Time and Distance

The topic of Time and Distance is a fundamental concept in quantitative aptitude, crucial for competitive exams. It deals with the relationship between three key variables: speed, distance, and time. Understanding these relationships and their applications in various scenarios is essential for solving problems efficiently.

Basic Concepts

The core formula that links speed, distance, and time is: Distance = Speed × Time

From this, we can derive the other two relationships: Speed = Distance / Time Time = Distance / Speed

It's vital to ensure that the units are consistent. If distance is in kilometers (km) and time is in hours (hr), then speed will be in kilometers per hour (km/hr). If distance is in meters (m) and time is in seconds (s), then speed will be in meters per second (m/s).

Unit Conversions

Often, problems involve different units, requiring conversion. The most common conversions are between km/hr and m/s.

To convert km/hr to m/s: 1 km/hr = 1000 meters / 3600 seconds = 5/18 m/s. So, multiply the speed in km/hr by 5/18 to get the speed in m/s.

To convert m/s to km/hr: 1 m/s = 18/5 km/hr. So, multiply the speed in m/s by 18/5 to get the speed in km/hr.

Average Speed

Average speed is not simply the average of the speeds. It is defined as the total distance traveled divided by the total time taken. Average Speed = Total Distance / Total Time

Consider a scenario where a person travels a distance 'd' at a speed 's1' and returns the same distance 'd' at a speed 's2'. The average speed for the entire journey is not (s1 + s2)/2.

Total Distance = d + d = 2d Time taken for the first part = d / s1 Time taken for the second part = d / s2 Total Time = (d / s1) + (d / s2) = d * (1/s1 + 1/s2) = d * ((s1 + s2) / (s1 * s2)) Average Speed = (2d) / [d * ((s1 + s2) / (s1 * s2))] Average Speed = 2 / [(s1 + s2) / (s1 * s2)] Average Speed = 2 * s1 * s2 / (s1 + s2)

If an object travels three equal distances at speeds s1, s2, and s3, the average speed is 3*s1*s2*s3 / (s1*s2 + s2*s3 + s3*s1).

Relative Speed

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

• Objects Moving in the Same Direction: If two objects move in the same direction with speeds s1 and s2 (where s1 > s2), their relative speed is (s1 - s2). This is the speed at which the faster object gains on or overtakes the slower object.
• Objects Moving in Opposite Directions: If two objects move towards each other or away from each other (in opposite directions) with speeds s1 and s2, their relative speed is (s1 + s2). This is the speed at which the distance between them decreases (if moving towards each other) or increases (if moving away from each other).

Problems on Trains

These problems typically involve a train moving past a stationary object (like a pole or a person) or a moving object (like another train, a platform, or a bridge).

• Train Passing a Point Object (Pole, Man): When a train of length 'L' passes a pole or a man, the distance covered by the train is its own length 'L'. The time taken is L / Speed of the train.
• Train Passing a Platform or Bridge: When a train of length 'L1' passes a platform or a bridge of length 'L2', the total distance the train needs to cover is the sum of its own length and the length of the platform/bridge (L1 + L2). The time taken is (L1 + L2) / Speed of the train.
• Two Trains Moving Towards Each Other: If two trains of lengths L1 and L2 are moving towards each other with speeds s1 and s2, the time taken for them to cross each other is (L1 + L2) / (s1 + s2).
• Two Trains Moving in the Same Direction: If two trains of lengths L1 and L2 are moving in the same direction with speeds s1 and s2 (s1 > s2), the time taken for the faster train to cross the slower train is (L1 + L2) / (s1 - s2).

Circular Motion

Problems involving circular tracks often ask when two people or objects starting from the same point and moving at different speeds will meet again at the starting point or at any point on the track.

Let the length of the circular track be 'D'. Let two people A and B move with speeds s1 and s2 respectively.

• Meeting at the Starting Point: They will meet at the starting point after a time equal to the Least Common Multiple (LCM) of the times taken by each to complete one round. Time for A = D/s1, Time for B = D/s2. LCM(D/s1, D/s2).
• Meeting at Any Point on the Track (Moving in Opposite Directions): The time taken for them to meet for the first time is D / (s1 + s2). They will meet 'k' times in a given time 'T' where k = T / (D / (s1 + s2)).
• Meeting at Any Point on the Track (Moving in the Same Direction): The time taken for the faster person to gain one full lap on the slower person is D / (s1 - s2). This is the time interval between consecutive meetings.

Example Problems

Example 1: A car travels from city A to city B at 40 km/hr and returns from city B to city A at 60 km/hr. What is the average speed of the car for the entire journey? Solution: Since the distance is the same in both directions, we use the shortcut for equal distances. Average Speed = 2 * s1 * s2 / (s1 + s2) = 2 * 40 * 60 / (40 + 60) = 2 * 40 * 60 / 100 = 4800 / 100 = 48 km/hr.

