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Chapter 8: Rational Expressions

8.8 Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .

[latex]r\cdot t=d[/latex]

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.

Example 8.8.1

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \\ 3r&+&3r&+&6&=&30 \\ &&&-&6&&-6 \\ \hline &&&&\dfrac{6r}{6}&=&\dfrac{24}{6} \\ \\ &&&&r&=&4 \text{ km/h} \end{array}[/latex]

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Example 8.8.2

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

[latex]\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \hline \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}[/latex]

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Example 8.8.3

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

The distance travelled by both is the same. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \hline \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}[/latex]

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

Example 8.8.4

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

[latex]\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \hline &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}[/latex]

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

  • A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
  • Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
  • Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
  • Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
  • A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
  • Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
  • A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
  • A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

  • A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
  • A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
  • A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
  • As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
  • Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
  • A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
  • A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
  • A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
  • Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
  • Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
  • Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
  • Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
  • On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
  • Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answer Key 8.8

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Need Help Solving Those Dreaded Word Problems Involving Quadratic Equations?

Yes, I know it's tough. You've finally mastered factoring and using the quadratic formula and now you are asked to solve more problems!

Except these are even more tough. Now you have to figure out what the problem even means before trying to solve it. I completely understand and here's where I am going to try to help!

There are many types of problems that can easily be solved using your knowledge of quadratic equations. You may come across problems that deal with money and predicted incomes (financial) or problems that deal with physics such as projectiles. You may also come across construction type problems that deal with area or geometry problems that deal with right triangles.

Lucky for you, you can solve the quadratic equations, now you just have to learn how to apply this useful skill.

On this particular page, we are going to take a look at a physics "projectile problem".

Projectiles - Example 1

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. Find the maximum height attained by the ball.

Let's first take a minute to understand this problem and what it means. We know that a ball is being shot from a cannon. So, in your mind, imagine a cannon firing a ball. We know that the ball is going to shoot from the cannon, go into the air, and then fall to the ground.

So, here's a mathematical picture that I see in my head.

Now let's talk about what each part of this problem means. In our equation that we are given we must be given the value for the force of gravity (coefficient of t 2 ). We must also use our upward velocity (coefficient of t) and our original height of the cannon/ball (the constant or 1.5). Take a look...

Now that you have a mental picture of what's happening and you understand the formula given, we can go ahead and solve the problem.

  • First, ask yourself, "What am I solving for?" "What do I need to find?" You are asked to find the maximum height (go back and take a look at the diagram). What part of the parabola is this? Yes, it's the vertex! We will need to use the vertex formula and I will need to know the y coordinate of the vertex because it's asking for the height.
  • Next Step: Solve! Now that I know that I need to use the vertex formula, I can get to work.

Just as simple as that, this problem is solved.

Let's not stop here. Let's take this same problem and put a twist on it. There are many other things that we could find out about this ball!

Projectiles - Example 2

Same problem - different question. Take a look...

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long did it take for the ball to reach the ground?

Now, we've changed the question and we want to know how long did it take the ball to reach the ground.

What ground, you may ask. The problem didn't mention anything about a ground. Let's take a look at the picture "in our mind" again.

Do you see where the ball must fall to the ground. The x-axis is our "ground" in this problem. What do we know about points on the x-axis when we are dealing with quadratic equations and parabolas?

Yes, the points on the x-axis are our "zeros" or x-intercepts. This means that we must solve the quadratic equation in order to find the x-intercept.

Let's do it! Let's solve this equation. I'm thinking that this may not be a factorable equation. Do you agree? So, what's our solution?

Hopefully, you agree that we can use the quadratic formula to solve this equation.

The first time doesn't make sense because it's negative. This is the calculation for when the ball was on the ground initially before it was shot.

This actually never really occurred because the ball was shot from the cannon and was never shot from the ground. Therefore, we will disregard this answer.

The other answer was 2.54 seconds which is when the ball reached the ground (x-axis) after it was shot. Therefore, this is the only correct answer to this problem.

Ok, one more spin on this problem. What would you do in this case?

