Profit and Loss, Time and Work, Time Speed and Distance

Profit and Loss

Profit and Loss is a fundamental concept in quantitative aptitude that deals with the calculation of profit earned or loss incurred on a transaction. Understanding these concepts is crucial for analyzing business scenarios and making informed financial decisions. We will explore the key terms, formulas, and problem-solving techniques associated with profit and loss.

Key Terms and Definitions:

• Cost Price (CP): The price at which an article is purchased.
• Selling Price (SP): The price at which an article is sold.
• Profit: The excess of Selling Price over Cost Price (SP > CP).
• Loss: The excess of Cost Price over Selling Price (CP > SP).
• Marked Price (MP) / List Price (LP): The price marked on the article, from which discounts are offered.
• Discount: The reduction offered on the Marked Price.

Formulas:

• Profit = Selling Price (SP) - Cost Price (CP)
• Loss = Cost Price (CP) - Selling Price (SP)
• Profit Percentage = (Profit / CP) × 100
• Loss Percentage = (Loss / CP) × 100
• SP = CP + Profit = CP × (100 + Profit%) / 100
• SP = CP - Loss = CP × (100 - Loss%) / 100
• CP = SP / (1 + Profit%/100)
• CP = SP / (1 - Loss%/100)
• Discount = Marked Price (MP) - Selling Price (SP)
• Discount Percentage = (Discount / MP) × 100
• SP = MP - Discount = MP × (100 - Discount%) / 100
• MP = SP / (1 - Discount%/100)

Example:

A shopkeeper buys a refrigerator for ₹20,000 and sells it for ₹24,000. Find the profit percentage.

Cost Price (CP) = ₹20,000

Selling Price (SP) = ₹24,000

Profit = SP - CP = ₹24,000 - ₹20,000 = ₹4,000

Profit Percentage = (Profit / CP) × 100 = (4000 / 20000) × 100 = (1/5) × 100 = 20%

Calculations involving Multiple Discounts:

When successive discounts are offered, they are applied one after another on the successively reduced prices. If two successive discounts are x% and y%, the net discount is given by:

Net Discount % = (x + y - xy/100)%

Example with Successive Discounts:

A shopkeeper marks a shirt at ₹1000 and offers two successive discounts of 10% and 20%. What is the final selling price?

First Discount = 10% of ₹1000 = ₹100

Price after first discount = ₹1000 - ₹100 = ₹900

Second Discount = 20% of ₹900 = ₹180

Final Selling Price = ₹900 - ₹180 = ₹720

Alternatively, using the net discount formula:

Net Discount % = (10 + 20 - (10 × 20)/100)% = (30 - 2)% = 28%

Final Selling Price = Marked Price × (100 - Net Discount%) / 100 = 1000 × (100 - 28) / 100 = 1000 × 72 / 100 = ₹720

Profit/Loss on Selling Equal Number of Articles:

If a person sells two articles at the same selling price, and makes a profit of x% on one and a loss of x% on the other, then there is always a loss of (x/10)%².

Example:

A dealer sells two tables for ₹5000 each. On one, he gains 25% and on the other, he loses 25%. Find his overall profit or loss.

Since the selling price is the same and the profit and loss percentages are the same, there will be a loss.

Loss % = (25/10)² = (2.5)² = 6.25%

Profit/Loss when Cost Prices are Equal:

If a person buys two articles at the same cost price, and sells one at a profit of x% and the other at a loss of x%, then there is neither profit nor loss in the overall transaction.

Example:

A man buys two watches for ₹1000 each. He sells the first at a profit of 20% and the second at a loss of 20%. Find the overall profit or loss.

CP of first watch = ₹1000. SP of first watch = 1000 × (100 + 20)/100 = ₹1200.

CP of second watch = ₹1000. SP of second watch = 1000 × (100 - 20)/100 = ₹800.

Total CP = ₹1000 + ₹1000 = ₹2000

Total SP = ₹1200 + ₹800 = ₹2000

Total Profit/Loss = Total SP - Total CP = ₹2000 - ₹2000 = ₹0. No profit, no loss.

