Quantitative Aptitude PYQs

Shortcut Trick

Identify Total Distance = 90 + 120 + 150 = 360 km.

Calculate Total Time = (90 ÷ 45) + (120 ÷ 60) + (150 ÷ 75) = 2 + 2 + 2 = 6 hours.

Average Speed = Total Distance ÷ Total Time = 360 ÷ 6 = 60 km/h.

∴ The correct answer is 60 km/h.

Alternate Method

Given:

Distance₁ (d₁) = 90 km, Speed₁ (s₁) = 45 km/h

Distance₂ (d₂) = 120 km, Speed₂ (s₂) = 60 km/h

Distance₃ (d₃) = 150 km, Speed₃ (s₃) = 75 km/h

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Time taken for first part (t₁) = 90 ÷ 45 = 2 hours

⇒ Time taken for second part (t₂) = 120 ÷ 60 = 2 hours

⇒ Time taken for third part (t₃) = 150 ÷ 75 = 2 hours

⇒ Total Distance = 90 + 120 + 150 = 360 km

⇒ Total Time = 2 + 2 + 2 = 6 hours

⇒ Average Speed = 360 ÷ 6 = 60 km/h

∴ The correct answer is 60 km/h.

Additional Information

Average Speed for Equal Distances

If a journey is divided into two equal distances with speeds x and y, the Average Speed = 2xy ÷ (x + y).

Unit Conversion

To convert speed from km/h to m/s, multiply by 5/18. To convert from m/s to km/h, multiply by 18/5.

Relative Speed

When two objects move in the same direction, relative speed is s₁ − s₂. In opposite directions, it is s₁ + s₂.

Shortcut Trick

Sum both equations to find the cost of 8 chairs and 8 tables.

(5C + 3T) + (3C + 5T) = 12,500 + 15,500 ⇒ 8C + 8T = ₹28,000.

To find the cost of 2 chairs and 2 tables, divide the total sum by 4.

Cost = 28,000 ÷ 4 = 7,000.

∴ The correct answer is ₹7,000.

Alternate Method

Given:

Cost of 5 chairs and 3 tables = ₹12,500

Cost of 3 chairs and 5 tables = ₹15,500

Formula Used:

Total Cost = (Cost per unit × Quantity)

Calculations:

Let the cost of one chair be C and one table be T.

⇒ 5C + 3T = 12,500 — (Equation 1)

⇒ 3C + 5T = 15,500 — (Equation 2)

Adding Equation 1 and Equation 2:

⇒ (5C + 3C) + (3T + 5T) = 12,500 + 15,500

⇒ 8C + 8T = 28,000

⇒ 8 × (C + T) = 28,000

⇒ C + T = 28,000 ÷ 8

⇒ C + T = 3,500

We need the cost of 2 chairs and 2 tables:

⇒ 2 × (C + T) = 2 × 3,500

⇒ 2C + 2T = 7,000

∴ The correct answer is ₹7,000.

Additional Information

Linear Equations in Two Variables

Equations in the form ax + by = c can be solved using elimination, substitution, or cross-multiplication methods.

Symmetric Equation Property

When coefficients are swapped (ax + by and bx + ay), always add and subtract the equations to simplify the system quickly.

Unitary Concept

Finding the combined value of one unit of each item (C + T) often allows solving for any multiple (nC + nT) without finding individual prices.

Given:

(((16/21))) / (((16/4))) ÷ ((4/5 × 10/12) + 4/7) + 4/8 ÷ (26/14 of 14/8)

Formula used:

The VBODMAS rule: Brackets, Of, Division, Multiplication, Addition, Subtraction.

