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Problems on Calculating Speed | Speed Questions and Answers

Solve different types of problems on calculating speed and get acquainted with various models of questions asked in your exams. Be aware of the Formula to Calculate and Relationship between Speed Time and Distance. Practice Speed Problems on a regular basis so that you can be confident while attempting the exams. We even provided solutions for all the Questions provided and explained everything in detail for better understanding. Try to solve the Speed Questions on your own and then cross-check where you are lagging.

We know the Speed of the Object is nothing but the distance traveled by the object in unit time.

Formula to find out Speed is given by Speed = Distance/Time

Word Problems on Calculating Speed

1.  A man walks 25 km in 6 hours. Find the speed of the man?

Solution: Distance traveled = 25 km Time taken to travel = 6 hours Speed of Man = Distance traveled/Time taken = 25km/6hr = 4.16 km/hr Therefore, a man travels at a speed of 4.16 km/hr

2. A car covers a distance of 420 m in 1 minute whereas a train covers 70 km in 30 minutes. Find the ratio of their speeds?

Solution: Speed of the Car = Distance Traveled/Time Taken = 420m/60 sec = 7 m/sec

Speed of the Train = Distance Traveled/Time Taken = 70 km/1/2 hr = 140 km/hr

To convert it into m/sec multiply with 5/18 = 140*5/18 = 38.8 m/sec = 39 m/sec (Approx) Ratio of Speeds = 7:39

3. A car moves from A to B at a speed of 70 km/hr and comes back from B to A at a speed of 40 km/hr. Find its average speed during the journey?

Solution: Since the distance traveled is the same the Average Speed= (x+y)/2 where x, y are two different speeds Substitute the Speeds in the given formula Average Speed = (70+40)/2 = 110/2 = 55 km/hr The Average Speed of the Car is 55 km/hr

4. A bus covers a certain distance in 45 minutes if it runs at a speed of 50 km/hr. What must be the speed of the bus in order to reduce the time of journey by 20 minutes?

Solution: Speed = Distance/Time 50 = x/3/4 50 = 4x/3 4x = 150 x = 150/4 = 37.5 km

Now by applying the same formula we can find the speed

Now, time = 40 mins or 0.66 hr since the journey is reduced by 20 mins

S = Distance/Time = 37.5/0.66 = 56.81 km/hr

5. Ram traveled 200 km in 3 hours by train and then traveled 140 km in 3 hours by car and 5 km in 1/2 hour by cycle. What is the average speed during the whole journey?

Solution: Distance traveled by Train is 200 km in 3 hours Distance Traveled by Car is 140 km in 3 hours Distance Traveled by Cycle is 5 km in 1/2 hour Average Speed = Total Distance/Total Time = (200+140+5)/(3+3+1/2) = 345/6 1/2 = 345/(13/2)” = 345*2/13 = 53.07 km/hr

6. A train covers 150 km in 3 hours. Find its speed?

Solution: Speed = Distance/Time = 150 km/3 hr = 50 km/hr Therefore, Speed of the Train is 50 km/hr.

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Speed Formula – Definition, Examples, Practice Problems, FAQs

What is the formula for speed in math, formula for speed, how to calculate speed in math, speed, distance, and time: formulas, solved examples on speed formula, practice problems on speed formula, frequently asked questions about the speed formula.

The formula for speed in math is determined by dividing the distance traveled by the time taken to cover that distance . 

Speed $= \frac{Distance}{Time}$

Speed is a measure of how quickly an object moves or the rate at which an object covers a certain distance. It is a scalar quantity and is typically expressed in units such as meters per second (m/s), kilometers per hour (km/h), or miles per hour (mph).

Example: Suppose that a car travels from point A to point B in half an hour. If the distance between A and B is 2 miles, what was the speed of the car?

Distance = 2 miles

Time taken = 0.5 hour

Speed $= \frac{Distance}{Time} = \frac{2}{0.5} = 4$ miles per hour

A mathematical equation called the speed formula is used to determine how fast an object is moving. It is determined by dividing the total distance traveled by the time needed to complete that distance. 

The formula to find speed is

Speed$ = \frac{Distance}{Time}$

S $= \frac{D}{T}$

D: Distance

The unit of the speed calculated by this formula is dependent on the units of time and distance. 

In order to determine the speed in mathematics, it is necessary to know both the distance traveled and the amount of time taken to cover that distance. Follow this step-by-step method to calculate speed.

Step 1: Determine the total distance traveled

Note that if the problem mentions a round journey, it means that the same distance is covered twice.

Step 2: Measure or determine the time taken to cover the distance

Step 3: Use the formula for speed

To find the speed, divide the distance by the amount of time taken.

Step 4: Write the answer using the correct unit of measurement.

The speed will be given in meters per second (m/s), for instance, if the distance is in meters and the time is in seconds. The speed will be expressed in miles per hour (mph) if the distance is measured in miles and the time in hours.

To calculate one quantity, one must have information about the other two quantities involved.

The picture given above shows three different triangles showing rearrangements of the formula of speed. The first triangle shows the formula for speed. The second and third triangles depict the formulae for distance and time, respectively.

Facts on Speed Formula

• Speed is a scalar quantity. Speed does not consider the direction of motion, only the magnitude.
• Distance is the measure of how much ground an object has covered or the total movement of an object on ground.
• Time is the measure of the duration or the interval between two events.
• Speed is inversely proportional to time. As the speed increases, the time required to travel a certain distance decreases.

In this article, we learned about the concept of speed and its formula, which calculates the rate of motion. Speed is a scalar quantity that considers only magnitude. To reinforce our understanding, let’s now apply the speed formula through solving examples and attempting MCQs for better comprehension.

1. What is the average speed of a car that travels a distance of 120 miles in 2 hours?

Distance = 120 miles

Time = 2 hours

Use the speed formula to calculate the speed.

Speed $= \frac{120 \;miles}{2 \;hours} = 60$mph

Therefore, the average speed of a car that travels a distance of 120 miles in 2 hours is 60 mph.

2. What is the average speed of a cyclist who covers a distance of 30 miles in 1.5 hours?

Distance = 30 miles

Time = 1.5 hours

Speed $= \frac{30 \;miles}{1.5 \;hours} = 20$mph

Therefore, the average speed of a cyclist who covers a distance of 30 miles in 1.5 hours is 20 mph.

3. How much distance does a train cover when traveling at a constant speed of 80 miles for 3 hours?

Speed = 80 mph

Time = 3 hours

Use the rearranged speed formula to calculate the distance.

Distance $= Speed \times Time$

Distance $= 80\; mph \times 3\; hours = 240$ miles

Therefore, the train, when traveling at a constant speed of 80 miles for 3 hours, is 240 miles.

4. What is the time required by a car that travels a distance of 400 miles at a speed of 60 mph?

Distance $= 400$ miles

Speed $= 60$ mph

Use the rearranged speed formula to calculate the time.

Time $= \frac{Distance}{Speed}$

Time = \frac{400 \;miles}{60 \;mph} ≈ 6.67$ hours

Therefore, for the car to cover a distance of 400 miles, it takes approximately 6.67 hours (or 6 hours and 40 minutes).

5. What will be the time a car takes at a speed of 60 miles per hour to move a distance of 180 miles?

Solution: 

Distance = 180 miles

Speed = 60 miles per hour

Using the formula of time, we can calculate the time-

Time$= \frac{Distance}{Speed}$

Time $= \frac{180 \;miles}{60 \;miles\; per\; hour} = 3$ hours

Therefore, the car takes approximately 3 hours to move a distance of 180 miles at a speed of 60 miles per hour.

Speed Formula - Definition, Examples, Practice Problems, FAQs

Attend this quiz & Test your knowledge.

What is the formula to calculate speed?

What is the speed of a car that travels a distance of 120 miles in 2 hours, how much time does it take for a cyclist to travel a distance of 45 miles at a speed of 15 mph, what is a runner’s average speed for completing a marathon at a distance of 42.195 miles in 3.5 hours, how far did a train travel at a speed of 80 mph for 2.5 hours.

What is the difference between speed and average speed?

Speed is a one-time calculation. Speed represents the rate of travel at a specific moment, but the average speed takes into account the different speeds maintained throughout the trip and provides an overall average rate of travel.

How do you calculate speed by using the speed formula?

Typically, the formula for speed is written as:

In this equation, “Speed” denotes the rate at which an object goes, “Distance” denotes the total distance covered by the object, and “Time” denotes how long it takes for the thing to travel that distance.

How is speed different from velocity?

Speed and velocity have distinct meanings in physics. Speed is a scalar quantity that refers to the magnitude of the rate at which an object covers a distance, while velocity is a vector quantity that includes both the magnitude and direction of the object’s motion.

What is the difference between speed from acceleration?

An object’s speed is how quickly it travels, whereas its acceleration is how quickly its speed varies. Positive (speeding up) or negative (slowing down) acceleration are both possible.

What is the mileage of a vehicle?

The mileage of a vehicle refers to the distance it can travel on a certain amount of fuel. It is a measure of the vehicle’s fuel efficiency.

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In order to access this I need to be confident with:

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This topic is relevant for:

Speed Distance Time

Speed Distance Time Triangle

Here we will learn about the speed distance time triangle including how they relate to each other, how to calculate each one and how to solve problems involving them.

There are also speed distance time triangle worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is speed distance time?

