Problem Solving

Anthony covers a certain distance on a bike. Had he moved 3mph faster, he would have taken 40 minutes less. Had he moved 2mph slower, he would have taken 40 minutes more. The distance (in miles) is?

A. 30
B. 35
C. 36
D. 37.5
E. 40

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I stuck here,

I know that Time = Dist / Speed, then I have 3 different values of time,
$$T_1=\frac{D}{S}$$
$$T_2=\frac{D}{S+3}$$
$$T_3=\frac{D}{S-2}$$
I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.

swerve wrote:Anthony covers a certain distance on a bike. Had he moved 3mph faster, he would have taken 40 minutes less. Had he moved 2mph slower, he would have taken 40 minutes more. The distance (in miles) is?

A. 30
B. 35
C. 36
D. 37.5
E. 40

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I stuck here,

I know that Time = Dist / Speed, then I have 3 different values of time,
$$T_1=\frac{D}{S}$$
$$T_2=\frac{D}{S+3}$$
$$T_3=\frac{D}{S-2}$$
I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.

Call the original rate 'r' and the original time 't.' The distance Anthony covers is rt.

If he moved 3 mph faster, his new rate is r + 3. If he took 40 minutes, or 2/3 of an hour less, his new time is (t - 2/3). So his distance is (r + 3)(t - 2/3).

Of course, the distance hasn't changed, so (r + 3)(t - 2/3) = rt
rt -(2/3)r + 3t - 2 = rt
-(2/3)r + 3t = 2 --> call this equation 1.

If he moved 2 mph slower, his new rate is r - 2. If he took 40 minutes, or 2/3 of an hour more, his new time is (t + 2/3). So his distance is (r - 2)(t + 2/3).

Again the distance hasn't changed, so (r - 2)(t + 2/3) = rt.
rt + (2/3)r -2t -4/3 = rt
(2/3)r - 2t = 4/3 --> call this equation 2.

If we add equation 1 to equation 2, the 'r' terms will cancel, and we'll get: t = 2 + 4/3 = 10/3
Plug this value for t back into equation 1 and we get -(2/3)r + 10 = 2; r = 12

If r = 12 and t = 10/3, then rt = 12 * 10/3 = 120/3 = 40. The answer is E

swerve wrote:Anthony covers a certain distance on a bike. Had he moved 3mph faster, he would have taken 40 minutes less. Had he moved 2mph slower, he would have taken 40 minutes more. The distance (in miles) is?

A. 30
B. 35
C. 36
D. 37.5
E. 40

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I stuck here,

I know that Time = Dist / Speed, then I have 3 different values of time,
$$T_1=\frac{D}{S}$$
$$T_2=\frac{D}{S+3}$$
$$T_3=\frac{D}{S-2}$$
I would like to know how to solve it in less than 2 minutes. I need your help. Thanks.

Note: this question appears to have been modeled on this official problem: https://www.beatthegmat.com/speed-dista ... tml#788767