No need for long algebra here. Use t = d/s. Let s be the speed of the slower sled. The times are 300/s and 300/(s+5) for teams B and A respectively. These times differ by 3 hours:

300/s - 300/(s+5) = 3
1/s - 1/(s+5) = 1/100

Lots of ways to finish this, including the conventional common denominator approach. I prefer number theory: from the answer choices, s is a whole number. From the above equation, it's clear s and s+5 are divisors of 100 (indeed 100 must be their LCM), so the only possible answer is s=20.

gmataspirant wrote:I am asking a basic question

300 = (r-5)(t+3)

Time should be t-3 and Rate should be r+5 right ?

The problem says " 3 fewer hours than Team B" and Team's A speed 5 miles greater than "Team B".

I just trying to fix my basics ? Your help is appreciated.

It all depends on what t and r mean in your solution:

If t and r represent the time and speed for Team A, then the time and speed for Team B would be t+3 (B takes three hours longer than A) and r-5 (B travels 5 mph slower than Team A). Now, if you solve for t, that will tell you how long it takes A to finish the race, and r will represent the speed at which A traveled.

You could instead make t represent: 'the time it takes Team B to finish', and you could let r mean: 'the speed at which Team B travels'. Then A's time will be t-3, and A's speed with be r+5. Now, if you solve for t, that will tell you how long it takes B to finish the race. You'd need to subtract 3 to find how long it takes A to finish the race. Similarly, here r represents B's speed, and you'd need to add 5 to find A's speed.

That's a long way of saying that you're right, but so is the text you've quoted: it just depends on what the letters represent.