Speed distance time is the formula used to explain the relationship between speed, distance and time. That is speed = distance ÷ time . Or to put it another way distance divided by speed will give you the time. Provided you know two of the inputs you can work out the third.

For example if a car travels for 2 hours and covers 120 miles we can work out speed as 120 ÷ 2 = 60 miles per hour.

The units of the the distance and time tell you the units for the speed.

What is the speed distance time triangle?

The speed distance time triangle is a way to describe the relationship between speed, distance and time as shown by the formula below.

\textbf{Speed } \bf{=} \textbf{ distance } \bf{\div} \textbf{ time}

“Speed equals distance divided by time”

Let’s look at an example to calculate speed.

If a car travels 66km in 1.5 hours then we can use this formula to calculate the speed.

This formula can also be rearranged to calculate distance or calculate time given the other two measures. An easy way to remember the formula and the different rearrangements is to use this speed distance time triangle. 

From this triangle we can work out how to calculate each measure: We can ‘cover up’ what we are trying to find and the formula triangle tells us what calculation to do.

Let’s look at an example to calculate time.

How long does it take for a car to travel 34 miles at a speed of 68 miles per hour?

Let’s look at an example to calculate distance.

What distance does a bike cover if it travels at a speed of 7 metres per second for 50 seconds?

What is the speed distance time formula?

The speed distance time formula is just another way of referring to the speed distance time triangle or calculation you can use to determine speed, time or distance.

• speed = distance ÷ time
• time = distance ÷ speed
• distance = speed x time

Time problem

We can solve problems involving time by remembering the formula for speed , distance and time . 

Calculate the time if a car travels at 15 miles at a speed of 36 mph.

Time = distance ÷ speed

Time = 15 ÷ 36 = 0.42 hours

0.42 ✕ 60 = 25.2 minutes

A train travels 42km between two stops at an average speed of 36 km/h. 

If the train departs at 4 pm, when does the train arrive?

Time = 42 ÷ 36 = 1.17 hours

1.17 ✕ 60= 70 minutes = 1 hour 10 minutes.

The average speed of a scooter is 18 km/h and the average speed of a cycle is 10 km/h. 

When both have travelled 99 km what is the difference in the time taken?

Time A = 99 ÷ 18 = 5.5 hours

Time B= 99 ÷ 10 = 9.9 hours

Difference in time = 9.9 – 5.5 = 4.4 hours

4.4 hours = 4 hours and 24 minutes

Units of speed, distance and time

• The speed of an object is the magnitude of its velocity. We measure speed most commonly in metres per second (m/s), miles per hour (mph) and kilometres per hour (km/hr).

The average speed of a small plane is 124mph.

The average walking speed of a person is 1.4m/s.

• We measure the distance an object has travelled most commonly in millimetres (mm), centimetres (cm), metres (m) and kilometres (km).

The distance from London to Birmingham is 162.54km.

• We measure time taken in milliseconds, seconds, minutes, hours, days, weeks, months and years. 

The time taken for the Earth to orbit the sun is 1 year or 365 days. We don’t measure this in smaller units like minutes of hours.

A short bus journey however, would be measured in minutes.

Speed, distance and time are proportional.

If we know two of the measurements we can find the other.

A car drives 150 miles in 3 hours.

Calculate the average speed, in mph, of the car.

Distance = 150 miles

Time = 3 hours

Speed = 150÷ 3= 50mph

Speed, distance, time and units of measure

It is very important to be aware of the units being used when calculating speed, distance and time.

• Examples of units of distance: mm, \ cm, \ m, \ km, \ miles
• Examples of units of time: seconds (sec), minutes, (mins) hours (hrs), days
• Examples of units of speed: metres per second (m/s), miles per hour (mph)

Note that speed is a compound measure and therefore involves two units; a combination of a distance in relation to a time.

When you use the speed distance time formula you must check that each measure  is in the appropriate unit before you carry out the calculation. Sometimes you will need to convert a measure into different units. Here are some useful conversions to remember.

Units of length

Units of time

1 minute = 60 seconds

1 hour = 60 minutes

1 day = 24 hours

Let’s look at an example.

What distance does a bike cover if it travels at a speed of 5 metres per second for 3 minutes?

Note here that the speed involves seconds, but the time given is in minutes. So before using the formula you must change 3 minutes into seconds.

3 minutes = 3\times 60 =180 seconds

Note also that sometimes you may need to convert an answer into different units at the end of a calculation.

Constant speed / average speed

For the GCSE course you will be asked to calculate either a constant speed or an average speed . Both of these can be calculated using the same formula as shown above.

However, this terminology is used because in real life speed varies throughout a journey. You should also be familiar with the terms acceleration (getting faster) and deceleration (getting slower).

Constant speed

A part of a journey where the speed stays the same.

Average speed

A journey might involve a variety of different constant speeds and some acceleration and deceleration. We can use the formula for speed to calculate the average speed over the course of the whole journey.

Average Speed Formula

Average speed is the total distance travelled by an object divided by the total time taken. To do this we can use the formula

Average speed =\frac{Total\, distance}{Total\, time}

If we are calculating an average speed in mph or km/h, we will need to ensure we have decimalised the time before we divide.

How to calculate speed distance time

In order to calculate speed, distance or time:

Write down the values of the measures you know with the units.

Check that the units are compatible with each other, converting them if necessary.

Substitute the values into the selected formula and carry out the resulting calculation.

Write your final answer with the required units.

Explain how to calculate speed distance time

Speed distance time triangle worksheet

Get your free speed distance time triangle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on   compound measures

Speed distance time  is part of our series of lessons to support revision on  compound measures . You may find it helpful to start with the main compound measures lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Compound measures
• Mass density volume
• Pressure force area
• Formula for speed
• Average speed formula

Speed distance time triangle examples

Example 1: calculating average speed.

Calculate the average speed of a car which travels 68 miles in 2 hours.

Speed: unknown

Distance: 68 miles

Time: 2 hours

2 Write down the formula you need to use from the speed, distance, time triangle.

3 Check that the units are compatible with each other, converting them if necessary.

The distance is in miles and the time is in hours. These units are compatible to give the speed in miles per hour.

4 Substitute the values into the formula and carry out the resulting calculation.

5 Write your final answer with the required units.

Example 2: calculating time

A golden eagle can fly at a speed of 55 kilometres per hour. Calculate the time taken for a golden eagle to fly 66 \ km, giving your answer in hours.

Speed: 55 \ km/hour

Distance: 66 \ km

Time: unknown

Write down the formula  you need to use from the speed, distance, time  triangle.

T=\frac{D}{S}

Time = distance \div speed

Speed is in km per hour and the distance is in km , so these are compatible to give an answer for time in hours.

Example 3: calculating distance

Calculate the distance covered by a train travelling at a constant speed of 112 miles per hour for 4 hours.

Speed: 112 \ mph

Distance: unknown

Time: 4 hours

D= S \times T

Distance = speed \times time

Speed is in miles per hour. The time is in hours. These units are compatible to find the distance in miles.

Example 4: calculating speed with unit conversion

A car travels for 1 hour and 45 minutes, covering a distance of 63 miles. Calculate the average speed of the car giving your answer in miles per hour (mph) .

Distance: 63 miles

Time: 1 hour and 45 minutes

S = \frac{D}{T}

Speed = distance \div time

The distance is in miles . The time is in hours and minutes. To calculate the speed in miles per hour , the time needs to be converted into hours only.

1 hour 45 minutes = 1\frac{3}{4} hours = 1.75 hours

Example 5: calculating time with unit conversion

A small plane can travel at an average speed of 120 miles per hour. Calculate the time taken for this plane to fly 80 miles giving your answer in minutes.

Speed: 120 \ mph

Distance: 80 \ miles

T = \frac{D}{S}

Speed is in miles per hour and the distance is in miles . These units are compatible to find the time in hours.

\frac{2}{3} hours in minutes

\frac{2}{3} \times 60 = 40

Example 6: calculating distance with unit conversion

A train travels at a constant speed of 96 miles per hour for 135 minutes. Calculate the distance covered giving your answer in miles.

Speed: 96 \ mph

Time: 135 minutes

D = S \times T

The speed is in miles per hour , but the time is in minutes. To make these compatible the time needs changing into hours and then the calculation will give the distance in miles .

135 minutes

135 \div 60 = \frac{9}{4} = 2\frac{1}{4} = 2.25

Common misconceptions

• Incorrectly rearranging the formula Speed = distance \div time

Make sure you rearrange the formula correctly. One of the simplest ways of doing this is to use the formula triangle. In the triangle you cover up the measure you want to find out and then the triangle shows you what calculation to do with the other two measures.

