CUET UG Applied Mathematics Booster Test 2 - Quantification and Mixture Applications -

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CUET UG Applied Mathematics Booster Test 2 - 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 2 - Quantification and Mixture Applications

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QUESTION 1 OF 20

An application of alligation is to find the ratio of quantities. If rice at Rs 69 per kg is mixed with rice at Rs 100 per kg to form a mixture worth Rs 80 per kg, what is the ratio of cheaper to dearer ingredients?

Answer Review

1 An application of alligation is to find the ratio of quantities. If rice at Rs 69 per kg is mixed with rice at Rs 100 per kg to form a mixture worth Rs 80 per kg, what is the ratio of cheaper to dearer ingredients?

A 11:20

B 20:11

C 1:2

D 10:9

💡 Short Explanation

�� Apply the rule of alligation. �� Ratio = (d − m) : (m − c). �� Substituting values gives 20:11.

✅ Correct Answer Explanation

Using alligation: Cheaper price (c) = 69 Dearer price (d) = 100 Mean price (m) = 80 Ratio of cheaper : dearer Hence, the ratio is 20:11. Option B is correct because alligation always gives the ratio inversely proportional to the price differences. Option A reverses the ratio. Option C and D do not satisfy the price balance condition.

❓ Why Other Options Are Incorrect

�� Option A → Ratio is reversed incorrectly.
�� Option C → Does not satisfy weighted average condition.
�� Option D → Incorrect proportional difference calculation.

🎯 Strategy

Used: Substitution

Application: Substitute directly into the alligation formula to avoid calculation confusion.

Final Logic: (100 − 80):(80 − 69) = 20:11.

🧠 Memory Trick

"Difference opposite gives ratio."

📚 Class XI – Mathematics – Chapter: Linear Inequalities & Applications – Topic: Weighted Average/Ratio Logic

2 Match the specific mixture variables and parameters with their algebraic substitutions.

List 1	List 2
1. Quantity of pure liquid after n replacements	a. (d - m) / (m - c)
2. Ratio of cheaper to dearer ingredients	b. x(1 - y/x)^n
3. Speed of boat in still water	c. 1/2(a + b)
4. Combined tank fill portion in 1 hour	d. 1/x - 1/y

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

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

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

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

💡 Short Explanation

�� Each formula belongs to a standard aptitude concept. �� Replacement uses exponential reduction. �� Boat speed in still water is the average of upstream and downstream speeds.

✅ Correct Answer Explanation

Match each variable with its standard formula. • Quantity of pure liquid after n replacements: Remaining quantity = x(1 - y/x)^n. Therefore, 1 → b. • Ratio of cheaper to dearer ingredients: According to the rule of alligation, Quantity of Cheaper / Quantity of Dearer = (d - m) / (m - c). Therefore, 2 → a. • Speed of boat in still water: Boat speed = (Upstream Speed + Downstream Speed) / 2 = 1/2(a + b). Therefore, 3 → c. • Combined tank fill portion in 1 hour: For one pipe filling and another emptying, Net work done in one hour = 1/x - 1/y. Therefore, 4 → d. Thus, the correct matching is 1 → b 2 → a 3 → c 4 → d Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because the replacement and alligation formulas are interchanged.
�� Option C → Incorrect because the boat speed and tank-work formulas are interchanged.
�� Option D → Incorrect because all four formulas are mismatched.

🎯 Strategy

Used: Option Grouping

Application:

Identify each standard formula independently and compare the complete mapping.

Final Logic:

Only Option A correctly matches all four concepts.

🧠 Memory Trick

Replacement → Power formula.

📚 Class XI – Mathematics – Chapter: Mathematical Reasoning – Topic: Formula-Based Logical Matching

3 Which of the following boundary conditions MUST hold true for the Mean Price (m) when mixing a cheaper ingredient (c) and a dearer ingredient (d)?
1. m > c
2. m < d
3. m = c + d
4. m > d

A 1, 2

B 1, 3

C 2, 4

D 1, 2, 3

💡 Short Explanation

�� Mean price always lies between the cheaper and dearer prices. �� It can never exceed the dearer price. �� It cannot equal the sum of the two prices.

✅ Correct Answer Explanation

In alligation, c < m < d, where c = Cost of the cheaper ingredient, m = Mean price, d = Cost of the dearer ingredient. Evaluate each statement. • Statement 1: m > c. This is true. • Statement 2: m < d. This is true. • Statement 3: m = c + d. This is false because the mean price lies between c and d, not equal to their sum. • Statement 4: m > d. This is false because the mean price cannot exceed the dearer price. Therefore, only Statements 1 and 2 are correct. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because Statement 3 is false.
�� Option C → Incorrect because Statement 4 is false.
�� Option D → Incorrect because Statement 3 is false.

