Sunday, December 4, 2011

Plane Geometric Figure Problem 3 Solution

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 

Saturday, December 3, 2011

Number Problem 14 Solution

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

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

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..

Friday, October 21, 2011

Number Problem 13 Solution

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

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;

Sunday, October 9, 2011

Number Problem 12 Solution

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....


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