• Using incompatible units in a calculation

When using the speed distance time formula you must ensure that the units of the measures are compatible. For example, if a car travels at 80 \ km per hour for 30 minutes and you are asked to calculate the distance, a common error is to substitute the values straight into the formula and do the following calculation. Distance = speed \times time = 80 \times 30 = 2400 \ km The correct way is to notice that the speed uses hours but the time given is in minutes. Therefore you must change 30 minutes into 0.5 hours and substitute these compatible values into the formula and do the following calculation. Distance = speed \times time = 80 \times 0.5 = 40 \ km

Practice speed distance time triangle questions

1. A car drives 120 miles in 3 hours. Calculate its average speed.

2. A cyclist travels 100 miles at an average speed of 20 \ mph. Calculate how long the journey takes.

3. An eagle flies for 30 minutes at a speed of 66 \ km per hour. Calculate the total distance the bird has flown.

30 minutes = 0.5 hours

4. Calculate the average speed of a lorry travelling 54 miles in 90 minutes. Give your answer in miles per hour (mph).

Firstly convert 90 minutes to hours. 90 minutes = 1.5 hours

5. Calculate the time taken for a plane to fly 90 miles at an average speed of 120 \ mph. Give your answer in minutes.

180 minutes

Convert 0.75 hours to minutes

6. A helicopter flies 18 \ km in 20 minutes. Calculate its average speed in km/h .

Firstly convert 20 minutes to hours. 20 minutes is a third of an hour or \frac{1}{3} hours. \begin{aligned} &Speed = distance \div time \\\\ &Speed =18 \div \frac{1}{3} \\\\ &Speed = 54 \\\\ &54 \ km/h \end{aligned}

Speed distance time triangle GCSE questions

1. A commercial aircraft travels from its origin to its destination in a time of 2 hours and 15 minutes. The journey is 1462.5 \ km.

What is the average speed of the plane in km/hour?

2 hours 15 minutes = 2\frac{15}{60} = 2\frac{1}{4} = 2.25

2. John travelled 30 \ km in 90 minutes.

Nadine travelled 52.5 \ km in 2.5 hours.

Who had the greater average speed?

You must show your working.

90 minutes = 1.5 hours

John = 30 \div 1.5 = 20 \ km/h

Nadine = 52.5 \div 2.5 = 21 \ km/h

Nadine has the greater average speed.

3. The distance from Birmingham to Rugby is 40 miles.

Omar drives from Rugby to Birmingham at 60 \ mph.

Ayushi drives from Rugby to Birmingham at 50 \ mph.

How much longer was Ayushi’s journey compared to Omar’s journey? Give your answer in minutes.

For calculating time in hours for Omar or Ayushi.

For converting hours into minutes for Omar or Ayushi.

For correct final answer of 8 minutes.

Learning checklist

You have now learned how to:

• Use compound units such as speed
• Solve simple kinematic problem involving distance and speed
• Change freely between related standard units (e.g. time, length) and compound units (e.g. speed) in numerical contexts
• Work with compound units in a numerical context

The next lessons are

• Best buy maths
• Scale maths

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Video transcript

Distance, Time and Speed Word Problems | GMAT GRE Maths

Before you get into distance, time and speed word problems, check out how you can add a little power to your resume by getting our mini-MBA certificate.  

Problems involving Time, Distance and Speed are solved based on one simple formula.

Distance = Speed * Time

Which implies →

Speed = Distance / Time   and

Time = Distance / Speed

Let us take a look at some simple examples of distance, time and speed problems.   Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?

Time = Distance / speed = 20/4 = 5 hours.   Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.

Speed = Distance/time = 15/2 = 7.5 miles per hour.   Example 3. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?

Distance covered = 4*40 = 160 miles

Speed required to cover the same distance in 1.5 hours = 160/1.5 = 106.66 mph   Now, take a look at the following example:

Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?

Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.

Let us see how this question can be solved.

For these kinds of questions, a table like this might make it easier to solve.

  Let the distance covered by that person be ‘d’.

Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’

IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.

He does this in a time of (d+7.5)/9.

Since the time is same in both the cases →

d/4 = (d+7.5)/9            →        9d = 4(d+7.5)   →        9d=4d+30        →        d = 6.

So, he covered a distance of 6 miles in 1.5 hours.   Example 5. A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

Here, we see that the distance is same.

Let us assume that its usual speed is ‘s’ and time is ‘t’, then

  s*t = (1/3)s*(t+30)      →        t = t/3 + 10      →        t = 15.

So the actual time taken to cover the distance is 15 minutes.

Note: Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.

Solved Questions on Trains

Example 1. X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?

Let the time after which they meet be ‘t’ hours.

Then the time travelled by second train becomes ‘t-2’.

Distance covered by first train+Distance covered by second train = 320 miles

70t+20(t-2) = 320

Solving this gives t = 4.

So the two trains meet after 4 hours.   Example 2. A train leaves from a station and moves at a certain speed. After 2 hours, another train leaves from the same station and moves in the same direction at a speed of 60 mph. If it catches up with the first train in 4 hours, what is the speed of the first train?

Let the speed of the first train be ‘s’.

Distance covered by the first train in (2+4) hours = Distance covered by second train in 4 hours

Therefore, 6s = 60*4

Solving which gives s=40.

So the slower train is moving at the rate of 40 mph.  

Questions on Boats/Airplanes

For problems with boats and streams,

Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream

[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]

Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream

[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]

Similarly, for airplanes travelling with/against the wind,

Speed of the plane with the wind = speed of the plane + speed of the wind

Speed of the plane against the wind = speed of the plane – speed of the wind

Let us look at some examples.

Example 1. A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?

Let the distance be ‘d’ miles.

Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3

Speed upstream = 3-1 = 2 mph

Speed downstream = 3+1 = 4 mph

So, our equation would be d/2 + d/4 = 3 → solving which, we get d = 4 miles.   Example 2. With the wind, an airplane covers a distance of 2400 kms in 4 hours and against the wind in 6 hours. What is the speed of the plane and that of the wind?

Let the speed of the plane be ‘a’ and that of the wind be ‘w’.

Our table looks like this:  

  4(a+w) = 2400 and 6(a-w) = 2400

Expressing one unknown variable in terms of the other makes it easier to solve, which means

a+w = 600 → w=600-a

Substituting the value of w in the second equation,

a-(600-a) = 400 → a = 500

The speed of the plane is 500 kmph and that of the wind is 100 kmph.  

More solved examples on Speed, Distance and Time

Example 1. A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.

Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

Solving which gives t=2.

Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d/30 + (100-d)/40 = 3

Solving which gives d = 60, which is the distance travelled by train. 100-60 = 40 miles is the distance travelled by bus.   Example 2. A plane covered a distance of 630 miles in 6 hours. For the first part of the trip, the average speed was 100 mph and for the second part of the trip, the average speed was 110 mph. what is the time it flew at each speed?

Our table looks like this.

Assuming the distance covered in the 1 st part of journey to be ‘d’, the distance covered in the second half becomes ‘630-d’.

Assuming the time taken for the first part of the journey to be ‘t’, the time taken for the second half becomes ‘6-t’.

From the first equation, d=100t

The second equation is 630-d = 110(6-t).

Substituting the value of d from the first equation, we get

630-100t = 110(6-t)

Solving this gives t=3.

So the plane flew the first part of the journey in 3 hours and the second part in 3 hours.   Example 2. Two persons are walking towards each other on a walking path that is 20 miles long. One is walking at the rate of 3 mph and the other at 4 mph. After how much time will they meet each other?

  Assuming the distance travelled by the first person to be ‘d’, the distance travelled by the second person is ’20-d’.

The time is ‘t’ for both of them because when they meet, they would have walked for the same time.

Since time is same, we can equate as

d/3 = (20-d)/4

Solving this gives d=60/7 miles (8.5 miles approximately)

Then t = 20/7 hours

So the two persons meet after 2 6/7 hours.  

Practice Questions for you to solve

Problem 1: Click here

A boat covers a certain distance in 2 hours, while it comes back in 3 hours. If the speed of the stream is 4 kmph, what is the boat’s speed in still water?

A) 30 kmph B) 20 kmph C) 15 kmph D) 40 kmph

Answer 1: Click here

Explanation

Let the speed of the boat be ‘s’ kmph.

Then, 2(s+4) = 3(s-4) → s = 20

Problem 2: Click here

A cyclist travels for 3 hours, travelling for the first half of the journey at 12 mph and the second half at 15 mph. Find the total distance he covered.

A) 30 miles B) 35 miles C) 40 miles D) 180 miles

Answer 2: Click here

Since it is mentioned, that the first ‘half’ of the journey is covered in 12 mph and the second in 15, the equation looks like

(d/2)/12 + (d/2)/15 = 3

Solving this gives d = 40 miles

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17 thoughts on “Distance, Time and Speed Word Problems | GMAT GRE Maths”

Meera walked to school at a speed of 3 miles per hour. Once she reached the school, she realized that she forgot to bring her books, so rushed back home at a speed of 6 miles per hour. She then walked back to school at a speed of 4 miles per hour. All the times, she walked in the same route. please explain above problem

When she walks faster the time she takes to reach her home and school is lower. There is nothing wrong with the statement. They never mentioned how long she took every time.

a man covers a distance on a toy train.if the train moved 4km/hr faster,it would take 30 min. less. if it moved 2km/hr slower, it would have taken 20 min. more .find the distance.

Let the speed be x. and time be y. A.T.Q, (x+4)(y-1/2)=d and (x-2)(y+1/3)=d. Equate these two and get the answer

Could you explain how ? you have two equations and there are 3 variables.

The 3rd equation is d=xy. Now, you have 3 equations with 3 unknowns. The variables x and y represent the usual speed and usual time to travel distance d.

Speed comes out to be 20 km/hr and the time taken is 3 hrs. The distance traveled is 60 km.

