# Time, Work and Distance

## Time And Distance

Decision making for planning, policy, and management relies increasingly on quantitative reasoning, which entails the collection, analysis and interpretation of quantitative data. This course is designed to introduce principles and techniques to solve time and distance related problems.
Develop Logical Reasoning in a Problem-Solving Framework. One goal is to develop a disciplined logical analysis of “word” problems. Such reasoning is the foundation for building simple mathematical models of problems – models implicit in counting time and distance. A good way to learn this logical thinking is by working many exercises. Students will use this reasoning often (consciously or unconsciously) in computer science, operations research, and probability and statistics. However, a logical mind will serve a person well in ANY field.

At the end of these course students:
To be understood the types of formula are used to measure time and Distance.
To be able to calculate time and distance while an object that moves at a constant rate is said to be in uniform motion, Uniform motion problems may involve objects going the same direction, opposite directions, or round trips.
To be solving many problems related to time and distance and which can be useful to students to face interviews.

Prerequisities (Important Formulas):

Speed = Distance/Time

Time = Distance/speed

Distance = speed*time

1km/hr = 5/18 m/s

1 m/s = 18/5 Km/hr

If the ratio of the speed of A and B is a:b,then the ratio of the time taken by them to cover the same distance is 1/a : 1/b or b:a

Suppose a man covers a distance at x kmph and an equal distance at y kmph. then the average speed during the whole journey is (2xy/x+y) kmph

An object that moves at a constant rate is said to be in uniform motion. The formula d = rt gives the relationship between distance d, rater, and time t. Uniform motion problems may involve objects going the same direction, opposite directions, or round trips.

## Time, Work and Distance Solved Examples

Example 1.) Walking at the rate of 4kmph a man cover certain distance in 2hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.

Answer: 40min
Explanation: Distance = Speed * time
4*11/4 = 11km
New Speed = 16.5 kmph
Therefore time = D/S = 11/16.5 = 40min

Example 2.) A train covers a distance in 50 min, if it runs at a speed of 48kmph on an average. The speed at which the train must run to reduce the time of journey to 40min will be.

Answer: 60 min
Explanation
: Time = 50/60 hr = 5/6hr
Speed = 48mph
distance = S*T = 48 * 5/6 = 40km
time = 40/60hr = 2/3hr
New speed = 40* 3/2 kmph = 60kmph

Example 3.) Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph respectively. What time will they take to be 8.5km apart, if they walk in the same direction?

Answer: 17 hrs
Explanation: The relative speed of
the boys = 5.5kmph – 5kmp = 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs

Example 4.) In covering a distance of 30 km, Sachin takes 2 hours more than Rohit. If Sachin doubles his speed, then he would take 1 hour less than Rohit. Sachin speed is:

Answer: 5 kmph
Explanation: Let Sachin’s speed be X km/hr.
Then, 30/x - 30/2x = 3
6x = 30
x = 5 km/hr.

Example 5.) A train 140m long running at 60kmph. In how much time will it pass a platform 260m long?

Answer: 5 kmph
Explanation: Distance travelled = 140 + 260m = 400m,
Speed = 60 * 5/8 = 50/3m
Time = 400 * 3/50 = 24 Seconds

## Time and Work

Work is the quantity of energy transferred from one system to another but for question based on this topic. OR Problem on work are based on the application of concept of ratio of time and speed.

Above mentioned definition of work throws light on three important points.
1.Work = 1 ( as it is always measured as a whole) = Distance
2.Rate at which work is done = speed
3.Number of days required to do the work = Time

Formulas:
1
.Work from Days: If A can do a piece of work in n days, then A's 1 day's work =$\frac{1}{n}$
2.Days from Work: If A's 1 day's work =$\frac{1}{n}$ then A can finish the work in n days.

3.Ratio:
(a) If A is thrice as good a workman as B, then:
(b) Ratio of work done by A and B = 3 : 1.
(c) Ratio of times taken by A and B to finish a work = 1 : 3.

4.If A is 'x' times as good a workman as B, then he will take $\frac{1}{x}th$of the time by B to do the same work.

5.A and B can do a piece of work in 'a' days and 'b' days respectively, then working together, they will take$\frac{xy}{x+y}$ days to finish the work and in one day, they will finish$\frac{x+y}{xy}th$ part of work

Problem 1: If 5 women or 8 girls can do a work in 84 days. In how many days can 10 women and 5 girls can do the same work?

Solution: Given that 5 women is equal to 8 girls to complete a work. So, 10 women = 16 girls. Therefore 10 women + 5 girls = 16 girls + 5 girls = 21 girls.
8 girls can do a work in 84 days then 21 girls can do a work in (8*$\frac{84}{21}$) = 32 days.
Therefore 10 women and 5 girls can a work in 32 days

Problem 2: Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long it take both A & B, working together but independently,to do the same job?

Solution: A's one hour work = $\frac{1}{8}$. B's one hour work = $\frac{1}{10}$. (A+B)'s one hour work
= $\frac{1}{8}$+$\frac{1}{10}$ =$\frac{9}{40}$. Both A & B can finish the work in $\frac{40}{9}$ days

Problem 3: A can finish a work in 18 days and B can do the same work in half the time taken by A. Then, working together, what part of the same work they can finish in a day?

Solution: Given that B alone can complete the same work in days = half the time taken by A = 9 days
A's one day work = $\frac{1}{18}$
B's one day work = $\frac{1}{9}$
(A+B)'s one day work = $\frac{1}{18}$+$\frac{1}{9}$ = $\frac{1}{6}$

Problem 4: A is twice as good a workman as B and together they finish a piece of work in 18 days.In how many days will A alone finish the work.

Solution: if A takes x days to do a work then B takes 2x days to do the same work
= > $\frac{1}{x}$+$\frac{1}{2x}$ = $\frac{1}{18}$
= > $\frac{3}{2x}$ = $\frac{1}{18}$
= > x = 27 days.
Hence, A alone can finish the work in 27 days.

Problem 5: A can do a certain work in 12 days. B is 60% more efficient than A.
How many days does B alone take to do the same job?

Solution: Ratio of time taken by A & B = 160:100 = 8:5
Suppose B alone takes x days to do the job.
Then, 8:5::12:x
= > 8x = 5*12
= > x = $\frac{15}{2}$ days.

## Time, Work and Distance Questions from Previous Year Exams

IBPS bank clerk CWE

## Time, Work and Distance Video Lecture

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