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A hiker walked for two days. On the second day the hiker wal

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Re: A hiker walked for two days [#permalink]
10 sec approach is great
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Re: A hiker walked for two days [#permalink]
yep, liked the 10 sec app as well...
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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
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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?

Hiker walked for two days for a total of 18 hours. He spent two more hours walking on the second day than the first: h1+h2=18 ===> h+h+2 = 18 ===> 2h = 16 ===> h=8.
He spent 8 hours the first day and 10 hours the second day.
Day 1 = 8h
Day 2 = 10h

Average speed = total distance/combined rate
Average speed = 64/r1+r2

rate = distance/time
r = d/8
r+1 = d/10 ===> r = (d/10)-1

64/(d/8) + (d/10)-1

Why wouldn't I use that formula? Is it because I would be trying to figure out average speed when I don't know the speed of the hiker on day one or two?

distance=rate*time
d1=r*8
d2=(r+1)*10

d1+d2=total distance
d1+d2=64
(r8)+(10r+10)=64
18r=64
r=3

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A hiker walked for two days. On the second day the hiker wal [#permalink]
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

1-st day: vt=d1(v:velocity, t:time taken, d1:distance)
2-nd day: (v+1)(t+2)=d2 and d1+d2=64

t+t+2=18, 2t=16, t=8 and 8v=d1 <-- equation 1), 10(v+1)=d2, 10v+10=d2 <-- equation 2)

equation 1)+ equation 2), then 8v+10v+10=d1+d2=64, 18v=54, v=3

Therefore the answer is B. thank you!!
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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
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Hi All,

This question requires a certain degree of note-taking and organization. Once you've got everything on the pad, you can then approach the math in a couple of ways:

Since we're dealing with distance, rate and time, I'm going to make two equations based on the two days of travel:

Day 1: Distance = (R mph)(T hours)
Day 2: Distance#2 = (R+1 mph)(T+2 hours)

Since the TOTAL TIME for both days was 18 hours, we can solve for T...

T + (T+2) = 18
2T = 16
T = 8

Since the TOTAL DISTANCE for both days was 64 miles, we can now set up 1 gigantic equation....

64 miles = (R)(8) + (R+1)(10)

At this point, we have 1 variable and 1 equation, so we can solve for R....

64 = 8R + 10R + 10
54 = 18R
3 = R

The prompt asks for the speed on DAY 1; since the hiker traveled R mph on Day 1, the answer is 3...

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A hiker walked for two days. On the second day the hiker wal [#permalink]
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ajit257 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

If we let x = the hiker's walking speed on day 1, then x+1 = the hiker's walking speed on day 2 (since the hiker walked 1 mph faster on the second day)
So, the hiker's AVERAGE speed will be BETWEEN x and x+1 mph

The hiker walked 64 miles and 18 hours.
Average speed = distance/time = 64/18 ≈ 3.5 mph

So, according to the above property, it must be the case that x = 3 mph, and x+1 = 4 mph

Cheers,
Brent
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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
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ajit257 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

We can let t = the number of hours and r = the speed he walked on the first day. We can create the equations:

t + t + 2 = 18

and

tr + (t +2)(r + 1) = 64

Solving the first equation, we have:

2t = 16

t = 8

Now, substituting t = 8 into the second equation, we have:

8r + 10(r + 1) = 64

8r + 10r + 10 = 64

18r = 54

r = 3

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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
Rate Time Distance
Day 1: R T RT
Day 2: R+1 T+2 (R+1)*(T+2)
Total 18 64

2T +2 =18
2T=16
T=8

sub in

Rate Time Distance
Day 1: R 8 8R
Day 2: R+1 10 (R+1)*(10)
Total 18 64

8R+ 10R+10 = 64
18R = 54
R=3
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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
Top Contributor
ajit257 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

