Rate, Time, Distance- Velocity

[Guest]

A fires at a target 500 yards away and hears the bullet strike the target 4 1/3 seconds after gun is fired. B is 400 yards from target and 650 yards from A, hears the bullet strike the target 2 1/3 seconds after he hears the gun fired. Calculate velocity (or speed) of sound and bullet. Both sound and bullet are uniform.

Formula used:

Rate x Time = Distance Distance / Time = Rate Distance / Rate = Time

How do I proceed ?

[Guest]

You are told MAN-A fires the bullet, hits the target 500 yards away, sound travels back from target to MAN-A 4 1/3 seconds later.
Make up an equation of these facts..

Distance / speed = times taken...

Let B equal the speed of the bullet.
Let S equal the speed of sound in air.

Eqn... (bullet time) + (sound back time)= 4 1/3 ... delay until MAN-A hears target sound.

500/B + 500/S = 4 1/3 .....eqn1

Similar for MAN-B....

Told MAN-B is 650 yards from MAN-A (gun sound travels 650 yards from MAN-A to MAN-B)
Told MAN-B is 400 yards from target (target sound travels 400 yards from Target to MAN-B)
Told delay for MAN-B on hearing gun until hears target is 2 1/3 seconds.

Eqn...((time bullet fired to target) + (time target sound travel to MAN-B)) - (time Gun Sound travel to MAN-B) = 2 1/3 delay for MAN-B to hear target snd.

(500/B + 400/S) - 650/S = 2 1/3 ....eqn2

You now have 2 simultaneous equations, solve for B and S

[Guest]

I can't solve them.

[Guest]

500/B + 500/S = 4 1/3 .....eqn1

(500/B + 400/S) - 650/S = 2 1/3 ....eqn2

500/B - 250/S = 2 1/3 ....eqn2a..collecting like terms

Subtract eqn2a from eqn1 .....750/s = 2

So..... 2S = 750 ..... S = 375 yards per second for sound

Substitute S in eqn1 ..... 500/B + 500/375 = 4 1/3

Working out gives ......... 500/B + 4/3 = 4 1/3

and .........................500/B = (4 1/3) - (1 1/3)

and......................... 500/B = 3

So..............................3B = 500

So............................ B = 166.67 yards per second for bullet