Category: Mathematics

Time Speed and Distance is a major concept in Mathematics. Time and Distance are used extensively for questions relating to topics like circular motion, boats, and streams, motion in a straight line, clocks, races, etc. This article gives you a complete idea of the Relationship Between Time Speed and Distance, Units, Conversions, etc. Get Formula for Time and Distance, Solved Examples explaining the concept in detail.

Time Speed and Distance – Definition

Speed is a concept in motion that is all about how slow or fast an object travels. Speed is defined as the distance divided by Time. Speed, distance, and time are given to solve for one of the three variables when a piece of certain information is known.

Relationship between Time Speed and Distance

Speed = Distance/Time

Speed tells us how fast or slow an object travels and describes the distance traveled divided by the time taken to cover the distance.

From the above formula, Speed is directly proportional to Distance and inversely proportional to Time.

Units of Time Speed and Distance

Speed Distance and Time has different units for each of them and they are given as under

Time: seconds(s), minutes (min), hours (hr)

Distance: (meters (m), kilometers (km), miles, feet

Speed: m/s, km/hr

If Distance and Time Units are known Speed Units can be derived easily.

Time Speed and Distance Conversions

• To change between km/hour to m/sec, we multiply by 5/18. So, 1 km/hour = 5/18 m/sec
• To change between m/sec to km / hour, we multiply by 18/5. So, 1m /sec = 18/5 km/hour = 3.6 km/hour
• Similarly, 1 km/hr = 5/8 miles/hour
• 1 yard = 3 feet
• 1 mile= 1.609 kilometer
• 1 kilometer= 1000 meters = 0.6214 mile
• 1 mile = 1760 yards
• 1 mile = 5280 feet
• 1 hour= 60 minutes= 60*60 seconds= 3600 seconds
• 1 yard = 3 feet
• 1 mph = (1 x 5280) / (1 x 3600) = 22/15 ft/sec
• 1 mph = (1 x 1760) / (1 x 3600) = 22/45 yards/sec
• For a certain distance, if the ratio of speeds is a : b, then the ratio of times taken to cover the distance is given by b : a and vice versa.

Applications of Time Speed and Distance

Check out different models of questions asked on the concept of Time Speed and Distance. They are as under

Average Speed

Average Speed = Total Distance Travelled/Total Time Taken

Case 1: When Distance is Constant, Average Speed is given by = 2xy/x+y where x, y are two speeds at which the same distance is covered.

Case 2: When Time taken is Constant, Average Speed = (x+y)/2 where x, y are two speeds at which we traveled for the same time.

Examples

1. A person travels from one place to another at 40 km/hr and returns at 160 km/hr. If the total time taken is 5 hours, then find the Distance?

Solution:

Here the Distance is constant, so the Time taken will be inversely proportional to the Speed.

Ratio of Speed = 40:160

= 1:4

Ratio of Time = 4:1

Time taken = 5 hours

Therefore, time taken while going is 4 hours and while returning is 1 hour

Distance = Speed* Time

= 40*4

= 160 Km

Therefore, Distance Travelled is 160 Km.

2. Traveling at 3/2 nd of the original Speed a train is 20 minutes late. Find the usual Time taken by the train to complete the journey?

Solution:

Let the usual Speed be S1 and the usual Time is T₁. As the Distance to be covered in both cases is the same,

The ratio of usual Time to the Time taken when he is late will be the inverse of the usual Speed and the Speed when he is late

If the Speed is S₂ = S₁ then the Time taken T₂ = 3/2 T₁ Given T₂ – T₁ = 20 =>3/2 T1 – T1 = 20 => T1 = 40 minutes.

Inverse Proportionality of Speed & Time

Speed is inversely proportional to time when the distance is constant. If Speeds are in the ratio of m:n then the time taken will be n:m They are two methods of solving questions.

Using Inverse Proportionality
Using Constant Product Rule

Example

After traveling 60km, a train is meeting with an accident and travels at (2/3)rd of the usual Speed and reaches 30 min late. Had the accident happened 15km further on it would have reached 20 min late. Find the usual Speed?