(s + 4) (t – 1/2)= st 1…new equotion = -1/2s + 4t = 2

(s – 2) (t + 1/3)= st 2…new equotion = 1/3s – 2t = 2/3

Multiply all by 6 1… -3s + 24t = 12 2… 2s – 12t = 4 Next, use elimination t= 3 Find s: -3s + 24t = 12 -3s + 24(3) = 12 -3s = -60 s= 20

st or distance = 3 x 20 = 60 km/h

It’s probably the average speed that we are looking for here. Ave. Speed= total distance/ total time. Since it’s harder to look for one variable since both are absent, you can use, 3d/ d( V2V3 + V1V3 + V1V2/ V1V2V3)

2 girls meenu and priya start at the same time to ride from madurai to manamadurai, 60 km away.meenu travels 4kmph slower than priya. priya reaches manamadurai and at turns back meeting meenu 12km from manamaduai. find meenu’s speed?

Hi, when the two girls meet, they have taken equal time to travel their respective distance. So, we just need to equate their time equations

Distance travelled by Meenu = 60 -12 = 48 Distance travelled by Priya = 60 + 12 = 72 Let ‘s’ be the speed of Meenu

Time taken by Meenu => t1 = 48/s Time taken by Priya => t2 = 72/(s+4)

t1=t2 Thus, 48/s = 72/(s+4) => 24s = 192 => s = 8Km/hr

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 KMS away from A at the same time. On the way, however the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Let speed of the CAR BE x kmph.

Then, speed of the train = 3/2(x) .’. 75/x – 75/(3/2)x= 125/(10*60) — subtracting the times travelled by two them hence trains wastage time

therefore x= 120 kmph

A cyclist completes a distance of 60 km at the same speed throughout. She travels 10 km in one hour. She stops every 20 km for one hour to have a break. What are the two variables involved in this situation?

For the answer, not variables: 60km divided by 10km/h=6 hours 60 divided by 20= 3 hours 3 hours+6 hours= 9 hours Answer: 9 hours

Let the length of the train to prod past a point be the intrinsic distance (D) of the train and its speed be S. Its speed, S in passing the electric pole of negligible length is = D/12. The length of the platform added to the intrinsic length of the train. So, the total distance = D + 200. The time = 20 secs. The Speed, S = (D + 200)/20 At constant speed, D/12 = (D + 200)/20 Cross-multiplying, 20D = 12D + 200*12 20D – 12D = 200*12; 8D = 200*12 D = 200*12/8 = 300m. 4th Aug, 2018

Can anyone solve this? Nathan and Philip agree to meet up at the park at 5:00 pm. Nathan lives 300 m due north of the park, and Philip lives 500 m due west of the park. Philip leaves his house at 4:54 pm and walks towards the park at a pace of 1.5 m/s, but Nathan loses track of time and doesn’t leave until 4:59 pm. Trying to avoid being too late, he jogs towards the park at 2.5 m/s. At what rate is the distance between the two friends changing 30 seconds after Nathan has departed?

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Math Word Problems and Solutions - Distance, Speed, Time

Problem 1 A salesman sold twice as much pears in the afternoon than in the morning. If he sold 360 kilograms of pears that day, how many kilograms did he sell in the morning and how many in the afternoon? Click to see solution Solution: Let $x$ be the number of kilograms he sold in the morning.Then in the afternoon he sold $2x$ kilograms. So, the total is $x + 2x = 3x$. This must be equal to 360. $3x = 360$ $x = \frac{360}{3}$ $x = 120$ Therefore, the salesman sold 120 kg in the morning and $2\cdot 120 = 240$ kg in the afternoon.

Problem 2 Mary, Peter, and Lucy were picking chestnuts. Mary picked twice as much chestnuts than Peter. Lucy picked 2 kg more than Peter. Together the three of them picked 26 kg of chestnuts. How many kilograms did each of them pick? Click to see solution Solution: Let $x$ be the amount Peter picked. Then Mary and Lucy picked $2x$ and $x+2$, respectively. So $x+2x+x+2=26$ $4x=24$ $x=6$ Therefore, Peter, Mary, and Lucy picked 6, 12, and 8 kg, respectively.

Problem 3 Sophia finished $\frac{2}{3}$ of a book. She calculated that she finished 90 more pages than she has yet to read. How long is her book? Click to see solution Solution: Let $x$ be the total number of pages in the book, then she finished $\frac{2}{3}\cdot x$ pages. Then she has $x-\frac{2}{3}\cdot x=\frac{1}{3}\cdot x$ pages left. $\frac{2}{3}\cdot x-\frac{1}{3}\cdot x=90$ $\frac{1}{3}\cdot x=90$ $x=270$ So the book is 270 pages long.

Problem 4 A farming field can be ploughed by 6 tractors in 4 days. When 6 tractors work together, each of them ploughs 120 hectares a day. If two of the tractors were moved to another field, then the remaining 4 tractors could plough the same field in 5 days. How many hectares a day would one tractor plough then? Click to see solution Solution: If each of $6$ tractors ploughed $120$ hectares a day and they finished the work in $4$ days, then the whole field is: $120\cdot 6 \cdot 4 = 720 \cdot 4 = 2880$ hectares. Let's suppose that each of the four tractors ploughed $x$ hectares a day. Therefore in 5 days they ploughed $5 \cdot 4 \cdot x = 20 \cdot x$ hectares, which equals the area of the whole field, 2880 hectares. So, we get $20x = 2880$ $ x = \frac{2880}{20} = 144$. Hence, each of the four tractors would plough 144 hectares a day.

Problem 5 A student chose a number, multiplied it by 2, then subtracted 138 from the result and got 102. What was the number he chose? Click to see solution Solution: Let $x$ be the number he chose, then $2\cdot x - 138 = 102$ $2x = 240$ $x = 120$

Problem 6 I chose a number and divide it by 5. Then I subtracted 154 from the result and got 6. What was the number I chose? Click to see solution Solution: Let $x$ be the number I chose, then $\frac{x}{5}-154=6$ $\frac{x}{5}=160$ $x=800$

Problem 8 One side of a rectangle is 3 cm shorter than the other side. If we increase the length of each side by 1 cm, then the area of the rectangle will increase by 18 cm 2 . Find the lengths of all sides. Click to see solution Solution: Let $x$ be the length of the longer side $x \gt 3$, then the other side's length is $x-3$ cm. Then the area is S 1 = x(x - 3) cm 2 . After we increase the lengths of the sides they will become $(x +1)$ and $(x - 3 + 1) = (x - 2)$ cm long. Hence the area of the new rectangle will be $A_2 = (x + 1)\cdot(x - 2)$ cm 2 , which is 18 cm 2 more than the first area. Therefore $A_1 +18 = A_2$ $x(x - 3) + 18 = (x + 1)(x - 2)$ $x^2 - 3x + 18 = x^2 + x - 2x - 2$ $2x = 20$ $x = 10$. So, the sides of the rectangle are $10$ cm and $(10 - 3) = 7$ cm long.

Problem 9 The first year, two cows produced 8100 litres of milk. The second year their production increased by 15% and 10% respectively, and the total amount of milk increased to 9100 litres a year. How many litres were milked from each cow each year? Click to see solution Solution: Let x be the amount of milk the first cow produced during the first year. Then the second cow produced $(8100 - x)$ litres of milk that year. The second year, each cow produced the same amount of milk as they did the first year plus the increase of $15\%$ or $10\%$. So $8100 + \frac{15}{100}\cdot x + \frac{10}{100} \cdot (8100 - x) = 9100$ Therefore $8100 + \frac{3}{20}x + \frac{1}{10}(8100 - x) = 9100$ $\frac{1}{20}x = 190$ $x = 3800$ Therefore, the cows produced 3800 and 4300 litres of milk the first year, and $4370$ and $4730$ litres of milk the second year, respectively.

Problem 10 The distance between stations A and B is 148 km. An express train left station A towards station B with the speed of 80 km/hr. At the same time, a freight train left station B towards station A with the speed of 36 km/hr. They met at station C at 12 pm, and by that time the express train stopped at at intermediate station for 10 min and the freight train stopped for 5 min. Find: a) The distance between stations C and B. b) The time when the freight train left station B. Click to see solution Solution a) Let x be the distance between stations B and C. Then the distance from station C to station A is $(148 - x)$ km. By the time of the meeting at station C, the express train travelled for $\frac{148-x}{80}+\frac{10}{60}$ hours and the freight train travelled for $\frac{x}{36}+\frac{5}{60}$ hours. The trains left at the same time, so: $\frac{148 - x}{80} + \frac{1}{6} = \frac{x}{36} + \frac{1}{12}$. The common denominator for 6, 12, 36, 80 is 720. Then $9(148 - x) +120 = 20x +60$ $1332 - 9x + 120 = 20x + 60$ $29x = 1392$ $x = 48$. Therefore the distance between stations B and C is 48 km. b) By the time of the meeting at station C the freight train rode for $\frac{48}{36} + \frac{5}{60}$ hours, i.e. $1$ hour and $25$ min. Therefore it left station B at $12 - (1 + \frac{25}{60}) = 10 + \frac{35}{60}$ hours, i.e. at 10:35 am.

