Two soldiers started racing from point P to point Q, a distance of 10

Question

Two soldiers started racing from point P to point Q, a distance of 10 miles. Soldier 1 ran the first half of the distance at a constant speed of X miles per hour and the second half at a constant speed of Y miles per hour. Soldier 2 ran the entire race at a constant speed of Z miles per hour. X and Y are distinct multiples of 10. If neither of them stopped before reaching point Q, which soldier reached point Q first?

(1) Y is 50% more than X.

(1) Z is the average of X and Y.

(1) Y is 50% more than X.

(1) Z is the average of X and Y.

SSM700plus · Answer

Statement 1:- No information about Z. Not sufficient.

Statement 2: - Suppose X = 10 and Y= 20. Avg of X & Y = Z = 15.

Soldier 1 covers first 5 miles at a speed of 10mph. It takes 30 minutes to complete.
Soldier 1 covers last 5 miles at a speed of 20mph. It takes 15 minutes to complete.
Total Time for soldier 1 = 45 minutes.
 
Soldier 2 covers first 10 miles at a speed of 15 mph. It takes 40 minutes to complete.
Total Time for soldier 2 = 40 minutes.

We can also try with other values of X & Y. The result will be same.

It answers our question - which soldier reached point Q first? . It is soldier 2.

Hence Sufficient.

Answer B.

push12345 · Answer

I) Y=3X/2

But no info about z so Insufficient

II) Z=(x+y)/2

Time by S1= 10/ 
= 2xy/x+y

Time by S2= 10/z= 20/x+y

As we know that x and y are multiple of 10
So time of S1 I would be= 
2*10k*10k1/x+y

where k and k1 are positive integers

So time for S1 = 200k*k1/x+y

So time of S1> time of S2

So soldier 2 will reach early

B is ANSWER

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thefibonacci · Answer

Time taken by S1 = 10/  = 2XY/(X+Y)
Time taken by S2 = 10/Z

1) Nothing is given about Z 
Insufficient

2) Time taken by S2 = 10/(X+Y) 
And time taken by S1 is 2XY/(X+Y)

now we look at above equations, we are already told that X & Y are multiple of 10. Look at time taken by S2, i can write it as 1/(x+y) by taking 10 common and cancelling. 
time taken by S1 can be represented as 2xY/(x+y) in this case i can take 10 common either from X or Y in numerator.

Clearly time taken by S1 is more as it has a larger numerator. Hence, B is sufficient.

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Two soldiers started racing from point P to point Q, a distance of 10

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Two soldiers started racing from point P to point Q, a distance of 10

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