This is the solution to the algebra work word problem # 4 which read: "Joe and Jim paint a house. Joe can paint it alone in 5 days, Jim in 8 days. They start to paint it together but after 2 days Jim stops. How long will Joe finish it alone?"

Joe takes 5 days

Jim 8

to paint a house

So Joe can do 1/5 of his job in one day and Jim can do 1/8th of his job in 1 day. If you just take Joe for example you can see where I'm heading on solving this problem. Joe can do 1/5th of his job in one day. So how long would he finish his job -- well besides knowing the answer already -- you would set it up as x/5 = 1. 1 being when the job is complete not just a "fraction" of his work being done. So the answer is 5 of course 5/5 = 1. You use this exact same method to solve the problem for both Joe and Jim.

**********Correction************ I realized I solved the problem before if they both continued working the job, but this problem states JIM stops working in 2 days. So Joe does x/5 part of his job in 2 days? 2/5 and Jim does x/8 part of his job in 2 days? 2/8, and Joe keeps working. So the equation would be

2/5 + 2/8 + x/5 = 1 multiply by 40 and get

16 + 10 + 8x = 40

26 + 8x = 40

8x = 40 - 26

8x = 14

x = 14/8 or 1 and 3/4ths so it would take 1 and 3/4ths days more for Joe to finish the job alone..

## Saturday, November 14, 2009

### Work Problem 4 Solution

## Wednesday, October 7, 2009

### Mixture Problem # 6 Solution

This is the solution to the algebra mixture word problem # 7 which read:

"If a merchant has two types of tea, one worth $2.70 per kilogram and the other worth $3.00 per kilogram, how many kilograms of each type should the merchant use in order to produce 30 kilograms of a blend that is worth $2.95 per kilogram?"

Even though this involves money we treat it like a mixture problem.

You got to unknowns but if you look at it as having x = 1 amount and y = 30 -x then you just have x, and 30 -x so...

Its the x amount * the 270 cent worth tea plus the 30 - x amount * the 300 cent worth tea is going to equal 30 * 295 cent blend

270 * x + 300(30-x) = 30(295)

270x + 9000 - 300x = 8850

-30x = -150

30x = 150

x = 150/30 = 5

so 5 kilograms of the 270 cent or $2.70 tea and 30 -5 or 25 kilograms of the 300 cent or $3.00 tea would need to be mixed together to make 30 kilograms of the $2.95 cent tea.

check 270 * 5 + 300 * 25 = 30*295

1350 + 7500 = 8850

8850 = 8850

### Number Problem 7 Solution

This is the solution to the algebra number word problem 7 which asked "If the Numerator and Denominator of a certain fraction are both increased by 3, the resulting fraction equals 2/3. If the Numerator and Denominator are both decreased by 2, the resulting fraction equals 1/2. Determine the fraction."

Ok so you have 2 unknowns --> x the numerator and y the denominator. You can solve 2 unknowns with 2 equations. They give you 2 equations.

(x+3)/(y+3) = 2/3 and

(x-2)/(y-2) = 1/2

take the (x+3)/(y+3) = 2/3 and cross multiply and get:

2(y+3) = 3(x+3) which is

2y + 6 = 3x + 9

Now take the second equation: (x-2)/(y-2) = 1/2 and cross multiply

2(x-2) = 1(y-2) which makes

2x - 4 = y - 2 now solve for y

y = 2x - 2

ok now plug this back into the 1st equation for y which we broke down to :

2y + 6 = 3x + 9 so you get 2(2x-2) + 6 = 3x + 9 solve for x

4x - 4 + 6 = 3x+ 9

x= 7 ok now plug 7 into y = 2x - 2 and get y = 2(7) -2 = 12

so the numerator is 7 and the denominator is 12 = 7/12

to check add 3 to the top and bottom and get 10/15 = 2/3

and subtract 2 from the top and bottom and get 5/10 = 1/2

## Saturday, September 12, 2009

### Time and Distance Problem 10 Solution

This is the solution to the algebra time and distance word problem # 10 which was submitted under the ask a question post anonymously which read..

"Junior's boat will go 15 miles per hour in still water. If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream then what is the speed of the current?"

distance = rate X time ... d = r*t we know the speed of the boat in still water 15 mph

we know he can go 12 miles downstream in the same time he can go 9 miles upstream-- so times are equal --- so since time is equal we can set the distances and rates equal to one another...

d1/r1 = d2/r2

The rate downstream will be 15 + x and upstream will be 15 - x so we have:

12/(15+x) = 9 /(15-x)

cross multiply and you get 12(15-x) = 9 (15+x)

180 - 12x = 135 + 9x

45 = 21x

x = 45/21 = 15/7 This is the answer the current is 15/7 or 2 and 1/7ths mph..