Problem 11 Susan drives from city A to city B. After two hours of driving she noticed that she covered 80 km and calculated that, if she continued driving at the same speed, she would end up been 15 minutes late. So she increased her speed by 10 km/hr and she arrived at city B 36 minutes earlier than she planned. Find the distance between cities A and B. Click to see solution Solution: Let $x$ be the distance between A and B. Since Susan covered 80 km in 2 hours, her speed was $V = \frac{80}{2} = 40$ km/hr. If she continued at the same speed she would be $15$ minutes late, i.e. the planned time on the road is $\frac{x}{40} - \frac{15}{60}$ hr. The rest of the distance is $(x - 80)$ km. $V = 40 + 10 = 50$ km/hr. So, she covered the distance between A and B in $2 +\frac{x - 80}{50}$ hr, and it was 36 min less than planned. Therefore, the planned time was $2 + \frac{x -80}{50} + \frac{36}{60}$. When we equalize the expressions for the scheduled time, we get the equation: $\frac{x}{40} - \frac{15}{60} = 2 + \frac{x -80}{50} + \frac{36}{60}$ $\frac{x - 10}{40} = \frac{100 + x - 80 + 30}{50}$ $\frac{x - 10}{4} = \frac{x +50}{5}$ $5x - 50 = 4x + 200$ $x = 250$ So, the distance between cities A and B is 250 km.

Problem 12 To deliver an order on time, a company has to make 25 parts a day. After making 25 parts per day for 3 days, the company started to produce 5 more parts per day, and by the last day of work 100 more parts than planned were produced. Find how many parts the company made and how many days this took. Click to see solution Solution: Let $x$ be the number of days the company worked. Then 25x is the number of parts they planned to make. At the new production rate they made: $3\cdot 25 + (x - 3)\cdot 30 = 75 + 30(x - 3)$ Therefore: $25 x = 75 + 30(x -3) - 100$ $25x = 75 +30x -90 - 100$ $190 -75 = 30x -25$ $115 = 5x$ $x = 23$ So the company worked 23 days and they made $23\cdot 25+100 = 675$ pieces.

Problem 13 There are 24 students in a seventh grade class. They decided to plant birches and roses at the school's backyard. While each girl planted 3 roses, every three boys planted 1 birch. By the end of the day they planted $24$ plants. How many birches and roses were planted? Click to see solution Solution: Let $x$ be the number of roses. Then the number of birches is $24 - x$, and the number of boys is $3\times (24-x)$. If each girl planted 3 roses, there are $\frac{x}{3}$ girls in the class. We know that there are 24 students in the class. Therefore $\frac{x}{3} + 3(24 - x) = 24$ $x + 9(24 - x) = 3\cdot 24$ $x +216 - 9x = 72$ $216 - 72 = 8x$ $\frac{144}{8} = x$ $x = 18$ So, students planted 18 roses and 24 - x = 24 - 18 = 6 birches.

Problem 14 A car left town A towards town B driving at a speed of V = 32 km/hr. After 3 hours on the road the driver stopped for 15 min in town C. Because of a closed road he had to change his route, making the trip 28 km longer. He increased his speed to V = 40 km/hr but still he was 30 min late. Find: a) The distance the car has covered. b) The time that took it to get from C to B. Click to see solution Solution: From the statement of the problem we don't know if the 15 min stop in town C was planned or it was unexpected. So we have to consider both cases. A The stop was planned. Let us consider only the trip from C to B, and let $x$ be the number of hours the driver spent on this trip. Then the distance from C to B is $S = 40\cdot x$ km. If the driver could use the initial route, it would take him $x - \frac{30}{60} = x - \frac{1}{2}$ hours to drive from C to B. The distance from C to B according to the initially itinerary was $(x - \frac{1}{2})\cdot 32$ km, and this distance is $28$ km shorter than $40\cdot x$ km. Then we have the equation $(x - 1/2)\cdot 32 + 28 = 40x$ $32x -16 +28 = 40x$ $-8x = -12$ $8x = 12$ $x = \frac{12}{8}$ $x = 1 \frac{4}{8} = 1 \frac{1}{2} = 1 \frac{30}{60} =$ 1 hr 30 min. So, the car covered the distance between C and B in 1 hour and 30 min. The distance from A to B is $3\cdot 32 + \frac{12}{8}\cdot 40 = 96 + 60 = 156$ km. B Suppose it took $x$ hours for him to get from C to B. Then the distance is $S = 40\cdot x$ km. The driver did not plan the stop at C. Let we accept that he stopped because he had to change the route. It took $x - \frac{30}{60} + \frac{15}{60} = x - \frac{15}{60} = x - \frac{1}{4}$ h to drive from C to B. The distance from C to B is $32(x - \frac{1}{4})$ km, which is $28$ km shorter than $40\cdot x$, i.e. $32(x - \frac{1}{4}) + 28 = 40x$ $32x - 8 +28 = 40x$ $20= 8x$ $x = \frac{20}{8} = \frac{5}{2} = 2 \text{hr } 30 \text{min}.$ The distance covered equals $ 40 \times 2.5 = 100 km$.

Problem 15 If a farmer wants to plough a farm field on time, he must plough 120 hectares a day. For technical reasons he ploughed only 85 hectares a day, hence he had to plough 2 more days than he planned and he still has 40 hectares left. What is the area of the farm field and how many days the farmer planned to work initially? Click to see solution Solution: Let $x$ be the number of days in the initial plan. Therefore, the whole field is $120\cdot x$ hectares. The farmer had to work for $x + 2$ days, and he ploughed $85(x + 2)$ hectares, leaving $40$ hectares unploughed. Then we have the equation: $120x = 85(x + 2) + 40$ $35x = 210$ $x = 6$ So the farmer planned to have the work done in 6 days, and the area of the farm field is $120\cdot 6 = 720$ hectares.

Problem 16 A woodworker normally makes a certain number of parts in 24 days. But he was able to increase his productivity by 5 parts per day, and so he not only finished the job in only 22 days but also he made 80 extra parts. How many parts does the woodworker normally makes per day and how many pieces does he make in 24 days? Click to see solution Solution: Let $x$ be the number of parts the woodworker normally makes daily. In 24 days he makes $24\cdot x$ pieces. His new daily production rate is $x + 5$ pieces and in $22$ days he made $22 \cdot (x + 5)$ parts. This is 80 more than $24\cdot x$. Therefore the equation is: $24\cdot x + 80 = 22(x +5)$ $30 = 2x$ $x = 15$ Normally he makes 15 parts a day and in 24 days he makes $15 \cdot 24 = 360$ parts.

Problem 17 A biker covered half the distance between two towns in 2 hr 30 min. After that he increased his speed by 2 km/hr. He covered the second half of the distance in 2 hr 20 min. Find the distance between the two towns and the initial speed of the biker. Click to see solution Solution: Let x km/hr be the initial speed of the biker, then his speed during the second part of the trip is x + 2 km/hr. Half the distance between two cities equals $2\frac{30}{60} \cdot x$ km and $2\frac{20}{60} \cdot (x + 2)$ km. From the equation: $2\frac{30}{60} \cdot x = 2\frac{20}{60} \cdot (x+2)$ we get $x = 28$ km/hr. The intial speed of the biker is 28 km/h. Half the distance between the two towns is $2 h 30 min \times 28 = 2.5 \times 28 = 70$. So the distance is $2 \times 70 = 140$ km.

Problem 18 A train covered half of the distance between stations A and B at the speed of 48 km/hr, but then it had to stop for 15 min. To make up for the delay, it increased its speed by $\frac{5}{3}$ m/sec and it arrived to station B on time. Find the distance between the two stations and the speed of the train after the stop. Click to see solution Solution: First let us determine the speed of the train after the stop. The speed was increased by $\frac{5}{3}$ m/sec $= \frac{5\cdot 60\cdot 60}{\frac{3}{1000}}$ km/hr = $6$ km/hr. Therefore, the new speed is $48 + 6 = 54$ km/hr. If it took $x$ hours to cover the first half of the distance, then it took $x - \frac{15}{60} = x - 0.25$ hr to cover the second part. So the equation is: $48 \cdot x = 54 \cdot (x - 0.25)$ $48 \cdot x = 54 \cdot x - 54\cdot 0.25$ $48 \cdot x - 54 \cdot x = - 13.5$ $-6x = - 13.5$ $x = 2.25$ h. The whole distance is $2 \times 48 \times 2.25 = 216$ km.

Problem 19 Elizabeth can get a certain job done in 15 days, and Tony can finish only 75% of that job within the same time. Tony worked alone for several days and then Elizabeth joined him, so they finished the rest of the job in 6 days, working together. For how many days have each of them worked and what percentage of the job have each of them completed? Click to see solution Solution: First we will find the daily productivity of every worker. If we consider the whole job as unit (1), Elizabeth does $\frac{1}{15}$ of the job per day and Tony does $75\%$ of $\frac{1}{15}$, i.e. $\frac{75}{100}\cdot \frac{1}{15} = \frac{1}{20}$. Suppose that Tony worked alone for $x$ days. Then he finished $\frac{x}{20}$ of the total job alone. Working together for 6 days, the two workers finished $6\cdot (\frac{1}{15}+\frac{1}{20}) = 6\cdot \frac{7}{60} = \frac{7}{10}$ of the job. The sum of $\frac{x}{20}$ and $\frac{7}{10}$ gives us the whole job, i.e. $1$. So we get the equation: $\frac{x}{20}+\frac{7}{10}=1$ $\frac{x}{20} = \frac{3}{10}$ $x = 6$. Tony worked for 6 + 6 = 12 days and Elizabeth worked for $6$ days. The part of job done is $12\cdot \frac{1}{20} = \frac{60}{100} = 60\%$ for Tony, and $6\cdot \frac{1}{15} = \frac{40}{100} = 40\%$ for Elizabeth.