False Weights and Measures:

In problems involving false weights, a shopkeeper uses a faulty weighing scale to cheat customers. For example, a shopkeeper claims to sell goods at CP but uses a faulty weight.

Example:

A dishonest shopkeeper professes to sell his goods at Cost Price, but uses a weight of 900 grams instead of 1000 grams. What is his gain percentage?

The shopkeeper claims to sell 1000 grams for a certain price. Let's assume the price of 1000 grams is ₹100.

He sells 900 grams for ₹100 (as he claims to sell at CP).

So, the Cost Price of 900 grams = (90/100) × ₹100 = ₹90.

The shopkeeper is selling goods worth ₹90 for ₹100.

Profit = ₹100 - ₹90 = ₹10.

Gain Percentage = (Profit / CP) × 100 = (10 / 90) × 100 = (1/9) × 100 = 11.11%

Profit/Loss on Goods Bought and Sold:

When a person buys goods at one price and sells them at another price, we calculate the overall profit or loss by comparing the total cost price and total selling price.

Example:

A merchant buys 10 kg of rice for ₹80 per kg and 15 kg of rice for ₹90 per kg. He mixes them and sells the mixture at ₹95 per kg. Find the profit or loss percentage.

Cost of 10 kg rice = 10 × 80 = ₹800

Cost of 15 kg rice = 15 × 90 = ₹1350

Total Cost Price = ₹800 + ₹1350 = ₹2150

Total quantity of rice = 10 kg + 15 kg = 25 kg

Selling Price of 25 kg rice = 25 × 95 = ₹2375

Total Profit = ₹2375 - ₹2150 = ₹225

Profit Percentage = (Profit / Total CP) × 100 = (225 / 2150) × 100 ≈ 10.47%

This section covers the core concepts of Profit and Loss. Practice various types of problems to gain proficiency.

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Time and Work

Time and Work is a crucial topic that assesses a candidate's ability to calculate the time taken to complete a task by individuals or groups, considering their individual efficiencies. This topic is widely tested in competitive exams.

Basic Concepts:

• The total amount of work is considered as a unit (i.e., 1 unit of work).
• Efficiency of a person is the amount of work done by that person in one unit of time.
• If a person A can complete a work in 'x' days, then A's efficiency (work done in 1 day) is 1/x.
• If A's efficiency is 1/x, then the time taken by A to complete the work is 'x' days.
• Work Done = Efficiency × Time
• Efficiency = Work Done / Time
• Time = Work Done / Efficiency

Formulas:

• If A can do a piece of work in x days and B can do the same work in y days, then together they can finish the work in: 1 / (1/x + 1/y) = xy / (x + y) days.
• If A can do a piece of work in x days and B can do the same work in y days, and A leaves after a days, then the remaining work will be completed by B in: (Total Work - Work done by A) / Efficiency of B days.
• If A can do a piece of work in x days and B can do the same work in y days, and they work together for a days, then the remaining work is: 1 - a × (1/x + 1/y).

Example 1: Basic Calculation

A can complete a work in 10 days and B can complete the same work in 15 days. In how many days can they together complete the work?

A's efficiency = 1/10 work per day.

B's efficiency = 1/15 work per day.

Combined efficiency = 1/10 + 1/15 = (3 + 2) / 30 = 5/30 = 1/6 work per day.

Time taken together = 1 / (Combined efficiency) = 1 / (1/6) = 6 days.

Using the formula: Time = (10 × 15) / (10 + 15) = 150 / 25 = 6 days.

Example 2: Work and Wages

A can do a work in 8 days and B can do it in 10 days. They together complete the work and get ₹4000 for it. The share of A is?

Time taken by A = 8 days. Time taken by B = 10 days.

Ratio of efficiencies of A and B = (1/8) : (1/10) = 10 : 8 = 5 : 4.

Total ratio parts = 5 + 4 = 9.

A's share = (A's ratio part / Total ratio parts) × Total wages

A's share = (5 / 9) × 4000 = ₹20000 / 9 ≈ ₹2222.22

Example 3: Men, Women, and Children

12 men can complete a work in 8 days. 16 women can complete the same work in 12 days. 10 men and 10 women start working together. In how many days will they complete 50% of the work?

