CUET UG Applied Mathematics Booster Test 3 - Quantification and Mixture Applications -

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CUET UG Applied Mathematics Booster Test 3 - Quantification and Mixture Applications

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 3 - Quantification and Mixture Applications

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QUESTION 1 OF 20

When mixing a 40% jaggery syrup with pure jaggery to achieve a 50% jaggery syrup, what is the concentration percentage of the "dearer ingredient" (pure jaggery) used in the alligation calculation?

Answer Review

1 When mixing a 40% jaggery syrup with pure jaggery to achieve a 50% jaggery syrup, what is the concentration percentage of the "dearer ingredient" (pure jaggery) used in the alligation calculation?

A 50%

B 60%

C 100%

D 0%

💡 Short Explanation

�� Pure jaggery means complete concentration. �� Pure substance always has 100% concentration. �� Hence it acts as the dearer ingredient.

✅ Correct Answer Explanation

In alligation: • Cheaper ingredient = 40% jaggery syrup • Dearer ingredient = pure jaggery Pure jaggery means: Therefore, the dearer ingredient concentration is 100%. Option A is the mean concentration, not the dearer ingredient. Option B is unrelated to the given mixture. Option D represents water or absence of jaggery.

❓ Why Other Options Are Incorrect

�� Option A → Represents target concentration, not pure jaggery.
�� Option B → No such concentration is mentioned.
�� Option D → Indicates zero jaggery content.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret the meaning of "pure jaggery" directly from concentration terminology.

Final Logic: Pure substance always means 100% concentration.

🧠 Memory Trick

"Pure means 100%."

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Mixture Concentration

2 Match the problem scenarios to their corresponding formula applications.

List 1	List 2
1. Diluting honey costing Rs 240/L with water	a. x(1 - y/x)^n
2. Replacing 5 L juice with water from a 50 L container 5 times	b. (1/A + 1/B - 1/C)
3. A boat moving with the flow of the river	c. (x + y)
4. Two pipes filling a tank and one emptying it	d. c = 0, d = 240

A 1-b, 2-a, 3-d, 4-c

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-c, 3-b, 4-a

D 1-d, 2-a, 3-c, 4-b

💡 Short Explanation

�� Each aptitude problem has a standard formula. �� Replacement problems use the exponential reduction formula. �� Boats and streams use addition or subtraction of speeds.

✅ Correct Answer Explanation

Match each situation with its appropriate formula. • Diluting honey costing Rs 240/L with water: In alligation, water is treated as the cheaper component with cost c = 0, while honey costs d = 240. Therefore, 1 → d. • Replacing 5 L of juice with water repeatedly: The quantity of the original liquid left after repeated replacement is x(1 - y/x)^n. Therefore, 2 → a. • Boat moving with the flow of the river: Downstream speed is Boat speed + Stream speed = x + y. Therefore, 3 → c. • Two pipes filling a tank and one emptying it: The combined rate is 1/A + 1/B - 1/C. Therefore, 4 → b. Thus, the correct matching is 1 → d 2 → a 3 → c 4 → b Hence, Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because it assigns the pipe formula to the honey dilution problem.
�� Option B → Incorrect because the honey dilution scenario does not use the replacement formula.
�� Option C → Incorrect because it interchanges the boat and replacement formulas.

🎯 Strategy

Used: Option Grouping

Application:

Identify each standard aptitude formula first, then match it with the corresponding problem.

Final Logic:

Only Option D correctly matches all four scenarios.

🧠 Memory Trick

Replacement → Power formula.

📚 Class XI – Mathematics – Chapter: Mathematical Reasoning – Topic: Formula-Based Problem Solving

3 Which of the following mathematical properties fundamentally validate the mean price (m) in the rule of alligation?
1. Quantity of Cheaper / Quantity of Dearer = (d - m) / (m - c)
2. m = (Quantity of Cheaper × c + Quantity of Dearer × d) / Total Quantity
3. m must lie strictly between c and d.
4. m is independent of the quantities mixed.

