Speed time Distance

[somu]

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

Please provide explaination

SPOILER: answer B

[Dynamo]

IMO D .... I guess i am going wrong somewhere?

let the original speed be s and time be t and distance be d

st = x

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

5t+st+5+s-st=70
5t+s = 65

(t+2)(10+s) = 10t+st+2s+20-st=2(5t +s)+20

2*65 +20 = 150 miles

[singhs04]

Hello,

I'm getting 150 as answer. Could you please re-confirm the OA?

[singhs04]

I happend to see Dyanmo's post after mine. Here is what I did:

Extra distance is due to extra speed in the alloted time + distance covered in extra time. Lets assume time and speed were to be t and s. So:

70 = 5*t + 1* (s+5). This equation is for extra distance covered in t hours due to doing 5mph extra and then 1 extra hour of travel at s+5 mph. This will solve to 5t + s = 65

Now for the second case, the extra distance is:

x = 10t + 2 (s+10)
x = 10t + 2s + 20
x = 2( 5t + s) + 20
x = 2* 65 + 20 (65 from equation above)
x = 150

[somu]

Actually I believe the official answer may be wrong .I too got the same answer(D) and then thought I must be doing something wrong .Thanks guys ..that was really fast !

[somu]

One more reason why I thought the answer has to be wrong is that if a person driving at 5 miles/hour and for 1 hour longer gains a distance of 70 miles ,then at 10 miles/hour and 1 hour longer he will atleast gain 140 miles .So answer had to be between D and E .But then I thought I may be going wrong somewhere ..so posted it on the forum ...

[Praso]

agreew with D