage of profit. A sleeping partner only provides funds and gets a share solely based on capital and time ratio.

Question 206

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

What is the average of all three-digit numbers divisible by 23?

  1. AOption (a)

    752

  2. BOption (b)

    552

  3. COption (c)

    652

  4. DOption (d)

    452

View answer & explanation

Correct answer

Answer: (b)B.

552

Why this answer is correct

Shortcut Trick

The sequence of multiples forms an Arithmetic Progression (A.P.).

Average of an A.P. = (First Term + Last Term) ÷ 2

First 3-digit multiple of 23 = 115

Last 3-digit multiple of 23 = 989

Average = (115 + 989) ÷ 2 = 552

∴ The correct answer is 552.

Alternate Method

Given:

We need to find the average of all three-digit numbers divisible by 23.

Formula Used:

For an Arithmetic Progression (A.P.), Average = (a + l) ÷ 2

Where 'a' is the first term and 'l' is the last term.

115 First Term

989 Last Term

552 Average (Midpoint)

Average = (First Term + Last Term) ÷ 2

Calculations:

⇒ Smallest three-digit number = 100

⇒ 100 ÷ 23 = 4 with a remainder of 8

⇒ First 3-digit multiple of 23 (a) = 23 × 5 = 115

⇒ Largest three-digit number = 999

⇒ 999 ÷ 23 = 43 with a remainder of 10

⇒ Last 3-digit multiple of 23 (l) = 23 × 43 = 989

⇒ The sequence of numbers forms an A.P.: 115, 138, 161, ..., 989

⇒ Average of the A.P. = (115 + 989) ÷ 2

⇒ Average = 1104 ÷ 2 = 552

∴ The correct answer is 552.

Additional Information

Arithmetic Progression (A.P.)

A sequence of numbers where the difference between consecutive terms is constant. General form: a, a+d, a+2d, ...

Average of an A.P.

The average (mean) of an A.P. is strictly equal to the average of its first and last terms: Average = (a + l) ÷ 2.

Number of Terms in an A.P.

The total number of terms (n) in a finite arithmetic sequence can be calculated using the formula: n = [(l − a) ÷ d] + 1.

Question 207

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

Two containers of the same volume are 40% and 50% full of water, respectively. They are then filled completely with milk. If the contents of both containers are mixed in a larger vessel, what is the percentage of milk in the final mixture?

  1. AOption (a)

    0.45

  2. BOption (b)

    0.5

  3. COption (c)

    0.55

  4. DOption (d)

    0.6

View answer & explanation

Correct answer

Answer: (c)C.

0.55

Why this answer is correct

Shortcut Trick

Let the volume of each container be 100 units.

Container 1 has 40 units water ⇒ Milk filled = 100 − 40 = 60 units.

Container 2 has 50 units water ⇒ Milk filled = 100 − 50 = 50 units.

Total milk in mixture = 60 + 50 = 110 units; Total volume = 100 + 100 = 200 units.

Fraction of milk in final mixture = 110 ÷ 200 = 0.55.

∴ The correct answer is 0.55 (or 55%).

Alternate Method

Given:

⇒ Volume of Container 1 = Volume of Container 2

⇒ Water in Container 1 = 40%

⇒ Water in Container 2 = 50%

⇒ Remaining capacity of both containers is completely filled with milk.

Formula Used:

⇒ Fraction of Milk = (Total Volume of Milk) ÷ (Total Volume of Mixture)

Milk: 60Water: 40Vol: 100+Milk: 50Water: 50Vol: 100Total Milk: 110(55%)Total Water: 90(45%)Final Vol: 200Mixture Process

Calculations:

⇒ Let the capacity of each container be 100 units.

⇒ Volume of water in Container 1 = 40% of 100 = 40 units

⇒ Volume of milk added in Container 1 = 100 − 40 = 60 units

⇒ Volume of water in Container 2 = 50% of 100 = 50 units

⇒ Volume of milk added in Container 2 = 100 − 50 = 50 units

⇒ Total volume of milk in final mixture = 60 + 50 = 110 units

⇒ Total volume of the final mixture = 100 + 100 = 200 units

⇒ Fraction of milk in final mixture = 110 ÷ 200

⇒ Fraction = 11 ÷ 20 = 0.55

∴ The correct answer is 0.55 (or 55%).

Additional Information

Mixtures & Alligations

When mixing two quantities, the ratio of their volumes directly determines the weighted average of any concentration present in them.

