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

Question

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.

Bunuel · Accepted Answer

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.

RJ7X0DefiningMyX · Answer

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.

Kinshook · Answer

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

hiranmay · Answer

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

firas92 · Answer

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)

shridhar786 · Answer

let D1 = x km
then D2 = 3x (given D2 is thrice of D1)
so we need to find the average speed

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

speed of Jim while running D1 (S1) = 1 km/hr
speed of Jim while running D2 (s2) = 6 km/hr
total distance traveled by Jim = x+3x = 4x

total time taken =  \frac{x}{1} + \frac{3x}{6}  =  \frac{3x}{2} 
average speed =  \frac{total distance}{total time} 
average speed =  \frac{4x}{3x/2}  =  \frac{8}{3} 
so from statement 1 we can find average speed SUFFICIENT

statement (2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.
let S1 be speed while traveling D1
S2 while traveling D2

then  \frac{D1}{S1}  = 2 \frac{D2}{S2}

\frac{x}{S1}  = 2 \frac{3x}{S2}  ------  \frac{S1}{S2}  =  \frac{1}{6} 
from ratio let S1 = y
S2 = 6y
average speed =  \frac{total distance}{total time}

average speed =  \frac{4x}{x/y+3x/6y}  =  \frac{8y}{3}

since we don't know the value of y so INSUFFICIENT

A is the answer

Sayon · Answer

Given, D2 = 3*D1 ->    
Let, S1= Speed while running D1; S2= Speed while running D2
Average Speed =  \frac{Total Distance}{Total Time}  ->    
=  \frac{(D1 + D2)}{\frac{D1}{S1}+\frac{D2}{S2}}  Substituting     into this we get
=  \frac{(D1 + 3D1)}{\frac{D1}{S1} + \frac{3D1}{S2}} 
=  \frac{4}{\frac{1}{S1} + \frac{3}{S2}}

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.   
Given, S1 = 1 km/hour; S2 = 6 km/hour
Therefore, Average speed =  \frac{4}{\frac{1}{1} + \frac{3}{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.   
If T1= time taken to cover D1 and T2= time taken to cover D2 then,
It is given that T1 = 2 T2. Putting this in     we get
Average Speed =  \frac{D1 + 3D1}{T1+ T\frac{1}{2}} = \frac{8D1}{3T1} = \frac{8}{3}*S1 
We do not know the value of S1

Not Sufficient

Answer A

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