Tracy Said:

2 math word problems?

We Answered:

Let x = quantity to be replaced with pure solution
0.30(10-x) + 1.00*x = 0.50*10
3 - 0.30x + x = 5
0.70x = 2
x = 2.86 liters

Let x = original number of people
90,000/x = original share

(x+3) * ( 90,000/x - 2,500 ) = 90,000
(x+3) * ( 90,000 - 2,500x ) = 90,000x
90,000x + 270,000 - 2,500x^2 - 7,500x = 90,000x
2,500x^2 + 7,500x - 270,000 = 0
x^2 + 3x - 108 = 0
( x + 12 ) ( x - 9 ) = 0

x = 9 persons

Cory Said:

help with math word problems.?

We Answered:

These are all WORK Problems.

Let's start with the first Question, then you should able to answer the rest if you understand it.

Since we don't know how big and the measurement of the house, and there is no way to calculate it. So we will just set the whole painting job as a WHOLE ~ 1 (100%)

Time Spent(hr) Fraction Done in 1 Hour

Jack 5 1/5
Joan 10 1/10
______________________________________…

Together let it be X 1/X


(The table above will not shown correctly after posted, try to break it into 3 columns and 4 rows)


We know that each of them can paint 1/5 and 1/10 of the house respectively in 1 hour by themselves.

Then 1/X = 1/5 + 1/10
1/X = 3/10
X = 10/3 Hrs
X = 3 1/3 Hrs
X = 3 Hours 20 Min.
Then it will take them 3 Hours and 20 min to paint the house together.

When you get use to it, you just need to set the formula as:

1/a + 1/b = 1/x

Where a and b are the time working alone and x is the time spent working together!

It can also be applied for more than 2 persons or even pipes in Q4 where there is an OUTLET pipe! That means you MUST SUBTRACT the OUTLET pipe... Hope this make sense: I'll lay down the answer for you, then you can check after you try them out.


Here are the answers:

Q2. 1 hour

Q3. This one is a little tricky. Remember the 1/x? the 1 on top is the part of work that is left over. In this case:

Ray had completed 3/14 of the work, then 11/14 was left

Let time alone for Dodi = b, then

1/14 + 1/b = (11/14) / 4
1/14 + 1/b = 11/56
1/b = 11/56 - 4/56
1/b = 7/56
b = 8 hours

Q4. 72 hours

Q5. 1/J + 1/A = 1/1.5
1/J + 1/A = 2/3 and A = 2J - 3
1/J + 1/(2J - 3) = 2/3
3(2J-3) + 3J = 2J(2J-3)
6J - 9 + 3J = 4J^2 - 6J
4J^2 -15J + 9 = 0
(4J - 3) (J - 3) = 0

then J = 3/4 hr or J = 3 hrs.
If J = 3/4
A = 2(3/4) - 3
= negative!!!
(reject!)

Then J =3
A = 2(3) - 3
A = 3 hors.

Then both spend 3 hrs when working alone.

Hope this will help!


Q1.
Let the numbers be x and x+2

Then x^2 + (x+2) = 32
x^2 + x - 30 = 0
(x + 6) (x - 5) = 0
since x can't be negative, then x = 5
the numbers are 5, 7

Q2. (x+1)^2 + (x+2)^2 = 61
(x^2 + 2x+1) + (x^2+4x+4) - 61 = 0
2x^2 + 6x - 56 = 0
x^2 + 3x - 28 = 0
(x+7) (x-4) = 0
since x can't be negative, x = 4
the numbers are 4, 5, 6, 7

Q3. x(x+2) = 4(x+4) -1
x^2 + 2x = 4x + 16 - 1
x^2 -2x - 15 = 0
(x+3) (x-5) = 0
since x can't be negative, x = 5
the numbers are 5, 7, 9

find three consecutive odd integers such that the producy of the second and third decreased by 4 times the second is 5 more than 5 times the first.

(x+2)(x+4) - 4(x+2) = 5x + 5
x^2 +6x + 8 - 4x - 8 - 5x - 5 = 0
x^2 -3x -5 = 0
This can be worked out.. can't be factorized....

Dale Said:

Some word problems in math, help?

We Answered:

1) yes it's a parabola
Since the coefficient of x^2 is positive, it opens upward.

The vertex is at x = -3 / 2.6
and the x-intercepts are between (-3 and -4) and between (1 and 2).

2)
Since any part of a day incurs a whole day's charge, this is a step function.
At every integer value for x, it jumps higher, then stays flat until the next integer.
The graph looks like a staircase, viewed from the side:

.................__|
...........___|
......___|
___|

It starts at (0,0), then (1,10) and (2, 18), (3,26), (4, 34) and so on.

Cost of 13/2 days = 6 1/2 but that is same as 7 = 10 + (7-1) * 8 = 10 + 48 = $58.

The function would be defined as:
0 for x <= 0
10 for 0 < x <= 1
10 + (ceiling(x)-1) * 8 for 1 < x
ceiling(x) = next higher integer
ceiling (13/2) = 7