A 1, 2

B 2, 3

C 1, 2, 3

D 1, 3, 4

💡 Short Explanation

�� Alligation is based on weighted averages. �� The mean price always lies between the cheaper and dearer prices. �� The quantities mixed directly determine the mean price.

✅ Correct Answer Explanation

Evaluate each statement. • Statement 1: The rule of alligation states Quantity of Cheaper / Quantity of Dearer = (d - m) / (m - c). Hence, Statement 1 is correct. • Statement 2: The mean price is the weighted average: m = (Quantity of Cheaper × c + Quantity of Dearer × d) / (Quantity of Cheaper + Quantity of Dearer). Hence, Statement 2 is correct. • Statement 3: For a valid mixture, c < m < d. Therefore, the mean price always lies between the cheaper and dearer prices. Hence, Statement 3 is correct. • Statement 4: The mean price depends on the quantities of the two ingredients mixed. Therefore, it is not independent of the quantities. Hence, Statement 4 is incorrect. Thus, Statements 1, 2 and 3 are correct. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because it omits the valid Statement 3.
�� Option B → Incorrect because it omits the valid Statement 1.
�� Option D → Incorrect because Statement 4 is false.

🎯 Strategy

Used: Elimination

Application:

Identify the statement that contradicts the weighted average principle and eliminate all options containing it.

Final Logic:

Since the mean price depends on the quantities mixed, Statement 4 is false, leaving Option C.

🧠 Memory Trick

Mean always lies between the two prices.

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Weighted Mean

4 Identify the INCORRECT statement regarding the inverse proportion formula in alligation.

A The ratio of the quantities is directly proportional to the actual cost prices.

B Quantity of dearer ingredient corresponds to the cross difference (m - c).

C Quantity of cheaper ingredient corresponds to the cross difference (d - m).

D The diagrammatic method visually represents this inverse relationship via cross-subtraction.

💡 Short Explanation

�� Alligation follows inverse proportionality. �� Cross differences determine the quantity ratio. �� Direct proportionality is not used.

✅ Correct Answer Explanation

In the rule of alligation, Quantity of Cheaper / Quantity of Dearer = (d - m) / (m - c), where c = Cost of the cheaper ingredient, d = Cost of the dearer ingredient, m = Mean cost. This shows that the quantities are determined by the cross differences, not by the actual prices themselves. • Option A is incorrect because the ratio of quantities is not directly proportional to the cost prices. It depends on the cross differences and represents an inverse relationship. • Option B is correct because the quantity of the dearer ingredient corresponds to the cross difference (m - c). • Option C is correct because the quantity of the cheaper ingredient corresponds to the cross difference (d - m). • Option D is correct because the alligation diagram uses cross-subtraction to represent this inverse relationship visually. Hence, Option A is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option B → Correct because the quantity of the dearer ingredient is represented by the cross difference (m - c).
�� Option C → Correct because the quantity of the cheaper ingredient is represented by the cross difference (d - m).
�� Option D → Correct because the alligation diagram is based on the cross-subtraction method.

🎯 Strategy

Used: Extreme Word Filter

Application:

The phrase "directly proportional" contradicts the basic principle of alligation, which uses inverse proportionality.

Final Logic:

Alligation always determines quantities through inverse proportionality using cross differences.

🧠 Memory Trick

Higher price → Lower quantity.

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Alligation Principle

5 A shopkeeper has 1 quintal of wheat, part of which she sells at 18% gain and the rest at 28% gain. In total, she gains 24%. What is the quantity of wheat sold at 18% gain?

A 40 kg

B 60 kg

C 50 kg

D 30 kg

💡 Short Explanation

�� Use the alligation rule on the profit percentages. �� The total quantity is 100 kg. �� Divide the wheat according to the alligation ratio.

