Yan is somewhere between his home and the stadium. To get to the stadi

Question

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan’s distance from his home to his distance from the stadium?

(A) \(\frac{2}{3}\)

(B) \(\frac{3}{4}\)

(C) \(\frac{4}{5}\)

(D) \(\frac{5}{6}\)

(E) \(\frac{6}{7}\)

(A)  \frac{2}{3}

(B)  \frac{3}{4}

(C)  \frac{4}{5}

(D)  \frac{5}{6}

(E)  \frac{6}{7}

ElninoEffect · Answer

Let the distance of Yan from the stadium be = a
&
Let the distance of Yan from the home be = b

Let speed by walk be = x
So, through cycling will be = 7x

Time taken by walk to go the stadium = a/x
Time taken by cycling to go to the stadium = b/x (time taken by walk to go to the home) + (a+b)/7x (Time taken by bicycle from home to stadium)

The two times are same so equating both the equation:
a/x = (7b + a + b)/ 7x
7a = 8b + a
6a = 8b

We need to find the ratio of b/a
  So B is the answer.

CrackverbalGMAT · Answer

Home ...(x)............. YAN ..........(y)........... Stadium

Let the distance between home and Yan's initial position be "x" and that of Yan's initial position to stadium be "y".

We have to find  Yan’s distance from his home to his distance from the stadium or  x / y

Yan rides 7 times as fast as he walks

y = x + (x + y ) / 7

=> 7y = 8x + y

=>6y = 8x or x/y =6/8 = 3/4

(option b)

Alternately,
Let the distance from home to the stadium be 1, and the distance from Yan to home is x

Let speed of walking be 1, and hence the speed of riding the bike is 7

So 1 - x = x + (1/7)

x = (3/7)

We have to find x / (1-x)

= (3/7) / (4/7)

=3/4

(option b)

D.S
GMAT SME

Fdambro294 · Answer

Let distance from H (home) to Y(Yan) = A

Let distance from Y (Yan) to S (stadium) = B

Question is asking for the ratio of:

A / B = ?

Speed walking = R

Time 1 = (distance from Y to S) / (Speed of Walking) = B/R

Speed riding bicycle = 7R

Time 2 = (distance from Y to H) / (speed walking) + (distance from H to S) / (speed of riding bicycle)

Time 2 = (A / R) + (A + B)/(7R)

And we are told that: Time 1 = Time 2

B/R = (A/R) + (A + B)/(7R)

—> multiply both sides of the equation by: * (7R)

7B = (7A) + (A + B)

6B = 8A

A/B = 6/8 = 3/4

D

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