Calculations:

Simplify the first complex fraction term:

⇒ (16/21) / (16/4) = 16/21 × 4/16 = 4/21

Simplify the expression inside the brackets:

⇒ (4/5 × 10/12 + 4/7) = (2/3 + 4/7) = (14 + 12) / 21 = 26/21

Solve the Of operation (which takes priority over division):

⇒ 26/14 of 14/8 = (26/14) × (14/8) = 26/8 = 13/4

Now, substitute these simplified values back into the full expression:

⇒ (4/21 ÷ 26/21) + (4/8 ÷ 13/4)

⇒ (4/21 × 21/26) + (4/8 × 4/13)

⇒ 4/26 + 16/104

⇒ 2/13 + 2/13

⇒ 4/13

∴ The correct answer is option (4).

Shortcut Trick

The numbers are in continued proportion, so the ratio 16 : 24 must equal the ratio 24 : third number.

Simplify the first ratio: 16 : 24 = 2 : 3.

This means the second ratio must also be 2 : 3.

If 2 units = 24, then 1 unit = 12.

Therefore, 3 units = 12 × 3 = 36.

∴ The correct answer is 36.

Alternate Method

Given: First number = 16, Second number = 24.

Formula Used: For three numbers a, b, and c in continued proportion, a ÷ b = b ÷ c.

Calculations:

⇒ Let the third number be c.

⇒ 16 ÷ 24 = 24 ÷ c

⇒ (16 × c) = (24 × 24)

⇒ 16c = 576

⇒ c = 576 ÷ 16

⇒ c = 36

∴ The correct answer is 36.

Additional Information

Mean Proportional

If a, b, and c are in continued proportion, then b is called the mean proportional between a and c, where b = √(a × c).

Third Proportional

In the proportion a : b = b : c, the value c is known as the third proportional to a and b, calculated as c = b² ÷ a.

Fourth Proportional

For four numbers a, b, c, and d in proportion (a : b = c : d), d is the fourth proportional, where d = (b × c) ÷ a.

Shortcut Trick

The Mean Proportional between a and b is √(a × b).

Given Mean Proportional = 12, numbers are x and 36.

Square of Mean = Product of numbers ⇒ 12² = x × 36.

144 = 36 × x ⇒ x = 144 ÷ 36 = 4.

∴ The correct answer is 4.

Alternate Method

Given:

First term = x

Mean proportional (Middle term) = 12

Third term = 36

Formula Used:

In a Continued Proportion (a : b :: b : c), the relation is b² = a × c.

Calculations:

⇒ x / 12 = 12 / 36

⇒ x × 36 = 12 × 12

⇒ 36x = 144

⇒ x = 144 / 36

⇒ x = 4

∴ The correct answer is 4.

Additional Information

Mean Proportional

For any two numbers a and c, the mean proportional b is calculated as b = √(ac).

Third Proportional

In a continued proportion a : b = b : c, the term c is the third proportional, where c = b² / a.

Fourth Proportional

For four quantities a, b, c, d in proportion, a/b = c/d, the fourth proportional is d = (b × c) / a.

Shortcut Trick

Let the Original Price be 100.

Step 1: 20% increase → 100 + 20 = 120.

Step 3: Final price becomes 100 after a 10% increase of the Original Price (+10).

This means the price before Step 3 was 100 − 10 = 90.

Step 2: Value of x is the percentage decrease from 120 to 90 ⇒ (30 ÷ 120) × 100 = 25.

∴ The correct answer is 25.

Alternate Method

Given:

Initial Price = 100

First Increase = 20%

Final Increase = 10% of Original Price

Final Price = Original Price = 100

Calculations:

⇒ Let original price = 100

⇒ Price after 20% increase = 100 + 20 = 120

⇒ Price after x% decrease = 120 − (x/100 × 120) = 120 − 1.2x

⇒ Final price after 10% of original price increase = (120 − 1.2x) + 10

⇒ Given final price = 100

⇒ 130 − 1.2x = 100

⇒ 1.2x = 30

⇒ x = 30 ÷ 1.2 = 25

∴ The correct answer is 25.

