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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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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Lever Problem 4 Solution

In this Algebra Lever problem solution, you again will have a problem where the distance times the weights have to equal on both sides. Also since Steve sits twice as far, he is 8 feet away since Samuel is 4 feet away and Steve sits twice as far as Samuel.

Here's the algebra equation:

65(4) + 85(8) = 200x

260 + 680 = 200x

940 = 200x

040/200 = x

94/20 = x = 4 and 14/20 or 4 and 7/10 or 4.7 feet is how far Big Billy would have to sit on the opposite side of the teeter totter across from Samuel and Steve.

Work Problem 1 Solution

In this Algebra work word problem solution the way to figure out the answer is to realize that they work a fractional part of work each hour. So in 1 hours time Steve does 1/9 of his work, so in 9 hours he does 9/9 or his whole job. Same thing goes for Joe except he gets 1/10th of his job done in an hour. For just one guy, take Steve for instance, the algebra for his work by himself would be (1/9)x = 1 or x/9 = 1 which makes x = 9 or 9 hours. for Joe it would be x/10 = 1 or 10 hours. Together they would take x/9 + x/10 = 1 hours(x) to get the job done.

x/9 + x/10 = 1
multiply by 90
10x + 9x + 90
19x = 90
x = 90/19

90/19 = 4.74 or approximately 4 hours and 45 minutes (.75 would 3/4ths of an hour or 45 minutes). If you want it exact just take the fractional part(.7368) and multiply by 60. You'll end up with 44.2 which is 44 minutes and .2*60 seconds or 12 seconds. The detailed answer would be 4 hours 44 minutes and 12 seconds. So if he hires both of these guys he'll get the lawn mowed before 6 hours.

Time Distance Problem 6 Solution

In this Algebra word problem solution, an non-algebraic approach would be looking at far Bob goes in 1 hour, since he has an hour head start. He's going 60 mi/hr so he went 60 miles in 1 hour. Steve is going 90 mi/hr or 30 mi/hr faster. So how long would it take someone going 30 mi/hr to go 60 miles? Well the 1st hour he went 30 miles, the 2nd hour he goes the other 30 or 2 hours total until he catches up to Bob. If your going 90 miles an hour for 2 hours-- you have gone 90 * 2 or 180 miles.

Algebraic approach:

x = time

rate * time = distance

60 = # of miles per hour Bob is going

90 = # of miles per hour Steve is going

x + 1 = the hour head start Bob had

Bob = 60(x+1)

Steve = 90x

Since Bob & Steve are going the same distance we can set their respective distance * time equal to one another.

60(x+1) = 90x
60x + 60 = 90x
60 = 30x
30x = 60
x =60/30
x = 2 or 2 hours

check
60(2+1) = 90(2)
180 = 180 --- rate times time = distance which is the answer to the 2nd part of the problem they went 180 miles total.....

Lever Problem 3 Solution

1st piece of information that's important in the Algebra lever problem is that the teeter totter was 12 feet long and the balance point was directly in the middle(which most teeter totter's are). So each side is 6 feet. This means if Bob and Susan sat on opposite ends then they were 6 feet from the fulcrum. They weigh 70 and 50 lbs respectively. On the same side as Susan 5 feet away from the center 40 lb Chrissy sat. So we got the weights and distances for everyone but Steve who weighs 35 lbs but is an unknown (x) distance from the center on the same side as Bob.

As we said before the distance times the weight of one side has to equal the other. In this case since there's more than 1 person on each side we add the respective weights and distance and set them equal to one another...

This side:

Susan = 50(6)
Chrissy = 40(5)

is equal to this side:

Bob = 70(6)
Steve = 35(x)

300 + 200 = 420 + 35x
500 = 420 + 35x

80 = 35x
x = 80/35
2 & 10/35 feet or 80/35 from the center is where Steve is located on the same side as Bob.

Just to check 500 = 420 + 35(80/35) = 420 + 80 == correct

Lever Problem 2 Solution

In lever problems you are trying to get the weight * the distance of one side to equal the weight times the distance of the other. The fulcrum would be considered the balance point, and since a weight of 100 lbs is on that side we can see that the weight of 100 * 2 = 200 has to equal the distance of 8 (10 - 2) times a certain amount of weight which were trying to figure out for the other side. So algebraically we have:

100 * 2 = 8 * x

200 = 8x

x = 200/8

x = 25

This means that a weight of 25 lbs would have to be applied to one side of the lever to lift the 100 lbs. As far as levers are concerned this makes sense. You definitely want to exert less force on one end that the weight of object your trying to move!