✅ Correct Answer Explanation

Given: Lower gain = 18% Higher gain = 28% Mean gain = 24% Using the rule of alligation, Quantity at 18% gain : Quantity at 28% gain = (28 - 24) : (24 - 18) = 4 : 6 = 2 : 3. The total quantity of wheat is 1 quintal = 100 kg. Therefore, Quantity sold at 18% gain = (2 / (2 + 3)) × 100 = (2 / 5) × 100 = 40 kg. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because 60 kg is the quantity sold at 28% gain.
�� Option C → Incorrect because it assumes equal quantities, which does not satisfy the overall gain of 24%.
�� Option D → Incorrect because it does not satisfy the alligation ratio of 2 : 3.

🎯 Strategy

Used: Substitution

Application:

Use the alligation formula to obtain the ratio and then divide the total quantity accordingly.

Final Logic:

The ratio 2 : 3 gives 40 kg at 18% gain and 60 kg at 28% gain.

🧠 Memory Trick

Cross differences give the quantity ratio.

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Weighted Average and Alligation

6 In a mixture weighing 600 gm containing 40% jaggery, pure jaggery is added to achieve 50% concentration. Evaluating the constraints (m=50, c=40, d=100), what ratio is formulated for cheaper to dearer ingredients?

A 5:1

B 1:5

C 50:10

D 4:1

💡 Short Explanation

�� Apply alligation formula. �� Use concentration values directly. �� Simplify obtained ratio.

✅ Correct Answer Explanation

Using alligation: Thus cheaper : dearer = 5:1. Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverse ratio.
�� Option C → Unsimplified form.
�� Option D → Incorrect simplification.

🎯 Strategy

Used: Substitution

Application: Direct substitution into alligation formula avoids confusion.

Final Logic: 50:10 simplifies to 5:1.

🧠 Memory Trick

"Opposite differences form ratio."

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Concentration Mixtures

7 Equating the components of 600 gm of 40% jaggery syrup mixed with pure jaggery (ratio 5:1), how much pure jaggery must be added?

A 100 g

B 120 g

C 150 g

D 200 g

💡 Short Explanation

�� Cheaper to dearer ratio = 5:1. �� 600 g corresponds to 5 parts. �� One part equals 120 g.

✅ Correct Answer Explanation

From previous ratio: 600 g corresponds to 5 parts. Therefore: Thus pure jaggery added = 120 g. Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect part division.
�� Option C → Overestimation.
�� Option D → Does not satisfy ratio condition.

🎯 Strategy

Used: Substitution

Application: Use ratio relation directly with total quantity.

Final Logic: One ratio part equals 120 g.

🧠 Memory Trick

"Total ÷ parts = one part."

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Mixture Ratio Problems

8 Moving to a profit scenario: A retailer has 250 kg of rice. He sells a part at 10% profit and the rest at 5% loss, averaging 7% profit overall. The ratio of selling is 1:4 (loss to profit). What quantity was sold at 5% loss?

A 50 kg

B 200 kg

C 100 kg

D 150 kg

💡 Short Explanation

�� Loss to profit ratio = 1:4. �� Total quantity = 250 kg. �� One-fifth quantity sold at loss.

✅ Correct Answer Explanation

Ratio of quantities: Total parts = 5. Quantity sold at 5% loss: Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents quantity sold at profit.
�� Option C → Assumes wrong ratio.
�� Option D → Incorrect proportional split.

🎯 Strategy

Used: Substitution

Application: Divide total quantity according to given ratio.

Final Logic: One-fifth of 250 equals 50 kg.

🧠 Memory Trick

"Ratio part × total."

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Profit Mixtures

9 Utilizing the probability of remaining juice: A container has 50 L of juice. 5 L is taken out and replaced by water. This process is repeated 4 more times. What is the approximate final amount of juice?

A 32.50 litres

B 29.5 litres

C 25.0 litres

D 35.0 litres

💡 Short Explanation

�� Replacement process repeated 5 times total. �� Remaining fraction each time = 45/50. �� Apply exponential reduction formula.

✅ Correct Answer Explanation

Formula: Here: x = 50, y = 5, n = 5 Thus approximate juice left = 29.5 litres.