## Tuesday, August 25, 2009

### Time and Distance Problem 9 Solution

This is the solution for the time and distance algebra word problem #9 that read:

"if I walked 9 laps around a track. One lap is 1312 ft. I walked at a rate of 4 miles an hour. how many minutes did it take me to walk 9 laps?"

The easiest way to think about this problem is to think about an easier problem and then copy the steps. What if the problem read:

if I walked 9 laps around a track -- 1 lap is 1 mile and i walked at a rate of 4.5 miles and hour. how many minutes did take me to walk 9 laps?

well the total distance would be the length of 1 lap: 1 mile * the # of laps which is 9 --> 9 * 1 = 9 miles

the rate was 4.5 miles an hour -- distance = rate * time

so 9 miles = 4.5miles/hour* time

9 miles/4.5 miles/hour = time or 2 hours would be how long it would take--->

So this is not a hard math problem the arithmetic is what's hard in this problem, but if you just follow the same steps you will get the answer easily enough. So, again we have:

"if I walked 9 laps around a track. One lap is 1312 ft. I walked at a rate of 4 miles an hour. how many minutes did it take me to walk 9 laps"

ok so take 9 * 1312 1st to get the total distance in feet which is : 11808

4 miles an hour would be (5280 ft in a mile) 5280 * 4 = 21120 ft/hour

But we dont want hours we want minutes-- there are 60 minutes in an hour

so (21120 ft*1hour)/(hour*60minutes) --> you're multiplying by 1hour/60 minutes. so divide by 60..

362ft/minute --- the hours(1 hour = hour) cancel out..

*** **Correction** -- user Donaldson pointed out I made a division error -- I said 362 feet should be 352 feet/min...*** Thanks!

ok so now we have a total distance in feet of 11808 feet -- divided by the rate of 352 ft/minute

11808 ft/(352 ft/minute) = 11808 ft *minute/ 352 ft

the ft cancel out you divide 11808 by 352 and get 33.54545 = 33.55 minutes --> which is a good enough answer but if you want it more exact:

.545454 of 60 seconds would be 60 *.5454 = 32.72 seconds == approx 33 seconds --- so 32 minutes 33 seconds would be a more recognizable (as far as common usage of time) answer.

You can kind of double check the answer by making sure its reasonable when you see that the total distance of 11808 feet that you walked is roughly 2 miles and your dividing by 4 miles an hour=== 1/2 an hour or 30 minutes approximately

## Sunday, August 2, 2009

### Plane Geometric Figure Problem 1 Solution

This is the solution to the Plane Geometric Figure Problem 1 which asked you to "find the dimension of a rectangle whose length is 9 less than twice its width if the perimeter is 120 cm."

In a rectangle 2 times the width plus to times the length equals the total perimeter.

Let the width equal x

Then the length is equal to 2x - 9 which means "9 less than twice its width"

Remember 2 times the width plus to times the length equals the perimeter so..

2(x) + 2(2x - 9) = 120

which is..

2x + 4x - 18 = 120

6x - 18 = 120

6x = 120 + 18

6x = 138

x = 138/6

x = 23

Since we let x stand for the width then the width of the rectangle is 23 cm and the length is 2(23) - 9 = 46 - 9 = 37

So the width is 23 cm and the length is 37 cm

Check 2(23) + 2(37) = 120

46 + 74 = 120

120 = 120

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### Age Problem 4 Solution

This is the solution to the algebra age problem # 4... "A man is 27 years older than his son and 10 years from now, he will be twice as old as his son. how old is each now?"

Let x = the son and x + 27 = the man(father)

in 10 years the son will be x + 10 and the father x + 27 + 10 or x + 37 years old. So if he will be twice as old as his son then "twice" the sons age then will "equal" each other.

So, 2(x+10) = x + 37

multiply by 2 on the left side

2x + 20 = x + 37

x = 17 This makes sense since 10 + 17 = 27 and 17 + 37 = 54 which is twice the sons age 10 years from now. So the son is 17 and the father is 17 + 27 or 44 years old.

## Tuesday, March 17, 2009

### Number Problem 6 Solution

This is the solution to the number problem # 6.

