Graham walked to school at an average speed of 3 miles an hour and jogged back along the same route at 5 miles an hour. If his total traveling time was 1 hour, what was the total number of miles in the round trip?
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Why not just take the distance to be x (in miles), so the total time is 3 x + 5 x hours and it is 1
This gives 1 5 8 x = 1 ⟹ x = 8 1 5
and total distance travelled is twice of this ,giving 4 1 5
Your solution looks bit bigger !
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Yeah you're right, but how would you explain the expression
3 x + 5 x = 1 ?
Why is that the travel time and how does it relate to the distance?
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Distance is x , speeds for going and coming are 3 and 5 respectively, giving t i m e for each part being 3 x and 5 x respectively, and addition will be 1 hour.
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@Aditya Raut – Just add that 3 x and 5 x come from the fact that t i m e = s p e e d d i s t a n c e , and that should do it. The fractions look alien to a non-math person (that makes most of the SAT takers), and you need to be more clear on where you derive your expressions from.
Cheers
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@John M. – Yep, right! Hey, btw, would you like to try my set mistakes give rise to problems , Crazy Fibonacci , Vegetable Combinatorics and Bashing Unavailable .
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@Aditya Raut – Hey, those are some wonderful sets you've got there. But right now my school started and I'm not doing any math besides AP Statistics and Multivariable Calculus. The only time I'll use Brilliant will be for my weekly Conceptual Physics problems and occasional randoms, but I don't think I'll be able to do much else.
But I think I'll give those a look over break. Thanks!
total time = t1 + t2 = s/3 + s/5 s= 3.75
let d be the distance of the whole route
d/2 be half the distance of the whole route
s=d/t
t=d/s
(d/2)/3 + (d/2)/5 = 1
d/6 + d/10 = 1
(5d + 3d)/30 = 1
8d/30 = 1
8d=30
d=3.75 or 3 3/4
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Short way:
Realize that t i m e = s p e e d d i s t a n c e
So,
5 m p h X m i l e s + 3 m p h X m i l e s = 1 h o u r
Thus, solving for the distance ( X ),
1 5 m p h 8 X m i l e s = 1 h (get the common denominator)
X = 8 1 5 m i l e s (multiply both sides by 8 m i l e s 1 5 m p h )
But X is the one-way distance. So for back and forth,
D i s t a n c e t o t a l = 2 X = 4 1 5 m i l e s = 3 4 3 m i l e s
Long way:
Three key ingredients for cooking this problem:
Speed, Distance, and Time.
From basic mechanics, we know that
s p e e d = t i m e d i s t a n c e , i.e. s = t d .
From the problem given, we know that:
Graham's total travel time was 1 hour. On his trip to school, he traveled at an average speed of 3mph.
On his way back, he traveled at an average speed of 5mph.
His total travel time was 1 hour.
Putting all of this together mathematically, we get the following:
⎩ ⎪ ⎨ ⎪ ⎧ t s c h o o l + t b a c k = 1 h r t s c h o o l = s s c h o o l d t b a c k = s b a c k d
Adding 2nd and 3rd equations together, we get
t s c h o o l + t b a c k = s s c h o o l d + s b a c k d
Combining the right side, we get
t s c h o o l + t b a c k = s s c h o o l s b a c k d ( s s c h o o l + s b a c k )
Now all we have to do is plug in the values we already know:
t s c h o o l + t b a c k = 1 h r , s s c h o o l = 3 m p h , s b a c k = 5 m p h .
And so,
t s c h o o l + t b a c k = s s c h o o l s b a c k d ( s s c h o o l + s b a c k )
⇒ 1 h r = 3 m p h 5 m p h d ( 3 m p h + 5 m p h )
⇒ 1 h r = 1 5 ( m p h ) 2 d ( 8 m p h )
Note that mph stands for "miles per hour," i.e. h o u r m i l e s = h r m i .
Thus,
1 h r = 1 5 h r 2 m i 2 d ( 8 h r m i )
⇒ 1 h r = 1 5 h r m i d ( 8 )
⇒ 1 h r ( 1 5 h r m i ) = 8 d
⇒ 1 5 m i = 8 d
⇒ d = 8 1 5 m i .
But we're not done yet. Remember from the original system we've set up, the d represented the distance it took Graham to travel from one place to another. Since this distance is the same to and fro, d is the same to school and back. Thus, d denotes HALF the distance.
Hence,
d = 2 1 d t o t a l = 8 1 5 m i ,
d t o t a l = 8 3 0 m i = 4 1 5 m i = 3 4 3 m i .
NOTE: Alternative solutions exist.