🎯 Strategy

Used: Elimination

Application:

Remove statements that violate the basic property of averages.

Final Logic:

The mean price must always remain between the cheaper and dearer prices.

🧠 Memory Trick

Average always stays between the extremes.

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

4 Consider the cost price of a unit quantity in practical alligation mixtures. Identify the INCORRECT statement.

A The mean price establishes the new cost price of the mixed unit.

B The ratio of the quantity of cheaper to dearer ingredient relies strictly on the mean price.

C If profit margin is applied to the mixture, the selling price equals the mean price.

D (d - m) yields the proportional ratio metric for the cheaper quantity.

💡 Short Explanation

�� Mean price represents the cost price of the mixture. �� Selling price changes when profit or loss is applied. �� Cost price and selling price are not the same after profit.

✅ Correct Answer Explanation

In alligation, the mean price represents the effective cost price of the final mixture. If a profit is applied, Selling Price = Cost Price + Profit. Therefore, the selling price becomes greater than the mean cost price. Evaluate each option. • Option A is correct because the mean price becomes the effective cost price of the mixture. • Option B is correct because the alligation ratio depends on the mean price and the cross differences. • Option C is incorrect because after adding a profit margin, the selling price is no longer equal to the mean cost price. • Option D is correct because the cross difference (d - m) determines the proportional quantity of the cheaper ingredient. Hence, Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct because the mean price is the effective cost price of the mixture.
�� Option B → Correct because the alligation ratio is based on the mean price.
�� Option D → Correct because (d - m) corresponds to the quantity of the cheaper ingredient.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "equals" becomes incorrect once a profit margin is introduced.

Final Logic:

Profit changes the selling price, while the mean price remains the cost price.

🧠 Memory Trick

Profit changes SP, not CP.

📚 Class XI – Mathematics – Chapter: Commercial Mathematics – Topic: Cost Price and Selling Price

5 Cost of two types of pulses is Rs 55 per kg and Rs 90 per kg. If both pulses are mixed together in the ratio 2:3, what is the mean price of the mixed variety of pulses per kg?

A Rs 76

B Rs 70

C Rs 80

D Rs 72

💡 Short Explanation

�� Use the weighted average formula. �� Multiply each price by its corresponding quantity. �� Divide the total cost by the total quantity.

✅ Correct Answer Explanation

The weighted average (mean price) is Mean Price = (55 × 2 + 90 × 3) / (2 + 3) = (110 + 270) / 5 = 380 / 5 = Rs 76 per kg. Therefore, the mean price of the mixture is Rs 76 per kg. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because it does not use the correct weighted average calculation.
�� Option C → Incorrect because it overestimates the average price.
�� Option D → Incorrect due to an arithmetic error while calculating the weighted average.

🎯 Strategy

Used: Substitution

Application:

Substitute the given prices and quantities into the weighted average formula.

Final Logic:

Total cost divided by total quantity equals Rs 76 per kg.

🧠 Memory Trick

Weighted average = Total Cost ÷ Total Quantity.

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

6 Based on the inverse proportion to cost constraint region, if the mean price is exactly halfway between the cost of the cheaper and dearer ingredients, what will be the ratio of their quantities?

A 2:1

B 1:1

C 3:2

D 1:2

💡 Short Explanation

�� Equal price differences produce equal quantities. �� Cross differences become equal. �� Therefore, the ratio is 1:1.

✅ Correct Answer Explanation

Using the rule of alligation, Quantity of Cheaper : Quantity of Dearer = (d - m) : (m - c). If the mean price lies exactly midway between the cheaper and dearer prices, then d - m = m - c. Therefore, Quantity of Cheaper : Quantity of Dearer = 1 : 1. Hence, equal quantities of both ingredients are mixed. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because it requires unequal price differences.
�� Option C → Incorrect because it does not represent equal cross differences.
�� Option D → Incorrect because it also requires unequal price differences.

🎯 Strategy

Used: Elimination

Application:

Recognize that a midpoint implies equal cross differences.

Final Logic:

Equal price gaps always produce a 1:1 quantity ratio.

🧠 Memory Trick

Midpoint means equal mix.

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

7 In formulating the equated alligation balance, if honey costs Rs 240 per litre and is diluted with water to make a syrup worth Rs 200 per litre, what is the dearer ingredient and its cost?

