Algebra Word Problem Solutions

Plane Geometric Figure Problem 3 Solution

2011-12-04T08:07:00.000-08:00

This is the solution for the Algebra plane geometric figure problem as asked by an anonymous visitor which read, "catrina and tom want to buy a rug for a room that s 14 by 15 feet. They want to leave an even strip of flooring uncovered around the edges of the room. How wide a strip if they buy a rug with an area of 110 square feet?"

It's easier to picture this. Draw a rectangle that's almost a square 14 x 15 feet. Within in draw another such figure but smaller. The space between that figure and the 14 x 15 figure is x. Now the smaller square has an area of 110 square feet. We now have an equation :

(15 -x ) (14 -x ) = 110

210 + x^2 -14x - 15x = 110

x^2 - 29x + 210 = 110

x^2 -29x + 100 = 0

(x - 25)(x - 4) = 0

x = 4 or 25 however using 25 would result in negative dimensions which does not make sense. So 4 is the answer

Check: (15 - 4) (14 - 4) = 110 = 11*10 = 110

Number Problem 14 Solution

2011-12-03T15:46:00.000-08:00

This is the solution for the Algebra number word problem which was asked by an anonymous visitor that read, "55 bowls of the same size capacity of food.
1. everyone gets their own bowl of soup
2. every two gets one bowl of spaghetti to share
3. every three will get one bowl of salad
4. all r required to have their own helping of salad, spaghetti, and soup."

Ok you know the total number of people is x. Since '" every two gets one bowl of spaghetti to share" then x/2 of the people will have those bowls and since "every three will get one bowl of salad" then x/3 of the people will have those bowls in which they will total 55 bowls.
So we have:

x + x/2 + x/3 = 55 ** multiply by 6

6x + 3x + 2x = 330

11x = 330

x = 30

So there are 30 bowls for soup, 30/2 = 15 bowls for spaghetti , and 30/3 = 10 bowls for salad.

Check 30 + 15 + 10 = 55

Finance Problem 4 Solution

2011-12-03T14:37:00.000-08:00

This is the solution to the Algebra finance word problem a anonymous user asked, "An administrative assistant orders cellular phones for people in her department. The brand A phones cost $89.95 and the brand b phones $34.95.If she ordered 3 times as many brand b phones as brand a phones at a total cost of $584.40, how many of each did she order?"

Ok to make things easier lets make it cents. Just remember to divide by 100 when done to get the answer in dollars and cents...

So brand A is 8995 and b is 3495 which had 3 times as many sold as a and all the combinations of phones totalled 58440.

So 8995x + 3495(3x) = 58440

8995x + 10485x = 58440

19480x = 58440

x = 3 So there were 3 brand A's and 3(3) = 9 brand B's sold.

Finance Problem 3 Solution

2011-12-03T14:29:00.000-08:00

This is the solution to the Algebra finance word problem asked by an anonymous user which read, "Alex has made 42 of the 48 payments he owes on his car but is having trouble continuing to make the payment in full. He has made an arrangement with the bank to pay 2/3rds of his monthly payment, rather than the entire payment, every month. How many months will it take Alex to pay off his car loan with this new payment arrangement?"

Ok, 1st find the remaining months : 48 - 42 = 6 So there are 6 months at full payments. But, you are no longer doing full payments you are doing 2/3s of a payment. So 2/3 of what would equal out 6 ? Or

2x/3 = 6

2x = 18

x = 18/2 = 9 .. So it will take Alex 9 months to pay off his car loan with the new payment arrangement..

Number Problem 13 Solution

2011-10-21T17:30:00.000-07:00

This is the solution to algebra number problem 13 as asked by an anonymous user: "A maths test contains 10 questions. Ten points are given for each correct answer and three points deducted for an incorrect answer. If Ralph scored 61, how many did he get correct?"

Ok on this problem you can come up with a solution faster by just quick trial an error. You know that if it only counts 3 points off for each wrong answer and they got a 61 then they had to have at least a 70 before the points were taken off. If they got a 70 that means they missed 3 problems (3 * 3 = 9) 70 - 9 = 61 . So the answer is they got 7 problems right and 3 wrong.

