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.
Saturday, January 23, 2010
Finance Problem 1 Solution
Sunday, January 10, 2010
Time and Distance Problem 11 Solution
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
Saturday, November 14, 2009
Work Problem 4 Solution
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..
Wednesday, October 7, 2009
Mixture Problem # 6 Solution
This is the solution to the algebra mixture word problem # 7 which read:
"If a merchant has two types of tea, one worth $2.70 per kilogram and the other worth $3.00 per kilogram, how many kilograms of each type should the merchant use in order to produce 30 kilograms of a blend that is worth $2.95 per kilogram?"
Even though this involves money we treat it like a mixture problem.
You got to unknowns but if you look at it as having x = 1 amount and y = 30 -x then you just have x, and 30 -x so...
Its the x amount * the 270 cent worth tea plus the 30 - x amount * the 300 cent worth tea is going to equal 30 * 295 cent blend
270 * x + 300(30-x) = 30(295)
270x + 9000 - 300x = 8850
-30x = -150
30x = 150
x = 150/30 = 5
so 5 kilograms of the 270 cent or $2.70 tea and 30 -5 or 25 kilograms of the 300 cent or $3.00 tea would need to be mixed together to make 30 kilograms of the $2.95 cent tea.
check 270 * 5 + 300 * 25 = 30*295
1350 + 7500 = 8850
8850 = 8850
Number Problem 7 Solution
This is the solution to the algebra number word problem 7 which asked "If the Numerator and Denominator of a certain fraction are both increased by 3, the resulting fraction equals 2/3. If the Numerator and Denominator are both decreased by 2, the resulting fraction equals 1/2. Determine the fraction."
Ok so you have 2 unknowns --> x the numerator and y the denominator. You can solve 2 unknowns with 2 equations. They give you 2 equations.
(x+3)/(y+3) = 2/3 and
(x-2)/(y-2) = 1/2
take the (x+3)/(y+3) = 2/3 and cross multiply and get:
2(y+3) = 3(x+3) which is
2y + 6 = 3x + 9
Now take the second equation: (x-2)/(y-2) = 1/2 and cross multiply
2(x-2) = 1(y-2) which makes
2x - 4 = y - 2 now solve for y
y = 2x - 2
ok now plug this back into the 1st equation for y which we broke down to :
2y + 6 = 3x + 9 so you get 2(2x-2) + 6 = 3x + 9 solve for x
4x - 4 + 6 = 3x+ 9
x= 7 ok now plug 7 into y = 2x - 2 and get y = 2(7) -2 = 12
so the numerator is 7 and the denominator is 12 = 7/12
to check add 3 to the top and bottom and get 10/15 = 2/3
and subtract 2 from the top and bottom and get 5/10 = 1/2
Saturday, September 12, 2009
Time and Distance Problem 10 Solution
This is the solution to the algebra time and distance word problem # 10 which was submitted under the ask a question post anonymously which read..
"Junior's boat will go 15 miles per hour in still water. If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream then what is the speed of the current?"
distance = rate X time ... d = r*t we know the speed of the boat in still water 15 mph
we know he can go 12 miles downstream in the same time he can go 9 miles upstream-- so times are equal --- so since time is equal we can set the distances and rates equal to one another...
d1/r1 = d2/r2
The rate downstream will be 15 + x and upstream will be 15 - x so we have:
12/(15+x) = 9 /(15-x)
cross multiply and you get 12(15-x) = 9 (15+x)
180 - 12x = 135 + 9x
45 = 21x
x = 45/21 = 15/7 This is the answer the current is 15/7 or 2 and 1/7ths mph..
Tuesday, August 25, 2009
Time and Distance Problem 9 Solution
This is the solution for the time and distance algebra word problem #9 that read:
"if I walked 9 laps around a track. One lap is 1312 ft. I walked at a rate of 4 miles an hour. how many minutes did it take me to walk 9 laps?"
The easiest way to think about this problem is to think about an easier problem and then copy the steps. What if the problem read:
if I walked 9 laps around a track -- 1 lap is 1 mile and i walked at a rate of 4.5 miles and hour. how many minutes did take me to walk 9 laps?
well the total distance would be the length of 1 lap: 1 mile * the # of laps which is 9 --> 9 * 1 = 9 miles
the rate was 4.5 miles an hour -- distance = rate * time
so 9 miles = 4.5miles/hour* time
9 miles/4.5 miles/hour = time or 2 hours would be how long it would take--->
So this is not a hard math problem the arithmetic is what's hard in this problem, but if you just follow the same steps you will get the answer easily enough. So, again we have:
"if I walked 9 laps around a track. One lap is 1312 ft. I walked at a rate of 4 miles an hour. how many minutes did it take me to walk 9 laps"
ok so take 9 * 1312 1st to get the total distance in feet which is : 11808
4 miles an hour would be (5280 ft in a mile) 5280 * 4 = 21120 ft/hour
But we dont want hours we want minutes-- there are 60 minutes in an hour
so (21120 ft*1hour)/(hour*60minutes) --> you're multiplying by 1hour/60 minutes. so divide by 60..
362ft/minute --- the hours(1 hour = hour) cancel out..
ok so now we have a total distance in feet of 11808 feet -- divided by the rate of 362 ft/minute
11808 ft/(362 ft/minute) = 11808 ft *minute/ 362 ft
the ft cancel out you divide 11808 by 362 and get 32.619 minutes --> which is a good enough answer but if you want it more exact:
.619 of 60 seconds would be 60 *.619 = 37 seconds --- so 32 minutes 37 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
