Quantitative Aptitude PYQs

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.

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.

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.

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.

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 = (Capital_A × Time_A) : (Capital_B × Time_B)

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.

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.

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.

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.

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.

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.

Shortcut Trick

Initial Ratio (Acid : Water) = 5 : 3

Final Ratio (Acid : Water) = 5 : 4

Since only water is added, the ratio of acid (5 parts) correctly remains unchanged.

Increase in water ratio = 4 − 3 = 1 part.

1 part = 4 litres ⇒ Original Acid = 5 parts = 5 × 4 = 20 litres.

∴ The correct answer is 20 litres.

Alternate Method

Initial Acid = 5u Water = 3u

Final Acid = 5u Water = 4u

• 4 L Water

Water Change: 4u − 3u = 1u = 4 Litres Original Acid: 5u = 5 × 4 = 20 Litres

Given:

Initial ratio of acid to water = 5 : 3

Water added = 4 litres

New ratio of acid to water = 5 : 4

Calculations:

⇒ Let initial acid = 5x and initial water = 3x.

⇒ After adding 4 litres of water, new quantity of water = 3x + 4.

⇒ New ratio = 5x / (3x + 4) = 5 / 4

⇒ 20x = 15x + 20

⇒ 20x − 15x = 20

⇒ 5x = 20 ⇒ x = 4

⇒ Original quantity of acid = 5x = 5 × 4 = 20 litres.

∴ The correct answer is 20 litres.

Additional Information

Constant Component Concept

When one substance is added to a mixture, the quantity of the other substance remains strictly constant. Equating the ratio parts of the undisturbed component simplifies calculations directly without full equations.

Cross-Multiplication Method

If the initial ratio is a:b and final ratio becomes c:d after adding x and y respectively, the relation is: (a × d) − (b × c) = (y × c) − (x × d). Here, y = 4 and x = 0.

Replacement in Mixtures

When a portion of a mixture is withdrawn without adding anything, the ratio of its components remains identical to the original mixture. The ratio alters only upon adding a new quantity.

Shortcut Trick

Total work = LCM(12, 18, 5) = 180 units.

Efficiency of A = 180 ÷ 12 = 15, B = 180 ÷ 18 = 10, A+B+C = 180 ÷ 5 = 36.

Efficiency of C = 36 − (15 + 10) = 11.

Wage ratio A : B : C = 15 : 10 : 11.

C's share = (11 ÷ 36) × 7200 = ₹2,200.

∴ The correct answer is ₹2,200.

Alternate Method

Given:

Time taken by A alone = 12 days

Time taken by B alone = 18 days

Time taken by A, B, and C together = 5 days

Total wages = ₹ 7,200

Formula Used:

Total Work = LCM of time taken

Efficiency = Total Work ÷ Time

Wages are distributed in the ratio of efficiencies when workers work for the same total time.

Calculations:

⇒ Total efficiency of A, B, and C = 36 units/day

⇒ Efficiency of C = Total efficiency − (Efficiency of A + Efficiency of B)

⇒ Efficiency of C = 36 − (15 + 10) = 11 units/day

⇒ Since they worked together for all 5 days, wages are shared directly in the ratio of their individual efficiencies.

⇒ Ratio of wages A : B : C = 15 : 10 : 11

⇒ Total ratio parts = 15 + 10 + 11 = 36 parts

⇒ 36 parts = ₹ 7,200

⇒ 1 part = 7200 ÷ 36 = ₹ 200

⇒ C's share = 11 parts = 11 × 200 = ₹ 2,200

∴ The correct answer is ₹2,200.

Additional Information

Work and Wages

Wages are strictly proportional to the actual amount of work done by each individual. If workers work together for the same duration, their wages are distributed directly in the ratio of their daily efficiencies.

Different Working Days

If workers work for different numbers of days, their respective wage shares are calculated by multiplying their efficiency with the number of days they worked: Share = Efficiency × Days Worked.

Pipes and Cisterns

This concept mirrors Time and Work exactly. Filling pipes have a positive efficiency (doing work), while emptying pipes or leaks have a negative efficiency (undoing work).

Shortcut Trick

Let remaining mixture components A, B, and C be 7k, 8k, and 5k (Total = 20k).

Final quantity of A = 7k + 15 L and Final quantity of B = 8k.