Projectiles - Example 3

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long does it take the ball to reach a height of 20 feet?

Yes, this problem is a little trickier because the question is not asking for the maximum height (vertex) or the time it takes to reach the ground (zeros), instead it it asking for the time it takes to reach a height of 20 feet.

Since the ball reaches a maximum height of 26.5 ft, we know that it will reach a height of 20 feet on the way up and on the way down.

Let's just estimate on our graph and also make sure that we get this visual in our head.

From looking at this graph, I would estimate the times to be about 0.7 sec and 1.9 sec. Do you see how the ball will reach 20 feet on the way up and on the way down?

Now, let's find the actual values. Where will we substitute 20 feet?

Yes, we must substitute 20 feet for h(t) because this is the given height. We will now be solving for t using the quadratic formula. Take a look.

Our actual times were pretty close to our estimates. Just don't forget that when you solve a quadratic equation, you must have the equation set equal to 0. Therefore, we had to subtract 20 from both sides in order to have the equation set to 0.

You've now seen it all when it comes to projectiles!

Great Job! Hopefully you've been able to understand how to solve problems involving quadratic equations. I also hope that you better understand these common velocity equations and how to think about what this problem looks like graphically in order to help you to understand which process or formula to use in order to solve the problem.

  • Quadratic Equations
  • Projectile Problems

word problem quadratic equation speed

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Quadratic Equations Word Problems

These lessons, with videos, examples, and step-by-step solutions, help Algebra 1 students learn to solve geometry word problems using quadratic equations.

Related Pages Solving Quadratic Equations by Factoring Solving Quadratic Equations by Completing the Square More Lessons for Grade 9 Math Worksheets

Quadratic equations - Solving word problems using factoring of trinomials Question 1a: Find two consecutive integers that have a product of 42

Quadratic equations - Solving word problems using factoring of trinomials Question 1b: There are three consecutive integers. The product of the two larger integers is 30. Find the three integers.

Quadratic Equations - Solving Word problems by Factoring Question 1c: A rectangular building is to be placed on a lot that measures 30 m by 40 m. The building must be placed in the lot so that the width of the lawn is the same on all four sides of the building. Local restrictions state that the building cannot occupy any more than 50% of the property. What are the dimensions of the largest building that can be built on the property?

More Word Problems Using Quadratic Equations Example 1 Suppose the area of a rectangle is 114.4 m 2 and the length is 14 m longer than the width. Find the length and width of the rectangle.

More Word Problems Using Quadratic Equations Example 2 A manufacturer develops a formula to determine the demand for its product depending on the price in dollars. The formula is D = 2,000 + 100P - 6P 2 where P is the price per unit, and D is the number of units in demand. At what price will the demand drop to 1000 units?

More Word Problems Using Quadratic Equations Example 3 The length of a car’s skid mark in feet as a function of the car’s speed in miles per hour is given by l(s) = .046s 2 - .199s + 0.264 If the length of skid mark is 220 ft, find the speed in miles per hour the car was traveling.

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Lesson Using quadratic equations to solve word problems

Quadradic Equation Word Problems - Examples & Practice - Expii

Quadradic equation word problems - examples & practice, explanations (3), quadratic word problems.

Let's look at 2 pretty common types of word problems that use quadratic functions. Recall that quadratic expressions follow this general form: y=ax2+bx+c In a quadratic expression, a and b are coefficients (numbers in front of the variable x), and c is constant (a number by itself). It's important to remember that a≠0 . It's possible for b=0 and/or c=0, but they don't have to.

Example 1: Word problems involving area

Many quadratic word problems use the area of a rectangle. Here is a graphic to get you started

word problem quadratic equation speed

Image source: By Caroline Kulczycky

Here is another example.

Singh knows the total area of his garden is 50 m2, but he doesn't know the length or width. Use the diagram below to find the length and width in meters (m).

A rectangle with width (x - 5) and length x.

