A biker travels 60 miles in 2.5 hours. Determine the biker's average speed.

SolutionThe equation relating distance, velocity and time provides

60 = V(5/2)

Divide both sides by 5/2 to solve for V.

V = (60)2/5 = 24 miles/hour

 

Example 16

A car travels between two cities 400 miles apart in 7 hours. The return trip takes 9 hours. Find the average speed of the car.

SolutionThe total distance is 2(400) = 800 miles. The total time is 7 + 9 = 16 hours. The average speed is found from D = VT:

800 = V(16)

V = 800/16 = 50 miles/hour

Example 17

A police officer, traveling at 100 miles per hour, pursues Philip who has a 30 minute head start. The police officer overtakes Philip in two hours. Find Philip's speed.

SolutionLet x miles per hour be Philip's speed. The distance traveled by the officer equals the distance traveled by Philip:

2 × 100 = (2 + 30/60)x

200 = (2 + 0.5)x

200 = 2.5x therefore x = 80 mph

A 555-mile, 5-hour plane trip was flown at two speeds. For the first part of the trip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed?

First I'll set up a grid:   Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved

  d r tfirst part d 105 t

second part 555 – d 115 5 – ttotal 555 --- 5

Using "d = rt", the first row gives me d = 105t and the second row gives me:

555 – d = 115(5 – t)

Since the two distances add up to 555, I'll add the two distance expressions, and set their sum equal to the given total:

555 = 105t + 115(5 – t)

Then I'll solve:

555 = 105t + 575 – 115t 555 = 575 – 10t –20 = –10t 2 = t

According to my grid, "t" stands for the time spent on the first part of the trip, so my answer is "The plane flew for two hours at 105 mph and three hours at 115 mph."

You can add distances and you can add times, but you cannot add rates. Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph?

As you can see, the actual math involved is often quite simple. It's the set-up that's the hard part. So what follows are some more examples, but with just the set-up displayed.

An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.   

  d r tdriving d 30 tflying 150 – d 60 3 – ttotal 150 --- 3

The first row gives me the equation d = 30t. Since the first part of his trip accounted for d miles of the total 150-mile distance and t hours of the total 3-hour time, I am left with 150 – d miles and 3 – t hours for the second part. The second row gives me the equation:

150 – d = 60(3 – t)

Adding the two "distance" expressions and setting their sum equal to the given total distance, I get:

150 = 30t + 60(3 – t)

Solve for t; interpret the value; state the final answer.

A car and a bus set out at 2 p.m. from the same point, headed in the same direction. The average speed of the car is 30 mph slower than twice the speed of the bus. In two hours, the car is 20 miles ahead of the bus. Find the rate of the car.   

  d r tcar d + 20 2r – 30 2bus d r 2total --- --- ---

(As it turns out, I won't need the "total" row this time.) The first row gives me:

d + 20 = 2(2r – 30)

This is not terribly helpful. The second row gives me:

d = 2r

Use the second equation to simplify the first equation by substituting "2r" in for "d", and then solve for "r". Interpret this value within the context of the exercise, and state the final answer.

A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if the passenger train overtakes the freight train in three hours.   

  d r t

passenger train d r 3freight train d r – 20 3 + 2 = 5

total --- --- ---

(As it turns out, I won't need the "total" row this time.) Why is the distance just "d" for both trains? Partly, that's because the problem doesn't say how far the trains actually went. But mostly it's because they went the same distance as far as I'm concerned, because I'm only counting from the depot to wherever they met. After that meet, I don't care what happens. And how did I get those times? I know that the passenger train drove for three hours to catch up to the freight train; that's how I got the "3". But note that the freight train had a two-hour head start. That means that the freight train was going for five hours.

  d r tpassenger train d = 3r r 3

freight train d = 5(r – 20) r – 20 3 + 2 = 5total --- --- ---

Now that I have this information, I can try to find my equation. Using the fact that d = rt, the first row gives me d = 3r (note the revised table above). The second row gives me:

d = 5(r – 20)

Since the distances are equal, I will set the equations equal:

3r = 5(r – 20)

Solve for r; interpret the value within the context of the exercise, and state the final answer.

Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet?   

  d r tslow guy d 14 tfast guy 45 – d 16 t

total 45 --- ---

Why is t the same for both cyclists? Because I am measuring from the time they both started to the time they meet somewhere in the middle. And how did I get "d" and "45 – d" for the distances? Because once I'd assigned the slow guy as having covered d miles, that left 45 – d miles for the fast guy to cover: the two guys together covered the whole 45 miles.

Using "d = rt", I get d = 14t from the first row, and 45 – d = 16t from the second row. Since these distances add up to 45, I will add the distance expressions and set equal to the given total:

45 = 14t + 16t

Solve for t.

A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water? How far did the boat travel one way?   

  d r tdownstream d b + 3 3

upstream d b – 3 4total 2d --- 7

(It may turn out that I won't need the "total" row.)

I have used "b" to indicate the boat's speed. Why are the rates "b + 3" and "b – 3"? Because I actually have two speeds combined into one on each trip. The boat has a certain speed (the "speed in calm water" that I'm looking for; this is the speed that registers on the speedometer), and the water has a certain speed (this is the "current"). When the boat is going with the current, the water's speed is added to the boat's speed. This makes sense, if you think about it: even if you cut the engine, the boat would still be moving, because the water would be carrying it downstream. When the boat is going against the current, the water's speed is subtracted from the boat's speed. This makes sense, too: if the water is going fast enough, the boat will still be going downstream (a "negative" speed, because the boat would be going backwards at this point), because the water is more powerful than the boat. (Think of a boat in a cartoon heading toward a waterfall. The guy paddles like crazy, but he still goes over the edge.)

  d r tdownstream d = 3(b + 3) b + 3 3

upstream d = 4(b – 3) b – 3 4total 2d --- 7

Using "d = rt", the first row (of the revised table above) gives me:

d = 3(b + 3)

The second row gives me:

d = 4(b – 3)

Since these distances are the same, I will set them equal:

3(b + 3) = 4(b – 3)

Solve for b. Then back-solve for d.

In this case, I didn't need the "total" row.

With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight hours. Find the rate of the plane in still air and the velocity of the wind.   

  d r ttailwind 1120 p + w 7

headwind 1120 p – w 8total 2240 --- 15

(I probably won't need the "total" row.) Just as with the last problem, I am really dealing with two rates together: the plane's speedometer reading, and the wind speed. When the plane turns around, the wind is no longer pushing the plane to go faster, but is instead pushing against the plane to slow it down.   Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved

The first row gives me:

1120 = 7(p + w)

The second row gives me:

1120 = 8(p – w)

The temptation is to just set these equal, like I did with the last problem, but that just gives me:

7(p + w) = 8(p – w)

...which doesn't help much. I need to get rid of one of the variables.

I'll take that first equation:

1120 = 7(p + w)

...and divide through by 7:

160 = p + w

Then, subtracting w from either side, I get that p = 160 – w. I'll substitute "160 – w" for "p" in the second equation:

1120 = 8([160 – w] – w)

1120 = 8(160 – 2w)

...and solve for w. Then I'll back-solve to find p.

A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the sound of the hammer strike both through the air and through the rail itself. These sounds arrive at your point six seconds apart. You know that sound travels through air at 1100 feet per second and through steel at 16,500 feet per second. How far away is that spike?  

  d r tair d = 1100t 1100 t

steel d = 16,500(t – 6) 16,500 t – 6total --- --- 6

However long the sound took to travel through the air, it took six seconds less to propagate through the steel. (Since the speed through the steel is faster, then that travel-time must be shorter.) I multiply the rate by the time to get the values for the distance column. (Once again, I didn't need the "total" row.)