❓ Why Other Options Are Incorrect

�� Option A → Underestimates repeated reduction.
�� Option C → Excessive reduction assumption.
�� Option D → Too high after five operations.

🎯 Strategy

Used: Substitution

Application: Apply repeated replacement formula carefully.

Final Logic: Multiplying by (9/10)^5 gives approximately 29.5.

🧠 Memory Trick

"Replacement means repeated multiplication."

📚 Class XI – Mathematics – Chapter: Exponential Functions – Topic: Repeated Reduction

10 If the pure liquid vector equation is defined as x(1 - y/x)^n, what is the computed fraction multiplier (1 - y/x) in the case of a 70 L squash container where 7 L is repeatedly replaced?

A 9/10

B 1/10

C 8/10

D 7/10

💡 Short Explanation

�� Use replacement fraction formula. �� x = 70 and y = 7. �� Simplify remaining fraction.

✅ Correct Answer Explanation

Compute: Simplifying: Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents removed fraction only.
�� Option C → Incorrect subtraction.
�� Option D → Directly uses replacement fraction.

🎯 Strategy

Used: Substitution

Application: Substitute numerical values directly into formula.

Final Logic: Remaining fraction after each operation is 9/10.

🧠 Memory Trick

"Remaining = 1 − removed fraction."

📚 Class XI – Mathematics – Chapter: Exponential Functions – Topic: Successive Replacement

11 If evaluating the surface area of costs, what happens analytically when determining the mixture ratio if the cheaper ingredient is an entirely free resource like water (c = 0)?

A The ratio formula denominator simply becomes the mean price 'm'.

B The ratio cannot be determined.

C The dearer cost 'd' must also be zero.

D The mean price equates to the dearer price.

💡 Short Explanation

�� Water is treated as zero-cost ingredient. �� In alligation, c = 0. �� So (m − c) simplifies to m.

✅ Correct Answer Explanation

In the alligation formula: If the cheaper ingredient is free water: Then: Thus the denominator becomes the mean price itself. Option B is incorrect because the ratio remains determinable. Option C is incorrect because the dearer ingredient still has positive cost. Option D is incorrect because mean price lies between cheaper and dearer prices.

❓ Why Other Options Are Incorrect

�� Option B → Formula still works normally when c = 0.
�� Option C → Dearer ingredient need not be free.
�� Option D → Mean price cannot equal dearer price in a valid mixture.

🎯 Strategy

Used: Substitution

Application: Directly substitute c = 0 into the formula.

Final Logic: m − 0 simplifies to m.

🧠 Memory Trick

"Free water makes denominator mean."

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Alligation and Mixtures

12 When calculating integral profit and loss parameters natively into alligation, how is a 5% loss modeled against a 7% average profit when identifying the (m - c) term?

A 7 - 5 = 2%

B 7 - (-5) = 12%

C 5 - (-7) = 12%

D -5 - 7 = -12%

💡 Short Explanation

�� Loss is treated as negative percentage. �� Average profit is positive. �� Subtracting negative values increases the difference.

✅ Correct Answer Explanation

In alligation involving profit and loss: • Profit is positive. • Loss is negative. Thus: Hence option B is correct. Option A ignores negative sign convention. Option C reverses the meaning of average profit. Option D incorrectly subtracts both values negatively.

❓ Why Other Options Are Incorrect

�� Option A → Loss must be represented using negative sign.
�� Option C → Uses incorrect sign placement.
�� Option D → Produces meaningless negative difference.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Proper sign handling is essential in profit-loss calculations.

Final Logic: Subtracting a loss means adding its magnitude.

🧠 Memory Trick

"Minus loss becomes plus."

📚 Class XI – Mathematics – Chapter: Commercial Mathematics – Topic: Profit-Loss Representation

13 A boat covers 32 km upstream and 36 km downstream in 7 hours. It also covers 40 km upstream and 48 km downstream in 9 hours. What is the speed of the boat in still water?

