A and B are two diametrically opposite points on a circular road whose

Question

A and B are two diametrically opposite points on a circular road whose circumference is 6 km long. A cyclist started from point A and made two circles. He made the first circle with a certain uniform speed and then decreased his speed by 3 km/hr. The interval between his two circles through point B is 50 minutes. What is the speed at which the cyclist made the first circle?

A. 6 km/r
B. 7 km/r
C. 8 km/r
D. 9 km/r
E. 10 km/r

A. 6 km/r
B. 7 km/r
C. 8 km/r
D. 9 km/r
E. 10 km/r

yashikaaggarwal · Answer

Let say a cyclist completed distance = 6 in S speed for first round.
for reaching B in first round at speed S time taken will be 6/2s -------------(1)

for reaching B in second round time taken by cyclist = 6/2s+ 6/2(s-3)

where (6/2s)*60+ (6/2(s-3))*60 = 50 min approx

(3/s+3/s-3)*60 = 50

putting s = 10
(3/10+3/10-3)*60 = 50 => (3/10+3/7)*60 = 50 => (0.3+0.43)*60 = 50
=> 0.73*60 => 43.8 is not equal to 50

from this we can induced that value of s is less than 10 but not very far value

putting s = 9
(3/9+3/9-3)*60 = 50 => (3/9+3/6)*60 = 50 => (0.33+0.5)*60 = 50
=> 0.83*60 => 49.8 or 50 which is equal to 50

hence the speed of cyclist in first circle is 9km/hr

Answer is D

GMATWhizTeam · Answer

Solution

• Let us assume that the speed of the cyclist during his first circle is s km/hour
 o So, the speed of the cyclist during second circle  = (s – 3)  km/hour 
 https://www.dropbox.com/s/fcf8iavyrnuxq44/circle_A_B.jpg?dl=1 
• Now, time taken between his two circles through point B = time taken to travel B to A at a speed of s during his first circle + time taken to travel A to B at speed of (s- 3) during his second circle
 o Also, distance between A and B  = \frac{1}{2} *  circumference  = \frac{1}{2} * 6 = 3  km

• Thus,  \frac{50}{60} = \frac{3}{ s} + \frac{3}{s- 3}

⟹ \frac{1}{s} + \frac{1}{s-3} = \frac{5}{18}

⟹\frac{1}{s} + \frac{1}{s-3} = \frac{3 + 2}{18}

⟹\frac{1}{s} + \frac{1}{s-3} = \frac{3}{18} + \frac{2}{18}

⟹\frac{1}{s} + \frac{1}{s-3} = \frac{1}{6} + \frac{1}{9}

• On comparing both sides of the above equation, we get,  s = 9  
Thus, the correct answer is  Option D.

ScottTargetTestPrep · Answer

Solution:

The distance between A and B around the road is 3 km since the entire circumference of the circular road is 6 km.

We can let r = the speed of the cyclist in completing the first circle. Notice that the distance traveled between the first time he passes through point B and the second time he passes it is the circumference of the road, or 6 km. However, the first 3 km of this 6 km is the 2nd half of the first circle in which his speed is r km/hr and the last 3 km is the 1st half of the second circle, in which his speed is (r - 3) km/hr. Therefore, we can create the equation:

3/r + 3/(r - 3) = 5/6

Multiplying the equation by 6r(r - 3), we have:

18(r - 3) + 18r = 5r(r - 3)

18r - 54 + 18r = 5r^2 - 15r

5r^2 - 51r + 54 = 0

(5r - 6)(r - 9) = 0

r = 6/5 or r = 9

Although both values of r are positive, r can’t be 6/5, otherwise, r - 3 would be negative. Therefore, r = 9.

Answer: D

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