Since the distances are the same, I set the distance expressions equal to get:

1100t = 16,500(t – 6)

Solve for the time t, and then back-solve for the distance d by plugging t into either expression for the distance d.

Adding To The Solution

Mixture Problems: Example 1:

John has 20 ounces of a 20% of salt solution, How much salt should he add to make it a 25% solution?

Solution:

Step 1: Set up a table for salt.

  original added result

concentration      

amount      

Step 2: Fill in the table with information given in the question.

John has 20 ounces of a 20% of salt solution. How much salt should he add to make it a 25% solution?

The salt added is 100% salt, which is 1 in decimal.Change all the percent to decimals

Let x = amount of salt added. The result would be 20 + x.

  original added result

concentration 0.2 1 0.25

amount 20 x 20 + x

Step 3: Multiply down each column.

  original added result

concentration 0.2 1 0.25

amount 20 x 20 + x

multiply 0.2 × 20 1 × x 0.25(20 + x)

Step 4: original + added = result

0.2 × 20 + 1 × x = 0.25(20 + x)4 + x = 5 + 0.25x

Isolate variable xx – 0.25x = 5 – 40.75x = 1

Answer: He should add ounces of salt.

 

 

Removing From The Solution

Mixture Problems: Example 2:

John has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 30% solution?

Solution:

Step 1: Set up a table for water. The water is removed from the original.

  original removed result

concentration      

amount      

Step 2: Fill in the table with information given in the question.

John has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 30% solution?

The original concentration of water is 100% – 20% = 80%The resulted concentration of water is 100% – 30% = 70%The water evaporated is 100% water, which is 1 in decimal.

Change all the percent to decimals.

Let x = amount of water evaporated. The result would be 20 – x.

  original removed result

concentration 0.8 1 0.7

amount 20 x 20 – x

Step 3: Multiply down each column.

  original removed result

concentration 0.8 1 0.7

amount 20 x 20 – x

multiply 0.8 × 20 1 × x 0.70(20 – x)

Step 4: Since the water is removed, we need to subtract

original – removed = result0.8 × 20 – 1 × x = 0.70(20 – x)16 – x = 14 – 0.7x

Isolate variable xx – 0.7x = 16 – 140.3x = 2

Answer: He should evaporate 6.67 ounces of water.

 

 

Replacing The Solution

Mixture Problems: Example 3:

A tank has a capacity of 10 gallons. When it is full, it contains 15% alcohol. How many gallons must be replaced by an 80% alcohol solution to give 10 gallons of 70% solution?

Solution:

Step 1: Set up a table for alcohol. The alcohol is replaced i.e. removed and added.

  original removed added result

concentration        

amount        

Step 2: Fill in the table with information given in the question.

A tank has a capacity of 10 gallons. When it is full, it contains 15% alcohol. How many gallons must be replaced by an 80% alcohol solution to give 10 gallons of 70% solution?

Change all the percent to decimals.

Let x = amount of alcohol solution replaced.

  original removed added result

concentration 0.15 0.15 0.8 0.7

amount 10 x x 10

Step 3: Multiply down each column.

  original removed added result

concentration 0.15 0.15 0.8 0.7

amount 10 x x 10

multiply 0.15 × 10 0.15 × x 0.8 × x 0.7 × 10

Step 4: Since the alcohol solution is replaced, we need to subtract and add.

original – removed + added = result0.15 × 10 – 0.15 × x + 0.8 × x = 0.7 × 10 1.5 – 0.15x + 0.8x = 7

Isolate variable x0.8x – 0.15x = 7 – 1.50.65x = 5.5

Answer: 8.46 gallons of alcohol solution needs to be replaced.

 

 

Mixing Quantities Of Different Costs

Mixture Problems: Example 4:

How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to produce a mixture worth $1.00 a pound?

Solution:

Step 1: Set up a table for different types of chocolate.

  original added result

cost      

amount      

Step 2: Fill in the table with information given in the question.

How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to produce a mixture worth $1.00 a pound?