A 8 km/hr

B 12 km/hr

C 10 km/hr

D 14 km/hr

💡 Short Explanation

�� Form simultaneous equations using speed formulas. �� Upstream speed = b − s. �� Downstream speed = b + s.

✅ Correct Answer Explanation

Let: • Boat speed in still water = b • Stream speed = s Then: and Solving gives: Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Does not satisfy both equations.
�� Option B → Produces inconsistent travel times.
�� Option D → Too large for the given conditions.

🎯 Strategy

Used: Substitution

Application: Use simultaneous equations from upstream and downstream motions.

Final Logic: Solving equations yields boat speed = 10 km/hr.

🧠 Memory Trick

"Boat ± stream gives river speeds."

📚 Class XI – Physics – Chapter: Motion in a Plane – Topic: Relative Velocity

14 A man can row 7.5 km/h in still water. In a river running at 1.5 km/h, it takes him 50 minutes to row to a place and back. How far off is the place?

A 4 km

B 5 km

C 3 km

D 6 km

💡 Short Explanation

�� Downstream speed = 9 km/h. �� Upstream speed = 6 km/h. �� Use total round-trip time equation.

✅ Correct Answer Explanation

Still water speed = 7.5 km/h Current speed = 1.5 km/h Thus: Total time = 50 minutes = 5/6 hour. Let distance = x km. Then: Solving: Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Gives excess travel time.
�� Option B → Does not satisfy equation.
�� Option D → Produces much greater total time.

🎯 Strategy

Used: Substitution

Application: Convert river speeds and substitute into time equation.

Final Logic: Equation solution gives distance = 3 km.

🧠 Memory Trick

"Round trip uses sum of times."

📚 Class XI – Physics – Chapter: Motion in a Plane – Topic: Relative Motion

15 The speed of a motor boat and that of the current of water is 36:5. The boat goes along with the current in 5 hours 10 minutes. How much time will it take to come back?

A 6 hours 50 minutes

B 5 hours 50 minutes

C 6 hours

D 7 hours

💡 Short Explanation

�� Still water : current = 36 : 5. �� Downstream speed = 41 units. �� Upstream speed = 31 units.

✅ Correct Answer Explanation

Let: Boat speed = 36x Current speed = 5x Then: Time ratio is inverse of speed ratio: Downstream time = 5 hr 10 min = 310 min. Thus upstream time: 410 min = 6 hr 50 min. Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect proportional conversion.
�� Option C → Underestimates upstream time.
�� Option D → Excessively large value.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Use inverse relation between speed and time.

Final Logic: Lower upstream speed increases return time.

🧠 Memory Trick

"Time inversely follows speed."

📚 Class XI – Physics – Chapter: Motion in a Straight Line – Topic: Speed-Time Relation

16 A person rows 5 km/hr in still water. It takes him thrice as long to row upstream as to row downstream. Find the rate of the stream.

A 1.5 km/hr

B 2.0 km/hr

C 2.5 km/hr

D 3.0 km/hr

💡 Short Explanation

�� Time ratio is inverse of speed ratio. �� Upstream time is three times downstream time. �� Use boat ± stream relation.

✅ Correct Answer Explanation

Let stream speed = s. Then: Solving: Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Does not satisfy time ratio.
�� Option B → Gives incorrect proportionality.
�� Option D → Makes upstream speed too low.

🎯 Strategy

Used: Substitution

Application: Convert time relation into speed equation.

Final Logic: Solving equation gives stream speed = 2.5 km/hr.

🧠 Memory Trick

"More time means lower speed."

📚 Class XI – Physics – Chapter: Relative Motion – Topic: River Boat Problems

17 A pipe fills a cistern in 6 hours. Due to a leakage, it takes 9 hours to fill. How much time will the leakage take to empty the full tank?

A 15 hours

B 18 hours

C 12 hours

D 24 hours

💡 Short Explanation

�� Filling rate decreases due to leakage. �� Leakage rate = difference of rates. �� Inverse of leakage rate gives emptying time.

