when doing word problems, you just write what it says. then solve.

1. The numerator of a fraction is 1 more than the denominator. If the numerator and the denominator are both increased by 2, the new fraction will be one fourth less than the original fraction. Find the original fraction.

(N/(N-1)) = (N+2)/(N+1) * 4/3

2. Jan and Jean start at the same time from the same place and travel inn opposite directions. Jan averages 50 mi/h and Jean averages 55 mi/h. How long before they will be 630 miles apart? (6 hours)

630/(50+55)

3. Martin started a run at 9:00 AM and maintained an average speed of 12 km/hr. Carl started on the same course at 9:10 AM and averaged 16 km/hr. What distance had Martin run when caught be Carl?

12t = 16(t-1/6)
then t * 12 = Carl’s distance.

4. A rectangular garden is 12 m longer than it is wide. A second rectangular garden is planned so that it will be three times as wide and half as long as the first garden. Find the area of the first garden if the sum of the areas of both gardens will be 98 m^2.

w*(w+12) + 3w*(w+12)/2 = 98

etc. you just write what it says.

1st change the word problem to a mathematical equation, I will do just one of your questions to give u a hint.
Q#1. lets say the numerator is x , and the denominator is y.
x+1=y —– the numerator is 1 more than the denominator
(x+2)/(y+2) = 1/4(x/y)
so, lets solve for x first, in the 2nd equation we will replace y by (x+1) because we said those 2 e equal
(x+2)/((x+1)+2) = 1/4(x/(x+1))
(x+2)/(x+3)= x/4(x+1)
(x+2)(4(x+1))=(X+3)X
4x(X+1)+8x+x=[(x^2)+3x
4(x^2)+4x+9x=(x^2)+3x
4(x^2)-(x^2)=3x-13x
3x^2+13x-3x
this is quadratic equation solve for x using the formula and then once u get x value substitute that number in place of x and solve for y. If u still cant get the answer and if u have to have this questions answered email me at anav59@rocketmail.com coz this is too much to do it here on yahoo answers.

1. "The numerator of a fraction is 1 more than the denominator. If the numerator and the denominator are both increased by 2, the new fraction will be one fourth less than the original fraction. Find the original fraction."
Let x be the denominator of the original fraction. Then the original fraction is (x + 1)/x. If the numerator and denominator are both increased by 2, the fraction you get is (x + 3)/(x + 2). Set up an equation:
(x + 3)/(x + 2) = (x + 1)/x – (1/4)
Multiply both sides by 4x(x + 2):
4x(x + 3) = 4(x + 2)(x + 1) – x(x + 2)
Expand:
4x² + 12x = 4x² + 12x + 8 – x² – 2x
4x² + 12x = 3x² + 10x + 8
x² + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = –4 or 2
The original fraction is 3/2 or (–3)/(–4). The latter isn’t in simplest form, so the former is the answer.
Answer—->> 3/2

2. "Jan and Jean start at the same time from the same place and travel in opposite directions. Jan averages 50 mi/h and Jean averages 55 mi/h. How long before they will be 630 miles apart?"
Let t be time in hours. Remember D = RT where D is distance, R is rate, and T is time.
Jan’s distance traveled = 50t
Jean’s distance traveled = 55t
Distance apart = (Jan’s distance) + (Jean’s distance) = 105t
Set up an equation:
105t = 630
t = 6
Answer—->> 6 hours

3. "Martin started a run at 9:00 AM and maintained an average speed of 12 km/hr. Carl started on the same course at 9:10 AM and averaged 16 km/hr. What distance had Martin run when caught be Carl?" Let t be time in hours and let t = 0 represent 9:00 AM.
Martin’s distance ran = 12t, t ≥ 0
Carl’s distance ran = 16[t – (1/6)] = 16t – (8/3)…..(10 minutes = 1/6 hour), t ≥ 1/6
Set Martin’s distance equal to Carl’s distance:
12t = 16t – (8/3)
–4t = –8/3
t = 2/3…..(at 9:40 AM)
When t = 2/3, Martin’s distance ran = 12(2/3) = 8 km
Answer—->> 8 km