Let x = amount of chocolate added.

  original added result

cost 0.9 1.2 1

amount 10 x x + 10

Step 3: Multiply down each column.

  original added result

cost 0.9 1.2 1

amount 10 x x + 10

multiply 0.9 × 10 1.2 × x 1 × (x + 10)

Step 4: original + added = result

0.9 × 10 + 1.2 × x = 1 × (x + 10) 9 + 1.2x = x + 10

Isolate variable x1.2x – x = 10 - 90.2x = 1

Answer: 5 pounds of the $1.20 chocolate needs to be added.

Word Problems: Mixtures In order to solve problems involving mixture, it is necessary to 

solve one variable equations involving: one step, two steps, multi-steps analyze and understand the problem

write and solve an equation for the problem

 Even though there are two different types of mixes, the process for solving them is the same. 

dry mixture created from two differently priced components such as two types of candy or two types of nuts

liquid mixture of a specific strength made from two or more solutions having differing concentrations

 Suppose the owner of a candy store mixes two types of candies. She decides to create a 20-pound mixture of raspberry-flavored gumdrops and cherry-flavored jelly beans. The gumdrops sell for $0.95 per pound and the jelly beans sell for $1.20 per pound. She plans to sell the mix for $1.10 per pound. How many pounds of each candy should she use in her

mix? First, since two quantities are to be mixed together to produce one mixture, we need to recognize that we will set up an equation that shows the following: total cost of gumdrops plus total cost of jelly beans equals the total cost of mixture  To arrive at the equation, it is typically helpful to use a table illustrating the problem such as the following:

Type of Candy Cost of Candy(unit price)

Amount of Candy(in pounds)

Total Cost(in dollars)

gumdrops

jelly beans

mixture

 The first column shows the types of candy involved, cost is displayed in the second column, amount of each type of candy is listed in the third column, and the fourth column is the product of each cost and each amount for each type of candy. The total cost of the mixture is found by multiplying the cost of each type candy times the amount of each type of candy used in the mixture. The total cost column will be used to write the equation. The candy store owner knows that she wants the total amount of the mixture of candy to be 20 pounds. However, she does not know how many pounds of each type to mix. That is the objective of the problem. So in the “Amount of Candy” column we will use x to represent the amount of gumdrops. Then the “total pounds of candy minus x” will represent the amount of jelly beans: (20 – x). The last column demonstrates that the price of each type candy multiplied times the amount of each type candy represents the total cost of each type candy. The last column is what we use to write the equation. Remember: total cost of gumdrops plus total cost of jelly beans equals the total cost of mixture  So, using the information in the last column: 

 We are ready to solve the equation to find the amount of each type of candy the store owner should use in her mixture.

First distribute to remove parentheses

Now multiply the equation by 100

Solve for x

pounds

 Since x represents the amount of gumdrops to be used in the mixture the candy store owner will use 8 pounds of gumdrops. From column three of the table, you can see that the amount of jelly beans to be used is (20 - x). Substituting 8 for the x, we see that the store owner needs to use (20 - 8) which is 12 pounds of jelly beans to create the desired mix. We have now learned the candy store owner will mix 8 pounds of raspberry-flavored gumdrops that cost $0.95 per pound and 12 pounds of cherry-flavored jelly beans that cost $1.20 per pound to create a mixture of 20 pounds of candy that sells for $1.10 per pound.

Let's Practice

Question #1A local grocer has decided to mix 100 pounds of cashews and almonds for a holiday special. Cashews typically cost $8 per pound and almonds cost $3 per pound. How many pounds of each type of nut must he mix to obtain a mixture that will cost his customers $5 per pound?

Question #2A chemist needs a 40% solution of alcohol. He plans to mix 3 liters of a 60% solution with a 25% solution. How many liters of the 25% solution must we mix with the 3 liters of the 60% solution to obtain the desired 40% solution of alcohol?

