Bob and Wendy left home to walk together to a restaurant for

enigma123 · **Posted:** Thu Nov 10, 2011 4:34 pm


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Bob and Wendy left home to walk together to a restaurant for dinner. They started out walking at a constant pace of 3 mph. At precisely the halfway point, Bob realized he had forgotten to lock the front door of their home. Wendy continued on to the restaurant at the same constant pace. Meanwhile, Bob, traveling at a new constant speed on the same route, returned home to lock the door and then went to the restaurant to join Wendy. How long did Wendy have to wait for Bob at the restaurant?

(1) Bob’s average speed for the entire journey was 4 mph.
(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

Guys - any idea how to solve this? Also, can someone please explain how should I approach to solve these questions?

piyatiwari · **Posted:** Thu Nov 10, 2011 6:46 pm

Let

Bob's initial speed B = Wendy's initial speed W = 3 MPH

Total distance from home to restaurant = D

Bobs new speed NB

So,

1. Total time Wendy took to reach restaurant = D/3 hours.

2. Bob's travel = (D/2 miles With wendy at 3 MPH) + (1.5 D at NB)

So total time bob took to reach restaurant = (D/2)/3 + 1.5D/NB = D/6 + 1.5D/NB = D (1/6 + 1.5/NB)

What is asked : How long did Wendy have to wait for Bob at the restaurant?

= Time Bob took - Time wendy took.
= D (1/6 + 1.5/NB) - D/3

So if we know D and NB we will be able to find answer.

Given statements:

(1) Avg speed = 4 MPH

=> total distance / total speed = 2D / (3+NB) = 4

(2) (D/2)/3 + 32 = 1.5D/NB

Hence both statements together are necessary.

Winthrop02 · **Posted:** Sat Nov 12, 2011 2:39 pm

Thanks for sharing this how to process. The step by step process is very easy to follow. Specifically, the numbering part make everything more clearer.

Bunuel · **Posted:** Sat Feb 04, 2012 5:08 pm

enigma123 wrote:

Bob and Wendy left home to walk together to a restaurant for dinner. They started out walking at a constant pace of 3 mph. At precisely the halfway point, Bob realized he had forgotten to lock the front door of their home. Wendy continued on to the restaurant at the same constant pace. Meanwhile, Bob, traveling at a new constant speed on the same route, returned home to lock the door and then went to the restaurant to join Wendy. How long did Wendy have to wait for Bob at the restaurant?

(1) Bob’s average speed for the entire journey was 4 mph.
(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

Guys - any idea how to solve this? Also, can someone please explain how should I approach to solve these questions?

Check similar problem here: bob-and-wendy-planned-to-walk-from-their-home-to-a-89738.html#p679745

Hope it helps.

budablasta · **Posted:** Sat Mar 10, 2012 1:38 am

piyatiwari wrote:

Given statements:

(1) Avg speed = 4 MPH

=> total distance / total speed = 2D / (3+NB) = 4

Hold on. (3+NB) doesn't seem ok. You can't add speeds. You can add distances and times, but not speeds.

Bunuel · **Posted:** Sat Mar 10, 2012 1:54 am

budablasta wrote:

piyatiwari wrote:

Given statements:

(1) Avg speed = 4 MPH

=> total distance / total speed = 2D / (3+NB) = 4

Hold on. (3+NB) doesn't seem ok. You can't add speeds. You can add distances and times, but not speeds.

You can add or subtract rates to get relative rate.

For example if two cars are moving toward each other from A to B (AB=100 miles) with 10mph and 15mph respectively, then their relative (combined) rate is 10+15=25mph, and they'll meet in time=distance/rate=100/25=4 hours;

Similarly if car x is 100 miles ahead of car y and they are moving in the same direction with 10mph and 15mph respectively then their relative rate is 15-10=5mph, and y will catch up x in 100/5=20 hours.

Hope it's clear.

budablasta · **Posted:** Sat Mar 10, 2012 2:18 am

Oh yes, of course. Relative velocity problems involve adding and subtracting speeds. Thank you!

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