D-150

assuming t is the original travel time in hrs, and x is the original speed in miles per hour

from the first half - 5t+(x+5)=70 => 5t+x=65
Now need to find out 10t+2(x+10) = 10t+2x+20 = 2(5t+x)+20 = 150

70 additional miles came from 2 sources - 5 miles additional distance traveled for t hours (5t) + distance traveled in the additional hour. The later one is x+5 where in x is distance traveled in an hour with the original speed and 5 is the distance came from the additional speed (making total speed x+5 mph)

D1=100, V1=50, T1=2
T2=3, V2=55 -->D2= 165
T3=5, V3=65 --> D3=325
D3-D2=150
D
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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
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Hi sondenso,
My approach was very similar to yours except for a minor change in the 3rd step;

"...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?" can be rephrased as;

(t+2)(v+10) = ?
tv+10t+2v+20 = ?
10t+2v+20 = 2(5t+v)+20 = 2*65 (from 2) +20 = 150.

Let original speed be s, distance be d, and time t
d=st
Also, d+70=(s+5)(t+1)
d+70=st+s+5t+5
st+70=st+s+5t+5
s+5t=65

And, d1=(s+10)(t+2)
d1=st+2s+10t+20
so, d1-d=2s+10t+20
=2(s+5t)+20
=2(65)+20
=130+20=150

OA D.
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My dad once said to me: Son, nothing succeeds like success.

Lets take x as time and y as avg speed
If he has driven 1 hr longer and 5 km/hr faster the distance covered Is 70

(X+1)*(Y+5) – XY = 70
XY+Y+5+5X – XY = 70

Y+5X = 65
Scenario 2

(X+2)(Y+10)-XY = D
2x+10Y+20 = D
2(X+5Y)+20 = D
2*65 +20 = D
D = 150
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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.
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Solution :
s-speed.
t-time.
d-distance.

Two equations will be formed.
As per the conditions given we`ll get,

(s+5)(t+1) - st = 70.....(1)
on simplifying this,we get
s + 5t = 65....(a)

Another one will be,
(s+10)(t+2) - st = d
on simplifying this we`ll get,

2s + 10t +20 = d...(b)
substituting (a) in (b)

we get 65*2 + 20 = 150.

tm=(1+t)
dm=70+d
rm=(r+5)

Being r,t and d the actual rate, time and distance.

If you realize that the motorist has actually covered a distance of 70 miles in 1 hour, going at (r+5)speed, it is much easier...

70= (r+5)*1 then r=65mph

If the new rate is 10 mph more then it is r+10=75mph

Therefore in two hours the distance covered at this speed is 150miles