average speed

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average speed

by Ankitaverma » Wed Nov 20, 2013 12:18 pm
A hiker walked for two days. On the second day the hiker walked 2 hours longer and at an average speed 1 mile per hour faster than he walked on the first day. If during the two days he walked a total of 64 miles and spent a total of 18 hours walking, what was his average speed on the first day?
(A) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph[/b]
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by Mike@Magoosh » Wed Nov 20, 2013 1:06 pm
Ankitaverma wrote:A hiker walked for two days. On the second day the hiker walked 2 hours longer and at an average speed 1 mile per hour faster than he walked on the first day. If during the two days he walked a total of 64 miles and spent a total of 18 hours walking, what was his average speed on the first day?
(A) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph[/b]
Dear Ankitaverma
I'm happy to help. :-) Here's a blog on distance, rate, and time:
https://magoosh.com/gmat/2012/gmat-math-rate-questions/

For this question --- Day #2, hiker walks for 2 hours more than the first day, for a total of 18 hours of walking. That means 8 hr of walking the first day, and 10 hours of walking the second day. I think backsolving is much easier and more efficient than an algebraic approach
https://magoosh.com/gmat/2012/gmat-plugg ... -choice-c/
Starting with (C)
If he walks 4 mph for 8 hours on Day 1 = 32 mi
Then 5 mph for 10 hours on Day 2 = 50 mi
Total = 32 + 50 = 82
Too much distance. Choices (C) & (D) & (E) are incorrect.

Try (B).
Day 1: 3 mph for 8 hours = 24 mi
Day 2: 4 mph for 10 hours = 40 mi
Total = 64 mi.
That's it! Answer = (B).

Does all this make sense?
Mike :-)
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https://gmat.magoosh.com/

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by [email protected] » Thu Nov 21, 2013 12:01 am
Hi Ankitaverma,

These types of questions are usually easier to handle if you create a table to organize your information, although you're going to have to do some algebra no matter how you organize.

We're told about 2 days (Day 2 was 2 hours longer and at a speed that was 1 mph more than Day 1). The information implies that we'll use the Distance Formula:

Distance = Rate x Time

Day 1: D = R x T
Day 2: D = (R+1) x (T+2)

Then we're told that the TOTAL Distance = 64 miles and the TOTAL Time was 18 hours.

Since the 2 "times" add up to 18 hours, we know:

T + (T+2) = 18
2T = 16
T = 8
This gives us:

Day 1: D = R x 8
Day 2: D = (R+1) x 10

We know the total distance is sum of the two "distances":

64 miles = 8R + 10(R+1)
64 = 8R + 10R + 10
54 = 18R
3 = R

The question asks us for the rate on Day 1 (so, what is R?)

Final Answer: [spoiler]B: 3[/spoiler]

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by GMATGuruNY » Thu Nov 21, 2013 4:17 am
Ankitaverma wrote:A hiker walked for two days. On the second day the hiker walked 2 hours longer and at an average speed 1 mile per hour faster than he walked on the first day. If during the two days he walked a total of 64 miles and spent a total of 18 hours walking, what was his average speed on the first day?
(A) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph[/b]
The answer choices represent the slower speed.
The average speed for the entire trip = 64/18 ≈ 3.5 miles per hour.
Thus, the slower speed must be LESS than 3.5 miles per hour, while the faster speed must be GREATER than 3.5 miles per hour.
The difference between the two speeds is 1 mile per hour.
Only answer choice B -- which implies a slower speed of 3 miles per hour and a faster speed of 4 miles per hour -- is viable.

The correct answer is B.
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by Mathsbuddy » Thu Nov 21, 2013 6:40 am
If v = average speed on day 1
and t = total time on day 1
and using distance = V * T:

Total distance = 64 = [d1] + [d2]
64 = [vt] + [(v+1)(t+2)] EQUATION A

Also,total time = 18 = [t] + [t+2] = 2t + 2
So t =8

Substitute t=8 into equation A:
64 = 8v + (v+1)(8+2)
54 = 18v
v = 3