Now to set this up using algebra...

10x - 3(10-x) = 61 You are subtracting the wrong answers worth 3 points a piece from the right ones worth 10 a piece. You don't know how many of each so you are saying there are x amount of 10 valued answers and (10 - x) number of 3 valued wrong answers.

10x - (30 - 3x) = 61

10x -30 + 3x = 61

13x = 91 --> x = 91/13 = 7 --> so there are 7 right 10 valued problems and (10-7) or 3 wrong 3 point negative value ones.....

You could have also set it up like this 10(10 - x) - 3x = 61 and got the same answer....

Age Problem 8 Solution

2011-10-21T17:21:00.000-07:00

This is the solution to the algebra age problem asked by Joshua that read, "A man is 4 years older than his wife and 5 times as old as his son. When the son was born, the age of the wife was six-sevenths that of her husband's age. Find the age of each."
Ok so here we go...

"A man is 4 years older than his wife"
m = w + 4
Now we will subtract 4 from both sides to use this conveniently later ( you will see)
w = (m-4)

"and 5 times as old as his son."
m = 5s

"When the son was born, the age of the wife was six-sevenths that of her husband's age."
w - s = 6/7(m-s)

7(w-s) = 6(m-s) -->
7w - 7s = 6m - 6s --->
7w = 6m - 6s + 7s --->
7w = 6m + s
replace w with (m-4)
7(m-4) = 6m + s
7m - 28 = 6m + s
7m - 6m = s + 28
m = s + 28
replace m with 5s
5s = s + 28
5s - s = 28
4s = 28
s = 7 yrs is the son's age
then
5(7) = 35 yrs is Dad's age
and
35-4 = 31 yrs is Mom's age

Now check work...
"When the son was born, the age of the wife was six-sevenths that of her husband's age."
31 - 7 = 6/7(35 - 7)
24 = 6/7*28
24 = 24;

Number Problem 12 Solution

2011-10-09T17:43:00.000-07:00

This is the solution to a number algebra word problem that was asked by Zakeus who said , "The numerator and denominator of a fraction are together equal to 100. Increase the numerator by 18 and decrease the denominator by 16 and the fraction is doubled. What is the fraction?"

Ok you want to set up 2 equations the 1st one is easy. x + y = 100 so y = 100 - x

The other would be (x + 18)/(y - 16) = (x/y)*2

(x + 18) y = 2x (y - 16) Now substitute 100-x for y....

(x + 18)(100 - x) = 2x(100 - x - 16)

100 x + 1800 - x^2 -18x = 2x(84-x)

82x + 1800 -x^2 = 168x - 2x^2 Now combine like terms...

x^2 - 86x + 1800 = 0

(x - 50)(x - 36) = 0   Now 50/50 is 1 and that's not the fraction we are looking for ( y = 100 - 50 = 50)

So substituting 36 in for x we have y = 100 - 36 = 64     So the fraction is 36/64 Now to verify this is correct add 18 to the numerator and subtract 16 from the denominator and we get ..

54/48 ... Ok now to verify if this is double the other equation lets reduce the equations. 36/64 4 is the gcd so we get 9/16     and for 54/48 , 3 is the gcd and we get 18/16   and   9/16 * 2 does equal 18/16, so we have the correct answer....

Time and Distance Problem 13 Solution

2011-10-09T09:59:00.000-07:00

This is the solution for a algebra time and distance word problem asked by a anonymous user: "Two trains left a station at the same time. One traveled north at a certain speed and the other traveled south at twice that speed. After 3 hours , the trains were 400 miles apart. How fast was each train traveling?"

The speed of the 1st train is x and the 2nd train is twice that or 2x. Ok the trick here is lets say someone is going 50 mph north and the other person is going twice that 100 the opposite direction. This distance separating them would be the same thing as if you added their speeds and it was going one direction -- in other words if someone was just going 150mph they would cover the same ground. Also rate multiplied by time equals distance. We know the time is 3 hours and the distance is 400. So

(x + 2x) (3) = 400

3x(3) = 400

9x = 400

x = 400/9 = 44 and 4/9ths mph is the 1st train and 88 and 8/9ths (800/9) mph is the 2nd train.