On the second day the hiker walked 2 hours longer than he walked on the first day. During the two days he spent a total of 18 hours walking
Let h = # of hours walked on first day
So, h + 2 = # of hours walked on second day
We can write: h + (h + 2) = 18
Simplify: 2h + 2 = 18
Solve: h = 8
So, the hiker traveled for 8 hours on the first day and for 10 hours on the second day

On the second day the hiker walked at an average speed 1 mile per hour faster than he walked on the first day. During the two days he walked a total of 64 miles.
Let x = the hiker's speed (in kilometres per hour) on the first day
So, x + 1 = the hiker's speed (in kilometres per hour) on the second day

distance = (time)(speed)
Let's start with the following word equation: (distance traveled on the first day) + (distance traveled on the second day) = 64
Substitute to get: (8)(x) + (10)(x + 1) = 64
Simplify: 8x + 10x + 10 = 64
Simplify: 18x + 10 = 64
Subtract 10 from both sides: 18x = 54
Divide both sides by 18 to get: x = 3

So, the hiker's speed on day one was 3 kilometres per hour

Cheers,
Brent
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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
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ajit257 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

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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
Given: 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.
Asked: 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?

Let the time spent walking by the hiker on first day be x hours
The time spent walking by the hiker on second day = (x+2) hours
x + (x+2) = 18
x = 8

Let the average speed of the hiker be s mph
8*s + 10*(s+1) = 64
8s + 10s + 10 = 64
18s = 54
s = 3 mph

IMO B
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A hiker walked for two days. On the second day the hiker wal [#permalink]
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ajit257 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

He spent a total of 18 hours walking.
On the second day, the hiker walked 2 hours longer than he walked on the first day.

18 hours split evenly between two days = 9 hours on the first day, 9 hours on the second day.
Since a difference of 2 hours is needed between the two days, the first day must DECREASE by 1 hour, while the second day INCREASES by 1 hour, yielding a 2-hour difference between the two days:
8 hours on the first day, 10 hours on the second day

On the second day the hiker walked at an average speed 1 mile per hour faster than he walked on the first day.
He walked a total of 64 miles.
What was his average speed on the first day?
We can PLUG IN THE ANSWERS, which represent the speed on the first day.
When the correct answer is plugged in, the total distance = 64 miles.

B: 3 mph for 8 hours on the first day, implying 4 mph for 10 hours on the second day
Total distance = 3*8 + 4*10 = 64
Success!

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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
The 10 second approach is amazing!
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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
Let's assume that the hiker walked at an average speed of x miles per hour on the first day. Since the hiker walked for a total of 18 hours and the second day was 2 hours longer than the first day, we can set up the following equation:

First day: x mph * t1 hours = x * t1 miles
Second day: (x + 1) mph * (t1 + 2) hours = (x + 1) * (t1 + 2) miles

We are given that the total distance covered during the two days is 64 miles, so we can write:

x * t1 + (x + 1) * (t1 + 2) = 64

Simplifying this equation, we get:

xt1 + xt1 + 2x + t1 + 2 = 64
2xt1 + 2x + t1 + 2 = 64
2xt1 + t1 + 2x = 62

Now, we know that the hiker walked for a total of 18 hours, so we can write:

t1 + (t1 + 2) = 18
2t1 + 2 = 18
2t1 = 16
t1 = 8

Substituting the value of t1 back into the previous equation:

2x(8) + 8 + 2x = 62
16x + 8 + 2x = 62
18x + 8 = 62
18x = 54
x = 3

Therefore, the average speed of the hiker on the first day was 3 miles per hour.
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Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
You could literally use the answer choices to your advantage. The only two times that would work for day 1 and day 2 is 8 hours and 10 hours. Then you could test each answer choice to see which two speeds would get you a total of 64 miles across the two days.

Day 1 = 3*8
Day 2 = 4*10

24+40=64, therefore B is the answer. Keep it simple
Re: A hiker walked for two days. On the second day the hiker wal [#permalink]
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