Example 2: A train 150 meters long is running at 68 km/hr. In what time will it pass a man standing on the platform? Solution: Length of train (L) = 150 m Speed of train (S) = 68 km/hr. Convert to m/s: 68 * (5/18) m/s = 340/18 m/s = 170/9 m/s. The distance to be covered is the length of the train itself. Time = Distance / Speed = 150 m / (170/9 m/s) = 150 * 9 / 170 seconds = 1350 / 170 seconds = 135 / 17 seconds ≈ 7.94 seconds.

Example 3: Two trains start at the same time from points P and Q towards each other. The first train travels at 50 km/hr and the second train at 40 km/hr. They cross each other after 3 hours. What is the distance between P and Q? Solution: The trains are moving towards each other, so their relative speed is the sum of their speeds. Relative Speed = 50 km/hr + 40 km/hr = 90 km/hr. Time taken to cross = 3 hours. Distance = Relative Speed × Time = 90 km/hr × 3 hr = 270 km. The distance between P and Q is 270 km.

Time and Work

The Time and Work section in quantitative aptitude assesses your ability to calculate the time taken to complete a task when multiple individuals or entities work on it, possibly at different rates. This topic is closely related to Time and Distance, as it also involves the concepts of rate and total output.

Basic Concepts

The fundamental idea is that the amount of work done is proportional to the rate of work and the time spent working. Work = Rate × Time

The 'Rate' here refers to the amount of work an individual can complete in a unit of time (e.g., work done per day, per hour).

If a person can complete a whole task (let's say, 1 unit of work) in 'T' days, then their rate of work is 1/T per day.

Conversely, if a person's rate of work is 'R' per day, the time taken to complete 1 unit of work is 1/R days.

Example: If A can complete a piece of work in 10 days, then A's rate of work is 1/10 of the work per day.

Working Together

When two or more people work together, their individual rates of work add up to find their combined rate of work.

If person A can complete a work in 'a' days and person B can complete the same work in 'b' days: A's rate = 1/a work per day B's rate = 1/b work per day Combined rate of A and B = (1/a) + (1/b) = (a + b) / (a * b) work per day. The time taken by A and B together to complete the work is the reciprocal of their combined rate: Time taken together = 1 / [(a + b) / (a * b)] = (a * b) / (a + b) days.

If three people A, B, and C can complete a work in 'a', 'b', and 'c' days respectively, their combined rate is (1/a + 1/b + 1/c). The time taken together is 1 / (1/a + 1/b + 1/c).

Efficiency and Proportionality

Efficiency is directly related to the rate of work. A person who works faster has higher efficiency.

If A is 'x' times as efficient as B, it means A can do 'x' times the work B can do in the same amount of time. Or, A can complete the same work in 1/x times the time B takes. Efficiency is inversely proportional to the time taken. If Efficiency(A) / Efficiency(B) = x / y, then Time(A) / Time(B) = y / x.

Example: A is twice as efficient as B. If A can complete a work in 10 days, how many days will B take? Since A is twice as efficient, A takes half the time B takes. Time(A) = 10 days. Time(B) = 2 * Time(A) = 2 * 10 = 20 days. Alternatively, A's rate = 1/10. B's rate = 1/2 * A's rate = 1/2 * (1/10) = 1/20. So B takes 20 days.

Work Done by Multiple People with Different Efficiencies

Suppose A, B, and C work on a task. A's rate is rA, B's rate is rB, and C's rate is rC. Their combined rate is rA + rB + rC. If they work for 't' days, the total work done is (rA + rB + rC) * t.

If the total work is W, and A takes 'a' days, B takes 'b' days, and C takes 'c' days: rA = W/a, rB = W/b, rC = W/c. Combined Rate = W/a + W/b + W/c = W * (1/a + 1/b + 1/c). Time taken together = Total Work / Combined Rate = W / [W * (1/a + 1/b + 1/c)] = 1 / (1/a + 1/b + 1/c).

Work Done in Alternate Days

When individuals work on alternate days, we calculate the work done in one full cycle (e.g., 2 days if two people work alternately) and then determine how many cycles are needed to complete the work.

Example: A can do a work in 10 days, and B can do it in 20 days. They work on alternate days, starting with A. In how many days will the work be completed? A's rate = 1/10 work/day. B's rate = 1/20 work/day. In 2 days (1 cycle: A works day 1, B works day 2), work done = (1/10) + (1/20) = 3/20 work. Total work = 1. Number of cycles needed = Total Work / Work per cycle = 1 / (3/20) = 20/3 ≈ 6.67 cycles. This means 6 full cycles will be completed, and then some work will be done in the next cycle. Work done in 6 cycles (12 days) = 6 * (3/20) = 18/20 = 9/10 work. Remaining work = 1 - 9/10 = 1/10 work. On the 13th day, A starts again. A does 1/10 work per day. Time taken by A to complete the remaining 1/10 work = (1/10) / (1/10) = 1 day. Total time = 12 days (for 6 cycles) + 1 day (for remaining work) = 13 days.