Additional Information

Successive Percentage Change

When a value is changed by a% and then by b%, the net effective change is given by the formula: [a + b + (ab ÷ 100)]%.

Base Value Concept

Percentage calculations always depend on the base value. In this problem, x% was calculated on the new price (120), while the 10% increase was calculated on the original price (100).

Percentage to Fraction Conversion

Common fractions help in fast calculation: 25% = 1/4, 20% = 1/5, 10% = 1/10.

Shortcut Trick

Assume the car travelled all 9.5 hr at the slower speed of 40 km/h.

Hypothetical distance = 40 × 9.5 = 380 km.

Shortfall in distance = 560 − 380 = 180 km.

Difference in speeds = 160 − 40 = 120 km/h.

Time spent at 160 km/h = 180 ÷ 120 = 1.5 hr.

Distance covered at 160 km/h = 1.5 × 160 = 240 km.

∴ The correct answer is 240 km.

Alternate Method

Given: Total distance = 560 km, Total time = 9.5 hr, Speeds = 40 km/h and 160 km/h.

Formula Used: Time = Distance ÷ Speed

Let the distance travelled at 160 km/h be d km.

The distance travelled at 40 km/h will be (560 − d) km.

⇒ Total Time = Time₁ + Time₂

⇒ 9.5 = (560 − d) ÷ 40 + d ÷ 160

⇒ 9.5 = [4(560 − d) + d] ÷ 160

⇒ 9.5 × 160 = 2240 − 4d + d

⇒ 1520 = 2240 − 3d

⇒ 3d = 2240 − 1520

⇒ 3d = 720

⇒ d = 240 km

∴ The correct answer is 240 km.

Additional Information

Distance, Speed, and Time Relationship

The fundamental relation is Distance = Speed × Time. If speed is constant, distance is directly proportional to time.

Average Speed

Average Speed = Total Distance ÷ Total Time. It is not the simple arithmetic mean of the individual speeds unless the time spent at each speed is equal.

Unit Conversion

To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5.

Alligation Rule for Speed

When alligation is applied to speeds, the resulting ratio is the ratio of the time taken at those respective speeds.

Shortcut Trick

The total surface area of three spheres is given by the formula: 4π(r₁² + r₂² + r₃²).

Sum of squares of radii = 2² + 3² + 6² = 4 + 9 + 36 = 49 cm².

Total Area = 4 × (22/7) × 49 = 88 × 7 = 616 cm².

∴ The correct answer is 616 cm².

Alternate Method

Given:

Radius of first sphere (r₁) = 2 cm

Radius of second sphere (r₂) = 3 cm

Radius of third sphere (r₃) = 6 cm

Value of π = 22/7

Formula Used:

Surface area of a sphere = 4πr²

Calculations:

⇒ Surface area of first sphere = 4 × (22/7) × 2² = 4 × (22/7) × 4 = 352/7 cm²

⇒ Surface area of second sphere = 4 × (22/7) × 3² = 4 × (22/7) × 9 = 792/7 cm²

⇒ Surface area of third sphere = 4 × (22/7) × 6² = 4 × (22/7) × 36 = 3168/7 cm²

⇒ Sum of surface areas = (352/7) + (792/7) + (3168/7)

⇒ Sum of surface areas = (352 + 792 + 3168) ÷ 7

⇒ Sum of surface areas = 4312 ÷ 7

⇒ Sum of surface areas = 616 cm²

∴ The correct answer is 616 cm².

Additional Information

Volume of a Sphere

The volume of a sphere with radius r is given by the formula V = (4/3) × π × r³.

Surface Area of Hemisphere

Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr².

Volume of a Hemisphere

The volume of a hemisphere is exactly half of the sphere, given by V = (2/3) × π × r³.