Recall that the formula for area of a rectangle is length⋅width. So, let's plug the values from Singh's garden into that formula, and set it equal to 50 (the area). area=length⋅width50=x⋅(x−5)

Now, let's multiply out this equation and get it in the standard quadratic form (written at the top of this explanation):

Once we multiply the equation out and rearrange it, what does it look like (in quadratic form)?

−x2−5x−50=0

−x2−5x+50=0

Related Lessons

(videos) solving geometry word problems using quadratic equations.

by PatrickJMT

word problem quadratic equation speed

This video by Patrick JMT covers a quadratic word problem based on area of a rectangle.

Patrick walks us through how to solve the following quadratic geometry problem:

“A picture inside a frame is 2 inches longer than it is wide. The picture is in a frame that has width 3 inches on each side of the picture. If the area of the pic, including the frame is 195 in.2, find the dimensions of the frame.”

He goes on to explain how we get the equations for the sides of the frame, so the whole problem looks like:

A picture within a picture frame.  The picture has dimensions of x and x+2, and the picture frame has dimensions of x+6 and x+8

We know the area of the picture frame, 195 in2. We can set the two sides of the frame equal to that and find a value for x.

(x+6)(x+8)=195

Using the FOIL method , we get an equation that looks like

x2+14x+48=195

We want the left side to equal 0, so subtracting 195 from both sides will give us

x2+14x–147=0

This looks pretty hard to factor. Let’s use our handy dandy quadratic formula .

Our values are going to be a=1,b=14, and c=−147

x=−b±√b2−4ac2ax=−14±√142–4(1)(−147)2(1)x=−14±√196+5882x=−14±√7842x=−14±282x=−7±14

*Note: be careful with the square root!

This leaves us two values for x, x=7 or x=−21. We cannot have a negative measurement, so our final answer is x=7.

(Videos) Solve Projectile Problems Using Quadratics

word problem quadratic equation speed

This video by Patrick JMT works through a word problem based on projectile motion .

Solving a Projectile Problem Using Quadratics

Patrick uses the following problem as an example of projectile motion:

“Larry throws a rock in the air. The height, h, in feet above the ground of the rock is given by: h=−16t2+123t+40. How long is the rock in the air?”

From this equation, at time t=0, we gather that h=40. This means Larry is on a cliff with height 40. We want to know the time at which the rock reaches a height of 0. So, plugging in 0 for h, we get an equation:

0=−16t2+123t+40

This is now a quadratic equation, which we can solve for using the quadratic formula! We have values of a=16,b=123, and c=40. We can plug these into the quadratic formula: t=−b±√b2−4ac2at=−123±√(123)2−4(−16)(40)2(−16)t=−123±√17689−32t=123±√1768932t=123±13332t=8,−516

We get two solutions: 8 and −516. t is time, which can't be negative, so our answer is t=8.

Real World Examples of Quadratic Equations

A Quadratic Equation looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

  • Factoring Quadratics
  • Completing the Square
  • Graphing Quadratic Equations
  • The Quadratic Formula
  • Online Quadratic Equation Solver

Each example follows three general stages:

  • Take the real world description and make some equations
  • Use your common sense to interpret the results

ball throw

Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

Example: Throwing a Ball

A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. when does it hit the ground.

Ignoring air resistance, we can work out its height by adding up these three things: (Note: t is time in seconds)

Add them up and the height h at any time t is:

h = 3 + 14t − 5t 2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t 2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t 2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1 :

5t 2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give ac , and add to give b " method in Factoring Quadratics :

ac = −15 , and b = −14 .

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

The "t = −0.2" is a negative time, impossible in our case.

The "t = 3" is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = −5t 2 + 14t + 3

It shows you the height of the ball vs time

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes nearly 13 meters high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations , and has two steps:

Find where (along the horizontal axis) the top occurs using −b/2a :

  • t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

  • h = −5t 2 + 14t + 3 = −5(1.4) 2 + 14 × 1.4 + 3 = 12.8 meters

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

Example: New Sports Bike

bike

You have designed a new style of sports bicycle!

Now you want to make lots of them and sell them for profit.