In the following Algebra number word problem we have a 3 digit number which has a 100's digit 3 more than the 1's digit and its 10's digit is twice the 100's digit. When all 3 digits are added together, it is equal to 9. What is the number?

If you make the 1's digit x then you have:

100's digit = x + 3 and the

10's digit = 2(x + 3) or 2x + 6

Now when all these digits are added together you get 9 so...

x + x + 3 + 2x + 6 = 9

4x = 0

x = 0

so if x = 0(the 1's digit) then the 100's digit would be 0 +3 or just 3 and the 10's digit would be 0 + 6 or 6.

so the answer is 360

## Friday, March 13, 2009

### Time and Distance Problem 8 Solution

This is the solution to the algebra time and distance word problem # 8:

Distances in this problem are equal, and rate multiplied by time is equal to distance. You have what the times are--- 4 and 5 hours and and the rates are ( x + 15 ) for with the wind of 15 miles per an hour and ( x - 15) for against the wind at 15 miles per an hour.

So we have...

4( x + 15 ) = 5( x - 15 )

4x + 60 = 5x - 75

x = 135 --> This is how fast the vehicle is moving. But with the wind its 135 + 15 or 150 miles an hour going and 135 - 15 or 120 miles an hour coming back.

To check this 4(150) = 5(120) 600 = 600 ---> You also now know the distance was 600 miles, which was not asked in the problem.

### Time and Distance Problem 7 Solution

Here's is the answer to the algebra word time and distance problem # 7:

If Bill is averaging 60 miles per an hour and Craig doesn't leave until 2 hours at 90 mi/hour, then that means that Bill traveled 60 mi/hr * 2 hr = 120 miles before Craig even left. Now how much faster is 90 miles an hour than 60 ? 90 - 60 = 30(this is the net gain of Bill which is what were worried about since were trying to figure out when he will catch up). So the question then becomes how long would it take you going 30 miles per an hour to go 120 miles --- 120/30 =4 So it takes 4 hours hours for Craig to catch him and since he was going 90 miles an hour he went a total of 90 * 4 or 360 miles.

Algebraically>..

90x = 60x + 120

30x = 120

x = 4

90(4) = 360

## Thursday, February 5, 2009

### Work Problem 3 Solution

In this Algebra work problem solution we have a similar situation in dealing with hours and and figuring out how much of a job is done in 1 hour. The only difference is part of the job is being taken away -- negative --- by the leak. So just subtract this. That's the only difference between this work problem and the others.

So with Pump A in 1 hour it would 1/12th of a job Pump B would do 1/6th of a job and the leak would take away 1/24th of the job.

So algebraically we have:

x/12 + x/6 - x/24 = 1

multiply by 24

2x + 4x - x = 24

5x = 24

x = 24/5 or 4 and 4/5ths

4/5ths of an hour is 60*4/5 or 240/5 = 48 minutes

So the pool would be filled in 4 hours and 48 minutes.

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## Sunday, February 1, 2009

### Work Problem 2 Solution

This is the solution to the Algebra work problem # 2.

The key to solving any work problem is to figure out how much work can be done in 1 hour, minute, day- whatever time classification your using. In this problem it's talking hours. Since pump A takes 10 hours to finish then in 1 hours times 1/10th of the work would be complete and with the 8 hour pump B, 1/8th of the work would be done.

Now the way to understand this is to look at just 1 of the pumps. Take pump A for example. It takes 10 hours to fill the pool and 1/10 of the pool is finished in 1 hour. So ask yourself how many hours would it take a pump that fills 1/10th of a pool an hour to fill the entire pool? Or X/10 = 1 --- Without Algebra you know its 10 but using algebra you can see it works out the same way.

So all we do when were dealing with more than just 1 pump is to add their respective ratios together and set them equal to 1. We add because it's the added effect of both pumps that fills the pool.

So we have algebraically:

X/10 + X/8 = 1

multiply by 40 ----- the common denominator....

40X/10 + 40X/8 = 40

4X + 5X = 40

9X = 40

X = 40/9

X = 40/9 or 4 and 4/9ths

so almost 4 and a half hours. To get the exact minutes take 60 * 4/9 = 240/9 = 26 and 2/3rds

so we got 4 hours 26 minutes and if u want to be real anal 2/3's of a minute is 40 seconds (60 * 2/3 = 120/3 = 40)

So, it takes 4 hours 26 minutes and 40 seconds for pumps A and B to fill up the pool if they work together.

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