Given:

55 1/8 + 55 3/8 + 55 5/8 + 55 7/8

Formula used:

A Mixed Fraction is the sum of its Whole Number part and its Fractional Part:

x a/b = x + a/b

Calculations:

55 1/8 + 55 3/8 + 55 5/8 + 55 7/8

⇒ (55 + 1/8) + (55 + 3/8) + (55 + 5/8) + (55 + 7/8)

⇒ (55 + 55 + 55 + 55) + (1/8 + 3/8 + 5/8 + 7/8)

⇒ (55 × 4) + (1 + 3 + 5 + 7) / 8

⇒ 220 + 16 / 8

⇒ 220 + 2

⇒ 222

∴ The correct answer is option (1).

Shortcut Trick

The highest frequency is 13, which corresponds to the modal class 50−55.

Using the mode formula: Mode = l + [ (f₁ − f₀) ÷ (2f₁ − f₀ − f₂) ] × h

Here, l = 50, f₁ = 13, f₀ = 4, f₂ = 5, and h = 5.

Mode = 50 + [ (13 − 4) ÷ (26 − 9) ] × 5 = 50 + (9 ÷ 17) × 5 = 50 + 2.65 = 52.65.

∴ The correct answer is 52.65.

Alternate Method

Given Data:

Modal Class (Class with highest frequency) = 50−55

Lower limit of the modal class (l) = 50

Frequency of the modal class (f₁) = 13

Frequency of the class preceding the modal class (f₀) = 4

Frequency of the class succeeding the modal class (f₂) = 5

Class interval width (h) = 5

Calculations:

⇒ Mode = l + [ (f₁ − f₀) ÷ (2 × f₁ − f₀ − f₂) ] × h

⇒ Mode = 50 + [ (13 − 4) ÷ (2 × 13 − 4 − 5) ] × 5

⇒ Mode = 50 + [ 9 ÷ (26 − 9) ] × 5

⇒ Mode = 50 + [ 9 ÷ 17 ] × 5

⇒ Mode = 50 + (45 ÷ 17)

⇒ Mode = 50 + 2.64705...

⇒ Mode ≈ 52.65

∴ The correct answer is 52.65.

Additional Information

Empirical Relationship

For a moderately skewed distribution, the relationship is: Mode = 3 × Median − 2 × Mean.

Median for Grouped Data

Median = l + [ (N÷2 − cf) ÷ f ] × h, where cf is the cumulative frequency of the preceding class.

Modal Class Definition

It is the class interval which has the maximum frequency in a grouped frequency distribution.

polygon, the sum of all interior angles is given by (n − 2) × 180°ree;.

 

Cyclic Quadrilateral Property

If all vertices of a quadrilateral lie on a circle, the sum of opposite angles is 180°ree;.

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Shortcut Trick

In a standard 30°−60°−90° triangle, the side ratio is 1 : √3 : 2.

The tower height (side opposite to 60°) corresponds to √3 units = 18 m.

The horizontal distance (side opposite to 30°) corresponds to 1 unit.

1 unit = 18 ÷ √3 = (18 × √3) ÷ 3 = 6√3 m.

∴ The correct answer is 6√3 m.

Alternate Method

Given:

Height of the vertical tower (AB) = 18 m

Angle of elevation (θ) = 60°

Horizontal distance between person and tower = BC

Formula Used:

tan θ = Opposite Side / Adjacent Side

Calculations:

⇒ tan 60° = AB / BC

⇒ √3 = 18 / BC

⇒ BC = 18 / √3

⇒ BC = (18 × √3) / (√3 × √3)

⇒ BC = 18√3 / 3

⇒ BC = 6√3 m

∴ The correct answer is 6√3 m.

Additional Information

Trigonometric Values

Common values include tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3.

Angle of Elevation

The angle between the horizontal plane and the line of sight when an observer looks at an object above their eye level.

Rationalization

To simplify a fraction with a radical denominator, multiply the numerator and denominator by the radical: a/√b = (a√b)/b.

