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In a sprint race, a man has to run two distances, D1 and D2, to comple

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In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.

 

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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 19 Jul 2019, 02:25
Bunuel wrote:
In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.

 

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OFFICIAL EXPLANATION FROM Egmat



Correct Answer: A

Steps 1 & 2: Understand Question and Draw Inferences

Given:
* A man a man has to run two distances, D1 and D2, to complete a sprint race.
* Jim is participating in the race.
* Distance D2 is thrice of the distance D1.

To Find:
* Average speed of Jim. Average Speed = Total distance covered/Total time taken
* Total distance covered = D1 + D2 = D1 + 3 D1= 4D1
* Total time taken= Time taken to cover D1 + Time taken to cover D2
Time taken to cover \(D_1 = \frac{D_1}{S_1}\) , where S1 is the speed to cover D1.
Time taken to cover \(D_2 = \frac{D_2}{S_2} = \frac{(3D_1)}{S_2}\) , where S2 is the speed to cover D2.
* Average Speed = \(\frac{(4D_1)}{(Time \ taken \ to \ cover \ D_1+ Time \ taken \ to \ cover \ D_2 )} = \frac{4D_1}{\frac{D_1}{S_1} + \frac{D_2}{S_2}} = \frac{4D_1}{\frac{D_1}{S_1} + \frac{3D_1}{S_2}}= \frac{4}{\frac{1}{S_1} + 3S_2}\)
* We need S1 and S2 to find the answer.

We do not have enough information to find the answer, let us analyze the statements.

Step 3: Analyse Statement 1:
“Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.”
* S1 = 1 and S2 = 6
Therefore, statement 1 alone is sufficient.

Step 4: Analyse Statement 2:
“The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.”
* Time taken to cover D1 = 2 × Time taken to cover D2
* \(\frac{D_1}{S_1} = 2 × \frac{(3D_1)}{S_2}\)
* \(\frac{1}{S_1} =\frac{6}{S_2}\)
* \(S_2 = 6S_1\)
This statement gives a relation between S1 and S2. However, we can find the value of S1 and S2.
Therefore, statement 2 alone is NOT sufficient.

Step 5: Analyse both statements together:
This step is not required, since, we got the answer from statement 1

Hence, the correct answer is Option A.
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:15
1
Answer is A.
Say distance D1 = x then D2 = 3x
By option 1:
D1 is covered at 1km/hr. Hence time taken = x/1 = x hours
Similarly D2 is covered at 6km/hr. hence time taken = 3x/6 = x/2 hours.
Avg speed = dist/time = 4x/(x+x/2)
Sufficient

Option 2:
Let time taken to cover D1 = 2y
Time taken to cover D2 = y
So avg speed = 4x/3y

Not sufficient
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:30
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1
Refer attached image.

Ans. A
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In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post Updated on: 07 Jul 2019, 01:01
1
given
distance of D2=3*D1
#1
Speed D1= 1kmph and Speed D2 ; 6kmph
total distance/total time ; d1+3d1/d1/1+3d1/6 ; sufficient
#2
The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.
t1 =2t2
total distance/total time ; d1+3d1/2t2+t2 ; 4d1/3t2
insufficient
IMO A

In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.

Originally posted by Archit3110 on 04 Jul 2019, 08:33.
Last edited by Archit3110 on 07 Jul 2019, 01:01, edited 2 times in total.
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:35
2
IMO (A)

Statement (1) gives us clear and sufficient information from which we know that the time Jim took to cover the first distance (x) = x/1 and the time he too to cover the second distance (3x) = 3x/6 (or x/2)

Now average speed = total distance/total time = (x+3x)/((x/1)+(x/2)), solving which we easily get the average speed for Jim. HENCE, SUFFICIENT.