Problem 20 A farmer planned to plough a field by doing 120 hectares a day. After two days of work he increased his daily productivity by 25% and he finished the job two days ahead of schedule. a) What is the area of the field? b) In how many days did the farmer get the job done? c) In how many days did the farmer plan to get the job done? Click to see solution Solution: First of all we will find the new daily productivity of the farmer in hectares per day: 25% of 120 hectares is $\frac{25}{100} \cdot 120 = 30$ hectares, therefore $120 + 30 = 150$ hectares is the new daily productivity. Lets x be the planned number of days allotted for the job. Then the farm is $120\cdot x$ hectares. On the other hand, we get the same area if we add $120 \cdot 2$ hectares to $150(x -4)$ hectares. Then we get the equation $120x = 120\cdot 2 + 150(x -4)$ $x = 12$ So, the job was initially supposed to take 12 days, but actually the field was ploughed in 12 - 2 =10 days. The field's area is $120 \cdot 12 = 1440$ hectares.

Problem 21 To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by $33 \times \frac{1}{3}\%$, and finished the work 1 day earlier than it was planned. A) What is the area of the grass field? B) How many days did it take to mow the whole field? C) How many days were scheduled initially for this job? Hint : See problem 20 and solve by yourself. Answer: A) 120 hectares; B) 7 days; C) 8 days.

Problem 22 A train travels from station A to station B. If the train leaves station A and makes 75 km/hr, it arrives at station B 48 minutes ahead of scheduled. If it made 50 km/hr, then by the scheduled time of arrival it would still have 40 km more to go to station B. Find: A) The distance between the two stations; B) The time it takes the train to travel from A to B according to the schedule; C) The speed of the train when it's on schedule. Click to see solution Solution: Let $x$ be the scheduled time for the trip from A to B. Then the distance between A and B can be found in two ways. On one hand, this distance equals $75(x - \frac{48}{60})$ km. On the other hand, it is $50x + 40$ km. So we get the equation: $75(x - \frac{48}{60}) = 50x + 40$ $x = 4$ hr is the scheduled travel time. The distance between the two stations is $50\cdot 4 +40 = 240$ km. Then the speed the train must keep to be on schedule is $\frac{240}{4} = 60$ km/hr.

Problem 23 The distance between towns A and B is 300 km. One train departs from town A and another train departs from town B, both leaving at the same moment of time and heading towards each other. We know that one of them is 10 km/hr faster than the other. Find the speeds of both trains if 2 hours after their departure the distance between them is 40 km. Click to see solution Solution: Let the speed of the slower train be $x$ km/hr. Then the speed of the faster train is $(x + 10)$ km/hr. In 2 hours they cover $2x$ km and $2(x +10)$km, respectively. Therefore if they didn't meet yet, the whole distance from A to B is $2x + 2(x +10) +40 = 4x +60$ km. However, if they already met and continued to move, the distance would be $2x + 2(x + 10) - 40 = 4x - 20$km. So we get the following equations: $4x + 60 = 300$ $4x = 240$ $x = 60$ or $4x - 20 = 300$ $4x = 320$ $x = 80$ Hence the speed of the slower train is $60$ km/hr or $80$ km/hr and the speed of the faster train is $70$ km/hr or $90$ km/hr.

Problem 24 A bus travels from town A to town B. If the bus's speed is 50 km/hr, it will arrive in town B 42 min later than scheduled. If the bus increases its speed by $\frac{50}{9}$ m/sec, it will arrive in town B 30 min earlier than scheduled. Find: A) The distance between the two towns; B) The bus's scheduled time of arrival in B; C) The speed of the bus when it's on schedule. Click to see solution Solution: First we will determine the speed of the bus following its increase. The speed is increased by $\frac{50}{9}$ m/sec $= \frac{50\cdot60\cdot60}{\frac{9}{1000}}$ km/hr $= 20$ km/hr. Therefore, the new speed is $V = 50 + 20 = 70$ km/hr. If $x$ is the number of hours according to the schedule, then at the speed of 50 km/hr the bus travels from A to B within $(x +\frac{42}{60})$ hr. When the speed of the bus is $V = 70$ km/hr, the travel time is $x - \frac{30}{60}$ hr. Then $50(x +\frac{42}{60}) = 70(x-\frac{30}{60})$ $5(x+\frac{7}{10}) = 7(x-\frac{1}{2})$ $\frac{7}{2} + \frac{7}{2} = 7x -5x$ $2x = 7$ $x = \frac{7}{2}$ hr. So, the bus is scheduled to make the trip in $3$ hr $30$ min. The distance between the two towns is $70(\frac{7}{2} - \frac{1}{2}) = 70\cdot 3 = 210$ km and the scheduled speed is $\frac{210}{\frac{7}{2}} = 60$ km/hr.

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7 Problem-Solving Skills That Can Help You Be a More Successful Manager

Discover what problem-solving is, and why it's important for managers. Understand the steps of the process and learn about seven problem-solving skills.

1Managers oversee the day-to-day operations of a particular department, and sometimes a whole company, using their problem-solving skills regularly. Managers with good problem-solving skills can help ensure companies run smoothly and prosper.

If you're a current manager or are striving to become one, read this guide to discover what problem-solving skills are and why it's important for managers to have them. Learn the steps of the problem-solving process, and explore seven skills that can help make problem-solving easier and more effective.

What is problem-solving?

Problem-solving is both an ability and a process. As an ability, problem-solving can aid in resolving issues faced in different environments like home, school, abroad, and social situations, among others. As a process, problem-solving involves a series of steps for finding solutions to questions or concerns that arise throughout life.

The importance of problem-solving for managers

Managers deal with problems regularly, whether supervising a staff of two or 100. When people solve problems quickly and effectively, workplaces can benefit in a number of ways. These include:

Greater creativity

Higher productivity

Increased job fulfillment

Satisfied clients or customers

Better cooperation and cohesion

Improved environments for employees and customers

7 skills that make problem-solving easier

Companies depend on managers who can solve problems adeptly. Although problem-solving is a skill in its own right, a subset of seven skills can help make the process of problem-solving easier. These include analysis, communication, emotional intelligence, resilience, creativity, adaptability, and teamwork.

1. Analysis

As a manager , you'll solve each problem by assessing the situation first. Then, you’ll use analytical skills to distinguish between ineffective and effective solutions.

2. Communication

Effective communication plays a significant role in problem-solving, particularly when others are involved. Some skills that can help enhance communication at work include active listening, speaking with an even tone and volume, and supporting verbal information with written communication.

3. Emotional intelligence

Emotional intelligence is the ability to recognize and manage emotions in any situation. People with emotional intelligence usually solve problems calmly and systematically, which often yields better results.

4. Resilience

Emotional intelligence and resilience are closely related traits. Resiliency is the ability to cope with and bounce back quickly from difficult situations. Those who possess resilience are often capable of accurately interpreting people and situations, which can be incredibly advantageous when difficulties arise.

5. Creativity 

When brainstorming solutions to problems, creativity can help you to think outside the box. Problem-solving strategies can be enhanced with the application of creative techniques. You can use creativity to:

Approach problems from different angles

Improve your problem-solving process

Spark creativity in your employees and peers

6. Adaptability

Adaptability is the capacity to adjust to change. When a particular solution to an issue doesn't work, an adaptable person can revisit the concern to think up another one without getting frustrated.

7. Teamwork

Finding a solution to a problem regularly involves working in a team. Good teamwork requires being comfortable working with others and collaborating with them, which can result in better problem-solving overall.

Steps of the problem-solving process

Effective problem-solving involves five essential steps. One way to remember them is through the IDEAL model created in 1984 by psychology professors John D. Bransford and Barry S. Stein [ 1 ]. The steps to solving problems in this model include: identifying that there is a problem, defining the goals you hope to achieve, exploring potential solutions, choosing a solution and acting on it, and looking at (or evaluating) the outcome.

1. Identify that there is a problem and root out its cause.

To solve a problem, you must first admit that one exists to then find its root cause. Finding the cause of the problem may involve asking questions like:

Can the problem be solved?

How big of a problem is it?

Why do I think the problem is occurring?

What are some things I know about the situation?

What are some things I don't know about the situation?

Are there any people who contributed to the problem?

Are there materials or processes that contributed to the problem?

Are there any patterns I can identify?

2. Define the goals you hope to achieve.

Every problem is different. The goals you hope to achieve when problem-solving depend on the scope of the problem. Some examples of goals you might set include:

Gather as much factual information as possible.

Brainstorm many different strategies to come up with the best one.

Be flexible when considering other viewpoints.

Articulate clearly and encourage questions, so everyone involved is on the same page.

Be open to other strategies if the chosen strategy doesn't work.

Stay positive throughout the process.

3. Explore potential solutions.

Once you've defined the goals you hope to achieve when problem-solving , it's time to start the process. This involves steps that often include fact-finding, brainstorming, prioritizing solutions, and assessing the cost of top solutions in terms of time, labor, and money.

4. Choose a solution and act on it.

Evaluate the pros and cons of each potential solution, and choose the one most likely to solve the problem within your given budget, abilities, and resources. Once you choose a solution, it's important to make a commitment and see it through. Draw up a plan of action for implementation, and share it with all involved parties clearly and effectively, both verbally and in writing. Make sure everyone understands their role for a successful conclusion.

5. Look at (or evaluate) the outcome.

Evaluation offers insights into your current situation and future problem-solving. When evaluating the outcome, ask yourself questions like:

Did the solution work?

Will this solution work for other problems?