Given:

(2 + 3/2)² − [4 − (6/3)²] ÷ 1/2

Formula used:

The BODMAS rule is used to solve the expression: Brackets, Orders (powers), Division, Multiplication, Addition, and Subtraction.

Calculation:

⇒ (7/2)² − [4 − 2²] ÷ 1/2

⇒ 49/4 − [4 − 4] ÷ 1/2

⇒ 49/4 − 0 ÷ 1/2

⇒ 49/4 − 0

⇒ 49/4

∴ The correct answer is option (3).

Shortcut Trick

Numbers are in Ratio 9 : 7 : 5, which are co-prime to each other.

Shortcut formula: LCM = (Product of ratio parts) × HCF

2205 = (9 × 7 × 5) × HCF

2205 = 315 × HCF

HCF = 2205 ÷ 315 = 7

∴ The correct answer is 7.

Alternate Method

Given:

Ratio of three numbers = 9 : 7 : 5

LCM (Least Common Multiple) = 2205

Formula Used:

If numbers are ax, bx, and cx where a, b, and c are co-prime, then HCF = x and LCM = (a × b × c) × x.

Calculations:

⇒ Let the three numbers be 9x, 7x, and 5x.

⇒ Since 9, 7, and 5 have no common factors other than 1, they are co-prime.

⇒ HCF of (9x, 7x, 5x) = x

⇒ LCM of (9x, 7x, 5x) = 9 × 7 × 5 × x = 315x

⇒ 315x = 2205

⇒ x = 2205 ÷ 315

⇒ x = 7

⇒ HCF = 7

∴ The correct answer is 7.

Additional Information

Product of Two Numbers

For any two numbers A and B, A × B = HCF(A, B) × LCM(A, B).

Co-prime Numbers

Numbers that have only 1 as their common factor are called co-prime; their HCF is always 1.

LCM and HCF Relationship

The HCF of a set of numbers is always a factor of their LCM (LCM ÷ HCF is always an integer).

The solution is based on the relationship between Cost Price (CP), Selling Price (SP), and Profit Percentage.

Initial Case:

Let the Original Cost Price (CP₁) of the article be 100x.

Given the man sells it at a profit of 15%:

Original Selling Price (SP₁) = 100x + (15% of 100x) = 115x

New Scenario:

If the article was bought at 25% less cost price and sold for ₹294 less than the original selling price, the details change as follows:

In this new scenario, the Profit made is 30%.

We can express the New Selling Price in terms of the New Cost Price and Profit:

SP₂ = CP₂ + 30% of CP₂

SP₂ = 75x + (30/100) × 75x

SP₂ = 75x + 22.5x = 97.5x

Equating the New Selling Price values:

Now, we equate the two different expressions for SP₂ to find the value of x:

115x − 294 = 97.5x

115x − 97.5x = 294

17.5x = 294

x = 294 / 17.5

x = 16.8

Calculating the Original Cost Price:

The Original Cost Price (CP₁) was assumed to be 100x.

CP₁ = 100 × 16.8 = 1,680

The cost price of the article is ₹1,680.

The correct answer is option 4

Shortcut Trick

For divisibility by 2, the units digit Q must be an even number (0, 2, 4, 6, or 8).

For divisibility by 3, the sum of digits (3 + P + 2 + Q) = 5 + P + Q must be a multiple of 3.

The smallest multiple of 3 greater than or equal to 5 is 6.

Setting 5 + P + Q = 6 ⇒ P + Q = 1.

With Q = 0 (even) and P = 1, both conditions are satisfied, so the smallest value is 1.

∴ The correct answer is 1.

Alternate Method

Given: A four-digit number 3P2Q is divisible by both 2 and 3.

Formula Used:

1. Divisibility by 2: Last digit must be even (0, 2, 4, 6, 8).
2. Divisibility by 3: Sum of all digits must be divisible by 3.

Calculations:

⇒ For the number to be divisible by 2, Q ∈ {0, 2, 4, 6, 8}.