Try These

Question #1The owner of a coffee shop has decided to mix types of teas to create a new blend. He will mix a type of tea that sells for $4 per pound with a type that sells for $2.40 per pound to produce 80 pounds of mixture that he will sell for $3.60 per pound. How much of the tea that costs $2.40 per pound must he use in the mixture?

A. 60 pounds

B. 20 pounds

C. 30.625 pounds

D. -360 pounds

Question #2How many liters of a 92-octane gasoline should be mixed with 200 liters of a 98-octane gasoline to produce a mixture that is 96-octane gasoline?

A.

B.

C.

D.

This type of problem requires carefully setting up a table using the information given in the problem. Following the development of the table, a linear equation must be written and solved. Using the information written in the last column of the table, set up your equation in this format: total amount of item one plus total amount of item two equals total amount of the mixture  After writing the equation and solving for the unknown variable, always check to be certain that the answer obtained answers the question and/or question(s) asked in the problem. It is always advisable to substitute your answers back into the equation to check that your solution is correct.

Kuta Software - Infinite Algebra 1 Name___________________________________

Period___Mixture Word Problems Date________________ _1) 2 m³ of soil containing 35% sand was mixedinto 6 m³ of soil containing 15% sand. Whatis the sand content of the mixture?

2) 9 lbs. of mixed nuts containing 55% peanutswere mixed with 6 lbs. of another kind ofmixed nuts that contain 40% peanuts. Whatpercent of the new mixture is peanuts?3) 5 fl. oz. of a 2% alcohol solution was mixedwith 11 fl. oz. of a 66% alcohol solution.Find the concentration of the new mixture.4) 16 lb of Brand M Cinnamon was made bycombining 12 lb of Indonesian cinnamonwhich costs $19/lb with 4 lb of Thaicinnamon which costs $11/lb. Find the costper lb of the mixture.5) Emily mixed together 9 gal. of Brand A fruitdrink and 8 gal. of Brand B fruit drinkwhich contains 48% fruit juice. Find thepercent of fruit juice in Brand A if themixture contained 30% fruit juice.6) How many mg of a metal containing 45%nickel must be combined with 6 mg of purenickel to form an alloy containing 78%nickel?7) How much soil containing 45% sand do you needto add to 1 ft³ of soil containing 15% sand in orderto make a soil containing 35% sand?8) 9 gal. of a sugar solution was mixed with 6gal. of a 90% sugar solution to make a 84%sugar solution. Find the percentconcentration of the first solution.9) A metallurgist needs to make 12.4 lb. of analloy containing 50% gold. He is going tomelt and combine one metal that is 60%gold with another metal that is 40% gold.How much of each should he use?10) Brand X sells 21 oz. bags of mixed nuts thatcontain 29% peanuts. To make theirproduct they combine Brand A mixed nutswhich contain 35% peanuts and Brand Bmixed nuts which contain 25% peanuts.How much of each do they need to use?Kuta Software - Infinite Algebra 1 Name___________________________________

Period___Mixture Word Problems Date________________ _1) 2 m³ of soil containing 35% sand was mixedinto 6 m³ of soil containing 15% sand. Whatis the sand content of the mixture?20%

