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If a motorist had driven 1 hour longer on a certain day and

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Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 01 Jul 2012, 04:50
1)
(V+5)(T+1)-VT=70
(V+10)(T+2)-VT=X
============
2)
5T+V=65
10T+2V=X-20
2*65=X-20 =>X=150
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Re: Problem related to time and distance  [#permalink]

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New post 03 Jun 2013, 09:03
Bunuel wrote:
padmaranganathan wrote:
20. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160


Let \(t\) be the actual time and \(r\) be the actual rate.

"If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did" --> \((t+1)(r+5)-70=tr\) --> \(tr+5t+r+5-70=tr\) --> \(5t+r=65\);

"How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?" --> \((t+2)(r+10)-x=tr\) --> \(tr+10t+2r+20-x=tr\) --> \(2(5t+r)+20=x\) --> as from above \(5t+r=65\), then \(2(5t+r)+20=2*65+20=150=x\) --> so \(x=150\).

Answer: D.

OR another way:

70 miles of surplus in distance is composed of driving at 5 miles per hour faster for \(t\) hours plus driving for \(r+5\) miles per hour for additional 1 hour --> \(70=5t+(r+5)*1\) --> \(5t+r=65\);

With the same logic, surplus in distance generated by driving at 10 miles per hour faster for 2 hours longer will be composed of driving at 10 miles per hour faster for \(t\) hours plus driving for \(r+10\) miles per hour for additional 2 hour --> \(surplus=x=10t+(r+10)*2\) --> \(x=2(5t+r)+20\) --> as from above \(5t+r=65\), then \(x=2(5t+r)+20=150\).

Answer: D.

Note that the solutions proposed by dushver and dimitri92 are not correct (though correct answer was obtained). For this question we can not calculate neither \(t\) not \(r\) of the motorist.

Hope it helps.



Exactly, this is what I wanted to ask. Thanks again.
Their solution means that they were traveling at speed S for t hours and only in the last hour they traveled by S+5 speed, which is wrong.
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Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 10 Aug 2013, 12:15
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?


(A) 100
(B) 120
(C) 140
(D) 150
(E) 160
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Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 10 Aug 2013, 12:32
1
WholeLottaLove wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?


(A) 100
(B) 120
(C) 140
(D) 150
(E) 160


the error is in highlited part:
that should be:
\(X = (r+10) * (t+2)\) (d is the actual distance)....X==>distance coveres when travelling with 10 miles more average speed and 2 hours more
now you have to calculate X-d
hope its clear
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Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 10 Aug 2013, 13:54
el1981 wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A) 100
(B) 120
(C) 140
(D) 150
(E) 160

....
used 3times s=vt and input 65=v+5t ,then got 150miles
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Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 13 Aug 2013, 11:33
blueseas wrote:
WholeLottaLove wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?


(A) 100
(B) 120
(C) 140
(D) 150
(E) 160


the error is in highlited part:
that should be:
\(X = (r+10) * (t+2)\) (d is the actual distance)....X==>distance coveres when travelling with 10 miles more average speed and 2 hours more
now you have to calculate X-d
hope its clear


Thanks!

"How many more miles would he have covered than he actually did" doesn't refer to his actual distance but hypothetical miles. For example, if his actual distance (d) equaled 200 miles then his hypothetical distance (x) may = 50 miles which means his hypothetical total would be 200+50 = 250 miles.

We have two equations:
d+70 = (r+5) * (t+1)
d+70 = rt+r+5t+5
d+65 = rt+r+5t
d = rt-r-5t-65

x = (r+10)*(t+2)
x = rt+2r+10t+20
x - 2r - 10t - 20 = rt

Because we are looking for the difference between the hypothetical distance minus the actual distance we have to find x-d.
x-d
(rt+2r+10t+20) - (rt-r-5t-65)
(rt+2r+10t+20) - rt+r+5t+65
3r +15t + 85

But from here...I am stuck!
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Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 16 Oct 2013, 19:08
orignal speed = s ;

if you increase s by 5 m/hr, distance travelled is 70 miles in that extra 1 hour, therefore :

==> (s+5)*1=70 ; s=65

Now, in 2 more hours with an increased 10m/hr over original speed, distance traveled :

d=(65+10)*2 ; d=150. Ans (D)
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If a motorist had driven 1 hour longer on a certain day and at a  [#permalink]

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New post 24 Jan 2015, 15:18
I see thias as that, think is easier:
1) traveled 70 km in one additional hour, at an extra 5 m/h
2) so, in that un hour at that extra speed, advance 70 miles. s2 * t2 = d2 ------ s2 =(s1 + 5)-----s1=x ----t2=1;
Therefore: (x+5) * 1 = 70 ---- x =65
3) now 65 plus 10 m/h, new speed is 75 m/h; then, in 2 additional hours, he adnvaced 2 *75 = 150 miles.

