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!

Lever Problem 1 Solution

With any type of lever problem it's important to realize that the weight of object 1 TIMES the distance of that object from the center(or fulcrum) is going EQUAL the weight of object 2 TIMES the distance of that object from the center(fulcrum).

So in this problem you know the to weights are Sam at 100 lbs and James at 150 lbs, and the distance from the fulcrum is known for Sam which is 8(feet). However, you don't know James' distance so thats what were trying to find.

So set them equal

100 * 8 = 150 * x

800 = 150x

x = 800/150

x = 5 1/3 (feet) or 5 feet 4 inches (since 1(ft)/3 * 12 inches/ft = 4 inches)

Age Problem 3 Solution

Greg will be x and Greg's father will be x + 30. In 15 years Greg will be x + 15 and Greg's father will be x + 30 + 15 or x + 45, and the sum of their ages will be 130.

So,

x + 15 + x + 45 = 130

2x + 60 = 130

2x = 130 - 60

2x = 70

x = 70/2

x = 35 which is Greg's current age. Greg's father is 35 + 30 or 65 years old. Just to check add 15 years or 30 total to their ages. 35 + 65 + 30 = 130

Age Problem 2 Solution

Here's a non-algebraic approach to the problem:

If Bobs current age were 8 then 8 years ago he would be zero, so he has to be older than 8 or at least 9 years now. So assume 9 and 36 -- 8 years ago would make 1 and 28 which is 28 times and not what were looking for.. 10 and 40 ---- 8 years ago would make 2 and 32 which is 16 times not what were looking for... try 11 and 44-- 8 years ago would make 3 and 36-- bingo 12 times- So, Jason is 44 and Bob is 11 years old.

Algebraically we have:

x for Bobs age and

4x for Jason

So 4x = x

But 8 years ago Jason was 12 times older so

4x - 8 = 12(x-8)

4x - 8 = 12x - 96

88 = 8x

x = 11 So Bob is currently 11 years old and Jason is 4(11) or 44 years old.

Age Problem 1 Solution

For a non-algebraic approach:

If Steves current age was 20 then 20 years ago he would be zero, so he has to be at least 21 years old. This would make Bob 42 and Steve 21. 20 years ago they would be 22 and 1, which is 22 times as old and not 6 times. Were not that far off though. So go up by increments. Try Steve at 22 and Bob 44-- 20 years ago would be 2 and 24 which is 12 times-- at 23 and 46 we can see than 20 years ago they would be 3 and 26 which is over 8 times were close-- and at 24 and 48 20 years ago would make 4 and 28 which is 7 times and really close.. at 25 and 50 20 years ago we would have 5 and 30 which IS 6 times older and what were looking for. So, Steve is 25 and Bob is 50...


For the algebraic approach we know that if something is twice as old, big, fast, etc as something else it is 2x more than x. So in this problem Bob is 2x as old as x Steve is.

Twenty years ago would be :

2x - 20 for Bob

and

x - 20 for Steve

but, 20 years ago Bob was 6 times older so

2x - 20 = 6(x - 20)

2x - 20 = 6x - 120

4x = 100

x = 100/4

x = 25

So x or Steve is 25 and Bob is twice or 2(25): 50 years old.

Mixture Problem 5 Solution

This is sort of like the opposite of the last problem. Instead of diluting it your giving 100% pure alcohol to increase its level. As a fraction 100% is always just 1. anything else is always over a 100 so 30/100 = 3/10ths



So in this problem you would have:

(3/10)20 + 1(x) = (20 + x)(1/2)

6 + x = 10 + 1x/2 *multiply by 2

12 + 2x = 20 + x

x = 8

So 8 quarts of pure alcohol would need to be added to increase the % to 50% from 30%.

just to check:

(3/10)20 + 8 = (20 + 8) 1/2

6 + 8 = 14 is true

Mixture Problem 4 Solution

75% is the same thing as 3/4 ths, which makes an easier setup than using .75 in the problem.

(3/4) * 8 gallons = 24/4 = 6 gallons--- so there is 6 gallons of pure alcohol in the 8 gallons.

The problem asks how much water needs to be added to dilute it to only 50% or so only half of the total quarts is alcohol.

A non-algebra way to think of it is to treat it like an average.

((3/4)*8 + 0(8) ) / (8 +x) = 1/2 which is basically saying if you have 8 total cups which is 75% alcohol how many extra cups (x) would you need to add to 8 to make 1/2 or 50%

Keep in mind that water is the same thing as saying 0% alcohol and then treat it like the other mixture problems.