A Water, Rs 0

B Honey, Rs 200

C Honey, Rs 240

D Syrup, Rs 200

💡 Short Explanation

�� Honey costs more than water. �� Water is treated as having zero cost. �� Therefore, honey is the dearer ingredient.

✅ Correct Answer Explanation

Given, Cost of honey = Rs 240 per litre Cost of water = Rs 0 per litre Mean price of syrup = Rs 200 per litre Since 240 > 200 > 0, honey is the dearer ingredient and water is the cheaper ingredient. The syrup is the final mixture, not an ingredient. Therefore, the dearer ingredient is honey costing Rs 240 per litre. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because water is the cheaper ingredient.
�� Option B → Incorrect because Rs 200 is the mean price of the mixture, not the cost of honey.
�� Option D → Incorrect because syrup is the final mixture, not one of the ingredients.

🎯 Strategy

Used: Odd One Out

Application:

Identify the ingredient having the highest cost.

Final Logic:

The ingredient with the greatest cost is always the dearer ingredient.

🧠 Memory Trick

Higher price = Dearer ingredient.

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

8 Moving along the alligation diagrammatic method, calculate the ratio of water to honey added to dilute the Rs 240 honey to a moving average price of Rs 200.

A 1:5

B 5:1

C 1:4

D 4:1

💡 Short Explanation

�� Water costs Rs 0. �� Apply the alligation rule. �� Simplify the resulting ratio.

✅ Correct Answer Explanation

Using alligation, Cheaper ingredient = Water = Rs 0 Dearer ingredient = Honey = Rs 240 Mean price = Rs 200 Water : Honey = (240 - 200) : (200 - 0) = 40 : 200 = 1 : 5. Hence, the ratio of water to honey is 1 : 5. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because it reverses the required ratio.
�� Option C → Incorrect because 40 : 200 simplifies to 1 : 5, not 1 : 4.
�� Option D → Incorrect because it reverses the proportional relationship.

🎯 Strategy

Used: Substitution

Application:

Substitute the costs directly into the alligation formula.

Final Logic:

The ratio 40 : 200 simplifies to 1 : 5.

🧠 Memory Trick

Cross differences give the ratio.

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

9 A container contains 40 litres of milk. From this, 4 litres of milk is taken out and replaced with water. If this process is repeated two more times, calculate the final quantity of pure milk using the replacement formula.

A 32 litres

B 29.16 litres

C 28 litres

D 30.5 litres

💡 Short Explanation

�� Use the repeated replacement formula. �� The remaining fraction each time is 36/40. �� The process is performed three times in total.

✅ Correct Answer Explanation

Use the replacement formula: Remaining Quantity = x(1 - y/x)^n where x = 40 litres, y = 4 litres, n = 3. Substitute the values: Remaining Quantity = 40 × (1 - 4/40)^3 = 40 × (9/10)^3 = 40 × 729/1000 = 29.16 litres. Hence, the quantity of pure milk remaining is 29.16 litres. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because it assumes a simple subtraction instead of repeated replacement.
�� Option C → Incorrect because it does not apply exponential reduction.
�� Option D → Incorrect due to an arithmetic error.

🎯 Strategy

Used: Substitution

Application:

Apply the replacement formula using the given values.

Final Logic:

Repeated replacement gives 40 × (9/10)^3 = 29.16 litres.

🧠 Memory Trick

Replacement = Multiply by the remaining fraction each time.

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

10 A container holds 70 litres of orange squash. Seven litres of squash is removed and replaced with water. This process is repeated three times in total. How much orange squash is left?

A 51.03 litres

B 49.50 litres

C 55.00 litres

D 50.03 litres

💡 Short Explanation

�� Use the repeated replacement formula. �� The remaining fraction after each replacement is 63/70. �� The process is repeated three times.

✅ Correct Answer Explanation

Use the replacement formula: Remaining Quantity = x(1 - y/x)^n where x = 70 litres, y = 7 litres, n = 3. Substitute the values: Remaining Quantity = 70 × (1 - 7/70)^3 = 70 × (9/10)^3 = 70 × 729/1000 = 51.03 litres. Therefore, the quantity of orange squash remaining is 51.03 litres. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because it uses an incorrect repeated reduction.
�� Option C → Incorrect because it assumes only one replacement.
�� Option D → Incorrect because of an arithmetic error.

🎯 Strategy

Used: Substitution

Application:

Substitute the given values into the repeated replacement formula.

Final Logic:

70 × (9/10)^3 = 51.03 litres.

🧠 Memory Trick

Keep multiplying by the remaining fraction.