⇒ For the number to be divisible by 3, Sum = 3 + P + 2 + Q = 5 + P + Q must be a multiple of 3.

⇒ Possible multiples of 3 are 6, 9, 12, 15, ...

⇒ To minimize P + Q, we test the smallest possible sum: 5 + P + Q = 6.

⇒ P + Q = 6 − 5 = 1.

⇒ Now, check if there exist digits P and Q such that P + Q = 1 and Q is even.

⇒ If Q = 0 (even) and P = 1, then P + Q = 1.

⇒ The resulting number 3120 is divisible by both 2 and 3.

∴ The correct answer is 1.

Additional Information

Divisibility by 6

A number is divisible by 6 if it satisfies the divisibility rules for both 2 (even units digit) and 3 (sum of digits divisible by 3).

Divisibility by 4

A number is divisible by 4 if the number formed by its last two digits is divisible by 4 (e.g., in 3P2Q, the number 2Q must be divisible by 4).

Divisibility by 9

A number is divisible by 9 if the sum of all its digits is divisible by 9.

Shortcut Trick

Compare the Effective Discount for 5 toys in each scheme:

Scheme A: 24 + 33 − (24 × 33 ÷ 100) = 49.08%

Scheme B: Successive 39%, 37%, 23% → Obviously higher than 50% (approx 70.4%)

Scheme C: (2 × 24% + 3 × 12%) ÷ 5 = (48 + 36) ÷ 5 = 16.8%

Scheme D: 2 ÷ (5 + 2) × 100 ≈ 28.57%

Comparing all, 16.8% is the minimum discount.

∴ The correct answer is C.

Alternate Method

Assume the Marked Price (MP) of each toy = ₹100. For 5 toys, total MP = ₹500.

Scheme A (Successive 24% and 33%):

⇒ Net Discount % = 24 + 33 − (24 × 33)/100 = 57 − 7.92 = 49.08%

Scheme B (Successive 39%, 37% and 23%):

⇒ After 39% and 37%: 39 + 37 − (39 × 37)/100 = 76 − 14.43 = 61.57%

⇒ Net Discount % = 61.57 + 23 − (61.57 × 23)/100 = 84.57 − 14.16 = 70.41%

Scheme C (24% on first 2, 12% on rest):

⇒ Total Discount = (2 × 24) + (3 × 12) = 48 + 36 = ₹84

⇒ Effective Discount % = (84 ÷ 500) × 100 = 16.8%

Scheme D (2 toys free on 5 toys bought):

⇒ Total items = 5 + 2 = 7 toys

⇒ Discount % = (Free items ÷ Total items) × 100 = (2 ÷ 7) × 100 ≈ 28.57%

The Least Beneficial scheme for the customer is the one with the minimum discount.

∴ The correct answer is C.

Additional Information

Successive Discount Formula

For two successive discounts a% and b%, the single equivalent discount is [a + b − (a × b ÷ 100)]%.

Buy X Get Y Free

The effective discount percentage for such schemes is calculated as [Y ÷ (X + Y)] × 100.

Customer vs Shopkeeper Benefit

A customer seeks the maximum discount (Most Beneficial), while the shopkeeper seeks the minimum discount (Least Beneficial to customer).

Shortcut Trick

Use the direct algebraic identity: (Difference)² = (Sum)² − 4 × (Product)

Given Sum = 32 and Product = 220.

⇒ (Difference)² = 32² − 4 × 220

⇒ (Difference)² = 1024 − 880 = 144

⇒ Difference = √144 = 12

∴ The correct answer is 12.

Alternate Method

Let the two positive numbers be x and y.

Given:

1. x + y = 32
2. x × y = 220

Calculations:

⇒ (x + y)² = 32²

⇒ x² + y² + 2xy = 1024

⇒ x² + y² + 2(220) = 1024

⇒ x² + y² = 1024 − 440 = 584

⇒ (x − y)² = x² + y² − 2xy

⇒ (x − y)² = 584 − 440 = 144

⇒ x − y = √144 = 12

∴ The correct answer is 12.

