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;

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....

Time and Distance Problem 13 Solution

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

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

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

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