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## Alan and Peter are cycling at different constant rates on a

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### Alan and Peter are cycling at different constant rates on a

by vinni.k » Fri Nov 23, 2018 6:14 am

00:00

A

B

C

D

E

## Global Stats

Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.

OA is D

But i think OA should be B because in S(1) you can't find the speed of Alan. For Peter, i can get the speed of 4 mph as he covered 4 miles in 1 hr and now he is 3 miles behind Alan, but how to get Alan's speed.

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by [email protected] » Fri Nov 23, 2018 8:22 am

00:00

A

B

C

D

E

## Global Stats

vinni.k wrote:Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
Given: Alan is NOW 3 miles ahead of Peter

Target question: How many minutes from now will Peter be 1 mile ahead of Alan?

Statement 1: ONE HOUR AGO, Alan was 7 miles ahead of Peter.
Alan is NOW 3 miles ahead of Peter
So, in one hour, the GAP between Alan and Peter decreased by 4 miles
Another way to put it: Peter's speed is 4 mph greater than Alan's speed.
Another way to put it: in one hour, Peter traveled 4 miles more than Alan
So, in ONE HOUR FROM NOW, Peter will travel another 4 miles more than Alan. This means the Peter will not only close the 3-mile gap BUT ALSO travel 1 mile further than Alan travels.
So, the answer to the target question is in 60 MINUTES, Peter will be 1 mile ahead of Alan
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
Another way to put it: Peter's speed is 4 mph greater than Alan's speed.
Another way to put it: in one hour, Peter travels 4 miles more than Alan
At this point, we can apply the same logic we applied to statement 1 to conclude that statement 2 is SUFFICIENT

Cheers,
Brent

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by [email protected] » Fri Nov 23, 2018 3:03 pm

00:00

A

B

C

D

E

## Global Stats

vinni.k wrote:Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
Excellent opportunity for the relative velocity (speed) technique!

$$?\,\,\,:\,\,\,\min \,\,{\rm{for}}\,\,4\,\,{\rm{miles - }}\underline {{\rm{gaining}}} \,\,\,{\rm{Peter}}/{\rm{Alan}}\,\,$$

$$\left( 1 \right)\,\,{\text{RelativeSpee}}{{\text{d}}_{{\text{Peter/Alan}}}}\,\,\, = \,\,\,\,\,\frac{{4\,\,{\text{miles}}}}{{\boxed{1\,\,{\text{hour}}}}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \boxed{1\,{\text{h}}}\,\,\left( { = 60\min } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.$$
$$\left( 2 \right)\,{\text{RelativeSpee}}{{\text{d}}_{{\text{Peter/Alan}}}}\,\,\,\, = \,\,\,\,\frac{{4\,\,{\text{miles}}}}{{\boxed{1\,\,{\text{hour}}}}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \boxed{1\,{\text{h}}}\,\,\left( { = 60\min } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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