4. "A rectangular garden is 12 m longer than it is wide. A second rectangular garden is planned so that it will be three times as wide and half as long as the first garden. Find the area of the first garden if the sum of the areas of both gardens will be 98 m²."
Let x be the width in meters of the first garden. Then:
Area of garden 1 = (x + 12)x = x² + 12x
Area of garden 2 = (x + 12)/2 • 3x = 3x²/2 + 18x
Sum of areas = 5x²/2 + 30x = 98
I’ll complete the square to solve this equation:
x² + 12x = 39.2
x² + 12x + 36 = 75.2
(x + 6)² = 75.2
x + 6 = ±4√(4.7)
x = –6 ± 4√(4.7)
Negative answers don’t make sense, so:
x = –6 + 4√(4.7) ≈ 2.67179 m
Area of garden 1 = x² + 12x = …
= [–6 + 4√(4.7)]² + 12[–6 + 4√(4.7)]
= 36 – 48√(4.7) + 75.2 – 72 + 48√(4.7)
= 39.2 m²
Answer—->> 39.2 m²

5. "Find two numbers whose sum is 25 and the sum of whose reciprocals is 1/4."
x + y = 25
(1/x) + (1/y) = 1/4
Multiply the second equation by xy:
y + x = xy/4
Plug in x + y = 25:
25 = xy/4
100 = xy
100/y = x
Plug this into the first equation:
(100/y) + y = 25
Multiply by y:
100 + y² = 25y
y² – 25y + 100 = 0
(y – 20)(y – 5) = 0
y = 20 or 5
When y = 20, x = 5; when y = 5, x = 20.
Answer—->> 20 and 5

6. "A resort hotel has 200 rooms. Those with kitchen facilities rent for $100 per night and those without kitchen facilities rent for $80 per night. On a night when the hotel was completely occupied, revenues were $17,000. How many of each type of room does the hotel have?"
Let k be the number of rooms with kitchen facilities and n be the number of rooms without kitchen facilities.
k + n = 200…..the hotel was completely occupied, which means that all 200 rooms were used up.
100k + 80n = 17,000
Multiply the first equation by 5 and divide the second equation by 20:
5k + 5n = 1,000
5k + 4n = 850
Subtract the equations:
n = 150
k = 50
Answer—->> The hotel has 50 rooms with kitchen facilities and 150 rooms without kitchen facilities.

7. "Lou has only quarters and dimes in his coin bank. There are currently 26 coins in the bank. If he adds 7 quarters and 3 dimes, the total value of the coins will be $7.35. Find the number of quarters and the number of dimes Lou originally had in the bank." Let q and d be the number of quarters and dimes (respectively) that Lou originally had in the bank.
q + d = 26
25(q + 7) + 10(d + 3) = 735…..I put the money in cents; you can choose to put it in dollars if you want. For the second equation, put the variables all on one side and the constants all on the other:
25q + 175 + 10d + 30 = 735
25q + 10d + 205 = 735
25q + 10d = 530
5q + 2d = 106
Multiply the first equation by 5:
5q + 5d = 130
5q + 2d = 106
Subtract the equations:
3d = 24
d = 8
q = 18
Answer—->> Lou has 8 dimes and 18 quarters in his coin bank.

8. "A two-digit number is 3 times the sum of its digits. The number is also 45 less than the number formed by reversing the digits of the original number. What is the original number?"
Let x be the first digit and y be the second digit. Then the original number is 10x + y. Set up equations:
10x + y = 3(x + y)
10x + y = (10y + x) – 45
Put the variables all on one side and the constants on the other:
7x – 2y = 0
9x – 9y = –45
Multiply the second equation by –2/9:
7x – 2y = 0
–2x + 2y = 10
Add the equations:
5x = 10
x = 2
Plug in this value of x into either equation, then solve for y. I’ll choose the first:
7x – 2y = 0
14 – 2y = 0
14 = 2y
7 = y
Answer—->> 27

9. "Ken has 22 coins, some of which are dimes and the rest are quarters. Altogether, the coins are worth more than $3.40. At least how many of the coins are quarters? At most how many are dimes?" Let d be the number of dimes and q be the number of quarters.
d + q = 22
10d + 25q > 340…..again, I put the money in cents; you can put it in dollars if you want.
2d + 5q > 68
2d + 2q + 3q > 68
2(d + q) + 3q > 68
Plug in d + q = 22:
44 + 3q > 68
3q > 24
q > 8…there must be more than 8 quarters, so Ken must have at least 9 quarters since you can’t have a fractional number of quarters.
d + q = 22
d = 22 – q
q > 8
–q < –8
22 – q < 14
d = 22 – q < 14
d < 14…he has less than 14 dimes, so he has at most 13 dimes.
Answer—->> Ken has at least 9 quarters and at most 13 dimes.