Your costs are going to be:

  • $700,000 for manufacturing set-up costs, advertising, etc
  • $110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

Where "P" is the price.

For example, if you set the price:

  • at $0, you just give away 70,000 bikes
  • at $350, you won't sell any bikes at all
  • at $300 you might sell 70,000 − 200×300 = 10,000 bikes

So ... what is the best price? And how many should you make?

Let us make some equations!

How many you sell depends on price, so use "P" for Price as the variable

Profit = −200P 2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square .

Solve: −200P 2 + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2) 2 = (−460/2) 2 = (−230) 2 = 52900

Step 4 Take the square root on both sides of the equation:

Step 5 Subtract (-230) from both sides (in other words, add 230):

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

The best sale price is $230 , and you can expect:

  • Unit Sales = 70,000 − 200 x 230 = 24,000
  • Sales in Dollars = $230 x 24,000 = $5,520,000
  • Costs = 700,000 + $110 x 24,000 = $3,340,000
  • Profit = $5,520,000 − $3,340,000 = $2,180,000

A very profitable venture.

Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm 2

The inside of the frame has to be 11 cm by 6 cm

What should the width x of the metal be?

Area of steel before cutting:

Area of steel after cutting out the 11 × 6 middle:

Let us solve this one graphically !

Here is the graph of 4x 2 + 34x :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm 2 when:

x is about −9.3 or 0.8

The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

Example: River Cruise

A 3 hour river cruise goes 15 km upstream and then back again. the river has a current of 2 km an hour. what is the boat's speed and how long was the upstream journey.

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

  • Let x = the boat's speed in the water (km/h)
  • Let v = the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

  • when going upstream, v = x−2 (its speed is reduced by 2 km/h)
  • when going downstream, v = x+2 (its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by (x-2) (x+2) :

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x 2 −4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x 2 − 30x − 12 = 0

It is a Quadratic Equation!

Let us solve it using the Quadratic Formula :

Where a , b and c are from the Quadratic Equation in "Standard Form": ax 2 + bx + c = 0

Solve 3x 2 - 30x - 12 = 0

Answer: x = −0.39 or 10.39 (to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "R T " is:

1 R T   =   1 R 1 + 1 R 2

In this case, we have R T = 2 and R 2 = R 1 + 3

1 2   =   1 R 1 + 1 R 1 +3

To get rid of the fractions we can multiply all terms by 2R 1 (R 1 + 3) and then simplify:

Yes! A Quadratic Equation!

Let us solve it using our Quadratic Equation Solver .

  • Enter 1, −1 and −6
  • And you should get the answers −2 and 3

R 1 cannot be negative, so R 1 = 3 Ohms is the answer.

The two resistors are 3 ohms and 6 ohms.

Quadratic Equations are useful in many other areas:

parabolic dish

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Quadratic equations are also needed when studying lenses and curved mirrors.

And many questions involving time, distance and speed need quadratic equations.

  • RD Sharma Solutions
  • Chapter 8 Quadratic Equations
  • Exercise 8.8

RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.8

Quadratic equations have applications in many areas. One of these is solving problems on time and distance. In order to clear any conceptual doubts regarding these, refer to the RD Sharma Solutions Class 10 . More detailed solutions of the RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.8 are given in the PDF below.