Weighted Average Formula

Final Concentration = (C1V1 + C2V2) ÷ (V1 + V2), where C is concentration and V is volume.

Successive Replacement

If a fraction of a mixture is removed and replaced with a pure component iteratively, the remaining original component is calculated as Final = Initial × (1 − Removed/Total)n.

Question 208

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A stationery shop ordered 6 dozen red pens and some additional dozen blue pens. The price of the red pens per dozen was three times that of the blue ones. When the order was delivered, it was found that the number of dozens of the two colours had been interchanged. This increased the bill by 40%. Find the ratio of the number of dozens of red pens to the number of dozens of blue pens in the original order.

  1. AOption (a)

    1:1

  2. BOption (b)

    1:2

  3. COption (c)

    2:3

  4. DOption (d)

    3:4

View answer & explanation

Correct answer

Answer: (b)B.

1:2

Why this answer is correct

Shortcut Trick

Assume the price of Blue pens = 1 unit, Red pens = 3 units, and let Blue pens ordered = B dozens.

Original Bill = (6 × 3) + (B × 1) = 18 + B

Interchanged Bill = (B × 3) + (6 × 1) = 3B + 6

The new bill is 40% more, meaning it is 1.4 times the original bill.

⇒ 3B + 6 = 1.4 × (18 + B) ⇒ 1.6B = 19.2 ⇒ B = 12

∴ The correct answer is 1:2.

Alternate Method

Original BillRed : 6 × 3 = 18Blue : B × 1 = BTotal = 18 + BNew BillRed : B × 3 = 3BBlue : 6 × 1 = 6Total = 3B + 6+ 40% Increase

Given:

Number of red pens originally ordered = 6 dozens.

Let the number of blue pens originally ordered be B dozens.

Let the price of 1 dozen blue pens be x. Then, the price of 1 dozen red pens = 3x.

Calculations:

⇒ Original Bill = (6 × 3x) + (B × x) = 18x + Bx = x(18 + B)

⇒ When the quantities of the two colours are interchanged, Red = B dozens and Blue = 6 dozens.

⇒ New Bill = (B × 3x) + (6 × x) = 3Bx + 6x = x(3B + 6)

⇒ According to the problem, the new bill is 40% more than the original bill.

⇒ New Bill = Original Bill + 40% of Original Bill

⇒ x(3B + 6) = x(18 + B) × 1.4

⇒ Since price x is a positive value, we can divide both sides by x:

⇒ 3B + 6 = 1.4 × (18 + B)

⇒ 3B + 6 = 25.2 + 1.4B

⇒ 3B − 1.4B = 25.2 − 6

⇒ 1.6B = 19.2

⇒ B = 19.2 ÷ 1.6 = 12

⇒ The original order had 6 dozens of red pens and 12 dozens of blue pens.

⇒ Required Ratio = Red pens : Blue pens = 6 : 12 = 1 : 2.

∴ The correct answer is 1:2.

Additional Information

Percentage Increase Formula

If a value is increased by p%, the final value can directly be computed as Original Value × (1 + p/100).

Total Value Relationship

Total value of any item is given by Quantity × Unit Price. Any changes in quantities or unit prices directly impact the proportionate bill linearly.

Cross-Multiplication in Equations

To simplify decimal coefficients quickly, you can multiply the entire equation by a power of 10. For instance, 1.6B = 19.2 transforms immediately into 16B = 192.

Question 209

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A flower vendor bought 200 flowers for ₹800. 20 flowers withered and became unsellable. At what price per flower should he sell the remaining flowers to earn a profit of 30% on his total cost?

  1. AOption (a)

    ₹7.85

  2. BOption (b)

    ₹6.54

  3. COption (c)

    ₹5.78

  4. DOption (d)

    ₹3.82

View answer & explanation

Correct answer

Answer: (c)C.

₹5.78

Why this answer is correct

Shortcut Trick

Total Selling Price required for a 30% profit = 130% of ₹800 = ₹1040.

Total sellable flowers = 200 − 20 = 180 flowers.

Selling Price per flower = 1040 ÷ 180 = 52 ÷ 9 ≈ ₹5.78.

∴ The correct answer is ₹5.78.