12 men = 8 days => 1 man = 12 × 8 = 96 man-days.

16 women = 12 days => 1 woman = 16 × 12 = 192 woman-days.

So, 1 man = 1/96 work/day and 1 woman = 1/192 work/day.

Efficiency of 10 men = 10 × (1/96) = 10/96.

Efficiency of 10 women = 10 × (1/192) = 10/192.

Combined efficiency of 10 men and 10 women = 10/96 + 10/192 = (20 + 10) / 192 = 30/192.

Total work done by 10 men and 10 women in 1 day = 30/192.

Time to complete the whole work = 1 / (30/192) = 192/30 = 6.4 days.

Time to complete 50% of the work = 50% of 6.4 days = 0.5 × 6.4 = 3.2 days.

Example 4: Work Left and Joined Later

A can do a work in 20 days and B in 30 days. They begin the work together but A leaves after 5 days. In how many days will B alone complete the remaining work?

A's efficiency = 1/20. B's efficiency = 1/30.

Work done by A and B together in 1 day = 1/20 + 1/30 = (3 + 2) / 60 = 5/60 = 1/12.

Work done by A and B together in 5 days = 5 × (1/12) = 5/12.

Remaining work = 1 - 5/12 = 7/12.

Time taken by B to complete the remaining work = (Remaining Work) / (B's efficiency) = (7/12) / (1/30) = (7/12) × 30 = 7 × (30/12) = 7 × (5/2) = 35/2 = 17.5 days.

Example 5: Work Left and Left Earlier

A can do a work in 10 days. B can do it in 15 days. They start working together. After 2 days, A leaves the work. In how many days will B finish the work?

A's efficiency = 1/10. B's efficiency = 1/15.

Work done by A and B together in 1 day = 1/10 + 1/15 = (3 + 2) / 30 = 5/30 = 1/6.

Work done by A and B in 2 days = 2 × (1/6) = 2/6 = 1/3.

Remaining work = 1 - 1/3 = 2/3.

Time taken by B to finish the remaining work = (Remaining Work) / (B's efficiency) = (2/3) / (1/15) = (2/3) × 15 = 2 × 5 = 10 days.

Example 6: Work Started by One, Finished by Another

A can do a work in 20 days. B can do it in 30 days. A starts the work and works for 4 days. Then B takes over and finishes the work. For how many days did B work?

A's efficiency = 1/20. B's efficiency = 1/30.

Work done by A in 4 days = 4 × (1/20) = 4/20 = 1/5.

Remaining work = 1 - 1/5 = 4/5.

Time taken by B to finish the remaining work = (Remaining Work) / (B's efficiency) = (4/5) / (1/30) = (4/5) × 30 = 4 × 6 = 24 days.

Example 7: Efficiency Comparison

A is twice as good a workman as B. Together they can complete a work in 15 days. In how many days can A alone complete the work?

Let B's efficiency be 1 unit. Then A's efficiency is 2 units.

Combined efficiency = A's efficiency + B's efficiency = 2 + 1 = 3 units.

Total work = Combined efficiency × Time = 3 units × 15 days = 45 units.

Time taken by A alone = Total Work / A's efficiency = 45 units / 2 units = 22.5 days.

Example 8: Negative Work (Pipes and Cisterns Analogy** - though not explicitly pipes, the concept applies)**

A can do a work in 10 days. B can do it in 15 days. C can do it in 20 days. A and B work together for 4 days, and then C joins them. In how many more days will the work be completed?

A's efficiency = 1/10. B's efficiency = 1/15. C's efficiency = 1/20.

Combined efficiency of A and B = 1/10 + 1/15 = (3 + 2) / 30 = 5/30 = 1/6.

Work done by A and B in 4 days = 4 × (1/6) = 4/6 = 2/3.

Remaining work = 1 - 2/3 = 1/3.

Now C joins them. Combined efficiency of A, B, and C = 1/10 + 1/15 + 1/20 = (6 + 4 + 3) / 60 = 13/60.