Given, Final B − Final A = 25 ⇒ 8k − (7k + 15) = 25 ⇒ k = 40.

Total remaining mixture = 20 × 40 = 800 L.

Initial mixture = 800 L + 40 L (drained) = 840 L.

∴ The correct answer is 840 litres.

Alternate Method

Given:

⇒ Initial ratio of A : B : C = 7 : 8 : 5

⇒ Quantity of mixture drained = 40 litres

⇒ Added quantities: 15 litres of A and 5 litres of C

Remaining Mixture Components

7k

• 15 L Solution A

8k Solution B

5k

• 5 L Solution C

Final B − Final A = 8k − (7k + 15) = 25 L

Calculations:

⇒ Let the quantities of A, B, and C in the remaining mixture be 7k, 8k, and 5k respectively.

⇒ Total remaining mixture = 7k + 8k + 5k = 20k

⇒ After addition, Final quantity of A = 7k + 15

⇒ After addition, Final quantity of B = 8k

⇒ After addition, Final quantity of C = 5k + 5

⇒ We are given that Final A is 25 litres less than Final B:

⇒ 8k − (7k + 15) = 25

⇒ 8k − 7k − 15 = 25

⇒ k − 15 = 25

⇒ k = 40

⇒ Total remaining mixture = 20k = 20 × 40 = 800 litres

⇒ Initial quantity of mixture = Remaining mixture + Drained mixture

⇒ Initial quantity of mixture = 800 + 40 = 840 litres

∴ The correct answer is 840 litres.

Additional Information

Mixture Replacement Concept

When a part of a mixture is drained, the remaining mixture still retains the exact same ratio of ingredients as the original mixture before any subsequent pure addition is made.

Successive Replacement

If V is the initial volume of pure liquid, and x volume is replaced by water n times sequentially, the final volume of pure liquid is given by V × (1 − x/V)ⁿ.

Rule of Alligation

It allows calculating the required ratio in which two components of given prices or concentrations must be mixed to obtain a mixture of a desired average price or concentration.

Shortcut Trick

Time ratio of A : B = 3 : 2, so efficiency ratio of A : B = 2 : 3.

Total work = Efficiency of B × Time taken by B = 3 × 18 = 54 units.

Combined efficiency of A and B = 2 + 3 = 5 units/hour.

Time taken together = Total work ÷ Combined efficiency = 54 ÷ 5 = 10.8 hours.

∴ The correct answer is 10.8 hours.

Alternate Method

Given:

Ratio of time taken by A and B = 3 : 2

Time taken by B = 18 hours

Calculations:

⇒ Let the time taken by A be 3x and by B be 2x.

⇒ 2x = 18 hours

⇒ x = 9 hours

⇒ Time taken by A = 3 × 9 = 27 hours.

⇒ Total efficiency of A and B together = 2 + 3 = 5 units/hour.

⇒ Time taken by A and B working together = Total Work ÷ Total Efficiency

⇒ Time taken = 54 ÷ 5 = 10.8 hours.

∴ The correct answer is 10.8 hours.

Additional Information

Time & Efficiency Relationship

Efficiency is inversely proportional to the time taken. If the ratio of time taken by A and B is x : y, then the ratio of their efficiencies is y : x.

Work and Wages

Wages are generally distributed in the ratio of the total work done by each individual. If workers work for the same number of days, the wages are distributed directly in the ratio of their efficiencies.

Pipes & Cisterns

The concept of time and work directly applies to pipes. Inlet pipes have positive efficiency (filling work) while outlet pipes or leaks have negative efficiency (emptying work).

Shortcut Trick

Time taken for one round = Distance ÷ Speed.

T_x = 2/10 = 1/5 h, T_y = 2/12 = 1/6 h, T_z = 2/15 h.

Time to meet at start = LCM(1/5, 1/6, 2/15) = 2/1 = 2 hours.

Total time from 9:00 a.m. to 1:00 p.m. = 4 hours.

Number of meetings = Total time ÷ Meeting interval = 4 ÷ 2 = 2 times.

∴ The correct answer is 2 times.

