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23 Jun 2020, 10:59
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D
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58% (03:26) correct
42%
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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
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23 Jun 2020, 11:35
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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
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
24 Jun 2020, 00:41
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Bunuel
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
• 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, \(s = 9\)
