PS 4 # 20

Let original distance travelled be d and original time = t and original speed =s.

We know st=d

We also know that

(t+1)(s+5)=d+70
=> st+s+5t+5=d+70
Since st=d

s+5t=65

Now the question is asking for x in

(t+2)(s+10)=d+x
=> st+2s+10t+20=d+x
Since st=d

2s+10t+20=x
=> 2(s+5t)+20=x
=> 2(65)+20=x

x=150

Answer D

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

Suppose originally the motorist drove for @ "x" miles/hr in "y" hrs
Distance = xy (S*T)
Now speed = x+5 and time taken = y+1
So Distance = (x+5)(y+1)
(x+5)(y+1) - xy = 70
xy+x+5y+5 - xy = 70
x+5y = 65 ---- 1)
When speed is increased by 10 miles and hrs driven by 2...then distance covered = (x+10)(y+2) = xy+2x+10y+20 = xy + 2(x+5y) + 20
= xy + 2(65) + 20
= xy + 130 + 20
= xy + 150
Thus 150 more miles than one originally covered

Thanks so much