  • RD Sharma Solutions Class 10 Maths Chapter 1 Real Numbers
  • RD Sharma Solutions Class 10 Maths Chapter 2 Polynomials
  • RD Sharma Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
  • RD Sharma Solutions Class 10 Maths Chapter 4 Triangles
  • RD Sharma Solutions Class 10 Maths Chapter 5 Trigonometric Ratios
  • RD Sharma Solutions Class 10 Maths Chapter 6 Trigonometric Identities
  • RD Sharma Solutions Class 10 Maths Chapter 7 Statistics
  • RD Sharma Solutions Class 10 Maths Chapter 8 Quadratic Equations
  • RD Sharma Solutions Class 10 Maths Chapter 9 Arithmetic Progressions
  • RD Sharma Solutions Class 10 Maths Chapter 10 Circles
  • RD Sharma Solutions Class 10 Maths Chapter 11 Constructions
  • RD Sharma Solutions Class 10 Maths Chapter 12 Some Applications of Trigonometry
  • RD Sharma Solutions Class 10 Maths Chapter 13 Probability
  • RD Sharma Solutions Class 10 Maths Chapter 14 Co-ordinate Geometry
  • RD Sharma Solutions Class 10 Maths Chapter 15 Areas Related to Circles
  • RD Sharma Solutions Class 10 Maths Chapter 16 Surface Areas and Volumes
  • Exercise 8.1 Chapter 8 Quadratic Equations
  • Exercise 8.2 Chapter 8 Quadratic Equations
  • Exercise 8.3 Chapter 8 Quadratic Equations
  • Exercise 8.4 Chapter 8 Quadratic Equations
  • Exercise 8.5 Chapter 8 Quadratic Equations
  • Exercise 8.6 Chapter 8 Quadratic Equations
  • Exercise 8.7 Chapter 8 Quadratic Equations
  • Exercise 8.8 Chapter 8 Quadratic Equations
  • Exercise 8.9 Chapter 8 Quadratic Equations
  • Exercise 8.10 Chapter 8 Quadratic Equations
  • Exercise 8.11 Chapter 8 Quadratic Equations
  • Exercise 8.12 Chapter 8 Quadratic Equations
  • Exercise 8.13 Chapter 8 Quadratic Equations

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RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 57

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Access RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.8

1. The speed of a boat in still water is 8km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.

Let the speed of the stream be x km/hr

Given, speed of the boat in still water is 8km/hr.

So, speed of downstream = (8 + x) km/hr

And, speed of upstream = (8 – x) km/hr

Using, speed = distance/ time

Time taken by the boat to go 15 km upstream = 15/(8 – x)hr

And, time taken by the boat to return 22 km downstream = 22/(8 + x)hr

From the question, the boat returns to the same point in 5 hr.

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 1

5x 2  – 7x + 296 – 320 = 0

5x 2  – 7x – 24 = 0

5x 2  – 15x + 8x – 24 = 0 [by factorisation method]

5x(x – 3) + 8(x – 3) = 0

(x – 3)(5x + 8) = 0

∴ x = 3, x = – 8/5

As the speed of the stream can never be negative, only the positive solution is considered.

Therefore, the speed of the stream is 3 km/hr.

2. A train, traveling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

Let the original speed of the train be x km/hr

When increased by 5, speed of the train = (x + 5) km/hr

Time taken by the train for the original uniform speed to cover 360 km = 360/x hr.

And, time taken by the train for increased speed to cover 360 km = 360/(x + 5) hr.

Given that the difference in the times is 48 mins. ⇒ 48/60 hour

This can be expressed as below:

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 2

1800(5) = 4(x 2  + 5x)

9000 = 4x 2  + 20x

4x 2  + 20x – 9000 = 0

x 2  + 5x – 2250 = 0

x 2  + 50x – 45x – 2250 = 0 [by factorisation method]

x(x + 50) – 45(x + 50) = 0

(x + 50)(x – 45) = 0

∴ x = – 50 or x = 45

Since the speed of the train can never be negative x = -50 is not considered.

Therefore, the original speed of the train is 45 km/hr.

3. A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.

Let’s consider the speed of the fast train as x km/hr

Then, the speed of the slow train will be = (x -10) km/hr

Time taken by the fast train to cover 200 km = 200/x hr

And, time taken by the slow train to cover 200 km = 200/(x – 10) hr

Given that the difference in the times is 1 hour.

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 3

x 2 – 10x = – 2000

x 2  – 10x + 2000 = 0

x 2  – 50x + 40x + 2000 = 0 [by factorisation method]

x(x – 50) + 40(x – 50) = 0

(x – 50)(x + 40) = 0

x = 50 or x = – 40

As, the speed of a train can never be negative, we neglect x = -40

Thus, the speed of the fast train is 50 km/hr

And the speed of the slow train (50 – 10) = 40 km/hr

4. A passenger train takes one hour less for a journey of 150 km if its speed is increased 5 km/hr from its usual speed. Find the usual speed of the train.