So the problem would read:

(3/4)8 + 0(x) = (1/2)(8+x)

6 + 0 = 4 + 1/2(x) * multiply by 2
12 = 8 + x

x = 4

So 4 quarts of water would need to be added to dilute the alcohol solution to 50%.

Mixture Problem 3 Solution

This mixture problem deals with dollars instead of percentages. But, works the same way as before.

She has 10 pounds of $3 coffee for a total worth of $30 and x pounds of $5 coffee. The question asks of how much of the x $5 coffee is needed to mix with the 10 pounds of $3 coffee to make (10 + x) pounds of $4.50 coffee with a total value of $4.50 * (10 + x)

to set this up we would then have:

3(10) + 5x = (10 + x) * 4.50

30 + 5x = 45 + 4.50 x *multiply by 100 to get rid of decimal

3000 + 500x = 4500 + 450x

50x = 1500

x = 1500/50

x = 30

so 30 pounds of the $5 grade must be added to the 10 pounds of $3 grade to make $4.50 grade-- If you think about it that makes sense. If you had exactly the same amount of $5 coffee (10 pounds) as the $3 coffee-- then that would make an average grade of $4 coffee--- so you know it would have to be a lot more than just 10 pounds to increase the value close to $5-- which 30 pounds does.

But, just to check..

Does

30 + 5(30) = (10 + 30) * 4.50 ?

30 + 150 = 40(4.50)

180 = 180 correct

Coin Problem 5 Solution

Let the total amount in cents = 2251

let x = the # pennies and the value = 1(x) or just x

let 2x = the # of quarters and 25(2x) or 50x = the value in quarters

let 20(2x) or 40x equal the # of 50 cent pieces and 50(40x) or 2000x = the value of 50 cent pieces

So, we have:

x + 50x + 2000x = 2051
2051x = 2051
x = 2051/2051
x = 1

so we have (1) pennies

2(1) or 2 quarters

and 40(1) or 40 half dollars

Coin Problem 4 Solution

19.43 is 1943 cents.

In this problem, it helps to think about what the easiest setup would be before just setting x equal to the 1st thing you see. If you let x = pennies, well that would work, but you would end up worth a lot of fractions that you could avoid if you started with x = quarters instead.

So let x = the # of quarters and 25x the value

2x = the number of dimes and 10(2x) = the value or 20x

4x = the # of pennies and 1(4x) or 4x the value of pennies

4x + 1 = the # nickels and 5(4x+1) or 20x + 5 the value of nickels

and

4(4x+1) or 16x + 4 the # of half dollars and 50(16x + 4) or 800x + 200 the value

So altogether we have 25x + 20x + 4x + 20x + 5 + 800x + 200 = 1943

869x + 205 = 1943

869x = 1738

x = 1738/869

x = 2

So we have 2 quarters, (50 cents)

2(2) or 4 dimes (40 cents)

4(2) or 8 pennies (8 cents)

4(2) + 1 or 9 nickels (45 cents)

and 16(2) + 4 or 36 half dollars (1800 cents)

just to check if you add those cents up it will equal 1943 cents which is $19.43

Coin Problem 3 Solution

Again make $10.35 equal to cents --- 1035 cents 1st.

Algebraically we have

x = 32x 32 cent stamps

x + 2 = 25(x + 2) 25 cent stamps and

2(x + 2) or 2x + 4 = 50(2x + 4) 50 cent stamps

So, we then have..

32x + 25(x + 2) + 50(2x + 4) = 1035

32x + 25x + 50 + 100x + 200 = 1035

157x + 250 = 1035

157x = 1035 - 250

157x = 785

x = 785/157

x = 5

So we have 5 32 cent stamps, 5 + 2 or 7 25 cent stamps, and 2(5 + 2) or 14 50 cent stamps...

Coin Problem 2 Solution

1st of all on all these problems convert the total amount of money from dollar to cents-- $100 is equal to 10000 cents (just add 2 zeros), $7 is equal to 700 cents, and you already have 37 cents, so altogether you have 10000 + 600 + 87 = 10737 cents.

From the problem you can see that everything is relative to pennies so let that be 1(x) or just x

1 more than twice as many nickels as pennies is (2x + 1) and and since its nickels 5(2x +1) or 10x + 5

twice as many dimes as nickels 2(2x +1) = 4x + 2 and since its dimes 10(4x + 2) or 40x + 20

1 more than twice as many 50 cents pieces than dimes (this is tricky wording since you may expect the denomination to go to a quarter) which is 2(4x + 2) + 1 or 8x + 5 and since its 50 cent pieces 50(8x + 5) or 400x + 250

and twice as many quarters as 50 cent pieces or 2(8x + 5) or 16x + 10 and since its quarters 25(16x + 10) or 400x + 250

so algebraically we have
x + 10x + 5 + 40x + 20 + 400x + 250 + 400x + 250 = 10737 =
851x + 525 = 10737 =
851x = 10737 - 525 =
851x = 10212 =
x = 10212/851

x = 12

So you have 12 pennies, 2(12) + 1 or 25 nickels, 4(12) + 2 or 50 dimes, 8(12) + 5 or 101 Fifty cent pieces, and 16(12) + 10 or 202 quarters....

