Word Trans (Rates): John would have reduced the time it took

[II]

John would have reduced the time it took him to drive from his home to a certain store by 1/3 if he had increased his average speed by 15 miles per hour. What was John's actual average speed, in miles per hour, when he drove from his home to the store?

(A) 25 (B) 30 (C) 40 (D) 45 (E) 50

My approach:
This is a rates problem, so we will be using the RT=D formula.
So we know that original trip -> RT = D

"increased his average speed by 15 miles per hour"
New rate = R+15
"reduced time by 1/3"
New time = 2/3T

So using the RT=D formula, and using this new information, we can write:
(R+15)((2/3)t) = D

(R+15)((2/3)t) = RT (on right hand side we substitute the D with RT)
(R+15)(2/3) = R (divided both side by T)
(2/3)R + 10 = R (multiply out the left hand side of the equation)
10 = (1/3)R (subtract (2/3)R from both sides)
30 = R (multiply both sides by 3 to get R)

I was interested in learning other approaches to solving this questions ? Grateful for your input.

Thanks.
II

[Stuart@KaplanGMAT]

John would have reduced the time it took him to drive from his home to a certain store by 1/3 if he had increased his average speed by 15 miles per hour. What was John's actual average speed, in miles per hour, when he drove from his home to the store?

(A) 25 (B) 30 (C) 40 (D) 45 (E) 50


I was interested in learning other approaches to solving this questions ? Grateful for your input.

Thanks.
II

Sometimes we can take an intuitive approach to these questions.

His new time is 2/3 the old time. The time/rate changes will always be inverses of each other to keep D constant. Therefore, his new R must be 3/2 of the old R.

So, R has increased by 50% (from R to 3/2 R). 15 (the actual rate increase) = 50% of 30: choose (b).