VeritasPrepKarishma wrote:
SonyGmat wrote:
Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a start of 200m and beats him by 30 seconds. Next, Jerry gives Jim a start of 3mins and is beaten by 1000m. Find the time in minutes in which Jerry and Jim can run the race seperately?
1. 8,10
2. 4,5
3. 5,9
4. 6,9
5. 7,10
I find it rather challenging, any inputs??
Make diagrams in races. They help you understand the question better.
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Jerry gives Jim a head start of 200 m so Jim starts not from the starting point but from 200 m ahead. Jerry still beats him by 30 sec which means that Jerry completes the race while Jim takes another 30 sec to complete it.
In this race, Jerry covers 2000m. In the same time, Jim covers the distance shown by the red line. Since Jim needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance which is (1/2)*s where s is the speed of Jim. The distance Jim has actually covered in the same time as Jerry is [1800 - (1/2)*s] shown by the red line.
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Jerry gives Jim a start of 3 mins means Jim starts running first while Jerry sits at the starting point. After 3 mins, Jerry starts running too. Now, Jim beats Jerry by 1000 m which means that Jim reaches the end point while Jerry is still 1000 m away from the end.
In this race, Jerry covers a distance of 1000 m only. In that time, Jim covers the distance shown by the red line (the distance before that was covered by Jim in his first 3 mins). This distance shown by the red line is given by 2000 - 3s (3s is the distance covered by Jim in 3 minutes)
Now you see that in the first race, Jerry covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, Jim would also have covered half the previous distance i.e. the second red line will be half the first red line.
First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min
Time taken by Jim to run a 2000 m race = 2000/400 = 5 min
Answer (B)
Dear Experts, could you , please, evaluate my solution? I cannot find a flaw in it but not sure that my logic is correct. In my humble opinion, the solution is attractive because it's easier to understand.
I use Karishma's diagrams. ( in attachment below )
step 1 - finding Jim speed ; 1st Race : Let say \(S_m\)- Jim speed ;
Because 200 meters was given to Jim, we know that overall at the 1st race he made 2000 - 200 = 1800 meters before Jerry had finished .
Because Jim had lost the 1st race by 30 second, this means that if additional 30 second were given to Jim, he finished the 1st race simultaneously with Jerry.
To sum up all this facts we have : (2000meters - 200meters ) + \(S_m*\frac{1}{2}\)minute = 2000 meters
1800 + \(\frac{1}{2}min\)* \(S_m\) = 2000
\(S_m\)= 400 meters per min ;
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step 2 - finding Jerry speed; 2nd Race : \(S_m\)- Jim speed ;
\(S_r\) - Jerry speed ;
\(t_r\) - Jerry time from second race before Jim finished the race ;
We didn't know how many minutes it takes Jim to finish the 2d race. However, we know that he ran 3 minutes plus time that Jerry ran his 1000 meters before Jim ended the race.
So we have \(t_m = t_r + 3min\)
System of equations for 2nd Race :
(1) \(S_r*t_r = 1000\)
(2)\(S_m*(t_r + 3min ) = 2000\)
(3) \(S_m\)= 400 mpm ;
First solution : (2) \(400 *t_r + 1200meters = 2000meters\)
\(t_r\)= \(\frac{2000 - 1200}{400} = 2 minutes\);
put \(t_r\) in (1) and find \(S_r\)
\(S_r = 500\) mpm
Second solution (alternative) : From race 1, we can find that Jim can run 2000 in 5 minutes => \(\frac{2000 meters}{400 mpm}= 5 min\)
So 5 min - 3 min = 2 min
\(t_r\)= 2 minutes;
put \(t_r\) in (1) and find \(S_r\)
\(S_r = 500\) mpm
step 3 - finding the time in minutes in which Jerry and Jim can run the race separatelyJim: \(\frac{2000 meters}{400 mpm}= 5 min\)
Jerry: \(\frac{2000 meters}{500 mpm} = 4 min\)
Answer B
Attachments
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