2) 9 lbs. of mixed nuts containing 55% peanutswere mixed with 6 lbs. of another kind ofmixed nuts that contain 40% peanuts. Whatpercent of the new mixture is peanuts?49%3) 5 fl. oz. of a 2% alcohol solution was mixedwith 11 fl. oz. of a 66% alcohol solution.Find the concentration of the new mixture.46%4) 16 lb of Brand M Cinnamon was made bycombining 12 lb of Indonesian cinnamonwhich costs $19/lb with 4 lb of Thaicinnamon which costs $11/lb. Find the costper lb of the mixture.$17/lb5) Emily mixed together 9 gal. of Brand A fruitdrink and 8 gal. of Brand B fruit drinkwhich contains 48% fruit juice. Find thepercent of fruit juice in Brand A if themixture contained 30% fruit juice.14%6) How many mg of a metal containing 45%nickel must be combined with 6 mg of purenickel to form an alloy containing 78%nickel?4 mg7) How much soil containing 45% sand do you needto add to 1 ft³ of soil containing 15% sand in orderto make a soil containing 35% sand?2 ft³8) 9 gal. of a sugar solution was mixed with 6gal. of a 90% sugar solution to make a 84%sugar solution. Find the percentconcentration of the first solution.80%9) A metallurgist needs to make 12.4 lb. of analloy containing 50% gold. He is going tomelt and combine one metal that is 60%gold with another metal that is 40% gold.How much of each should he use?6.2 lb. of 60% gold, 6.2 lb. of 40% gold10) Brand X sells 21 oz. bags of mixed nuts thatcontain 29% peanuts. To make theirproduct they combine Brand A mixed nutswhich contain 35% peanuts and Brand Bmixed nuts which contain 25% peanuts.

How much of each do they need to use?8.4 oz. of Brand A, 12.6 oz. of Brand B

John has $20,000 to invest. He invests part of his money at an annual interest rate of 6%, the rest at 9% annual rate. The return on these two investments over one year is $1,440. How much does he invest at each rate?

Solution Paul made two investments totaling $15,000. The percentage return on the first investment was 7% annually, while the the percentage return on the second one was 10% annually. If the total return on the two investments over one year was $1,350, how much was invested at each rate?   Ben invested $30,000, part of which at 5% annual interest rate, the rest at 9% annual interest rate. The interest earned from the investments was $2,100 at the end of one year. How much did he invest at each rate?

Solution Jason invested $20,000 for one year, Part of his money was invested at an annual interest rate of 6%, the rest at an annual interest rate of 10%. If his total income from the two investments over one year was $1,700, how much was invested at each rate?  Jane had $20,000 to invest for one year. She deposited part of which into an account paying 5% annual interest. the rest into another account paying 8% annual interest. If the total interest earned at the end of one year was $1,390, how much was invested at each account?  A total of $18,000 was invested for 6 months, part at 4% annual interest rate and part at 7% annual interest rate. The total interest earned over the 6 month period was $450, how much was invested at each rate?

Solution $12,000 was invested for three months. Part of which was invested at 6% annual interest rate and the rest at 10% annual interest rate. If the total income for three months from the investments

was $240, how much was invested at each rate?  Sue has $15,000 to invest for 5 months, part at 6% annual rate, the rest at 10% annual rate. If the total interest earned at the end of five months is $450, how much was invested at each rate?

1. Lisa requires additional income to meet her everyday expenses. She has $30,000 to invest. To generate the required additional income, the annual return rate has to be 6%. She deposits part of her capital into an account paying 4% per year, and invest the rest in stocks earning 10% per year. How much does she need to invest at each rate?  

2. To balance risk and return on his investment, Ben invests part of his money in an low risk, low return bank saving account paying 5% per year, the rest in high risk, high return stocks earning 15% per year. To achieve the goal of a 12% annual return on $24,000 investment, how much does he need to invest at each rate? 

3. Sue has invested $12,000 at an annual interest rate of 6%. To realize an annual return of 8% on her investment, how much more funds must she invest at an annual rate of 12%? 

4. Paul deposited $21,000 into his bank saving account paying 5% per year. How much additional funds must he invest at 10% annually so that the annual return on his total investment is 7%? 

5. Joe made two investments. She earned 8% annually on her first investment, but lost 12% annually on her second investment. If her total investment was $15,000, and the total income was $240 for one year, how much money was allocated to each investment? 

6. Allan made two investments totaling $20,000. He earned 5% annually on his first investment, but lost 10% annually on his second investment. If the net loss was $140 for one year, how much money was allocated to each investment? 

7. Jan made two investments. She made a 7% profit on her first investment, but lost 10% annually on her second investment. If her total investment was $16,000, and the total income was $100, how much money was allocated to each investment? 