Time taken to complete the remaining work (1/3) with all three = (Remaining Work) / (Combined efficiency of A, B, C) = (1/3) / (13/60) = (1/3) × (60/13) = 60 / 39 = 20 / 13 days.

This topic requires careful understanding of individual and combined efficiencies. Practice is key to mastering different scenarios.

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Time Speed and Distance

Time, Speed, and Distance is a fundamental topic in quantitative aptitude that deals with the relationship between these three quantities. It is essential for solving problems related to vehicles, journeys, and relative motion.

Basic Relationship:

The core relationship is: Distance = Speed × Time.

From this, we can derive:

• Speed = Distance / Time
• Time = Distance / Speed

Units and Conversions:

It is crucial to maintain consistent units. The most common conversions are:

• To convert km/hr to m/s: Multiply by 5/18. (Since 1 km = 1000 m and 1 hr = 3600 s, km/hr = 1000/3600 m/s = 5/18 m/s).
• To convert m/s to km/hr: Multiply by 18/5.

Average Speed:

Average Speed is not simply the average of the speeds. It is defined as Total Distance covered divided by Total Time taken.

Average Speed = Total Distance / Total Time

Case 1: Equal Distances

If a person covers two equal distances at speeds s1 and s2 respectively, then the average speed is:

Average Speed = 2 / (1/s1 + 1/s2) = 2*s1*s2 / (s1 + s2) (Harmonic Mean)

Example 1: Equal Distances

A car travels from City A to City B at a speed of 40 km/hr and returns from City B to City A at a speed of 60 km/hr. What is the average speed for the entire journey?

Let the distance between City A and City B be D km.

Time taken to travel from A to B = D / 40 hours.

Time taken to travel from B to A = D / 60 hours.

Total Distance = D + D = 2D km.

Total Time = (D/40) + (D/60) = (3D + 2D) / 120 = 5D / 120 = D / 24 hours.

Average Speed = Total Distance / Total Time = 2D / (D/24) = 2D × (24/D) = 48 km/hr.

Using the formula for equal distances: Average Speed = (2 × 40 × 60) / (40 + 60) = (2 × 2400) / 100 = 4800 / 100 = 48 km/hr.

Case 2: Equal Times

If a person travels for two equal time intervals at speeds s1 and s2 respectively, then the average speed is:

Average Speed = (s1 + s2) / 2 (Arithmetic Mean)

Example 2: Equal Times

A train travels at 50 km/hr for the first hour and at 70 km/hr for the next hour. What is the average speed?

Since the time intervals are equal (1 hour each), we can use the arithmetic mean.

Average Speed = (50 + 70) / 2 = 120 / 2 = 60 km/hr.

Let's verify: Distance 1 = 50 km/hr × 1 hr = 50 km. Distance 2 = 70 km/hr × 1 hr = 70 km.

Total Distance = 50 + 70 = 120 km. Total Time = 1 + 1 = 2 hours.

Average Speed = 120 km / 2 hours = 60 km/hr.

Relative Speed:

Relative speed is used when two objects are moving. It is the speed of one object with respect to the other.

• Objects moving in the same direction: Relative Speed = |Speed1 - Speed2|. The faster object gains on the slower object at this speed.
• Objects moving in opposite directions: Relative Speed = Speed1 + Speed2. The objects are approaching or moving away from each other at this combined speed.

Example 3: Relative Speed (Same Direction)

Two trains, A and B, are running in the same direction with speeds 50 km/hr and 70 km/hr respectively. If train B is ahead of train A by 1 km, how long will it take for train A to catch up with train B?

Relative Speed of A with respect to B = 70 km/hr - 50 km/hr = 20 km/hr.

Distance to be covered by A to catch up = 1 km.

Time = Distance / Relative Speed = 1 km / 20 km/hr = 1/20 hours.

1/20 hours = (1/20) × 60 minutes = 3 minutes.

Example 4: Relative Speed (Opposite Direction)

Two trains start at the same time from two stations A and B, 300 km apart, and travel towards each other at speeds of 40 km/hr and 50 km/hr respectively. When will they meet?