Let’s assume the usual speed of the train as x km/hr

Then, the increased speed of the train = (x + 5) km/hr

Time taken by the train under usual speed to cover 150 km = 150/x hr

Time taken by the train under increased speed to cover 150 km = 150(x + 5)hr

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 4

750 = x 2  + 5x

x 2  + 5x – 750 = 0

x 2  – 25x + 30x -750 = 0 [by factorisation method]

x(x – 25) + 30(x – 25) = 0

(x – 25) (x + 30) = 0

x = 25  or x = -30 (neglected as the speed of the train can never be negative)

Hence, the usual speed of the train is x = 25 km/hr

5. The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at the speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?

Let the ongoing speed of person be x km/hr,

Then, the returning speed of the person is = (x + 10) km/hr (from the question)

Time taken by the person in the forward direction to cover 150 km = 150/x hr

And, time taken by the person in returning direction to cover 150 km = 150/(x + 10)hr

Given that the difference in the times is 2.5 hours ⇒ 5/2 hours

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 5

3000 = 5x 2  + 50x

5x 2  + 50x – 3000 = 0

5(x 2  + 10x – 600) = 0

x 2  + 10x – 600 = 0

x 2  – 20x + 30x – 600 = 0 [by factorisation method]

x(x – 20) + 30(x – 20) = 0

(x – 20)(x + 30) = 0

x = 20  or x = -30(neglected) As the speed of train can never be negative.

Thus, x = 20 Then, (x + 10) (20 + 10) = 30

Therefore, the ongoing speed of the person is 20km/hr.

And the returning speed of the person is 30 km/hr.

6. A plane left 40 minutes late due to bad weather and in order to reach the destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.

Let’s assume the usual speed of the plane to be x km/hr,

Then the increased speed of the plane is = (x + 4000) km/hr

Time taken by the plane under usual speed to cover 1600 km = 1600/x hr

Time taken by the plane under increased speed to cover 1600 km = 1600/(x + 400) hr

Given that the difference in the times is 40mins ⇒ 40/60 hours

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 6

1920000 = 2x 2  + 800x

2x 2  + 800x – 1920000 = 0

2(x 2 + 400x – 960000) = 0

x 2  + 400x – 960000 = 0

x 2  – 800x + 1200x – 960000 = 0 [by factorisation method]

x(x – 800) + 1200(x – 800) = 0

(x – 800)(x + 1200) = 0

x = 800 or x = -1200 (neglected)

As the speed of the train can never be negative.

Thus, the usual speed of the train is 800 km/hr.

7. An aero plane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed of the plane. Find its usual speed.

Let’s consider the usual speed of the plane as x km/hr,

Then, the increased speed of the plane is = (x + 100) km/hr

Time taken by the plane under usual speed to cover 1200 km = 1200/x hr

Time taken by the plane under increased speed to cover 1200 km = 1200/(x + 100)hr

So, this can be expressed as below:

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.8 - 7

120000 = x 2  + 100x

x 2  + 100x – 120000 = 0

x 2  – 300x + 400x – 120000 = 0 [by factorisation method]

x(x – 300) + 400(x – 300) = 0

x = 300 or x = – 400 neglected as the speed of the aeroplane can never be negative.

Therefore, the usual speed of the train is 300 km/hr.

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Quadratic Word Problems Worksheets

In mathematics, the term quadratic describes something that pertains to squares, to the operation of squaring, to terms of the second degree, or equations or formulas that involve such terms.

In the quadratic equations word problems, the equations wouldn't be given directly. In fact, you have to deduct the equation from the given facts within the equations. It can also include profit maximization or loss minimization questions in which you have to find either minimum or maximum value of the equation.

Download Quadratic Word Problem Worksheet PDFs

These math worksheets should be practiced regularly and are free to download in PDF formats.

IMAGES

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