Yeah, a little tedious arithmetic wise...

Coin Problem 1 Solution

On any kind of coin problem its important to turn any dollar amount (if given in dollars) to cents and to make sure you give the appropriate value of each denomination ---- a penny 1 or -- nickel 5--- dimes 10-- quarter 25-- sounds kind of obvious but a lot of problems are taken care of if you just know that information for sure.

Alright if you have 10 pennies you have 10(1) == 10 cents So,

If you have 12 nickels you would have 12(5) = 60 cents

If you had 5 dimes you would have 5(10) = 50 cents

and if you had 10 quarters you would have 10(25) = 250 cents or $2.50

To solve this problem you 1st set up the information given. You can see that everything is relative to the pennies (1 more nickel than pennies and 6 times as many dimes as pennies)
So let pennies equal 1x or just x since 1 times anything is just the number...

you have x + 1 nickels at 5 cents so 5(x+1) nickels

and 6 times as many dimes as pennies or 10(6x) or 60x

Theres no need to change the dollars to cents as the cents are already given 71 cents.

So the equation would be

x + 5(x+1) + 10(6x) = 71 =

x + 5x + 5 + 60x = 71 =

66x = 66 =
x = 66/66 = 1

So thers 1 penny, (1 + 1) or 2 nickels and 6(1) 6 dimes

To check if you add those up 1 + 10 + 60 you get 71 so the answer is correct.

Number Problem 5 Solution

This one's a lot easier to work algebraically. You could get it by trial an error or course, but its easy if you think of one number being just x and the other number being (40-x)

so for example if the number was 10 then the other number would be (40 - 10) or 30

*this is pretty close if you check 5(10) = 50 2(30) = 60 which is a 10 difference and not a 3 through trial and error you could get the right answer just switching out a new number

Algebraically:

So 2 times the 1st number lets use (40-x) since multiply by 2 with this is easier than by 5

2 (40 - x) is "equal" to

3 more than 5 times another number (x)

or 5x + 3

altogether we have 2(40-x) = 5x + 3

80 - 2x = 5x + 3
77 = 7x
x = 11

so 1 number is 11 and the other number is (40 - 11) or 29

check it--- 2(29) = 58 5(11) = 55 and indeed they are 3 apart

Number Problem 4 Solution

You might be able to figure this out in your head since it doesn't matter if the number was 900, 9 million etc if its a situation that add up to 9 and 1 number is twice as big that you're adding. So what numbers 1 through the number 9 could you add together to get 9? 1 and 8 , 2 and 7, 3 and 6, and 4 and 5. Now which one of these is twice as big as the 1st number? The 6 is twice as big as the 3. But the problem asked for what 2 numbers added up to 900. So just add 2 zeros onto the 6 and 3. 300 and 600 are the answers.

Algebraically,

x = 1st number
2x = 2nd number *2x means the same thing as 2 times x or 2 X x
900 is the number were looking for so..

x + 2x = 900
3x = 900
x = 900/3
x = 300
substituting back in x is 300 and 2x is 2(300) or 2 X 300 which is 600.

Time and Distance Problem 5 Solution

This is similar to the previous problem. The only difference is that they left at a certain time and you have to figure out what time they end up being 2100 miles apart.

Once again, they're both covering distance individually that adds to the total distance were looking for 2100, therefore we can just add their respective speeds to figure out how long it will take to cover that distance.

So with x being used as "time", 300 and 600 the respective speeds or rates to be added, and 2100 the distance --- rate X time = distance we have....

600x + 300x = 2100

900x = 2100

x = 2100/900

x = 2 and 1/3 hours or 7/3 of an hour

So to get the time in hours from when they left we can either figure out a what a third (1/3) of an hour is an add it to 2 (2 and 1/3 hours) or we can figure out how many minutes (7/3) X 60 would be..

1/3 of an hour is (1/3) X 60 or 60/3 which is 20-- so 20 minutes added to 2 is 2 hours and 20 minutes which is the answer but the alternate approach would be taking
(7/3) X 60 and you end up with 420/3 which is 140 or 140 minutes --- with 60 minutes in an hour you can see that 2 hours is 120 minutes with 20 minutes left over and once again 2 hours and 20 minutes

This is still not the answer the problem is looking for. It's asking what time were they that far apart. If they left at 12 (noon) , 2 hours and 20 minutes later would be 2:20 pm--- So 2:20 pm is the answer.