Solution:

Using Inverse Proportionality

Here there are 2 cases

Case 1: accident happens at 60 km

Case 2: accident happens at 75 km

The difference between the two cases is only for the 15 km between 60 and 75. The time difference of 10 minutes is only due to these 15 km.

In case 1, 15 km between 60 and 75 is covered at (2/3)rd Speed.

In case 2, 15 km between 60 and 75 is covered at the usual Speed.

So the usual Time “t” taken to cover 15 km, can be found out as below. 3/2 t – t = 10 mins = > t = 20 mins, d = 15 km

so Usual Speed = 15km/20min = 15km/0.33hr= 45Km/hr

Using Constant Product Rule Method

Let the actual Time taken be T

There is a (1/3)rd decrease in Speed, this will result in a (1/2)nd increase in Time taken as Speed and Time are inversely proportional

The delay due to this decrease is 10 minutes

Thus 1/2 T= 10 and T=20 minutes

Also, Distance = 15 km

Thus Speed = 15/20 minutes = 15km/0.33hr = 45km/hr

Meeting Point

If two people travel from points A and B towards each other they meet at point P. Distance covered by them on the meeting is AB. Time taken by both to meet is the same. Since Time is constant Distance AP and BP will be in the ratio of their Speeds.

Consider the distance between A and B is d.

If two people walking towards each other from A and B. when they meet for the first time they cover a distance of “d ” and when they meet for the second time they cover a distance of “3d” and when they meet for the third time they cover a distance of “5d” …..

Example

1. Amar and Arun have to travel from Delhi to Jaipur in their respective cars. Arun is driving at 45 kmph while Amar is driving at 60 kmph. Find the Time taken by Amar to reach Jaipur if Arun takes 6 hrs?

Solution:

Since the Distance covered is constant in both cases, the Time taken will be inversely proportional to the Speed.

From the given data, the Speed of Amar and Arun are in ratio 45:60 or 3:4.

So the ratio of the Time taken by Arun to that taken by Amar will be in the ratio 4:3. So if Arun takes 6 hrs, Amar will take 4.5 hrs.

Solved Examples on Time and Distance

1. Seetha is driving a car with a speed of 60 km/hr for 1.5hr. How much distance does she travel?

Solution:

Speed = 60 Km/hr

Time = 1.5 hr

Distance = Speed *Time

= 60*1.5

= 90 Km

Therefore, Sheela Travels a distance of 90km.

2. While going to an office, Ram travels at a speed of 35 kmph, and on his way back, he travels at a speed of 40 kmph. What is his average speed for the whole journey?

Solution:

In this case, Distance is constant

Average Speed = 2xy/x+y where x, y is the speeds at which the distance is covered

Substitute the Speeds from the given data

Average Speed = (2*35*40)/35+40

= 37.33 km/hr

The Average Speed of the Whole Journey is 37.33Kmph

3. Ramu and Somesh are standing at two ends of a room with a width of 40 m. They start walking towards each other along the width of the room with a Speed of 4 m/s and 3 m/s respectively. Find the total distance traveled by Ramu when he meets Somesh for the third time?

Solution:

When Ramu meets Somesh for the third time he would have covered a distance of 5d i.e. 5*40m = 200m

The ratio of Speed of Ramu and Somesh is 4:3 so the Distance traveled by both of them will also be in the ratio of 4:3

Probability is a branch of mathematics that deals with the occurrence of random events. It is expressed from zero to one and predicts how likely events are to happen. In general, Probability is basically the extent to which something is likely to happen. You will learn about Probability Distribution where you will learn the possibility of outcomes for a random experiment.

Probability Definition

Probability is the measure of the likelihood of an event to occur.  In the case of events, we can’t predict with total certainty. We can only predict the cancer of an event to occur i.e. how likely it is to happen. Probability ranges between 0 to 1 in which 0 indicates the event to be an impossible one and 1 indicates a certain event.

