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

Nice question. It's always worth trying to figure out a simpler way of approaching a problem.

I typically do these by creating a chart. One column of the chart is 'time'. The other columns tell me where each person in the problem is located at that time.

In this problem, it would look like this. I started counting up in 12-minute intervals, because all of the answer choices are multiples of 12. 12 minutes is 0.2 hours. In 0.2 hours, Tom will travel 1.2 miles and Linda will travel 0.4 miles.

0 hours:
Tom: 0, Linda: 0

1 hour:
Tom: 0, Linda: 2

1.2 hours:
Tom: 1.2
Linda: 2 + 0.4 = 2.4

1.4 hours:
Tom: 1.2 + 1.2 = 2.4
Linda: 2.4 + 0.4 = 2.8

1.6 hours:
Tom: 2.4 + 1.2 = 3.6
Linda: 2.8 + 0.4 = 3.2

1.8 hours:
Tom: 3.6 + 1.2 = 4.8
Linda: 3.2 + 0.4 = 3.6

2.0 hours:
Tom: 4.8 + 1.2 = 6.0
Linda: 3.6 + 0.4 = 4.0

2.2 hours:
Tom: 6.0 + 1.2 = 7.2
Linda: 4.0 + 0.4 = 4.4

2.4 hours:
Tom: 7.2 + 1.2 = 8.4
Linda: 4.4 + 0.4 = 4.8

2.6 hours:
Tom: 8.4 + 1.2 = 9.6
Linda: 4.8 + 0.4 = 5.2

2.8 hours:
Tom: 9.6 + 1.2 = 10.8
Linda: 5.2 + 0.4 = 5.6

3.0 hours:
Tom: 10.8 + 1.2 = 12.0
Linda: 5.6 + 0.4 = 6.0

Unfortunately this didn't give us an exact answer, but it's close enough that we can estimate. Tom and Linda covered the same distance at about 1.5 hours. Tom covered twice Linda's distance in exactly 3 hours. The difference is 3-1.5 = 1.5 hours, which is 90 minutes.

Here is the logical way to solve this..

Speed of Tom = 6 mph
Speed of Linda = 2 mph

Relative speed of Tom = 6-2 = 4 mph.

So Compared with Linda, Tom will cover 4 additional miles per hour.

Linda has a head start of 1 hour in which she has covered 2 miles.

Time taken by Tom to cover the same distance as Linda =\(\frac{(Lead Taken by Linda)}{(Relative Speed of Tom} = \frac{2}{4} = \frac{1}{2} = 30 Minutes\)

So, 30 Minutes after Tom started, he will have traveled the same distance as that by Linda, which is 3 miles.

Now, Following are the distances covered by Tom and Linda Henceforth.. (Back Solving..)

1 hour after they have traveled the same distance..

Distance by Tom = 3 + 6 = 9 miles
Distance by Linda = 3 + 2 = 5 miles.

Option A is out.
120 minutes is too large. E is out.

Lets Put in C as it will help us solve the question fastest now.

After 1 hour, distances stand at

Distance by Tom = 3 + 6 = 9 miles
Distance by Linda = 3 + 2 = 5 miles.

So, after 84 minutes = 60 + 24 minutes,

Distance by Tom = 9 + 2.4 = 11.4 miles
Distance by Linda = 5 + 0.8 = 5.8 miles.

5.8 * 2 = 11.6 - 11.4 = 0.2

C is out and since Tom hasn't covered twice the distance so far, D is answer.

1.
2+2x=6x
X=0.5
60(0.5+1)=90
2.
6x=2(2+2x)
X=2
60(2+1)=180
3.
180-90=90
So the answer is c
I am wondering why you use such a complex method

distance covered in 1st hour by both Linda and Tom
6t= 2(t+1)
t = 1/2 ; 30 mins
and time taken by Tom to cover twice distance covered by Linda
6t= 2*(2t+2)
t = 2 hrs ; 120 mins
∆ ; 120-30 ; 90 mins
option D

hey , everything is fine with the solution just need one clarification Tom started 1 hour late of linds , or you can say linda started one hour early.
so D=S*T ,
1. first solution must be 6*(t+1)=2t
i don't understand why you have added t+1 with respect to lind time ??

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

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

Assume that in \(T\) hours , Tom covers the exact same distance that Linda covered.
That means Linda has traveled for\( T + 1\) hours as Tom begins to jog after 1 hour

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

\(6* T = 2 *(T + 1) \)
\(4*T = 2\)
\(T= 2/4 = 1/2 =.5 hrs\)

we see that in 1.5 hours Linda has covered 2* 1.5 = 3 miles and in 0.5 hours Tom covered the same distance 0.5 * 6 = 3 miles
NOTE: Remember that tom started one hour later.

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,
\(6* T1 = 2*(2*(T1 + 1))\)
\(6* T1 = 4* T1 + 4\)
\(2* T1 = 4\)
\(T1= 2 hrs.\)

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 = 2 - .5 = 1.5 hours = 90 mins
Option D is the right answer.

Thanks,
Clifin J Francis,
GMAT SME

The formula to calculate distance is Distance = (Rate)(Time). So at any given moment Tom's distance (let's call it DT) can be expressed as DT = 6T. So, at any given moment, Linda's distance (let's call it DL) can be expressed as DL = 2(T + 1) (remember, Linda's time is one hour more than Tom's). The question asks us to find the positive difference between the amount of time it takes Tom to cover half of Linda's distance and the time it takes him to cover twice her distance. Let's find each time separately first.

When Tom has covered half of Linda's distance, the following equation will hold: 6T = (2(T + 1))/2. We can solve for T:
6T = (2(T + 1))/2
6T = (2T + 2)/2
6T = T +1
5T = 1
T = 1/5

So it will take Tom 1/5 hours, or 12 minutes, to cover half of Linda's distance. When Tom has covered twice Linda's distance, the following equation will hold: 6T = 2(2(T + 1). We can solve for T:
6T = 2(2(T + 1)
6T = 2(2T + 2)
6T = 4T + 4
2T = 4
T = 2

So it will take Tom 2 hours, or 120 minutes, to cover twice Linda's distance.
We need to find the positive difference between these times: 120 – 12 = 108.

The correct answer is E.

I have the same question

Linda travels for 1 hour more than Tom. So, if Tom travels for t hours, then Linda travels for t + 1 hours. It's as simple as that!