Alternate Method

Given:

Circumference of the track (D) = 2 km

Speeds: X = 10 km/h, Y = 12 km/h, Z = 15 km/h

Cycling duration = 9:00 a.m. to 1:00 p.m. = 4 hours

Formula Used:

Time = Distance ÷ Speed

Meeting time at start point = LCM of individual times taken for one round

LCM of fractions = LCM of Numerators ÷ HCF of Denominators

Start PointD = 2 kmSpeeds:X = 10 km/hY = 12 km/hZ = 15 km/hTime to meet = LCM of individual times

Calculations:

⇒ T_x = 2 ÷ 10 = 1/5 h

⇒ T_y = 2 ÷ 12 = 1/6 h

⇒ T_z = 2 ÷ 15 = 2/15 h

⇒ Meeting time = LCM(1/5, 1/6, 2/15)

⇒ LCM of numerators (1, 1, 2) = 2

⇒ HCF of denominators (5, 6, 15) = 1

⇒ LCM(1/5, 1/6, 2/15) = 2 ÷ 1 = 2 hours

⇒ They all meet at the starting point every 2 hours.

⇒ Total duration from 9:00 a.m. to 1:00 p.m. = 4 hours.

⇒ Number of meetings = 4 ÷ 2 = 2 times

∴ The correct answer is 2 times.

Additional Information

Meeting at the Starting Point

When multiple runners or objects move around a circular path starting at the same time, the time required for all of them to meet at the starting point simultaneously is always the LCM of the times taken by each to complete exactly one round.

First Meeting on a Circular Track

If two bodies move in the same direction, they will meet for the very first time anywhere on the track after a time = Distance ÷ Relative Speed (Difference of their speeds). If moving in opposite directions, relative speed is the sum of their speeds.

LCM & HCF of Fractions

LCM of fractions = LCM of Numerators ÷ HCF of Denominators.

HCF of fractions = HCF of Numerators ÷ LCM of Denominators.

Shortcut Trick

Area of a circular path = π × w × (2r + w)

Where w = 3 m (width of path) and r = 8 m (inner radius)

Area = π × 3 × (2 × 8 + 3) = π × 3 × 19

Area = 57π ≈ 57 × 3.14 ≈ 179 m²

∴ The correct answer is 179 m².

Alternate Method

Given:

Inner radius of the park (r) = 8 m

Width of the path (w) = 3 m

Outer radius of the park with path (R) = 8 m + 3 m = 11 m

8 m

3 m

11 m

Circular Park with Path

Formula Used:

Area of circular path = Area of outer circle − Area of inner circle

Area = πR² − πr² = π(R² − r²)

Calculations:

⇒ Area = π × (11² − 8²)

⇒ Area = π × (121 − 64)

⇒ Area = 57π

⇒ Area ≈ 57 × 3.14159

⇒ Area ≈ 179.07 m²

By comparing the options, the nearest value is 179 m².

∴ The correct answer is 179 m².

Additional Information

Area of a Circle

The total region enclosed by a circle is given by Area = πr², where r is the radius.

Area of a Circular Path

For a circular path of width w around a circle of radius r, the area is calculated as π(R² − r²) or πw(2r + w), where the outer radius R = r + w.

Circumference

The perimeter or length of the outer boundary of a circle is calculated as C = 2πr.

Shortcut Trick

The slope of the given line in the form y = mx + c is m = 1/4.

The slope of a perpendicular line is always the negative reciprocal of the original line's slope.

Perpendicular slope = −1 ÷ (1/4) = −4.

∴ The correct answer is -4.

Alternate Method

Given:

Equation of the line: y = 1/4x + 7

Formula Used:

For two perpendicular lines with slopes m₁ and m₂: m₁ × m₂ = −1

y = 1/4x + 7 m₁ = 1/4

Perpendicular Line m₂ = -4

Calculations:

⇒ The given line is structured in the standard form y = mx + c, where m represents the slope of the line.

⇒ By comparing the equations directly, the slope of the given line (m₁) = 1/4

⇒ Let the slope of the required perpendicular line be defined as m₂.

⇒ Since the properties of perpendicular lines state that their product is −1, we have m₁ × m₂ = −1

⇒ (1/4) × m₂ = −1

⇒ m₂ = −1 × 4

⇒ m₂ = −4

∴ The correct answer is -4.

Additional Information

Slope-Intercept Form

The equation of a straight line is typically expressed as y = mx + c, where m is the slope defining the steepness and c is the y-intercept.

Parallel Lines

If two straight lines are distinctly parallel to each other, their slopes will always be exactly equal (m₁ = m₂).