Example: For instance, when we toss a coin there are only two possibilities either head or tail(H,T). If we toss two coins in the air there are three possible outcomes that are both the coins show heads, both the coins show tails, one is head and the other is tail i.e. (H, H), (T, T), (H, T).

Formula for Probability

Probability is defined as the possibility of an event to occur. The formula for Probability is given as the ratio of the number of favorable events to the total number of possible outcomes.

Probability of an event to happen = No. of Favourable Outcomes/ Total Number of Outcomes

This is the basic formula for Probability.

Probability Tree

Tree Diagram helps to organize and visualize different possible outcomes. Branches and ends are the two main positions of the tree. Each branch Probability is written on the branch and the ends contain the final outcome. Tree Diagram helps you to figure out when to multiply and add.

Types of Probability

There are three types of major probabilities. They are

• Theoretical Probability
• Experimental Probability
• Axiomatic Probability

Theoretical Probability: It depends on the possible chances of something to happen. Theoretical Probability mainly depends on the reasoning behind probability.

Experimental Probability: This kind of Probability depends on the observation of the experiment. Experimental Probability can be calculated on the number of possible outcomes to the total number of trials.

Axiomatic Probability: A Set of Rules or Axioms are Set applies to all types. Using the axiomatic approach to probability, the chances of occurrence or non-occurrence of the events can be quantified.

Conditional Probability is nothing but the likelihood of an event or outcome occurring based on the occurrence of a previous event or outcome.

Probability of an Event

Let us consider an Event E that occurs in r ways out of n possible ways. The probability of happening an event or its success is given by

P(E) = r/n

The probability of an event or its failure is given by

P(E’) = (n-r)/n = 1-(r/n)

E’ represents the event will not occur.

Therefore, we can say that

P(E)+P(E’) = 1

What are Equally Likely Events?

If the events have the same theoretical probability of happening then they are called Equally Likely Events. Results of Sample Space are said to be equally likely if all of them have the same probability of occurring. Below are some examples of Equally Likely Events.

• Getting 2 or 3 on throwing a die.
• Getting 1, 3, 4 on throwing a die

are all Equally Likely Events since the Probabilities of Each Event are Equal.

Complementary Events

In the Case of Such Events, there will only be two outcomes that state whether an event will occur or not. The complement of an event occurring is the exact opposite that the probability of an event is not occurring.

• It may or may not rain today
• Winning a lottery or not.
• You win the lottery or you don’t.

Probability Density Function

It is a probability function that represents the density of continuous random variables lying between a certain range of values. Standard Normal Distribution is used to create a database or statistics that are used in science to represent the real-valued variables whose values are unknown.

Additive Law of Probability

If E₁ and E₂ be any two events (not necessarily mutually exclusive events), then P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)

Probability Terms

Some of the Important Probability Terms are discussed here

Sample Space: Set of all possible outcomes that occur in any trail.

Example:

Tossing a Coin, Sample Space (S) = {H, T}

When you Roll a Die Sample Space (S) = {1, 2, 3, 4, 5, 6}

Sample Point: It is one of the Possible Outcome.

Example:

In a deck of cards, 3 of hearts is a sample point.

Experiment or Trial: Series of Actions where outcomes are always uncertain.

Example: Tossing a Coin, Choosing a Card from a Deck of Cards, Throwing a Dice.

Event: Single outcome of an experiment.

Example: Getting Tails while Tossing a Coin is an Event.

Outcome: Possible Result of an Experiment.

Head is a possible outcome when a coin is tossed.

Impossible Event: The Event can’t happen

While Tossing a Coin it is impossible to get head and tail at the same time.

Solved Examples on Probability

1. Find the Probability of getting 2 on rolling a die?

Solution:

Sample Space = {1, 2, 3, 4, 5, 6}

Number of Favourable Events = 1 i.e. {2}

Total Number of Outcomes = 6

Probability P = 1/6

Therefore, Probability of getting 2 on rolling a die is 1/6.