✅ Correct Answer Explanation

Pipe filling rate: Effective filling rate: Leakage rate: Thus leakage alone empties tank in: Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect subtraction.
�� Option C → Assumes direct averaging.
�� Option D → Overestimates emptying time.

🎯 Strategy

Used: Substitution

Application: Convert times into work rates and subtract.

Final Logic: Leakage rate equals 1/18 tank per hour.

🧠 Memory Trick

"Leakage = original − effective rate."

📚 Class XI – Mathematics – Chapter: Rational Expressions – Topic: Pipes and Cisterns

18 A cistern can be filled by an inlet pipe in 20 hours and emptied by an outlet pipe in 25 hours. Both are opened. After 10 hours, the outlet pipe is closed. Find the total time taken to fill the tank.

A 15 hours

B 18 hours

C 20 hours

D 22 hours

💡 Short Explanation

�� Net filling rate works initially. �� After 10 hours, only inlet pipe works. �� Remaining tank filled separately.

✅ Correct Answer Explanation

Inlet rate: Outlet rate: Net rate: Filled in 10 hours: Remaining tank: Time to fill remaining using inlet alone: Total time: Therefore, the provided answer is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Ignores remaining work calculation.
�� Option B → Uses incomplete filling logic.
�� Option C → Assumes outlet effect negligible.
�� Option D → Underestimates remaining time.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Carefully compute work done in stages.

Final Logic: Stage-wise filling gives total time = 28 hours.

🧠 Memory Trick

"Net work first, remaining later."

📚 Class XI – Mathematics – Chapter: Rational Expressions – Topic: Pipes and Cisterns

📄 Passage

When a tank is connected to many pipes (inlets and outlets), then the difference between the sum of the work done by inlets and the sum of the work done by outlets gives the filled part of the tank. Consider a scenario where a cistern can be filled by two pipes A and B in 12 minutes and 15 minutes respectively. Another tap C can empty the full tank in 20 minutes. The tap C is opened 5 minutes after the pipes A and B are opened.

Question:
What is the total fraction of the cistern filled by pipes A and B alone in the first 5 minutes?

A 3/4

B 1/2

C 1/4

D 2/3

💡 Short Explanation

�� Add filling rates of pipes A and B. �� Multiply by 5 minutes. �� Obtain total filled fraction.

✅ Correct Answer Explanation

Pipe A rate: Pipe B rate: Combined rate: Work done in 5 minutes: Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Underestimates combined filling.
�� Option C → Incorrect multiplication.
�� Option D → Simplification error.

🎯 Strategy

Used: Substitution

Application: Convert individual times into rates and combine.

Final Logic: Combined filling after 5 minutes equals 3/4.

🧠 Memory Trick

"Add rates, then multiply time."

📚 Class XI – Mathematics – Chapter: Rational Expressions – Topic: Combined Work Rates

📄 Passage

Question:
After tap C is opened, what is the combined efficiency of pipes A, B, and C, and how much additional time is required to fill the remaining 1/4 of the cistern?

A 1/10 per min; 2 mins 30 secs

B 1/15 per min; 3 mins

C 1/10 per min; 5 mins

D 1/12 per min; 2 mins

💡 Short Explanation

�� Add inlet rates and subtract outlet rate. �� Remaining tank fraction = 1/4. �� Time = work ÷ rate.

✅ Correct Answer Explanation

Combined efficiency: LCM = 60: Remaining work: Required time: 2.5 minutes = 2 minutes 30 seconds. Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect combined efficiency.
�� Option C → Uses wrong remaining time.
�� Option D → Combined rate miscalculated.

🎯 Strategy

Used: Substitution

Application: Use combined work-rate formula directly.

Final Logic: Remaining quarter fills in 2.5 minutes at 1/10 rate.

🧠 Memory Trick

"Net rate = inlets − outlets."

📚 Class XI – Mathematics – Chapter: Rational Expressions – Topic: Pipes and Cisterns

CUET UG Applied Mathematics Booster Test 3 - Quantification and Mixture Applications

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