Were there any changes you would have made?

Would another solution have worked better?

As a current or future manager looking to build your problem-solving skills, it is often helpful to take a professional course. Consider Improving Communication Skills offered by the University of Pennsylvania on Coursera. You'll learn how to boost your ability to persuade, ask questions, negotiate, apologize, and more. 

You might also consider taking Emotional Intelligence: Cultivating Immensely Human Interactions , offered by the University of Michigan on Coursera. You'll explore the interpersonal and intrapersonal skills common to people with emotional intelligence, and you'll learn how emotional intelligence is connected to team success and leadership.

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Article sources

Tennessee Tech. “ The Ideal Problem Solver (2nd ed.) , https://www.tntech.edu/cat/pdf/useful_links/idealproblemsolver.pdf.” Accessed December 6, 2022.

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This content has been made available for informational purposes only. Learners are advised to conduct additional research to ensure that courses and other credentials pursued meet their personal, professional, and financial goals.

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Americans’ Dismal Views of the Nation’s Politics

1. the biggest problems and greatest strengths of the u.s. political system, table of contents.

• The impact of partisan polarization
• Persistent concerns over money in politics
• Views of the parties and possible changes to the two-party system
• Other important findings
• Explore chapters of this report
• In their own words: Americans on the political system’s biggest problems
• In their own words: Americans on the political system’s biggest strengths
• Are there clear solutions to the nation’s problems?
• Evaluations of the political system
• Trust in the federal government
• Feelings toward the federal government
• The relationship between the federal and state governments
• Americans’ ratings of their House member, governor and local officials
• Party favorability ratings
• Most characterize their party positively
• Quality of the parties’ ideas
• Influence in congressional decision-making
• Views on limiting the role of money in politics
• Views on what kinds of activities can change the country for the better
• How much can voting affect the future direction of the country?
• Views of members of Congress
• In their own words: Americans’ views of the major problems with today’s elected officials
• How much do elected officials care about people like me?
• What motivates people to run for office?
• Quality of recent political candidates
• In elections, is there usually at least one candidate who shares your views?
• What the public sees as most important in political candidates
• Impressions of the people who will be running for president in 2024
• Views about presidential campaigns
• How much of an impact does who is president have on your life?
• Whose priorities should the president focus on?
• How different are the Republican and Democratic parties?
• Views of how well the parties represent people’s interests
• What if there were more political parties?
• Would more parties make solving problems easier or harder?
• How likely is it that an independent candidate will become president?
• Americans who feel unrepresented by the parties have highly negative views of the political system
• Views of the Electoral College
• Should the size of the U.S. House of Representatives change?
• Senate seats and population size
• Younger adults more supportive of structural changes
• Politics in a single word or phrase: An outpouring of negative sentiments
• Negative emotions prevail when Americans think about politics
• Americans say the tone of political debate in the country has worsened
• Which political topics get too much – and too little – attention?
• Majority of Americans find it stressful to talk politics with people they disagree with
• Acknowledgments

The public sees a number of specific problems with American politics. Partisan fighting, the high cost of political campaigns, and the outsize influence of special interests and lobbyists are each seen as characteristic of the U.S. political system by at least 84% of Americans.

Yet 63% also say that “ordinary Americans care about making the political system work well” is a good description of U.S. politics today. Still, when asked to describe a strength of the political system in their own words, more than half either say “nothing” (22%) or decline to give an answer (34%).

Americans view negative statements as better descriptions of the political system than positive ones

More than eight-in-ten adults say that each of the following is at least a somewhat good description of the U.S. political system today:

• Republicans and Democrats are more focused on fighting each other than on solving problems (86%);
• The cost of political campaigns makes it hard for good people to run for office (85%);
• Special interest groups and lobbyists have too much say in what happens in politics (84%).

About six-in-ten (63%) think ordinary Americans want to make the political system work well. This is the rare positive sentiment that a majority views as a good descriptor of the political system.

Fewer than half of adults hold the view that the government deserves more credit than it gets: Majorities say that “the federal government does more for ordinary Americans than people give it credit for” (59%) and “Congress accomplishes more than people give it credit for” (65%) are both bad descriptions of the political system.

Nearly seven-in-ten adults express frustration with the availability of unbiased information about politics: 68% say the statement “it is easy to find unbiased information about what is happening in politics” is not a good description of the political system.

And just 22% of Americans say that political leaders facing consequences for acting unethically is a good description of the political system. They are more than three times as likely to say that this is a bad description (76% say this).

Many critiques of the political system are bipartisan

Partisans have similar views of many of the descriptions of the political system included in the survey.

Overwhelming majorities in both parties think there is too much partisan fighting, campaigns cost too much, and lobbyists and special interests have too much say in politics. And just 24% of Democrats and Democratic-leaning independents and 20% of Republicans and Republican leaners say that political leaders face consequences if they act unethically.

The widest partisan gap is over a description of the federal government. Democrats are roughly twice as likely as Republicans to say “the federal government does more for ordinary Americans than people give it credit for” (54% vs. 26%).

There is a narrower gap in views of Congress’ accomplishments: 37% of Democrats and 28% of Republicans say it accomplishes more than people give it credit for.

Democrats are also more likely to say, “It is easy to find unbiased information about what is happening in politics” (36% of Democrats and 25% of Republicans say this is a good description of the political system today), while Republicans are slightly more likely than Democrats to view ordinary Americans as wanting to make the political system work well (67% of Republicans and 61% of Democrats say this is a good description).

When asked to describe in their own words the biggest problem with the political system in the U.S. today, Americans point to a wide range of factors.

Negative characteristics attributed to politicians and political leaders are a common complaint: 31% of U.S. adults say politicians are the biggest problem with the system, including 15% who point to greed or corruption and 7% who cite dishonesty or a lack of trustworthiness.

The biggest problem, according to one woman in her 50s, is that politicians are “hiding the truth and fulfilling their own agendas.” Similarly, a man in his 30s says, “They don’t work for the people. They are too corrupt and busy filling their pockets.”

Explore more voices: The political system’s biggest problems

What do you see as the biggest problem with the political system in the U.S. today?

“An almost total lack of credibility and trust. Coupled with a media that’s so biased, that they’ve lost all objectivity.” –Man, 70s

“Lying about intentions or not following through with what elected officials said they would do.” –Woman, 20s

“Blind faith in political figures.” –Woman, 50s

“Our elected officials would rather play political games than serve the needs of their constituents.” –Woman, 50s

“Same politicians in office too long.” –Woman, 30s

“Extremism on both sides exploited by the mainstream media for ratings. It is making it impossible for both parties to work together.” –Man, 30s

“It has become too polarized. No one is willing to compromise or be moderate.” –Woman, 40s

“Too much money in politics coming from large corporations and special interest.” –Man, 30s

“The people have no say in important matters, we have NO representation at all. Our lawmakers are isolated and could care less what we want.” –Man, 60s

About two-in-ten adults cite deep divisions between the parties as the biggest problem with the U.S. political system, with respondents describing a lack of cooperation between the parties or among elected leaders in Washington.

“Both of the political parties are so busy trying to stop the other party, they are wasting their opportunities to solve the problems faced by our nation,” in the view of one man in his 70s.

Even as some blame polarization, others (10% of respondents) identify the other party as the system’s biggest problem. Some Republicans say that the biggest problem is “Democrats” while some Democrats simply say “Republicans.”

Smaller but substantial shares of adults name the media and political discourse (9%), the influence of money in politics (7%), government’s perceived failures (6%), specific policy areas and issues (6%) or problems with elections and voting (4%) as the biggest problem with the political system today.

Far fewer adults name a specific strength of the political system today when asked to describe the system’s biggest strength in their own words. More than half either say that the system lacks a biggest strength (22%) or decline to answer (34%). As one woman in her 60s writes, “I’m not seeing any strengths!”

Among those who do identify strengths of the U.S. political system, the structure of political institutions and the principles that define the constitutional order are named most frequently (by 12% of respondents). Many respondents specifically point to the Constitution itself or refer to the separation of powers or the checks and balances created by the Constitution.

A man in his 20s believes that the “separation of powers and federalism work pretty well,” while one in his 30s writes that the system’s greatest strength is “the checks and balances to make sure that monumental changes aren’t made unilaterally.”

Explore more voices: The political system’s biggest strengths

What do you see as the biggest strength of the U.S. political system today?

“Everyone getting a say; democracy.” –Woman, 40s

“The right to have your opinions heard.” –Man, 60s

“In spite of our differences, we are still a democracy, and I believe there are people within our government who still care and are interested in the betterment of our country.” –Woman, 50s

“The freedom of speech and religion” –Woman, 50s

“If we have fair, honest elections we can vote out the corruption and/or incompetent politicians.” –Man, 70s

“The Constitution.” –Man, 50s

“The checks and balances to control the power of any office. The voice of the people and the options to remove an official from office.” –Man, 60s

“New, younger voices in government.” –Woman, 40s

“If we can’t get more bipartisanship we’ll become weaker. Our biggest strength is our working together.” –Woman, 60s

“The way that every two years the people get to make their voice heard.” –Man, 30s

About one-in-ten (9%) refer to individual freedoms and related democratic values, while a similar share (8%) discuss the right to vote and the existence of free elections. A woman in her 70s echoes many similar comments when she points to “the possibility of change in upcoming elections.”