2. Two dice are rolled find the Probability that the Sum is

equal to 1

equal to 5

equal to 8

Solution:

In order to find the probability whose sum is equal to 5 we need to figure out the Sample Space

Practice the Questions based on the Uniform Rate of Growth and Depreciation from here. Learn about the Concept of Uniform Rate of Growth and Depreciation better by going through this entire article. You will find How to Apply Principal of Compound Interest on Combination of Uniform Rate of Growth and Depreciation. Check out Formula, Solved Examples on the concept of Uniform Rate of Increase or Decrease from this article. Get Step by Step Solutions for all the Problems provided and get a good hold on the concept.

How to find the Uniform Rate of Increase or Decrease?

Let us discuss how to find the Uniform Rate of Growth or Depreciation in detail in the below modules.

If a quantity P grows at the rate of r₁% in the first year and depreciates at r₂% in the second year and grows at r₃% in the third year then the quantity becomes Q after 3 years.

Take r/100 with a positive sign for each growth or appreciation of r % and r/100 with a negative sign for depreciation of r%.

Solved Examples on Uniform Rate of Growth or Depreciation

1. The current population of a town is 60,000. The population increases by 10 percent in the first year and decreases by 5% in the second year. Find the population after 2 years?

Solution:

Initial Population = 60,000

r₁ =10%, r₂ = 5%

Population after 2 Years Q = P(1+r₁/100)(1-r₂/100)

= 60,000(1+10/100)(1-5/100)

= 60,000(1+0.1)(1-0.05)

= 60,000(1.1)(0.95)

= 62,700

Therefore, the Population after 2 Years is 62,700.

2. The count of a certain breed of bacteria was found to increase at the rate of 4% per hour and then decrease by 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 2,00,000.

Solution:

Since the Population of bacteria increases and decreases we use the formula

Q = P(1+r₁/100)(1-r₂/100)

= 2,00,000(1+4/100)(1-2/100)

= 2,00,000(1+0.04)(1-0.02)

= 2,00,000(1.04)(0.98)

= 2,03,840

Bacteria at the end of 2 hours is 2,03,840

3. The price of a car is $ 2,50,000. The value of the car depreciates by 10% at the end of the first year and after that, it depreciates by 15%. What will be the value of the car after 2 years?

Solution:

Initial Price of the Car = $ 2,50,000

r₁  = 10% r₂ = 15%

Since the Price of a Car depreciates we use the formula

Q = P(1-r₁/100)(1-r₂/100)

= 2,50,000(1-10/100)(1-15/100)

= 2,50,000(90/100)(85/100)

= 2,50,000(0.9)(0.85)

= $1,91,250

Value of Car after 2 Years is $1,91,250.

Get to know about the Uniform Rate of Depreciation along with its Formula and Solved Examples. In this article, we will discuss how to apply the Principal of Compound Interest on Problems of Uniform Rate of Depreciation. If the Rate of Decrease is Uniform then we call it a Uniform Decrease or Depreciation. Refer to the Solved Examples on How to find Rate of Depreciation or Decrease. We have listed the Step by Step Solutions for each and every problem making it easy for you to understand.

How to Calculate the Rate of Uniform Decrease or Depreciation?

Let us discuss in detail how to find the Uniform Rate of Decrease or Depreciation in the coming sections.

If the Present Value P of a Quantity Decreases at the rate of r % per unit of time then Value Q of the Quantity after n units of time is given by

Q = P(1-r/100)ⁿ

Depreciation in Value = P – Q

= P – P(1-r/100)ⁿ

= P{1-(1-r/100)ⁿ}

Efficiency of a Machine after regular use, Decrease in the Value of Furniture, Buildings, Decrease in the Number of Diseases, etc. come under Uniform Rate of Decrease or Depreciation.

Solved Examples on Uniform Rate of Depreciation

1. The value of a residential flat constructed at a cost of Rs.1,20,000 is depreciating at the rate of 5% per annum. What will be its value 4 years after construction?