8. Paul made two investments totaling $25,000. He made a profit of 6% on his first investment, but made a loss of 10% on his second investment. If the net loss was $260, how much money was allocated to each investment? 

9. Ben invests 30% of his total funds at 5% annual rate, the rest at 8% annual rate. If his total income for one year is $625, how much does he invest at each rate? 

10. Ben invests 30% of his total funds at 5% annual rate, 40% at 6% annual rate, the rest at 8% annual rate. If his total income for one year is $560, how much does he invest at each rate?

11. A certain amount of money is to be invested for a period of one year. The amount of money invested at 6% per year is twice as much as the amount invested at 9% per year. If the income for one year is $1680, how much is invested at each rate?

Solution 

12. Jeff has some money to invest for one year. The amount of money invested at 5% per year is $5,000 more than that invested at 8% per year. The interest earned is $1,160. How much does he invest at each rate?

Solution 

13. One third of the funds is invested at an annual interest rate of 4%, the rest at 6% annual rate. If the income for one year is $480, how much is invested at each rate? 

14. The ratio of the the amount of money invested at 6% annual rate to the amount invested at 9% is 2 : 5. If the total income for one year is $570, how much is invested at each rate? 

15. A certain amount of money is to be invested for one year. The amount of money invested at 6% per year is $2,000 more than twice the amount invested at 9% per year. If the income for one year is $1,800, how much is invested at each rate? 

16. John has some money to invest, the amount of money invested at 5% per year is $3,000 less than three times the amount invested at 9% per year. If the income for one year is $1,050, how much is invested at each rate?

Worker A can produce 20 auto parts per hour, while worker B can produce 30 auto parts per hour. How long will it take for both of them, working together, to produce 1,000 auto parts?Detailed solution to this problem

2.  Relative rate of work

Example:

Working alone, worker A can complete a task in 3 hours, worker B can complete it in 6 hours. How long will it take for both of them, working together, to complete the task?          Detailed solution to this problem

More similar problems

3.  Man-hours

Example

If 20 workers can complete a task in 12 hours, how long would it take for 30 workers?

"Work" Problems: Two Persons

Example 1:

Peter can mow the lawn in 40 minutes and John can mow the lawn in 60 minutes. How long will it take for them to mow the lawn together?

Solution:

Step 1: Assign variables:

Let x = time to mow lawn together

Step 2: Use the formula:

Step 3: Solve the equation

The LCM of 40 and 60 is 120Multiply both sides with 120

Answer: The time taken for both of them to mow the lawn together is 24 minutes.

 

 

“Work” Problems: More than Two Persons

Example 1:

Jane, Paul and Peter can finish painting the fence in 2 hours. If Jane does the job alone she can finish it in 5 hours. If Paul does the job alone he can finish it in 6 hours. How long will it take for Peter to finish the job alone?

Solution:

Step 1: Assign variables:

Let x = time taken by Peter

Step 2: Use the formula:

Step 3: Solve the equation

Multiply both sides with 30x

Answer: The time taken for Peter to paint the fence alone is hours.

 

 

“Work” Problems: Pipes Filling up a Tank

Example 1:

A tank can be filled by pipe A in 3 hours and by pipe B in 5 hours. When the tank is full, it can be drained by pipe C in 4 hours. if the tank is initially empty and all three pipes are open, how many hours will it take to fill up the tank?

Solution:

Step 1: Assign variables:

Let x = time taken to fill up the tank

Step 2: Use the formula:

Since pipe C drains the water it is subtracted.

Step 3: Solve the equation

The LCM of 3, 4 and 5 is 60

Multiply both sides with 60

Answer: The time taken to fill the tank is hours.

Kuta Software - Infinite Algebra 1 Name___________________________________

Period___Work Word Problems Date________________ _Solve each question. Round your answer to the nearest hundredth.1) Working alone, Ryan can dig a 10 ft by 10 fthole in five hours. Castel can dig the samehole in six hours. How long would it takethem if they worked together?2) Shawna can pour a large concrete drivewayin six hours. Dan can pour the samedriveway in seven hours. Find how long itwould take them if they worked together.3) It takes Trevon ten hours to clean an attic.