Alternate Method

Given:

Total cost price of 200 flowers = ₹800

Number of withered flowers = 20

Desired profit percentage = 30%

Formula Used:

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

SP per unit = Total SP ÷ Total units available for sale

CP = ₹800 (200 flowers)

+30% Profit

Total SP = ₹1040 (130% of CP)

÷ 180 flowers

(200 − 20 withered)

SP / flower = ₹5.78

Calculations:

⇒ Total Selling Price = 800 × (100 + 30) ÷ 100

⇒ Total Selling Price = 800 × 130 ÷ 100 = ₹1040

⇒ Number of flowers available to sell = 200 − 20 = 180

⇒ Selling Price per flower = Total Selling Price ÷ 180

⇒ Selling Price per flower = 1040 ÷ 180

⇒ Selling Price per flower = 52 ÷ 9

⇒ Selling Price per flower = 5.777... ≈ ₹5.78

∴ The correct answer is ₹5.78.

Additional Information

Profit and Loss Percentage

Profit% or Loss% is strictly calculated based on the Total Cost Price (CP) of the initial quantity bought, unless stated otherwise. Formula: Profit% = (Profit ÷ CP) × 100.

Effective Cost Recovery

When a fraction of goods gets damaged or lost, the desired revenue (Total SP) must be fully recovered by selling only the remaining intact goods. This inherently increases the required Selling Price per unit.

Marked Price and Discount

If a vendor wants to offer discounts while maintaining a fixed profit rate, the Marked Price must be calculated above this calculated Selling Price (SP). Selling Price = Marked Price − Discount.

Question 210

SSC Combined Graduate Level (CGL) 2025• Quantitative Aptitude• Quantitative Aptitude

A furniture store sells a dining table for ₹Z, making a 25% profit. During a festive season, they increase the marked price of the same table to ₹1.6Z. They then offer a special discount of 20% on this increased marked price. What will be the percentage profit made by the store during the festive season?

  1. AOption (a)

    0.26

  2. BOption (b)

    0.56

  3. COption (c)

    0.62

  4. DOption (d)

    0.6

View answer & explanation

Correct answer

Answer: (d)D.

0.6

Why this answer is correct

Shortcut Trick

Given the initial Selling Price is ₹Z and the profit is 25%, the Cost Price (CP) = Z ÷ 1.25 = 0.8Z.

During the festive season, the new Marked Price is ₹1.6Z.

With a 20% discount on the Marked Price, the new SP = 1.6Z × (1 − 0.20) = 1.28Z.

Profit amount = New SP − CP = 1.28Z − 0.8Z = 0.48Z.

Percentage profit (as a fraction) = 0.48Z ÷ 0.8Z = 0.6.

∴ The correct answer is 60% (0.6).

Alternate Method

Given:

Initial Selling Price = ₹Z

Initial Profit = 25%

New Marked Price (MP) = ₹1.6Z

Festive Discount = 20%

Formula Used:

Cost Price (CP) = SP ÷ (1 + Profit/100)

Selling Price (SP) = MP × (1 − Discount/100)

Percentage Profit = [(SP − CP) ÷ CP] × 100

CP = 0.8ZSP = 1.28ZMP = 1.6Z−20% DiscountProfit = +0.48Z

Calculations:

⇒ First, calculate the original Cost Price (CP):

⇒ CP × (1 + 0.25) = Z

⇒ 1.25 × CP = Z

⇒ CP = Z ÷ 1.25 = 0.8Z

⇒ Next, calculate the new Selling Price after applying the 20% discount on the new MP:

⇒ New SP = 1.6Z × (1 − 0.20)

⇒ New SP = 1.6Z × 0.80 = 1.28Z

⇒ Calculate the profit earned during the festive season:

⇒ Profit = New SP − CP = 1.28Z − 0.8Z = 0.48Z

⇒ Calculate the profit fraction / percentage:

⇒ Profit fraction = Profit ÷ CP = 0.48Z ÷ 0.8Z = 0.6

⇒ This translates to exactly 60% profit.

∴ The correct answer is 60% (0.6).

Additional Information

Marked Price and Discount

A discount is strictly calculated on the Marked Price (MP), not the Cost Price. The Selling Price is derived directly by deducting the discount: SP = MP × (1 − Discount%).

Cost Price from Profit Percentage

When the final Selling Price and Profit% are provided, the Cost Price can be efficiently found using the formula: CP = SP × [100 ÷ (100 + Profit%)].

Successive Price Changes

When an article's price is marked up and then followed by a discount, a chain calculation or a successsive percentage formula [a + b + (a × b ÷ 100)] can be used to track the overall profit or loss.