However, even some of the descriptions of positive characteristics of the system are couched in respondents’ doubts about the way the system is working today. One woman in her 50s adds a qualification to what she views as the system’s biggest strength, saying, “Theoretically every voter has a say.”

Smaller shares of the public point to the positive characteristics of some politicians (4%) or the positive characteristics of the American people (4%) as reasons for optimism.

The public remains roughly evenly split over whether there are clear solutions to the biggest issues facing the country. Half of Americans today say there are clear solutions to most of the big issues facing the country, while about as many (48%) say most big issues don’t have clear solutions.

There are relatively modest demographic and political differences in perceptions of whether the solutions to the nations’ problems are clear or not.

While both men and women are relatively divided on this question, women are 6 percentage points more likely to think the big issues facing the country don’t have clear solutions.

Race and ethnicity

While 43% of Hispanic adults and about half of Black (50%) and White (48%) adults say there aren’t clear solutions for most big issues, that rises to 62% among Asian adults.

Age differences on this question are modest, but those under 30 are slightly more likely than those 30 and older to say most big issues have clear solutions.

Partisanship and political engagement

Both Republicans and Democrats are relatively split on this question, though Republicans are slightly more likely to say there are clear solutions to most big issues.

Those with higher levels of political engagement are more likely to say there are clear solutions to most big issues facing the country.

About six-in-ten adults with high levels of political engagement (61%) say there are clear solutions to big issues today, compared with half of those with medium levels of engagement and 41% of those with lower engagement.

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Research on Solving the Problem of Children’s Doctor-Patient Relationship from the Perspective of System Theory - The Example of Children’s Medical IP “Courageous Planet”

https://ror.org/03rc6as71Tongji University, Shanghai, China

Wuhan University of Technology, Wuhan, China

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HCI in Business, Government and Organizations: 11th International Conference, HCIBGO 2024, Held as Part of the 26th HCI International Conference, HCII 2024, Washington, DC, USA, June 29 – July 4, 2024, Proceedings, Part I

Due to children’s limited cognitive ability, children’s patients often suffer from fear and anxiety during treatment, which leads to resistance to medical treatment, and parents don’t know how to comfort their children, so the tension between doctors and patients in children’s hospitals has always been a thorny issue. Shanghai Children’s Medical Centre (SCMC), one of the best children’s medical institutions in Shanghai, was the first to set the goal of becoming a “no-cry children’s hospital”, and since 2014, the School of Design and Creativity in Tongji University has been establishing a design-driven collaboration with SCMC, aiming to improve children’s healthcare services and experience.

Hospitals are a systemic problem, made up of different roles, and different environments, and essentially a communication problem in connecting doctors, parents, and children. This paper is based on Richard Buchanan’s Definitions of system: The four modes of definitions of system: set, arrangement, group, and condition, of which set, and arrangement belong to the design thinking mode of problem-solving, which is also usually the way designers think about problems, but this design mode is not a good solution to the problem of communication between children, parents, and hospitals. This paper favors the thinking mode of CONDITION, which is the harmonious and orderly interaction of the whole, without the smallest part, close to the process of seeking truth, a mode of Assimilation, the process of mutual understanding and communication through natural life and extending to the social group. IP is such a bridge that it can communicate across media and at the same time assimilate each medium into a systematic ecology for its use. Through the children’s healthcare IP to solve the communication problem, to build a new concern of the three, so that people with different identities can be placed in it and understand each other.

Through researching 123 doctors and 121 families with children, we produced the “Courageous Planet” medical IP brand design based on children’s physiological and psychological characteristics and research in the medical field and then verified it to produce a business strategy model. The “Planet Courage” medical IP design aims to help doctors, parents, and children deliver messages and build emotional ties more effectively in the relatively special setting of a children’s hospital through the introduction of IP images. The name “Planet Courage” is centered on the word “courage”, hoping that every child is a little warrior who is brave enough to face up to illness and grow up healthy and happy.

References:

Recommendations, an inclusive design approach for developing video games for children with autism spectrum disorder.

The efficacy of therapeutic treatments for Autism Spectrum Disorder is mainly associated with the treatment's intensity in terms of weekly hours. This has led mental health professionals to explore the use of video games to complement traditional ...

A parental perspective on apps for young children

Touchscreen applications (apps) for young children have seen increasingly high rates of growth with more than a hundred thousand now available apps. As with other media, parents play a key role in young children's app selection and use. However, to date,...

Parent Problem Recognition and Service Use Willingness for Young Children

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Singapore Management University, Singapore, Singapore

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• Published: 29 June 2024

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• Children’s doctor-patient relationship
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GrayMatter scores $45M for robots that speed-up manufacturing with ‘physics-informed AI’

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Los Angeles-based GrayMatter , a startup addressing some of the hardest problems in manufacturing with AI-powered robots, today announced it has raised $45 million in a series B round of funding. The investment takes the total capital raised by the company to $70 million and has been led by Wellington Managemen with participation from multiple new and existing investors.

While robotic automation has been around for a long time, with companies like Apple using it in different functions of the assembly line, GrayMatter is pioneering what it describes as “physics-informed AI” — a technology that enables robots to self-program and handle high-mix, high-variability manufacturing environments. This is essentially the heart of the company, which has seen significant growth since its launch in 2020.

“There are so many parts, variations, and variabilities that a traditional robot cannot handle, so we’re bridging the gap with our technology for companies facing a minimum of two-year production backlogs,” Ariyan Kabir, co-founder and CEO of the company, told VentureBeat.

GrayMatter solving high-mix, high-variability manufacturing problems

The American manufacturing industry is worth $2.5 trillion , but companies are struggling with massive backlogs due to skilled worker shortages. There are as many as 3.8 million unfilled jobs across departments, keeping teams from meeting their delivery deadlines. Not to mention, in many cases, when there are enough workers, they fail to deliver the quality companies expect.

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Kabir, who was a part of the University of Southern California’s Center for Advanced Manufacturing, saw these problems while interfacing with several industry stakeholders. The situation was even worse for companies engaged in high-mix, high-variability manufacturing dealing with a variety of parts.

This led Kabir to launch GrayMatter, with a focus on building robotic solutions that could handle labor-intensive surface treatment and finishing jobs for all sorts of products being manufactured — from football helmets to aerospace equipment and everything in between. 

At the core, the company provides enterprises with smart robotic cells, a workspace of sorts where robots using its proprietary physics-informed AI, dubbed GMR-AI, perform tasks like sanding, buffing, polishing, spraying, coating, blasting and inspection. But here is the thing: unlike automation robots that are programmed to do one specific job (which takes weeks), these machines program themselves from a high-level task description. Their process parameters adapt based on observed performance to execute the desired task autonomously. 

The whole self-programming takes a matter of minutes. Once that’s done, the robots start producing highly consistent results at speed. This addresses the capacity and quality issues teams often face with manual efforts. On top of that, the cells can even monitor their health to reduce the risk of failure.

According to Kabir, GrayMatter’s physics-informed AI tries to augment existing manufacturing process models and knowledge with experimental data to deliver exactly what is expected from the robotic cell.

“It enforces known physics-based process models (or knowledge) as a constraint in the AI system to ensure that it does not learn anything that contradicts existing models/knowledge. For example, the system can enforce a constraint that increasing pressure on the sanding tool will increase the deflection of the part being sanded. We don’t need to conduct a large number of tests to learn this already-known fact. If the measured data contradicts this constraint, then it is highly likely either the sensor is malfunctioning, or the part/tool is not clamped properly,” he explained.

Adoption across different sectors

Since its launch, GrayMatter has deployed twenty custom-made smart robotic cells for enterprises in sectors such as aerospace & defense, specialty vehicles, marine & boats, metal fabrication, sports equipment and furniture & sanitary-ware.

The company did not share specific customer names, but noted these cells have cumulatively processed over 7.5 million square feet of product surface area for them.

“The work we’re doing at GrayMatter for companies…is becoming an integral part of their essential operations. It’s a big responsibility, and we’re seeing a generational shift in our lifetime. We’re in a fortunate position to be able to help millions of people elevate and improve their quality of life with our advanced AI-powered technology,” Kabir added. 

Generally, the CEO said the company’s solutions work 2-4 times faster than manual operators and cut down consumable waste by 30% or more.

In one case, an enterprise using its technology for sanding RV caps was able to bring the time taken to complete the task from one hour to six minutes per part. 

As the next step, GrayMatter plans to use this funding to scale its LA team and create next-gen AI robotic cells targeting more use cases.

“All of our current customers are asking us for adjacent products and applications because introducing our system to their production floor removes the bottleneck from that application and pushes it upstream or downstream. We have a strong product roadmap that we need to deliver. With the latest funding raise, we’re looking to create the next generation of AI robots as we continue to grow and expand our team in go-to-market, operations, product, and engineering to meet this growing customer demand,” Kabir said.

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Welcome to the daily solving of our PROBLEM OF THE DAY with Nitin Kaplas. We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Geometry but also build up problem-solving skills. Given three non-collinear points whose co-ordinates are p(p1, p2), q(q1, q2) and r(r1, r2) in the X-Y plane. Return the number of integral / lattice points strictly inside the triangle formed by these points. Note: - A point in the X-Y plane is said to be an integral / lattice point if both its coordinates are integral.

Input: p = (0,0), q = (0,5), r = (5,0) Output: 6

Give the problem a try before going through the video. All the best!!! Problem Link: https://practice.geeksforgeeks.org/problems/integral-points-in-triangle5026/1