Work Done by Men, Women, and Children

These problems often involve different groups of people (men, women, children) with different assumed work rates. The key is to establish a relationship between their efficiencies.

For example, if 3 men can do a work in 10 days, and 5 women can do the same work in 12 days, and 8 children can do it in 15 days. We need to find the time taken by, say, 1 man, 1 woman, and 1 child together. Total work done by 3 men in 1 day = 1/10. So, 1 man's work in 1 day = (1/10) / 3 = 1/30. Total work done by 5 women in 1 day = 1/12. So, 1 woman's work in 1 day = (1/12) / 5 = 1/60. Total work done by 8 children in 1 day = 1/15. So, 1 child's work in 1 day = (1/15) / 8 = 1/120. Now, we can find the combined rate of 1 man, 1 woman, and 1 child: Combined rate = 1/30 + 1/60 + 1/120 = (4 + 2 + 1) / 120 = 7/120 work/day. Time taken = 1 / (7/120) = 120/7 days.

Pipes and Cisterns

This is a variation of Time and Work where 'inlets' fill the cistern and 'outlets' empty it.

• An inlet pipe can fill a cistern in 'x' hours. Its rate of filling is +1/x of the cistern per hour.
• An outlet pipe can empty a cistern in 'y' hours. Its rate of emptying is -1/y of the cistern per hour.

If an inlet pipe fills a cistern in 'x' hours and an outlet pipe empties it in 'y' hours, and both are opened simultaneously: The net rate of filling = (1/x - 1/y) per hour. The time taken to fill the cistern is 1 / (1/x - 1/y) = xy / (y - x) hours (assuming x < y, i.e., filling rate is greater than emptying rate).

If there are multiple inlet pipes and outlet pipes, calculate the total filling rate and total emptying rate separately, then find the net rate.

Example Problems

Example 1: A can do a piece of work in 12 days and B can do it in 18 days. In how many days will they finish the work together? Solution: A's rate = 1/12. B's rate = 1/18. Combined rate = 1/12 + 1/18 = (3 + 2) / 36 = 5/36. Time taken together = 1 / (5/36) = 36/5 days = 7.2 days.

Example 2: P is twice as efficient as Q. Together, they can complete a work in 10 days. In how many days can P alone complete the work? Solution: Let Q's rate be 'r'. Then P's rate is '2r'. Combined rate = r + 2r = 3r. Time taken together = 10 days. So, 3r * 10 = 1 (total work). This means 30r = 1, or r = 1/30. P's rate = 2r = 2 * (1/30) = 1/15. So, P can complete the work alone in 15 days.

Example 3: An inlet pipe can fill a cistern in 6 hours. An outlet pipe can empty it in 8 hours. If both pipes are opened simultaneously, how long will it take to fill the cistern? Solution: Inlet pipe rate = +1/6 cistern/hour. Outlet pipe rate = -1/8 cistern/hour. Net rate = 1/6 - 1/8 = (4 - 3) / 24 = 1/24 cistern/hour. Time to fill = 1 / (1/24) = 24 hours.

Basic Algebraic Identities

Algebraic identities are fundamental equations that are true for all values of the variables involved. Mastering these identities is crucial for simplifying expressions, solving equations, and factoring polynomials in various branches of mathematics, including competitive exams.

Linear Expressions (Degree 1)

Identities involving linear expressions are straightforward and often used in basic equation manipulation.

Identity 1: (a + b)² = a² + 2ab + b² This identity states that the square of the sum of two terms is equal to the sum of the squares of each term plus twice their product. Example: (x + 3)² = x² + 2(x)(3) + 3² = x² + 6x + 9.

Identity 2: (a - b)² = a² - 2ab + b² This identity states that the square of the difference of two terms is equal to the sum of the squares of each term minus twice their product. Example: (2y - 5)² = (2y)² - 2(2y)(5) + 5² = 4y² - 20y + 25.

Identity 3: a² - b² = (a + b)(a - b) This is the "difference of squares" identity. It states that the difference between the squares of two terms is equal to the product of their sum and their difference. Example: 9x² - 16y² = (3x)² - (4y)² = (3x + 4y)(3x - 4y).

Identity 4: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca This identity extends the square of a binomial to a trinomial. It states that the square of the sum of three terms is the sum of the squares of each term plus twice the product of each pair of terms. Example: (x + y + 2)² = x² + y² + 2² + 2(x)(y) + 2(y)(2) + 2(2)(x) = x² + y² + 4 + 2xy + 4y + 4x.