Age Problem 7 Solution

2011-10-09T09:41:00.000-07:00

This is the solution to a algebra age word problem as asked by a anonymous user: "Nona is one-third as old as her mother. Five years ago, she was only one-fifth of the age of her mother. How old is Nona now? "

It is easier to set this up with two unknowns say x and y and have 2 equations.

The 1st equation would be: x = 1/3(y) or x = y/3   since Nona is currently 1/3 as old as her mother. Now 5 years ago she was only 1/5th the mothers age so the other equation would be..

x -5 = 1/5(y - 5)   multiply by 5..

5x - 25 = y - 5 so

y = 5x - 20 .. plug this back into the 1st equation and you get..

x = (5x - 20)/3 - -multiply by 3

3x = 5x - 20 so 2x = 20 and x = 10

So Nona is 10 years old currently and the mom is 30.   To check this go 5 years back... Nona would be 5 and the mother would be 25   ------   5/25 = 1/5ths of her mothers age...

Mixture Problem 7 Solution

2011-10-09T09:13:00.000-07:00

This is the solution to a new algebra mixture problem as asked by Califax who asked, "A 50-lb solution of acid and water is 20% (by weight). How much pure acid must be added to this solution to make it 30% acid?"

1/5 = 20 % ..... 3/10 = 30 % x = the unknown amount in lbs we want to add of 100% or 1/1 or 1 or just x of pure acid.

50(1/5) + x = 3/10 (50 + x) multiply by 10...

100 + 10x = 3(50 + x)

100 + 10x = 150 + 3x

7x = 50 = 50/7 = 7 and 1/7th lbs of pure acid needs to be added to make the solution 30% acid.

Number Problem 11 Solution

2011-10-07T12:32:00.000-07:00

This is a new problem as asked by a anonymous visitor which read: "a number added to five times its reciprocal is -6. find the number "

Ok non-algebraically you can think of what numbers added together will make 6(of course they will be negative to make negative 6) --- 1 and 5 and 4 and 2 , and 6 and 0... You can't have a reciprocal of 0 and a reciprocal of 6 : 1/6 when multiplied by 5 makes 5/6ths and you know those added together : -6 + - 5/6ths dont make -6.   Same thing goes for 2 and 3 --> 2 + 5/2 or 3 + 5/3 dont make 6.   1 and 5 however do work.   The reciprocal of 1 is 1 so 1 + 5/1 =6 and so does 5 + 5/5 = 6 (Of course negative again).   So there are 2 answers -1 and -5...

Algebraically...   -x + (1/-x)5 = -6 =

-x -5/x = -6 (multiply by -x)

x^2 + 5 = 6x

x^2 + 6x + 5 = 0

(x + 1) (x + 5 )    = 0

So the answers are x = -1 and -5

Age Problem 6 Solution

2010-12-14T11:57:00.000-08:00

This is the answer to the age algebra word problem that asked, "Bob is one third the age of his father. In 12 years he will be half the age of his father. How old is each now?"

Ok Bob being 1/3 the age of his father is the same thing as saying the father is 3 times as old as him. It makes it easier to work with. So let Bob be x and his father be 3x. And in 12 years bob will be half as old as his father. So we have..

x + 12 = (3x + 12) / 2

2x + 24 = 3x + 12

x = 12

So Bob is 12 years old and his father is 3(12) or 36 years old now. In 12 years Bob will be 12 + 12 or 24 years old and his father will be 36 + 12 or 48 years old which means Bob will be half as old so it checks out.

Age Problem 5 Solution

2010-11-25T22:49:00.000-08:00

This is the algebra age word problem that read Pol is 10 years younger than greg. In 7 years, he will be 10 years more than one halfas old as greg. Find their age at present. help me solve it.

