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

(t+1)(s+5) = d+70

expand to get ts+5t+s+5 = d+70, and recognize that ts = d (from standard equation speed = distance/time)

simplify down to 5t+s=65

now, (t+2)(s+10) = ts+10t+2s+20 = ts+2(5t+s)+20 = d+150

so your answer is 150

Interesting problem.

first we have (r+5)(t+1)=d+70 ---> rt+r+5t+5=d+70. Also notice that rt=d

thus we now have r+5t=65.

Second equation. (r+10)(t+2)=d+x

rt+2r+10t+20=d+x --> plug in rt and 2(r+5t)=2(65) ---->130 ---> rt+150=rt+x ---> x=150.

D

my approach was similar too than the posts above ... but i'll share it anyway

avg rate of 5mph faster means that the motorist drove 5miles more in each hour.
since he drove 5miles in the last (extra) hour too, so he drove 70-5=65 miles more in each of the hour earlier
65/5=13hrs
this is the actual time he drove

avg rate of 10mph more means that he drove 10*13=130miles more in the first 13 hrs
he covered 2*10=20miles more in the last 2 hours
130+20=150 more miles covered than he actually did

so D it is

The motorist covers 70 miles in an hour when he drives 5 miles/hr faster
=> He is currently driving at 65 miles/hr

If he drives 10 miles/hr faster for 2 more hours he would cover 75*2 = 150 miles

(D) it is.

despite the correct answer, the logic here is flawed, isn't it ?


M drives less than 70miles in one hour. Because M drove more than the original distance in the time before the extra hour starts. So that delta + another hour's worth of driving (at the elevated speed) is 70miles.

Yes, 70 miles is the sum of extra distance covered by increasing speed + extra distance covered in an hour.

But remember, this is a PS question. You do have the correct answer in the options and there is only one correct option.
I will just assume any case and whatever I get will be my answer.

Say the original speed was 60 miles/hr. The speed increased to 65.
So the motorist would have traveled 5 miles + 65 miles (= 70) extra.
(He travels for only one hour initially.)
If he instead increases speed by 10 miles/hr, his speed becomes 70. In the initial 1 hr, he will travel an extra 10 miles and then in additional 2 hrs he will travel an extra 70*2 = 140 miles.
So he will travel an extra 150 miles.

I could have just as well assumed the original speed to be 55 miles/hr. If the speed increases to 60 and extra distance covered is 70, it means that the motorist travels for 2 hrs initially.
70 = 5 + 5 + 60
If instead, the speed increases by 10 miles/hr, the speed becomes 65.
Extra distance covered = 10+10+2*65 = 150

In any case, the answer has to be the same since it is a PS question.

I could also have assumed the original speed to be 65 miles/hr (as done above) If the speed increases to 70 and I cover 70 miles extra, it means I didn't travel at all before.
If the speed instead becomes 75, I travel 2*75 = 150 miles extra.

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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

\(distance = rate * time \)

\(d = (r+5)(t+1)\)

\(d + 70 = rt + r + 5t + 5\)

\(70 = r + 5t + 5\)

\(65 = r + 5t\)

\(d + x = (r+10)(t+2)\)

\(d + x = rt + 2r + 10t + 20\)

\(d + x = rt + 130 + 20\)

\(d + x = rt + 150\)

\(x = 150\)

Answer is D.

Plug in a value for the RATE and solve for the TIME.

Let the actual rate = 10 mph and the actual time = t hours
Actual Distance = (actual rate)(actual time) = 10t

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
Since the rate here is 5 mph greater than the actual rate of 10 mph and the time 1 hour longer than the actual time of t hours, we get:
Distance 2 = (rate 2)(time 2) = (10+5)(t+1) = 15(t+1) = 15t+15

Since Distance 2 is 70 miles greater than the Actual Distance, we get:
Distance 2 - Actual Distance = 70
(15t+15) - 10t = 70
5t = 55
t = 11

In this case:
Actual Distance = (actual rate)(actual time) = 10*11 = 110 miles
Distance 2 = (rate 2)(time 2) = 15*12 = 180 miles --> 70 miles longer than the actual 110-mile distance

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?
Since the rate here is 10 mph greater than the actual rate of 10 mph and the time 2 hours longer than the actual time of 11 hours, we get:
Distance 3 = (rate 3)(time 3) = (10+10)(11+2) = 20*13 = 260 miles ----> 150 miles longer than the actual 110-mile distance



Note:
The actual rate, time and distance can be different values.