Additional Information

Algebraic Identity for Sum and Difference

The relation between sum, difference, and product is given by the identity: (a + b)² − (a − b)² = 4ab.

Quadratic Equation Roots

If the sum (S) and product (P) of two numbers are known, the numbers are roots of the equation: t² − St + P = 0.

Finding Individual Values

Once the sum (S) and difference (D) are known, the larger number is (S + D) ÷ 2 and the smaller number is (S − D) ÷ 2.

Shortcut Trick

When a value is first increased by x% and then decreased by x%, the net change is always a decrease given by the formula: (x/10)²%.

Here, x = 20.

Net percentage change = (20 / 10)² = 2² = 4% decrease.

Note: The original price (₹500) does not affect the net percentage change.

∴ The correct answer is 4% decrease.

Alternate Method

Given: Original price = ₹500, first increase = 20%, subsequent decrease = 20%.

Formula: New Value = Original Value × (100 + change%)/100

⇒ Original Price = ₹500

⇒ Price after 20% increase = 500 + (20/100) × 500 = 500 + 100 = ₹600

⇒ Price after 20% decrease = 600 − (20/100) × 600 = 600 − 120 = ₹480

⇒ Net change in price = 500 − 480 = ₹20

⇒ Net percentage change = (Net Change / Original Price) × 100

⇒ Net percentage change = (20 / 500) × 100 = 4%

⇒ Since the final price (₹480) is less than the original price (₹500), it is a decrease.

∴ The correct answer is 4% decrease.

Additional Information

Successive Percentage Change Formula

For two successive changes of a% and b%, the effective change is [a + b + (ab/100)]%. Use positive for increase and negative for decrease.

Order of Changes

The final net percentage change remains the same regardless of the order of the percentage increases or decreases (i.e., a 20% increase followed by a 20% decrease is the same as a 20% decrease followed by a 20% increase).

Constant Percentage Rule

If a quantity is increased by x% and then the resulting quantity is decreased by x%, the overall result is always a net decrease of (x/10)²% or x²/100%.

Shortcut Trick

Loss of 12% ⇒ SP is 88% of Cost Price (CP).

Desired Profit of 14% ⇒ New SP should be 114% of CP.

Using the ratio: (1786.40 ÷ 88) × 114

⇒ 20.3 × 114 = ₹2,314.20.

∴ The correct answer is ₹2,314.20.

Alternate Method

Given:

Initial Selling Price (SP₁) = ₹1,786.40

Initial Loss = 12%

Required Profit = 14%

Formula Used:

Cost Price (CP) = SP × [100 ÷ (100 − Loss%)]

New Selling Price (SP₂) = CP × [(100 + Profit%) ÷ 100]

Calculations:

⇒ CP = 1786.40 × [100 ÷ (100 − 12)]

⇒ CP = 1786.40 × (100 ÷ 88)

⇒ CP = 20.3 × 100 = ₹2,030

⇒ New SP = 2030 × [(100 + 14) ÷ 100]

⇒ New SP = 2030 × (114 ÷ 100)

⇒ New SP = 20.3 × 114

⇒ New SP = ₹2,314.20

∴ The correct answer is ₹2,314.20.

Additional Information

Relationship between CP and SP

Profit % = (Profit ÷ CP) × 100, where Profit = SP − CP. Similarly, Loss % = (Loss ÷ CP) × 100.

Direct SP Calculation

When CP is constant, SP₁ ÷ (100 ± Gain/Loss₁) = SP₂ ÷ (100 ± Gain/Loss₂).

Successive Changes

If an article is sold through multiple hands at a profit/loss of x%, y%, and z%, then Final SP = CP × (1 ± x/100) × (1 ± y/100) × (1 ± z/100).