Relative Speed = 40 km/hr + 50 km/hr = 90 km/hr.

Total Distance = 300 km.

Time to meet = Total Distance / Relative Speed = 300 km / 90 km/hr = 30/9 hours = 10/3 hours.

10/3 hours = 3 and 1/3 hours = 3 hours and (1/3) × 60 minutes = 3 hours and 20 minutes.

Problems on Trains:

These problems often involve trains crossing:

• A pole or a standing man: The distance covered is the length of the train. Time = Length of Train / Speed of Train.
• A platform or a bridge: The distance covered is the length of the train + length of the platform/bridge. Time = (Length of Train + Length of Platform) / Speed of Train.
• Another train:
  • Moving in the same direction: Distance = |Length of Train 1 - Length of Train 2|. Time = Distance / Relative Speed.
  • Moving in opposite directions: Distance = Length of Train 1 + Length of Train 2. Time = Distance / Relative Speed.

Example 5: Train Crossing a Platform

A train 150 meters long is running at 60 km/hr. In how many seconds will it cross a platform 250 meters long?

Length of train = 150 m.

Length of platform = 250 m.

Total distance to cover = 150 m + 250 m = 400 m.

Speed of train = 60 km/hr = 60 × (5/18) m/s = (10 × 5) / 3 m/s = 50/3 m/s.

Time = Total Distance / Speed = 400 m / (50/3 m/s) = 400 × (3/50) seconds = 8 × 3 = 24 seconds.

Boat and Stream Problems:

These problems involve a boat moving in a river. The speed of the river current affects the boat's speed.

• Speed Downstream (Speed of boat + Speed of stream): When the boat moves in the direction of the stream.
• Speed Upstream (Speed of boat - Speed of stream): When the boat moves against the direction of the stream.
• Let Speed of Boat in still water = u km/hr.
• Let Speed of Stream = v km/hr.
• Speed Downstream = u + v
• Speed Upstream = u - v

Example 6: Boat and Stream

A boat takes 3 hours to travel downstream a distance of 42 km. It takes 4 hours to travel upstream the same distance. Find the speed of the boat in still water and the speed of the stream.

Speed Downstream = Distance / Time Downstream = 42 km / 3 hours = 14 km/hr.

Speed Upstream = Distance / Time Upstream = 42 km / 4 hours = 10.5 km/hr.

We have: u + v = 14 and u - v = 10.5.

Adding the two equations: (u + v) + (u - v) = 14 + 10.5 => 2u = 24.5 => u = 12.25 km/hr.

Substituting u in the first equation: 12.25 + v = 14 => v = 14 - 12.25 = 1.75 km/hr.

Speed of boat in still water = 12.25 km/hr.

Speed of stream = 1.75 km/hr.

Circular Motion:

In problems involving circular tracks, we often need to find when participants will meet or when they will be at the starting point together.

• Meeting at the starting point: This happens after a time equal to the LCM of the individual times taken to complete one round.
• Meeting at any point on the track: This involves relative speed. If they move in opposite directions, they meet when the sum of distances covered equals the track length. If they move in the same direction, they meet when the difference in distances covered equals the track length.

Example 7: Circular Motion

Three runners A, B, and C run around a circular track of 1 km. They start from the same point at the same time. A runs at 10 km/hr, B at 8 km/hr, and C at 6 km/hr. When will they meet at the starting point for the first time?

Time taken by A to complete one round = Distance / Speed = 1 km / 10 km/hr = 1/10 hour.

Time taken by B to complete one round = 1 km / 8 km/hr = 1/8 hour.

Time taken by C to complete one round = 1 km / 6 km/hr = 1/6 hour.

They will meet at the starting point after a time equal to the LCM of (1/10, 1/8, 1/6) hours.

LCM of fractions = LCM of numerators / HCF of denominators.

LCM(1, 1, 1) = 1.

HCF(10, 8, 6) = 2.

LCM of times = 1 / 2 hour = 30 minutes.

They will meet at the starting point after 30 minutes.

Mastering Time, Speed, and Distance requires understanding the basic formula and practicing various scenarios involving average speed, relative speed, trains, boats, and circular motion.