📚 Class XI – Mathematics – Chapter: Exponential Functions – Topic: Successive Reduction Models

11 Taking into account the area of costs, how does the zero cost price of water impact the (m - c) term when water is the cheaper ingredient in a mixture?

A The term becomes negative.

B The term equals the mean price (m - 0 = m).

C The term equals the dearer price.

D The term is completely removed from the formula.

💡 Short Explanation

�� Water is treated as zero-cost ingredient. �� Cheaper price c = 0. �� So (m − c) becomes m.

✅ Correct Answer Explanation

In alligation, when water is the cheaper ingredient: Therefore: Hence option B is correct. Option A is incorrect because the value remains positive. Option C is incorrect because the term does not become the dearer price. Option D is incorrect because the term still exists in the formula.

❓ Why Other Options Are Incorrect

�� Option A → Mean price cannot become negative here.
�� Option C → Dearer price is unrelated to simplification.
�� Option D → Formula remains valid and unchanged structurally.

🎯 Strategy

Used: Substitution

Application: Substitute c = 0 directly into the expression.

Final Logic: m − 0 simplifies to m.

🧠 Memory Trick

"Water cost zero → term becomes mean."

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

12 Integrating profit and loss: A retailer has 250 kg of rice. He sells a part at 10% profit and the rest at a 5% loss, making an overall 7% profit. What represents 'c' and 'd' in the alligation calculation?

A c = 5, d = 10

B c = -5, d = 10

C c = 7, d = 10

D c = 0, d = 7

💡 Short Explanation

�� Loss is represented using negative sign. �� Profit is positive. �� So cheaper side becomes −5.

✅ Correct Answer Explanation

In profit-loss alligation: • Profit percentages are positive. • Loss percentages are negative. Thus: • 5% loss = −5 • 10% profit = +10 Therefore: Hence option B is correct. Option A ignores sign convention. Option C incorrectly uses overall profit as ingredient value. Option D does not represent the given conditions.

❓ Why Other Options Are Incorrect

�� Option A → Loss percentage must be negative.
�� Option C → Overall profit is not ingredient ratio input.
�� Option D → Values unrelated to given transaction.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Profit-loss percentages require proper sign interpretation.

Final Logic: Loss values are always treated as negative.

🧠 Memory Trick

"Loss carries minus sign."

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

13 A man rows 15 km upstream and 25 km downstream in 5 hours each time. What is his downstream speed?

A 3 km/hr

B 4 km/hr

C 5 km/hr

D 8 km/hr

💡 Short Explanation

�� Downstream speed = distance ÷ time. �� Distance = 25 km. �� Time = 5 hours.

✅ Correct Answer Explanation

Downstream speed: Thus downstream speed = 5 km/hr. Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to upstream speed.
�� Option B → Incorrect division.
�� Option D → Excessively high for given data.

🎯 Strategy

Used: Substitution

Application: Apply basic speed formula directly.

Final Logic: Speed = distance ÷ time = 5.

🧠 Memory Trick

"Downstream → larger distance per same time."

📚 Class XI – Mathematics – Chapter: Motion in a Straight Line – Topic: Speed Formula

14 A man rows 15 km upstream and 25 km downstream in 5 hours each time. What is his upstream speed?

A 3 km/hr

B 5 km/hr

C 4 km/hr

D 2 km/hr

💡 Short Explanation

�� Upstream speed = distance ÷ time. �� Distance = 15 km. �� Time = 5 hours.

✅ Correct Answer Explanation

Upstream speed: Hence upstream speed = 3 km/hr.

❓ Why Other Options Are Incorrect

�� Option B → This is downstream speed.
�� Option C → Incorrect calculation.
�� Option D → Does not satisfy given distance-time data.

🎯 Strategy

Used: Substitution

Application: Use speed = distance/time directly.

Final Logic: 15 ÷ 5 = 3 km/hr.

🧠 Memory Trick

"Upstream is slower."

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

15 Based on the downstream speed of 5 km/hr and upstream speed of 3 km/hr, what is the speed of the boat in still water?

A 1 km/hr

B 2 km/hr

C 4 km/hr

D 8 km/hr

💡 Short Explanation

�� Still water speed is average of upstream and downstream speeds. �� Add both speeds. �� Divide by 2.

✅ Correct Answer Explanation

Boat speed in still water: Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Too small for average.
�� Option B → Represents stream effect difference.
�� Option D → Sum instead of average.

🎯 Strategy

Used: Substitution

Application: Use standard boats-and-streams formula.

Final Logic: Average of 5 and 3 equals 4.

🧠 Memory Trick

"Still water = average speed."