Let Greg be X and Pol be X - 10 as far as current ages. In 7 years (thats x + 7) he will be 10 years more than 1/2 as old as greg (thats (x - 10 + 10)/2 or x/2). So set them equal..

X and x -10

x + 7 and x - 10 + 7 28 and 25

(x+7)/2 + 10 = x - 3

x + 17 = 2x - 6 ******** Correction 2 * 10 = 20 + 7 = 27 So it should be

x + 27 = 2x - 6

x =33 ---> Which is Greg's age so Pol's is x-10 or 33 -10 = 23

So Pol is 23 and Greg is 33

Work Problem 6 Solution

2010-11-25T14:31:00.000-08:00

This is the solution to Work Problem 6 which asked, "A mother can rake a yard in 90 minutes and her daughter can do it in 60 minutes.If the mother rakes for 15 minutes before her daughter joins her,how long will it take them to finish the work?"

Ok the mother can do 1/90th of her job in a minute so if you have x/90 in 90 minutes 90/90 = 1 she completes her job same thing goes for the the daughter .. 1/60 of her job done in a minute with x/60 in 60 minutes 60/60 = 1 she completes her job. Seems kind of silly that I'm pointing this out but this is how you solve the problem you just add their work together and set it equal to 1. Their is one additional stipulation in the problem though and thats the mother starting 15 minutes before the daughter joins the mother otherwise it would just be
x/90 + x/60 = 1... but the mother rakes for 15 minutes and completes 15/90 of her job so the way to write it out would be x/90 + 15/90 + x/60 = 1.

multiply everything by 180 and get 2x + 30 + 3x = 180.

5x + 30 = 180
5x = 150
x = 30 So it would take 30 minutes if they worked together. UPDATE: "how long will it take them to finish the work?" <-- Sounds to me like how long will they finish together thus 30 minutes the answer. But if you want to take this to mean how long did it take them to finish the total job then add the 15 minutes of initial work done by the mother which would be 30 + 15 = 45.

Plane Geometric Figure Problem 2 Solution

2010-11-25T14:09:00.000-08:00

This is the Plane Geometric Figure Problem 2 Solution which asked "The length of the second side of a triangle is four less than three times the length of the first side. The length of the third side is one more than the length of the first side. If the perimeter of the triangle is 37 feet, what is the length of the first side?"

Since everything is relative to the 1st side make the 1st side x.

The 2nd side is 3x - 4 (4 less than 3 times the length of the 1st side)

and the 3rd side is x + 1 ( 1 more than the length of the 1st side)

Now all the sides add up to 37 feet so set all this equal to 37.

x + 3x - 4 + x + 1 = 37

5x - 3 = 37

5x = 40

x = 8

So the 1st side is 8 the 2nd side is 3(8) - 4 or 20 and the 3rd side is 9 .. 8 + 20 + 9 = 37

Finance Problem 2 Solution

2010-11-25T13:50:00.000-08:00

This is the finance problem 2 solution which read: "A local furniture store is selling all $850 mattresses at 35% off. Alyssa is buying a mattress and has a coupon for an additional 10% off the sale price. What will Alyssa pay for her mattress? "

Ok you just have to take the discount off twice. 35% off of $850 would be $297.50 so 850 - 297.50 = $552.50 is the sale price. Now she has a coupon for another 10% off that sales price. So 10% of 552.50 is 55.25 and 552.50 - 55.25 = 497.25

So Alyssa will pay $497.25 for her mattress....

Coin Problem 6 Solution

2010-11-25T13:41:00.000-08:00

This is the coin problem 6 solution which read "a stack of pennies and dimes has a total value of $2.31. how many dimes are in the stack if there are twice as many dimes as pennies?"

make $2.31 into cents so 231 cents .. a dime is worth 10x and penny is worth x. So if you had 1 dime and 1 penny you would have 10(1) or 10 cents plus 1(1) or 1 cent which is 11 cents. Were trying to have it add up to 231 cents though and there are twice as many dimes 2(10x) or 20x as pennies.