Mixture Problem 2 Solution

A non-algebraic approach would see that in order to have x amount of quarts of boric acid solution you could use averages to figure out how many quarts at 30% boric acid solution you would need to mix with 2 quarts of 10% boric acid solution.

When I say averages I mean if you scored 10% on 2 tests in school how many 30% tests would you need to get you a 20% average?

Well you might have seen that 20% is exactly half way between 10% and 30%. You already know that if you had just 1 test at 10% you would need 1 test at 30% to get an average of 20%---- But, you have 2 tests at 10% - so you need 2 test at 30% to get the 20.

This is analogous to the boric acid solution. You would need 2 quarts of 30% and mix it with the 2 quarts of 10% in order to get 20% boric acid solution (4 quarts total).

Algebraically you can set it up like you would an average..

From earlier you had (10% + 10%+ 30% + 30%)/4 = 20%

It should be pointed out theres a couple ways to work with percents. You can either write it in decimal: 10% becoming .10 , 50% becoming .5
or you can just write your percentage (minus the percent sign) over 100:
10% becomes 10/100 , 50% becomes 50/100 for example

You can enter 10/100 and 50/100 into a calculator and you'll get the .10 and .5 respectively, which is the decimal version.

Once again from earlier we had (10% + 10%+ 30% + 30%)/4 = 20% which is the same thing as writing (10/100 + 10/100 + 30/100 + 30/100)/4 = 20/100 this equals...

(80/100)/4 = 20/100 =

(80/100) X (1/4) = 20/100 =

20/100 = 20/100 , which is correct since both sides of the "equals" are equal.


So to set this up with an unknown you have:

(10/100 + 10/100 + 30x/100)/(2+x) = 20/100

(20/100 + 30x/100) = (20/100) X (2+x)
multiply by 100 and the 100's on the denominator cancel out

20 + 30x = 20 X (2+x) =
20 + 30x = 40 + 20x =
10x = 20
x = 20/10 = 2

So there you have an algebraic way of solving it...

As a shortcut you can skip right to the chase algebraically by setting up the problem as (# of quarts from Jar A) X ((Percentage of Solution) + (x amount of quarts of Jar B)) X (Percentage of Solution) = (# of quarts from Jar A + x amount of quarts from Jar B) X (Needed Percentage of Solution)

in this case we had : 2 X (10/100) + x X (30/100) = (2+x) X (20/100)
= 20/100 + 30x/100 = 40/100 + 20x/100 multiply by 100

* could have just gotten rid of the 100's from the beginning but you can make sure of no mistakes this way especially when doing other problems

For all other mixture problems you can see that the percentages when written over 100 allow you to cancel the 100's -- so for now on problems like this you can start wrting it as
2(10) + 30x = 20(2+x) =

20 + 30x = 40 + 20x =
10x = 20
x = 2

OK I might have been long winded on this one, but I just wanted to say more than I had to on the 1st real mixture problem. I'll be less long winded on the next problem of this type.

Time & Distance Problem 4 Solution

Bob & Steve are both covering their respective distances (in this case the total distance to find is how far apart they are 1750 miles) individually with their given speeds. It's the same distance covered if they were going their added speeds, which is 500 mi/hr + 200 mi/hr = 700 mi/hr -- So, how long would it take to go 1750 miles at 700 mi/hour.. Distance = Rate X Time .... Time = Distance / Rate

So time = 1750 / 700 = 2.5 hours

Algebraically you would have

500x + 200x = 1750

700x = 1750
x = 1750/700
x = 2.5

Time & Distance Problem 3 Solution

In this problem its important to know that if you're moving a certain speed and someone else is moving another speed, the distance being covered between the both of you is the same as if you were going the added speeds yourself. In other words, if you're moving 20 miles per an hour and someone else moving towards you is going 60 miles an hour then the amount of distance being crosses is at the rate of 80 miles an hour.

Another way of looking at is if there's 800 miles between you and someone else, the 1st hour you went 20 miles (20 mi/hr) and they went 60 miles (60 mi/hr) or 80 miles total of the distance. Which you can see 2 things from. 1) It would take 10 of those hours 80 X 10 = 800 to go 800 miles and 2) You can just add the speeds 60 mi/hr + 20 mi/hr to get 80 mi/hr and figure out how long it would take you to go 800 miles.

