Tom and Linda stand at point A. Linda begins to walk in a straig

Question

Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover half of the distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

a)60
b)72
c)84
d)90
e)108

A. 60
B. 72
C. 84
D. 90
E. 108

gurpreetsingh · Accepted Answer

IMO E - 108

Case 1:  6*t1 = \frac{1}{2}* (t1+1) * 2 => t1 = \frac{1}{5}  hour = 12 minutes

Case 2:  6*t2 = 2* (t2+1) * 2 => t1 = 2 = 120 minutes

Difference = 120-12 = 108 minutes

amitjash · Answer

E is the answer....
D = TS where D=distance, T=Time and S=Speed
To travel half distance, (2+2T) = 6T ==> T = 1/5 ==> 12 minutes
To travel double distance, 2(2+2T) = 6T ==> 2 ==> 120 minutes
Difference, 108 minutes

mbah191 · Answer

Corrected....

D=RT
Linda's Distance: D = 2 + 2t, because she has a one hour advantage (of 2 miles) and moves at 2mph (or 2t).
Tom's Distance: D = 6t

To find the amount of time needed for Tom to reach half the distance Linda reached, we divide Linda's RT in half and set this equal to Tom's RT ----> (2+2t)/2 = 6t ----> t=1/5 ----> then multiply by 60 minutes and we get 12 minutes.

To find amount of time needed for Tom to reach double the distance Linda reached, we multiply Linda's RT by 2 and set this equal to Tom's RT ----> 2(2+2t) = 6t ----> t=2 ----> then multiply by 60 minutes and we get 120 minutes.

The difference between these two is 108 minutes. The answer is E.

CrackverbalGMAT · Answer

Case 1:   Time taken by Tom to cover half the distance that Linda has covered

Speed of Linda= 2 miles/hour.
Speed of Tom = 6 miles/hour

Assume that Tom travels  T  hours.
That means Linda has traveled for  T + 1  hours as Tom started to jog 1 hour after Linda.

Distance covered by Tom in T hours =  6* T 
Distance covered by Linda in T+ 1 hours =  2*(T + 1)

Since Tom covered half the distance covered by Linda,
 6* T = \frac{1}{2} *2*(T + 1)   
 6*T = T+1 
 T= 1/5 hours = 12 mins

Case 2:   Time taken by Tom to cover twice the distance that Linda has covered.

Assume that in  T1  hours, Tom covers twice the distance that Linda covered

Distance covered by Tom in T1 hours =  6* T1 
Distance covered by Linda in T1+1 hours =  2*(T1 + 1)

Since Tom covered twice the distance covered by Linda,
 6* T1 = 2*(2*(T1 + 1)) 
 6* T1 = 4* T1 + 4 
 2* T1 = 4 
 T1= 2 hrs= 120 mins.

That means Tom covered 2* 6 = 12 miles in 2 hrs that is double what Linda covered in 3 hours = 2* 3 = 6 miles.

Difference in time taken by Tom in both cases = 120 - 12 = 108 mins
 Option E is the right answer.

Thanks,
  Clifin J Francis,
GMAT SME

A2D2 · Answer

If instead of taking (t+1), we take Linda to be T and Tom to be T-1 why the ans is coming different….
Sorry if the question sounds dumb..

6(t-1)=2*t*1/2 ???

adewale223 · Answer

First case

Let x be distance Linda covers

x/2 = 2+x/2 X 1/6

x/2 = 2+x/12

12x = 4 + 2x

10x = 4

x = 0.4

Distance covered by Linda

2.4 miles

Distance covered by Tom

1.2 miles

Time it will take Tom

T = 1.2/6

0.2h to minutes = 12

Second case

Tom covers twice Linda's distance

Let distance covered by Linda in this case be y

y/2 = 2(2+y)/6

y/2 = 4+2y/6

6y = 8 + 4y

2y = 8

y = 4

Distance covered by Linda= 6 miles. Tom will then cover 12 miles.

Time Tom will take to cover 12 miles

T = 12/6
= 2hrs. To minutes = 120

120 - 12

= 108

Answer E

Posted from my mobile device

bumpbot · Answer

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Tom and Linda stand at point A. Linda begins to walk in a straig

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