So we have 20x + x = 231

21x = 231

x = 11 So there are 11 penny's or 11 cents and 22 dimes or 220 cents .. 220 cents plus 11 cents equals 231 cents or $2.31.

Number Problem 10 Solution

2010-11-25T13:25:00.000-08:00

This is number problem 10 solution which read:

"Find 3 consecutive even integers such that 4 times the first is decreased by the second is 12 more than twice the third. " Ok I'm not understading the "is decreased by the" part so I assume it was a typo and you just meant "Find 3 consecutive even integers such that 4 times the first is 12 more than twice the third. "

ok 3 consecutive even integers would b x, x + 2, and x + 4

4 times the 1st would be just 4x and is equal to 12 more than twice the 3rd which would be (x + 4) * 2 + 12

which is... 4x = (x+4)*2 + 12

4x = 2x + 8 + 12

2x = 20

x = 10 So the answer would be 10, 12, 14 to check 4(10) = (14)*2 + 12

40 = 28 + 12

40 = 40

Work Problem 5 Solution

2010-08-17T13:21:00.000-07:00

This is the solution to the work problem # 5 which read: Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?

Ok Tony takes 1 and 1/2 hours which is 90 minutes, and then we got jim at 30 minutes and Sue at 45 minutes. Convert everything to find out how much each can do in 1 minute. Tony does 1/90 th of his job in 1 minute Jim 1/30 th and Sue 1/45 th. So lets take just one individual persons job say Sue for example. If someone says Sue takes how long to finish her job if you know that she can do 1/45 of her job in a minute what would u say? 45 minutes. Or you can look at it as x/45 = 1 which means how many over 45 would = 1? The answer of course would be 45 again. So we set this problem up just like that only were ADDING the other peoples jobs together and setting it equal to 1.

So we got x/45 + x/90 + x/30 = 1 *Multiply both sides by 90 and get

2x + x + 3x = 90

6x = 90

x= 90/6 = 15

So it would take 15 minutes if all 3 worked together to finish the job.

Number Problem 9 Solution

2010-03-30T16:37:00.000-07:00

This is the solution to the algebra number word problem which was asked by a anonymous person in the ask question section which read, "One number is 5 more than another.their sum is 53.what are the two numbers?"

So we have one number is x and another is 5 more than it which is x + 5. they equal 53.

So we have x + x + 5 = 53

2x + 5 = 53

2x = 48

x = 48/2 = 24 for one number and 24 + 5 or 29 for the other.

So check 24 + 29 = 53

53 = 53

Number Problem 8 Solution

2010-03-05T12:13:00.001-08:00

This is the solution to the Algebra Number word problem # 8 which read "The ratio of 16 more than a number to 12 less than that number is 1 to 3. What is the number?"

Ok a number is x - the unknown

The problem says that ration of 16 more than a number --> which is x + 16 to 12 less than that number(the number is still x) or x - 12 is 1 to 3

So we have (x + 16)/(x -12) = 1/3
Cross multiply and you get 3*(x + 16) = 1*(x - 12)

3x + 48 = x - 12

2x = - 60

x = -30

Check:
(-30 + 16)/(-30 - 12) =
-14/-42 = 14/ 42 = 1/3

Finance Problem 1 Solution

2010-01-23T19:14:00.000-08:00

This is the solution to the algebra finance word problem that An anonymous poster recently asked which read:

"An employee's new salary is $19,110 after getting a 5% raise. What was the salary before the increase in pay?"

Ok easiest way to think about this is try something you do know for sure. Whats 5% of 100? 5 right--- Ok, so what if the problem said an employee's new salary was $105 after a 5% increase in pay what was the salary before... Ok you already know the answer 100.... So how would u come up with the new total? Well 5% of 100 is 5 and then you add it to the original amount of 100 which makes 105. So, you set up your problem like that:

5% can be written 5/100

so,

5/100 * x + x = 105 multiply by 100

5 * x + 100x = 10500

5x + 100x = 10500

105x = 10500
x = 10500/105 = 100 so 100 is the answer you were looking for and it checks out..

You solve this problem exactly the same way.