Cody can clean the same attic in sevenhours. Find how long it would take them ifthey worked together.4) Working alone, Carlos can oil the lanes in abowling alley in five hours. Jenny can oilthe same lanes in nine hours. If they workedtogether how long would it take them?5) Working together, Paul and Daniel can pickforty bushels of apples in 4.95 hours. Hadhe done it alone it would have taken Daniel9 hours. Find how long it would take Paulto do it alone.6) Working together, Jenny and Natalie canmop a warehouse in 5.14 hours. Had shedone it alone it would have taken Natalie 12hours. How long would it take Jenny to doit alone?7) Rob can tar a roof in nine hours. One dayhis friend Kayla helped him and it only took4.74 hours. How long would it take Kaylato do it alone?8) Working alone, it takes Kristin 11 hours toharvest a field. Kayla can harvest the samefield in 16 hours. Find how long it wouldtake them if they worked together.9) Krystal can wax a floor in 16 minutes. Oneday her friend Perry helped her and it onlytook 5.76 minutes. How long would it takePerry to do it alone?10) Working alone, Dan can sweep a porch in15 minutes. Alberto can sweep the sameporch in 11 minutes. If they workedtogether how long would it take them?11) Ryan can paint a fence in ten hours. Asanjican paint the same fence in eight hours. Ifthey worked together how long would it takethem?12) Working alone, it takes Asanji eight hours todig a 10 ft by 10 ft hole. Brenda can dig thesame hole in nine hours. How long would ittake them if they worked together?Kuta Software - Infinite Algebra 1 Name___________________________________

Period___Work Word Problems Date________________ _Solve each question. Round your answer to the nearest hundredth.1) Working alone, Ryan can dig a 10 ft by 10 ft

hole in five hours. Castel can dig the samehole in six hours. How long would it takethem if they worked together?2.73 hours2) Shawna can pour a large concrete drivewayin six hours. Dan can pour the samedriveway in seven hours. Find how long itwould take them if they worked together.3.23 hours3) It takes Trevon ten hours to clean an attic.Cody can clean the same attic in sevenhours. Find how long it would take them ifthey worked together.4.12 hours4) Working alone, Carlos can oil the lanes in abowling alley in five hours. Jenny can oilthe same lanes in nine hours. If they workedtogether how long would it take them?3.21 hours5) Working together, Paul and Daniel can pickforty bushels of apples in 4.95 hours. Hadhe done it alone it would have taken Daniel9 hours. Find how long it would take Paulto do it alone.11 hours6) Working together, Jenny and Natalie canmop a warehouse in 5.14 hours. Had shedone it alone it would have taken Natalie 12hours. How long would it take Jenny to doit alone?8.99 hours7) Rob can tar a roof in nine hours. One dayhis friend Kayla helped him and it only took4.74 hours. How long would it take Kaylato do it alone?10.01 hours8) Working alone, it takes Kristin 11 hours toharvest a field. Kayla can harvest the samefield in 16 hours. Find how long it wouldtake them if they worked together.6.52 hours9) Krystal can wax a floor in 16 minutes. Oneday her friend Perry helped her and it onlytook 5.76 minutes. How long would it takePerry to do it alone?9 minutes

10) Working alone, Dan can sweep a porch in15 minutes. Alberto can sweep the sameporch in 11 minutes. If they workedtogether how long would it take them?6.35 minutes11) Ryan can paint a fence in ten hours. Asanjican paint the same fence in eight hours. Ifthey worked together how long would it takethem?4.44 hours12) Working alone, it takes Asanji eight hours todig a 10 ft by 10 ft hole. Brenda can dig thesame hole in nine hours. How long would ittake them if they worked together?4.24 hours