Solution:

From the Given Data P = Rs. 1,20,000

r = 5%

n = 4 Years

We know the formula to find Q = P(1-r/100)ⁿ

Substitute the input values in the above formula and we have the equation as such

Q = 1,20,000(1-5/100)⁴

= 1,20,000(95/100)⁴

= 1,20,000(0.8145)

= Rs. 97,740

Therefore, the Value of a Residential Flat after 4 Years would be Rs. 97,740.

2. The price of a motor vehicle depreciates by 10% every year. By what percent will the price of the car reduce after 2 years?

Solution:

From the given data

Consider the Price be P

r = 10%

n = 2 Years

We know the formula to find Q = P(1-r/100)ⁿ

Substitute the input values in the above formula and we have the equation as such

Q = P(1-10/100)²

=P(90/100)²

= P(9/10)(9/10)

= 81P/100

Reduction in Price = P -81P/100

= 19P/100

Percent Reduction in Price = (19P/100)/P*100%

= 19%

3. The cost of a school bus depreciates by 8 % every year. If its present worth is $ 27,000. What will be its value after three years?

Solution:

From the given data

P = $27,000

r = 8%

n = 3 years

Formula to Calculate the Price of Depreciated Value Q = P(1-r/100)ⁿ

= 27,000(1-8/100)³

= 27,000(92/100)³

= 27,000(0.92)³

= 27,000(0.778)

= $ 21, 024

The Cost of a School Bus after the Depreciation is $ 21,024.

In our Day-Day Lives, we see entities such as Population of a City, Value of a Property, Weight of a Child, Height of a Tree that grow over a period of time. We call the Increase in Quantity as Growth and the Growth Per Unit Time is known as the Rate of Growth. If the Growth Rate occurs at the Same Rate then we call it Uniform Increase or Uniform Growth Rate. Check the Formulas for Uniform Rate of Growth and the Solved Problems for finding the Principal of Compound Interest in Uniform Rate of Growth explained step by step.

How to find the Uniform Rate of Growth?

Learn how to calculate the uniform growth rate by referring to the below modules.

If the Present Value P of a quantity increases at the rate of r % per unit of time then the value Q of the quantity after n units of time is as such

Q = P(1+r/100)ⁿ

Growth is obtained by subtracting the Increased Value from the Actual Value

Growth = Q – P

= P(1+r/100)ⁿ – P

= P{(1+r/100)ⁿ -1}

Solved Examples on Principal of Compound Interest in Uniform Rate of Growth

1. The population of the town increases by 8% every year. If the present population is 7000, what will be the population of the town after 2 years?

Solution:

Given Current Population = 7000

n = 2 Years

Rate of Interest = 8%

Q = P(1+r/100)ⁿ

Substitute the input values in the above formula

= 7000(1+8/100)²

= 7000(1+0.08)²

= 7000(1.08)²

= 7000(1.1664)

= 8164

Population of a Town after 2 years is 8164.

2. John buys a plot of land for $ 20,000. If the value of the land appreciates by 10% every year then find the profit that John will make by selling the plot after 3 years?

Solution:

P = $20,000

interest rate = 10%

n = 3 years

Q = P(1+r/100)ⁿ

= 20,000(1+10/100)³

= 20,000(1+0.1)³

= 20,000(1.1)³

= $26620

Profit made by John = $26620 – $20,000

= $6,620

3. Mike purchased a bike for Rs. 45,000. If the cost of his bike is appreciated at a rate of 5% per annum, then find the cost of the bike after 3 years?

Solution:

Initial Price = Rs. 45, 000

Rate of Appreciation = 5 % Per Annum

n = 3 years

Q = P(1+r/100)ⁿ

= 45,000(1+5/100)³

= 45,000(1+0.05)³

= 45,000(1.05)³

= Rs. 52093

Therefore, the Cost of the Bike after 3 years is Rs. 52093.