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Re: If a motorist had driven 1 hour longer on a certain day and at an aver  [#permalink]

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New post 24 Jan 2015, 18:58
Hi All,

This question can be solved by TESTing VALUES (as some of the explanations have noted). You can keep the values really small though and save some time on the calculations:

We're told that driving 1 extra hour AND at a speed that was 5 miles/hour faster (for the entire trip) would have increased TOTAL distance by 70 miles.

IF.....
We originally drove for 1 hour at 60 miles/hour, then we would have traveled 60 miles.

Adding 1 extra hour and increasing speed by 5 miles/hour would give us.....
2 hours at 65 miles/hour, which gives us a total distance of 130 miles (which is 70 miles MORE than originally traveled).

So now we we've established the starting time, speed and distance, so we can answer the given question:

How many MORE miles would be traveled if the original time was increased by 2 hours AND the original speed was increased by 10 miles/hour?

1+2 = 3 hours
60 + 10 = 70 miles/hour
3 hours at 70 miles/hour = 210 miles

Since we originally traveled 60 miles, the 210 total miles is 150 miles MORE than originally traveled.

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If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 18 Feb 2016, 16:24
let t=original time
r=original rate
m=additional miles
equation 1: 5t+r=65
equation 2: 10t+2r=m-20
multiply equation 1 by 2 and subtract equation 2
m=150 miles
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Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 04 Mar 2017, 12:43
Since he traveled an extra 70 miles at 5mph more plus one hour of travel, at 10mph more he would have covered double that amount (140 miles) in the same time, plus one more extra hour of travel (10 miles), for a total of 150 miles. Solved in seconds without any computations... is this logic correct?
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Re: If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 02 Sep 2017, 13:09
Let's assume original distance as x, speed as s and time as t

distance= speed *time
therefore,
x= st -----(1)
According to first scenario,
x+70= (s+5) * (t+1)
x+70= st+s+5t+5
x+70= x+s+5t+5 ---from 1

s+5t=65 -----(2)

Now consider the second scenario,
It is given that the speed would be 10 Miles/hr faster and total time would be 2 hrs longer
therefore ,
(s+10) * (t+2) will give us the total distance.
distance= st+2s+10t+20
distance= x+2(s+5t)+20 ---- from (1)
distance= x+2(65)+20 ------from (2)
distance= x+150.

hence in given scenario, extra 150 miles will be covered. Hence Answer is D

Kudos if it helps.
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If a motorist had driven 1 hour longer on a certain day and  [#permalink]

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New post 23 Sep 2018, 12:35
Let’s start with the basic D=RT equation. From there, the first theoretical trip can be represented as D + 70 = (R + 5)(T + 1), which expands to D + 70 = RT + R + 5T +5 and can be simplified to 65 = R + 5T.

The second theoretical trip can be represented as (R+10)(T+2), which expands to RT + 2R + 10T + 20 (note that you only have an expression since you don’t know what the distance is). The two middle terms (2R + 10T) can be factored to 2(R+5T), which allows us to use the second equation. RT + 2(R+5T) + 20 = RT + 2(65) + 20 = RT + 150. Since the original distance was RT, the additional distance is 150, or choice D.

Additionally, this problem can be solved with number picking. Imagine that the original speed was 50mph - then you know that the one extra hour would have resulted in 55 extra miles so there must have been 3hrs at the additional 5mph to create the extra 15 miles to get to 70. Now apply those same numbers in the second case: with an extra two hours you get 2(60mph) or 120 extra miles and then the 3hrs at the additional 10mph to get 150. This will yield 150 no matter what numbers you try for the original case!
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