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

16 Based on the downstream speed of 5 km/hr and upstream speed of 3 km/hr, what is the speed of the current (stream)?

A 2 km/hr

B 1 km/hr

C 4 km/hr

D 0.5 km/hr

💡 Short Explanation

�� Stream speed is half the difference. �� Difference = 5 − 3. �� Half of 2 equals 1.

✅ Correct Answer Explanation

Current speed: Thus stream speed = 1 km/hr.

❓ Why Other Options Are Incorrect

�� Option A → Uses full difference instead of half.
�� Option C → Equals still water speed.
�� Option D → Incorrect halving.

🎯 Strategy

Used: Substitution

Application: Apply stream speed formula directly.

Final Logic: Half of speed difference gives current speed.

🧠 Memory Trick

"Current = half the difference."

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

17 Pipe A can fill a tank in 30 hours. What portion of the tank does it fill in one hour functioning as an inlet pipe?

A 30

B 1/30

C 1/15

D 1

💡 Short Explanation

�� Work done per hour = reciprocal of total time. �� Time taken = 30 hours. �� So one-hour work = 1/30.

✅ Correct Answer Explanation

If a pipe fills a tank in x hours, then one-hour work: Here x = 30. Thus: Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents total hours, not work fraction.
�� Option C → Incorrect reciprocal.
�� Option D → Entire tank cannot fill in one hour.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Convert "hours to complete" into "portion per hour."

Final Logic: One-hour work = reciprocal of total time.

🧠 Memory Trick

"Work per hour = 1/time."

📚 Class XI – Mathematics – Chapter: Rational Expressions – Topic: Work Rate Concept

18 Pipe C can empty a full cistern in 20 minutes. What portion of the cistern does it empty per minute?

A 1/20

B 20

C 1/10

D -20

💡 Short Explanation

�� Emptying rate also uses reciprocal principle. �� Total time = 20 minutes. �� So one-minute work = 1/20 emptied.

✅ Correct Answer Explanation

Outlet pipe empties: of the cistern per minute. Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents time, not rate.
�� Option C → Incorrect reciprocal value.
�� Option D → Negative notation unnecessary here.

🎯 Strategy

Used: Substitution

Application: Apply reciprocal work-rate rule.

Final Logic: 1 ÷ 20 gives emptied portion per minute.

🧠 Memory Trick

"Rate is reciprocal of time."

📚 Class XI – Mathematics – Chapter: Rational Functions – Topic: Work and Efficiency

📄 Passage

A pipe connected to a tank or cistern which fills it is known as an inlet pipe and the pipe connected to the tank which drains or empties it is known as an outlet pipe. 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. Let a pipe fill a tank in x number of hours, then it can fill (1/x)th portion of the tank in one hour. If a pipe can empty a tank in y number of hours, then it can empty out (1/y)th portion of the tank in one hour. The portion of tank they can fill together in one hour = (1/x – 1/y)th.

Question:
Pipe A can fill a tank in 30 hours and pipe B in 45 hours. If both are opened in an empty tank, what portion of the tank is filled per hour?

A 1/15

B 1/18

C 1/75

D 5/90

💡 Short Explanation

�� Add individual filling rates. �� A fills 1/30 per hour. �� B fills 1/45 per hour.

✅ Correct Answer Explanation

Combined work: Thus combined filling rate = 1/18 per hour. Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Faster than actual combined rate.
�� Option C → Incorrect denominator handling.
�� Option D → Equivalent intermediate form but not simplified fully.

🎯 Strategy

Used: Substitution

Application: Convert times into hourly fractions and add.

Final Logic: Simplified combined rate equals 1/18.

🧠 Memory Trick

"Combined work = sum of rates."

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

📄 Passage

Question:
Based on the combined efficiency of 1/18 per hour calculated for Pipes A and B, how much total time will it take to fill the tank?

A 15 hours

B 18 hours

C 75 hours

D 90 hours

💡 Short Explanation

�� Time is reciprocal of rate. �� Combined rate = 1/18 per hour. �� So total time = 18 hours.

✅ Correct Answer Explanation

If combined filling rate: then required time: Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Faster than actual efficiency.
�� Option C → Ignores reciprocal relationship.
�� Option D → Uses LCM denominator incorrectly.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Convert work rate back into total time.

Final Logic: Time = reciprocal of combined rate.

🧠 Memory Trick

"Time and rate are reciprocals."

📚 Class XI – Mathematics – Chapter: Rational Expressions – Topic: Time–Work Reciprocal Relation

CUET UG Applied Mathematics Booster Test 2 - Quantification and Mixture Applications

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