So algebraically: 20x + 60x = 800
80x + 800
x = 800/80 = 10

So, in this problem it would take 10 hours for Bob and Steve to meet.

Mixture Problem 1 Solution

A easy way to set up percentages is to always put that percentage as a number over 100. So 15% percent is the same thing as 15/100 ...

So if a mixture of 40 quarts has 15% alcohol then there is 40 X (15/100) = 4 X (15/10) = 60/10 =

6 quarts of alcohol in the mixture.

Time and Distance Problem 2 Solution

 Here's the answer to the time and distance problem # 2 that read, "Bob leaves in his car driving at a constant speed of 40 miles/hour. 4 hours later Steve leaves going the same direction at 60 miles/hour. How long will it take until Steve catches up (is side by side) with Bob?"

A non algebra solution would be to realize that Steve is moving 20 miles/hour faster than Bob and has to cover 160 miles (40 * 4) to catch up to Bob (40 miles/hour X 4 hours = 160 miles) So how long until does it take to cover 160 miles going 20 miles/hour-- 8 hours

Algebraically you got 40(x+4) = 60x where x is time and x + 2 means 2 hours ahead

So, 40x + 160 = 60x which makes

160 = 20x ... x = 160/20 = 8 ----

So in 8 hours they meet.

Number Problem 3 Solution

Odd numbers are 5, 11, 17, 19, 21, etc and consecutive means one right after the other as in 3,4,5,6, etc . So 3 consecutive odd numbers means 45,47,49 for example.

You want 3 consecutive odd that add up to 51 so the 1st number is x, the next number is x +2, and the last number is (x + 2) + 2 or x + 4. So,

x + x + 2 + x + 4 = 51 which equals

3x + 6 = 51 =

3x = 45

x = 45/3 = 15

So substituting back in for x you have 15 + (15 + 2) + (15 +4) = 51

15 + 17 + 19 does = 51 which is the answer ...

Number Problem 2 Solution

Even numbers are 2, 6, 12, 18, etc- consecutive means one right after the other 1,2,3,4-- so consecutive even means 6,8,10,12, etc ...

So, the 1st number would be x, the 2nd number x + 2, and the 3rd number would be (x + 2) + 2 or x + 4 . The problem states when these 3 numbers are added together they equal 138.

We then have x + x + 2 + x + 4 = 138 which makes
3x + 6 = 138 which makes
3x = 132
which makes x = 132/3
which is x = 44

Substituting back in we have 44 + 44 + 2 + 44 + 4 = 138 and that equals
44 + 46 + 48 = 138 so 138 = 138

The 3 consecutive even numbers when added together that equal 138 are 44, 46, 48

Number Problem 1 Solution

Consecutive means 1 right after the other as in 3,4,5,6,7.... So in this problem the 1st number is x the next number is x + 1 and the last number is (x+1) +1 which is x + 2.

So we have x + x + 1 + x + 2 = 48
3x = 45 x = 15 ..... substituting back in the problem
15 + (15 + 1) + (15 + 2) = 48 = 15 + 16 + 17 = 48 ...
so the 3 consecutive numbers are 15,16,17.

Time and Distance Problem 1 Solution

I set this problem up so you can almost stumble upon the solution accidentally by playing with the numbers I gave. If you see that in 1 hour's time they both cover the same amount of miles as their speed in miles per in an hour. In this case Bill traveled 3.5 miles in 1 hour and Susie 1.5 miles in 1 hour. You might have also stumbled upon the fact that this number when added equals the distance of the track which was 5 miles. This means that if they together covered a distance of 5 miles then that's when they "meet"-- so, coincidentally this is the answer for the 1st & 2nd parts of the problem. Bill goes 3.5 miles (distance) and Susie 1.5 miles (distance) when they meet each other in the 1 hour (time). Whether or not they kept going after their argument is another story.

A pure algebraic solution would use this scenario and just add the respective speeds together so 3.5x + 1.5x = 5 , where x is the speed in miles/hour and 5 is the distance in miles. You can do this when you realize that if something is moving towards you at a given speed say 30 miles and your moving 60 miles/hour its as though you're moving towards an object standing still at 90 miles/hour. In other words, you can figure the time you cover a respective distance in relation to another person moving towards you just by adding the 2 speeds and figuring out the time it would take you to cover that distance with that new "added up" speed. So, 5x = 5 which means x = 1 hour*. Substituting back in the problem 3.5(1) + 1.5(1) = 5 =.... 5 = 5, which is correct. So, Bill went 3.5 miles and Susie 1.5 miles when they met in 1 hours time.



*When you divide miles by miles/hour the miles cancel and your left with the hours.....