5/100 * x + x = 19110 multiply by 100

5x + 100x = 1911000

105x = 1911000

1911000/105 = x

x = 18200 So check whats 5% of 18200 ? its 910 Now add 910 to 18200 and you get 19110. So it checks out.

Time and Distance Problem 11 Solution

2010-01-10T16:50:00.000-08:00

This is the answer to the algebra time and distance word problem #11 which was asked by another anonymous user on the "Algebra Word Problem Questions" post. The question read:

"Alexis left Miami and drove at a speed of 20 kph. Thomas left 3 hours later, from the same point, and drove at a speed of 30 kph. How long will it take Thomas to catch up to Alexis?"

Ok non-algebraically the easiest way to figure this one out is to figure how far kilometers Alexis has gone before Thomas even leaves. Since Thomas left 3 hours later and Alexis drove 20 kph then Alexis has gone 20(3) or 60 k or kilometers. Thomas is going faster but how much faster than Alexis? (30 - 20)kmh = 10 So, the question is if he's creeping up at 10 kmh how long would it take him to go the distance that Alexis is ahead which we figured out to be 60 kilometers. Well at 10 kph it would take 6 hours to to 60 k... 60 k / 10 kph = 6 h

For the algebra way you just set up the equation where you have :

20(3 + x) = 30x --> the time is x + 3 because she had a 3 hour head start

then you have 60 + 20x = 30x

60 = `10x

x = 60/10 = 6 --> 6 hours

Work Problem 4 Solution

2009-11-14T15:54:00.000-08:00

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..

Mixture Problem # 6 Solution

2009-10-07T09:10:00.000-07:00

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

2009-10-07T08:55:00.000-07:00

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

Time and Distance Problem 10 Solution

2009-09-12T19:40:00.001-07:00

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..

Time and Distance Problem 9 Solution

2009-08-25T19:15:00.001-07:00

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

Plane Geometric Figure Problem 1 Solution

2009-08-02T10:26:00.000-07:00

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

Age Problem 4 Solution

2009-08-02T08:10:00.000-07:00

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.

Number Problem 6 Solution

2009-03-17T13:17:00.001-07:00

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

Time and Distance Problem 8 Solution

2009-03-13T20:01:00.001-07:00

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

2009-03-13T19:43:00.000-07:00

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

Work Problem 3 Solution

2009-02-05T14:57:00.001-08:00

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.

Work Problem 2 Solution

2009-02-01T18:03:00.000-08:00

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.

Lever Problem 4 Solution

2008-10-09T18:36:00.000-07:00

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

2008-09-28T21:24:00.001-07:00

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

2008-09-28T21:05:00.000-07:00

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

2008-06-12T19:40:00.000-07:00

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

2008-05-12T14:52:00.000-07:00

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

2008-03-26T19:44:00.000-07:00

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

2008-03-20T20:07:00.000-07:00

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

2008-03-16T15:12:00.000-07:00

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

2008-03-15T15:35:00.000-07:00

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

2008-03-13T11:56:00.001-07:00

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

2008-03-13T11:43:00.000-07:00

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

2008-03-12T09:22:00.001-07:00

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

2008-03-12T09:15:00.000-07:00

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

2008-03-11T20:37:00.000-07:00

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

2008-03-10T21:21:00.000-07:00

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

2008-03-10T20:12:00.000-07:00

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

2008-03-10T18:59:00.000-07:00

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

2008-03-10T12:29:00.000-07:00

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

2008-03-10T12:02:00.000-07:00

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

2008-03-10T11:46:00.000-07:00

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

2008-03-09T21:10:00.001-07:00

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

2008-03-09T20:26:00.000-07:00

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

2008-03-09T18:05:00.000-07:00

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

2008-03-09T15:58:00.000-07:00

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

2008-03-09T14:24:00.000-07:00

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

2008-03-09T13:26:00.000-07:00

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

2008-03-08T22:56:00.000-08:00

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

2008-03-08T22:36:00.000-08:00

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

2008-03-08T21:53:00.000-08:00

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.....