Now, looking at Statement (B), though I can prove the insufficiency mathematically, I choose to apply logic here. I ONLY know the ratio of time taken during the first distance to that taken during the second distance. Now thinking LOGICALLY, Jim could run really fast to maintain and maintain the same ratio, or run really really slow (slower than a tortoise) and even still maintain the same ratio, so with the given information, we really can't know his average speed.

Hence, INSUFFICIENT.

(A) is the correct answer choice.
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:37
1
In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?

D2=3D1

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
time spend for the D1: D1 km/1km per hour
time spend for the D2: 3D1 km/6 km per hour

Weighed average speed is:

(D1/1 * 1 + 3D1*6/6)/(D1+3D1/6)
Sufficient to answer the question

(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.
time d1= 2*time d2
Not sufficient to answer the question

Answer: A
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:40
1
In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
D2 = 3D1
Time taken to cover D1 = D1/1 = D1 hours
Time taken to cover D2 = 3D1/6 = D1/2 hours
Average speed = (D1+D2)/(D1+D1/2) = 4D1/(3D1/2) = 8/3 kmh. SUFFICIENT
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.
Let us assume time taken to cover D1 = 2t = 2* time taken to cover D2 (t)
Average speed = 4D1/(2t+t) = 4D1/3t. NOT SUFFICIENT

IMO A
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:40
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avg speed will be : \((D1 +D2)/((D1/T1)+(D2/T2))\)or \((D1 +D2)/((v1)+(v2))\)

From1: avg speed is given and distance correlation is given
sufficient.

From 2:Not sufficient.

A
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:42
2
In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?
D2 = 3*D1

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour. --> average speed = (D1+D2) / (( D1/1) +( D2/6)) = (D1+3*D1) / (( D1/1) +( 3*D1/6)) = (4/1.5) km /h --> correct
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2. --> t1 = 2*t2 & D2 = 3*D1, two ratio can't give us a value i.e. two ratio can't gove us the vag speed

Answer: A
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In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post Updated on: 05 Jul 2019, 00:21
1
Let x and 3x be the distances D1 and D2.


(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
t1 (time to cover D1) = x/1,
t2 = 3x/6=x/2
Total time = x + x/2 = 3x/2
Average speed = total distance/total time = 4x/3x/2 = 8/3 hr
This statement is sufficient.

(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.
Let 2t, t be the time taken to cover distances D1 and D2.
Average speed = 4x/3t
This statement is not sufficient.

So first statement alone is sufficient.

I think Option A would be the answer.

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Originally posted by prashanths on 04 Jul 2019, 08:46.
Last edited by prashanths on 05 Jul 2019, 00:21, edited 4 times in total.
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In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post Updated on: 04 Jul 2019, 13:04
2
r1*t1=D1
r2*t2=D2

Since D2=3D1,

r2*t2=3D1

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.

We are given the average speed for first lap and the average speed for the second lap and we are asked to find the weighted average (say, x)

D1/3D1 = 6-x/x-1

We can arrive at a unique solution using the above equation

Sufficient

(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.

Using this statement we can say

r1*t1=D1
r2*t1/2=3D1

On dividing the two equations we get r1/r2=1/6

Here we have just the ratio of speeds but we can't be sure about their actual speed so we can't set up the weighted average equation

Insufficient

Answer is (A)

Originally posted by firas92 on 04 Jul 2019, 08:46.
Last edited by firas92 on 04 Jul 2019, 13:04, edited 2 times in total.
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In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:50
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Average speed = total distance/total speed

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.-

D2=3D1 ; T1=D1 and T2=D2/6=3D1/6=D1/2. since both distance and time is expressed in terms of one variable D1 average speed = 4D1/3D1/2=8/3 - Sufficient

(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2. - Time will be in one variable and Distance will be in another variable. Insufficient

IMO A
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In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:50
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In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.