Perpendicular Lines

When two non-vertical straight lines intersect perfectly at a right angle, the resulting product of their slopes evaluates to −1 (m₁ × m₂ = −1).

Shortcut Trick

We know the standard conversion where 1 radian is approximately equal to 57.3°.

To convert 2.5 radians to degrees, we simply multiply the value by 57.3°.

⇒ 2.5 × 57.3° = 143.25°

Looking at the given options, the closest exact value is 143.24°.

∴ The correct answer is 143.24°.

Alternate Method

Given:

Angle = 2.5 radians

Formula Used:

Angle in Degrees = Angle in Radians × (180° ÷ π)

2.5 rad 143.24°

Calculation:

⇒ Angle in Degrees = 2.5 × (180° ÷ π)

⇒ Angle in Degrees = 450° ÷ π

Using the precise value of π ≈ 3.14159:

⇒ Angle in Degrees = 450° ÷ 3.14159

⇒ Angle in Degrees ≈ 143.2395°

⇒ Rounding off to two decimal places, we get 143.24°.

∴ The correct answer is 143.24°.

Additional Information

Radian to Degree Conversion

To convert any given angle from radians to degrees, multiply the radian value by 180° ÷ π.

Degree to Radian Conversion

To convert an angle from degrees back to radians, multiply the degree value by π ÷ 180°.

Standard Conversions

Some common unit circle angles include: π radians = 180°, π ÷ 2 radians = 90°, and π ÷ 4 radians = 45°.

Shortcut Trick

Direct formula for the area of a sector in radians: Area = ½ × r² × θ

Substitute the given values: r = 10 cm and θ = π/6.

Area = ½ × (10)² × (π/6) = ½ × 100 × π/6 = 25π / 3

Using π ≈ 3.14, Area = (25 × 3.14) ÷ 3 = 78.5 ÷ 3 ≈ 26.17 cm².

∴ The correct answer is 26.17 cm².

Alternate Method

Given:

Radius of the sector (r) = 10 cm

Central angle (θ) = π/6 radians

O π/6 10 cm 10 cm Radius (r) = 10 cm Angle (θ) = π/6 rad Area = ?

Formula Used:

Area of a sector = ½ × r² × θ (where θ is in radians)

Calculations:

⇒ Area = ½ × (10)² × (π / 6)

⇒ Area = ½ × 100 × (π / 6)

⇒ Area = 50 × π / 6

⇒ Area = 25π / 3

Using π ≈ 3.14:

⇒ Area = (25 × 3.14) / 3

⇒ Area = 78.5 / 3

⇒ Area = 26.166... ≈ 26.17 cm²

∴ The correct answer is 26.17 cm².

Additional Information

Area of a Sector (in degrees)

If the central angle θ is given in degrees instead of radians, the formula to calculate the area becomes Area = (θ ÷ 360°) × πr².

Length of an Arc

The length of the arc (s) is calculated directly as s = rθ (if θ is in radians) or s = (θ ÷ 360°) × 2πr (if θ is in degrees).

Perimeter of a Sector

The total perimeter bounds the sector, including the curved arc length and the two straight radii. It is found using Perimeter = s + 2r.

Shortcut Trick

If tan x = cot y, then their angles are complementary: x + y = 90°

Given: tan A = cot(2A − 30°)

⇒ A + (2A − 30°) = 90°

⇒ 3A = 120°

⇒ A = 40°

∴ The correct answer is 40°.

Alternate Method

Given:

tan A = cot(2A − 30°)

Formula Used:

cot θ = tan(90° − θ)

A

2A − 30° Adjacent to A Opposite to A If tan x = cot y, then x + y = 90°

Calculations:

⇒ tan A = tan(90° − (2A − 30°))

Comparing the angles on both sides:

⇒ A = 90° − (2A − 30°)

⇒ A = 90° − 2A + 30°

⇒ A + 2A = 120°

⇒ 3A = 120°

⇒ A = 40°

∴ The correct answer is 40°.

Additional Information

Complementary Angle Identities

sin(90° − θ) = cos θ | tan(90° − θ) = cot θ | sec(90° − θ) = cosec θ

Supplementary Angle Identities

sin(180° − θ) = sin θ | cos(180° − θ) = −cos θ | tan(180° − θ) = −tan θ

Negative Angle Identities

sin(−θ) = −sin θ | cos(−θ) = cos θ | tan(−θ) = −tan θ