Problem Solving

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 that 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
d) 160

Motorist drives 1 hr longer and 5mph faster

(x+5)(t+1) = xt+70
xt+x+5t+5 = xt+70
x+5t = 65------[1]


If he drives 2 hrs longer and 10 mph faster. lets say he drove m miles more

(x+10)(t+2) = xt + m
xt+2x+10t+20 = xt+m
2x+10t+20 = m
m = 2(x+5t)+20
m = 130+20 = 150

Option D

Option D - 150

We can directly deduce from the question that in one hour extra the motorist completes 70 miles extra.

Therefore, he must have completed 70 miles in one hour.

This makes his increased Speed = 70 miles/ hour

Original Speed = 70- 5 = 65 miles / hour

Now, if we increase the speed by 10 miles / hr

New Speed = 75 miles per/ hour
Time = 2 hrs

Hence, Distance = Speed x Time = 75 x 2 = 150 miles.

As the question ask for the additional distance he would have travelled the answer will be 150.

gmatblood 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 that 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
d) 160

Not going into calculations.
had he gone 5 miles/hour faster for 2 hours he would have done 140 miles more right?
now for actually estimating the distance for 10 miles/hour faster for 2 hours, we just add the extra miles i.e (10-5) miles/ hour for 2 hours, which is 10 miles.

Adding them up =140+10=150 miles

IMO D

Let x be the # of hours he travels everyday then 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

=> 5(x+1) =70 => x =13

if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

=> 10(x+2) = ? 10*15 =150 !