let D1 = d
hence D2 = 3d
now average speed = \(\frac{(d+3d)}{[m][fraction]d/s1}\)+\(\frac{3d}{s2}\)[/fraction][/m] where s1 is the avg speed of first D1 distance and s2 is teh average speed of 3d distance
Stmt 1 gives clearly s1 = 1 and s2 =6
put in the values
\(\frac{(d+3d)}{[m][fraction]d/1}\)+\(\frac{3d}{6}\)[/fraction][/m]
=\(\frac{8}{2}\) hence sufficient
Stmt 2 says

\frac{d}{s1}= 2* \frac{3d}{s2}
so ti:t2 = 2:1 or 2k:k
put the values back in equation
\(\frac{(d+3d)}{2k+k}\)
we are left with still variables hence not sufficient
Thus A alone is sufficient
Hence A
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:53
1
D1=x and D2=3x
From 1, S1= 1km/hr and s2= 6km/hr,
Avg speed=total distance/total time
= 4x/((x/1)+(3x/6))
So, sufficient

From 2, t1=2t and t2=t
Shive we have two variables X and t, hence insufficient

So A

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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:53
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The answer is A.

In order to find the average rate of 2 distances and times, you need to divide the sum of 2 distances by the sum of 2 times.

In the Statement 1, the sum of 2 distances is 4*D1, while the sum of two times is 3/2*D1. Thus, by dividing 4*D1 by 3/2*D1 you get 8/3. Sufficient

Statement 2, on the other hand, doesn't offer numbers to work with since we are left with 4*D1 as the sum of distances and 3*T2 as the sum of times. Unfortunately, the division doesn't yield a number. Thus, it is not sufficient.


Thus, A is the answer.
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:54
Here is my approach to the above problem.


Average speed = total distance/total time

STAT1:

Average speed= (D1+3D1)/(D1/1+3D1/6)

Note: we don't need to solve Statement one as D1 will cancel off from Numerator and deno and so statement 1 is suffcient.

STAT2:


Let amount of time taken o cover distance D2 (that is 3D1) be t ,then amount of time to cover D1= 2t

Now

Average speed = total distance/total time = (D1+3D1)/ (2t+t) = 4D1/3t

Now we know

t= 3D1/6

So this we can substitute back in above equation and will get a define value of average speed again.

Hence,both statements are sufficient. :)
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:55
1
Average speed = total distance/total time
Also, D2 = 3D1

Considering statement (1) alone:
Total time taken = D1/1 + D2/6 = D1/1 + 3D1/6 = D1/1 + D1/2 = 1.5D1
SUFFICIENT

Considering statement (2) alone:
Time taken to cover D1 is 2 times the time taken to cover D2.
INSUFFICIENT

The answer is (A).
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:57
1
IMO A

In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.

Solution:


Avg time= (total distance)/(total time)

1.
if we only use d1 even without assuming any distance then

(d1+3d1)/(d1/1)+(3d1/6)

4d1/(2d1+d1)/2

8d1/3d1

Avg Speed=8/3

assuming d1=10 then

total dist=10+30=40;

total time =t1+t2=(10/1)+(30/6)=15

so 40/15=8/3 , same as above.

And A alone solves the problem.

2. let d1 be 100 and and d2 be 300

t1=100/10 and t2=300/5
then total time=10+60=70 hours

speed=400/70=40/7

also if we take d1=10 and d2=30 and
time for d1=10 hours and time for t2=5 hours then

40/15=8/3, this is inconsitent, so no B


A is the correct answer from above explaination
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple  [#permalink]

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New post 04 Jul 2019, 08:59
(1) not sufficient. this is because we still do not know the distances D1 and D2. the average speed will depend on the magnitude of the distances, and while we know the relative ratio, this is not sufficient as we don't know the magnitude. plugging in will show that if D1=1 and D2=3, or D1=2 and D2=6, the averages will be different.

(2) not sufficient. this is because we have no information about speed.

taken together, we still don't have any information about magnitude to solve for so E.
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Re: In a sprint race, a man has to run two distances, D1 and D2, to comple   [#permalink] 04 Jul 2019, 08:59

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