word problem help?

Question

The speed of train A is 16mph slower than the speed of train B. Train A travels 23o miles in the same time it takes train B to travel 310 miles. Find the speed of each train.

Answer 1

If speed of train A is x mph, speed of train B is (x +16) mph.
Then 230/x = 310/(x + 16)
=> 310x = 230(x + 16)
=> (31 -- 23)x = 23*16
=> 8x = 23*16
=> x = 46 and (x + 16) = 46 + 16 = 62
Speeds of trains are A = 46 mph
B = 62 mph

Answer 2

Let a be the speed of train A, and b be the speed of train B.

a=b-16

At the same time it train A travels 230 miles train B travels 310 miles:

Let t be the time:

a*t = 230 -> t= 230/a
b*t = 310 -> t=310/b

From this system of equation we get:
230/a = 310/b // cross multiply

(i) 230b = 310a

Let's plug a=b - 16 into (i)

310(b - 16) = 230b

310b - 310*16 = 230 b // - 230b

80b - 310*16 = 0

80b = 310 * 16 // divide by 10

8b = 31 * 16 // divide by 8

b = 31*2

b=62

a=62-16=46



Let's check the solution
230*62 = 14260
310*46 = 14260

Answer 3

Let the speed of train B be x.
Since distance = speed * time

For train A, 230 = (x - 16) * t => t = (230 / (x - 16))
For train B, 310 = (x)t => t = (310/x)

Solving x when t is the same.
(230) / (x-16)) = (310 / x)

Multiply both sides by x(x-16)
230x = 310(x-16)
230x = 310x - 4960
80x=4960
x=4960/80
x=62

Speed of train A is (62-16) = 46mph.
Speed of train B is 62mph.

Answer 4

A=B-16
AT=230
BT=310

T=230/A
T=310/B
230/A=310/B
230B=310A
23B=31A
B=31A/23

A=B-16
A=(31A/23)-16
23A=31A-368
8A=368
A=46
B=A+16=62

The speed of train A is 46mph and train B is 62mph.