Let the actual rate = 5 mph and the actual time = t hours
Actual Distance = (actual rate)(actual time) = 5t

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
Since the rate here is 5 mph greater than the actual rate of 5 mph and the time 1 hour longer than the actual time of t hours, we get:
Distance 2 = (rate 2)(time 2) = (5+5)(t+1) = 10(t+1) = 10t+10

Since Distance 2 is 70 miles greater than the Actual Distance, we get:
Distance 2 - Actual Distance = 70
(10t+10) - 5t = 70
5t = 60
t = 12

In this case:
Actual Distance = (actual rate)(actual time) = 5*12 = 60 miles
Distance 2 = (rate 2)(time 2) = 10*13 = 130 miles --> 70 miles longer than the actual 60-mile distance

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?
Since the rate here is 10 mph greater than the actual rate of 5 mph and the time 2 hours longer than the actual time of 12 hours, we get:
Distance 3 = (rate 3)(time 3) = (5+10)(12+2) = 15*14 = 210 miles --> 150 miles longer than the actual 60-mile distance

Another case:

Let the actual rate = 30 mph and the actual time = t hours
Actual Distance = (actual rate)(actual time) = 30t

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
Since the rate here is 5 mph greater than the actual rate of 30 mph and the time 1 hour longer than the actual time of t hours, we get:
Distance 2 = (rate 2)(time 2) = (30+5)(t+1) = 35(t+1) = 35t+35

Since Distance 2 is 70 miles greater than the Actual Distance, we get:
Distance 2 - Actual Distance = 70
(35t+35) - 30t = 70
5t = 35
t = 7

In this case:
Actual Distance = (actual rate)(actual time) = 30*7 = 210 miles
Distance 2 = (rate 2)(time 2) = 35*8 = 280 miles --> 70 miles longer than the actual 210-mile distance

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?
Since the rate here is 10 mph greater than the actual rate of 30 mph and the time 2 hours longer than the actual time of 7 hours, we get:
Distance 3 = (rate 3)(time 3) = (30+10)(7+2) = 40*9 = 360 miles --> 150 miles longer than the actual 210-mile distance

As the cases above illustrate:
The actual rate, time and distance cannot be determined.
In the first case, actual rate = 10 mph, actual time = 11 hours, actual distance = 110 miles.
In the second case, actual rate = 5 mph, actual time = 12 hours, actual distance = 60 miles.
In the third case, actual rate = 30 mph, actual time = 7 hours, actual distance = 210 miles.

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I received a PM with the following query:



As illustrated by the two cases in my earlier post, the answer is NO: the actual time and speed can be different values.

In this question, we're comparing a HYPOTHETICAL trip to an ACTUAL trip.
We're told that the hypothetical trip would have been 70 miles longer than the actual trip.

So, we can start with the following word equation: (distance of hypothetical trip) - (distance of actual trip) = 70

Let r = the speed during the ACTUAL trip
Let t = the total time of the ACTUAL trip

So, r + 5 = the speed during the HYPOTHETICAL trip
And t + 1 = the total time of the HYPOTHETICAL trip

Since distance = (rate)(time), we can substitute our values into the original word equation: ((r + 5)(t + 1)) - (rt) = 70
Expand to get: rt + r + 5t + 5 - rt = 70
Simplify: r + 5t + 5 = 70
Subtract 5 from both sides: r + 5t = 65 [we'll use this information shortly]

The question: 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?
So, we want to determine the following: (distance travelled during this new hypothetical trip) - (distance travelled during the actual trip)

In this case, r + 10 = the rate, and t + 2 = the travel time

Plug these values into the above expression to get: (r + 10)(t + 2) - (r)(t)
Expand: (rt + 2r + 10t + 20 - rt)
Simplify: 2r + 10t + 20
Rewrite as follows: 2(r + 5t) + 20

Since we already learned that r + 5t = 65, we can plug this into the above expression to get: 2(65) + 20, which simplifies to 150

Answer: D

Can you help point to error in my logic or what am i doing wrong?
Lets say you are driving 5mph faster for 1 hour longer.. u drive 70miles more.. now if you drive for 2 hrs...you would have driven 140 miles more. So you travel 140 miles more driving at 5mph more for 2 hrs. Now if the speed were 10mph more instead of 5mph more... wouldnt the difference from actual be double of 140 ie 280 miles? I know I am wrong but I need someone to help point the error in reasoning.

Bunuel KarishmaB

You cannot simply double 70. Think about it:
Say his original speed is s and he travels for t hrs. So he covers s miles every hour.

So total distance covered originally is st.

If he travels for another hour and at speed 5 extra, he covers extra distance of (s + 5 + 5t). This (s + 5 + 5t) = 70

Instead if he travels for 2 extra hours and at speed 10 extra, he covers extra distance of (s + 10 + s + 10 + 10t).
This is equal to 2s + 10 + 10t + 10 = 2(s + 5 + 5t) + 10 = 2*70 + 10 = 150 miles extra

Hey, your reasoning is flawed because you have assumed an increased speed of 5 miles/hr for the last extra hour only.

If D=R*T
Then Original R*T= (Current R * Current D= Current D) - 70 (The difference between the current D and original D)

R*T = (R+5)(T+1)-70
RT=RT + 5T + R +5 -70
5T + R = 65 (1)

Original R*T = (Forecasted R * Forecasted T) - The difference of distances X (the required value in the problem)

RT = (R+10)(T+2) - X
RT = RT + 10 T + 2R + 20 - X
X = 10 T + 2R + 20 (2)

From 1 & 2
X = 150

I got to the correct answer with a weird approach.

Since:
D+70 = R+5 * T+1
D+X= R+10 * T+2

I analogued the 2 equations with real numbers to spot a pattern
6*7 = 42
11*8 = 88
88 is 2,smth 42
so I multiplied the 70 with 2,smth and landed on D

Well, it worked!