Two Math problems.....

1) Two trains leave the train station at the same time but in opposite directions. The faster train travels at an average rate of 75 mph and the slower train travels at an average rate of 68 mph. In how many hours will they be 715 miles apart?

2) Mr. Richards, a cyclist, rode from his home to his office at an average of 18 mph. On is return home from his office, using the same route averaging 12 mph. If the total round trip took 5 hour, what was the distance from his home to his office?

Thanks in advance for any answers, and please include explanations.

1 ) sum of their speeds = 75+68 = 143 mph

distance = 715 miles

time = 715/143 = 5 hours

2) let the distance = d

time for going offce t1 = d/18

time for coming home t2= d/12

total trip = 5 hours

d/18 + d/12 = 5

d = 36 miles

1 )The sum of their speeds

= [75+68]mph
= 143 mph

the distance between these= 715 miles

Hence, time taken = 715/143 = 5 hours

2) let the required distance be p

required time for going offce = p/18..........[1]

So,time for returning back to home = p/12...[2]

Time taken for both jouneys = 5 hours

p/18 + p/12 = 5

p[12+18]/ [18x12] = 5

p= [5X18x12]/30 miles

p=36 miles

1) 75 + 68 = 143 mph

715 /143 = 5 hours ANSWER

2) Let the distance to office be x

home to office takes x / 18 hours

office to home takes x / 12 hours

x / 18 + x/12 = 5

MULTIPLY BY 36

2x + 3x = 180
5x = 180
x = 36 miles ANSWER
CHECK
36/18 = 2 hours
36/12 = 3 hours
total = 5 hours

1. 75 + 68 = 143

715/143 = 5 hrs

2. 18 + 12 = 30/2 = 15 mph average x 5 hours = 75 miles total/2 trips = 37.5 miles

75X+68X=715,so X=5,five hours will they be 715 miles apart

a:b=12:18 ,a:b=2:3. a+b=5,so a=2,b=3,then distance=2*18=36m