Cubic Expressions (Degree 3)

Identities involving cubic expressions are more complex and frequently tested in exams for simplification and factorization.

Identity 5: (a + b)³ = a³ + 3a²b + 3ab² + b³ This identity expands the cube of the sum of two terms. It can also be written as (a + b)³ = a³ + b³ + 3ab(a + b). Example: (x + 2)³ = x³ + 3(x²)(2) + 3(x)(2²) + 2³ = x³ + 6x² + 12x + 8. Using the alternative form: (x + 2)³ = x³ + 2³ + 3(x)(2)(x + 2) = x³ + 8 + 6x(x + 2) = x³ + 8 + 6x² + 12x.

Identity 6: (a - b)³ = a³ - 3a²b + 3ab² - b³ This identity expands the cube of the difference of two terms. It can also be written as (a - b)³ = a³ - b³ - 3ab(a - b). Example: (2y - 3)³ = (2y)³ - 3(2y)²(3) + 3(2y)(3²) - 3³ = 8y³ - 3(4y²)(3) + 3(2y)(9) - 27 = 8y³ - 36y² + 54y - 27.

Identity 7: a³ + b³ = (a + b)(a² - ab + b²) This is the sum of cubes factorization. It states that the sum of the cubes of two terms can be factored into the product of their sum and a trinomial (the square of the first term minus their product plus the square of the second term). Example: x³ + 27 = x³ + 3³ = (x + 3)(x² - 3x + 9).

Identity 8: a³ - b³ = (a - b)(a² + ab + b²) This is the difference of cubes factorization. It states that the difference of the cubes of two terms can be factored into the product of their difference and a trinomial (the square of the first term plus their product plus the square of the second term). Example: 8m³ - 125 = (2m)³ - 5³ = (2m - 5)((2m)² + (2m)(5) + 5²) = (2m - 5)(4m² + 10m + 25).

Identity 9: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) This is a very important identity involving three variables.

Special Case of Identity 9: If (a + b + c) = 0, then a³ + b³ + c³ - 3abc = 0, which implies a³ + b³ + c³ = 3abc. This special case is extremely useful for solving problems where the sum of three terms is zero. Example: If x + y + z = 0, then x³ + y³ + z³ = 3xyz. For instance, if x = 5, y = -3, z = -2, then x + y + z = 5 - 3 - 2 = 0. Therefore, 5³ + (-3)³ + (-2)³ = 3 * (5) * (-3) * (-2). 125 - 27 - 8 = 90. 90 = 90. The identity holds.

Other Useful Identities

Identity 10: (a + b)² + (a - b)² = 2(a² + b²) This identity is derived by adding Identity 1 and Identity 2. (a² + 2ab + b²) + (a² - 2ab + b²) = 2a² + 2b² = 2(a² + b²).

Identity 11: (a + b)² - (a - b)² = 4ab This identity is derived by subtracting Identity 2 from Identity 1. (a² + 2ab + b²) - (a² - 2ab + b²) = 4ab.

Applications of Algebraic Identities

Algebraic identities are used extensively in:

• Simplifying Expressions: Complex algebraic expressions can often be simplified into a more manageable form using identities. Example: Simplify [(x+y)² - (x-y)²] / xy. Using Identity 11, the numerator is 4xy. So, the expression becomes 4xy / xy = 4.
• Solving Equations: Identities help in transforming equations into simpler forms or in factoring them for easier solution. Example: Solve for x: x² - 10x + 25 = 0. Recognizing this as (x - 5)² = 0, we can easily find x = 5.
• Factorization: Identities provide standard patterns for factoring polynomials. Example: Factor 16a² + 24ab + 9b². This fits the pattern of (x + y)² = x² + 2xy + y², where x = 4a and y = 3b. So, 16a² + 24ab + 9b² = (4a + 3b)².
• Evaluating Numerical Expressions: Identities can simplify calculations involving large numbers. Example: Calculate 101². Using (a+b)² = a² + 2ab + b², let a = 100 and b = 1. 101² = (100 + 1)² = 100² + 2(100)(1) + 1² = 10000 + 200 + 1 = 10201. Example: Calculate 99². Using (a-b)² = a² - 2ab + b², let a = 100 and b = 1. 99² = (100 - 1)² = 100² - 2(100)(1) + 1² = 10000 - 200 + 1 = 9801. Example: Calculate 51² - 49². Using a² - b² = (a + b)(a - b), let a = 51 and b = 49. 51² - 49² = (51 + 49)(51 - 49) = (100)(2) = 200.

Practice is key to recognizing these patterns quickly and applying the correct identity. Regularly solving problems involving these identities will build speed and accuracy.