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17 Jul 2023, 08:00
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Difficulty:
95%
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Question Stats:
52% (02:48) correct
48%
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wrong
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The first cyclist from point A and the second cyclist from point B simultaneously started moving toward each other, at their respective constant speeds. They met each other after 36 minutes and continued moving without stopping. After the meeting, the first cyclist doubled his speed, and the second cyclist reduced his speed by 10%. The first cyclist reached point B 6 minutes after the second cyclist reached point A. How much time did the first cyclist take to travel from point A to point B ?
A. 1 hour
B. 1 hour and 4 minutes
C. 1 hour and 6 minutes
D. 1 hour and 10 minutes
E. 1 hour and 36 minutes
A. 1 hour
B. 1 hour and 4 minutes
C. 1 hour and 6 minutes
D. 1 hour and 10 minutes
E. 1 hour and 36 minutes
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18 Jul 2023, 01:57
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Let the speed of first cyclist be a m/min (taking miles/min just for sake of mentioning minutes. We can take km too, the working will be same)
and speed of second cyclist be b m/min
They meet after 36 min.
So, distance travelled by first cyclist = 36a
distance travelled by second cyclist = 36b
After the meet,
speed of first cyclist = 2a m/min
and speed of second cyclist = 0.9b m/min
Now the distance to be travelled by first cyclist = distance already travelled by second cyclist = 36b
Similarly the distance to be travelled by second cyclist = distance already travelled by first cyclist = 36a
Using the above information -
Time taken by first cyclist to travel remaining distance after meeting \(= \frac{36b}{2a} = \frac{18 b}{ a}\)
Similarly, time taken by second cyclist to travel remaining distance after meeting \(= \frac{36a}{0.9b} = \frac{40 a}{b}\)
Total time taken by first cyclist to travel from A to B \(= 36 + \frac{18 b}{ a}\)
Now , we are given that the first cyclist reached point B 6 minutes after the second cyclist reached point A
\(\frac{18 b}{ a} = \frac{40 a}{b} + 6\\
\\
Let \frac{b}{ a} = k\\
\\
18 k = \frac{40}{k }+ 6\\
\\
9k^2 - 3k - 20 = 0\\
\\
9k^2 + 12k - 15k - 20 = 0\\
\\
(3k+4) (3k-5)=0\)
Since ratio of two speeds will be positive ,\( k = \frac{5}{3}\)
Now, total time taken by first cyclist to travel from A to B \(= 36 + \frac{18 b}{ a} = 36 + 18k\)
\(= 36 + 18 * \frac{5}{3}\\
\\
=66 min\\
\\
\\
= 1 hour 6 min\)
and speed of second cyclist be b m/min
They meet after 36 min.
So, distance travelled by first cyclist = 36a
distance travelled by second cyclist = 36b
After the meet,
speed of first cyclist = 2a m/min
and speed of second cyclist = 0.9b m/min
Now the distance to be travelled by first cyclist = distance already travelled by second cyclist = 36b
Similarly the distance to be travelled by second cyclist = distance already travelled by first cyclist = 36a
Using the above information -
Time taken by first cyclist to travel remaining distance after meeting \(= \frac{36b}{2a} = \frac{18 b}{ a}\)
Similarly, time taken by second cyclist to travel remaining distance after meeting \(= \frac{36a}{0.9b} = \frac{40 a}{b}\)
Total time taken by first cyclist to travel from A to B \(= 36 + \frac{18 b}{ a}\)
Now , we are given that the first cyclist reached point B 6 minutes after the second cyclist reached point A
\(\frac{18 b}{ a} = \frac{40 a}{b} + 6\\
\\
Let \frac{b}{ a} = k\\
\\
18 k = \frac{40}{k }+ 6\\
\\
9k^2 - 3k - 20 = 0\\
\\
9k^2 + 12k - 15k - 20 = 0\\
\\
(3k+4) (3k-5)=0\)
Since ratio of two speeds will be positive ,\( k = \frac{5}{3}\)
Now, total time taken by first cyclist to travel from A to B \(= 36 + \frac{18 b}{ a} = 36 + 18k\)
\(= 36 + 18 * \frac{5}{3}\\
\\
=66 min\\
\\
\\
= 1 hour 6 min\)
17 Jul 2023, 08:30
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Suppose A and B are far D one from the other.
In 36 minutes, A goes to d, and B goes to D-d.
d = vA * (36 min)
D-d = vB * (36 min).
Furthermore: d = 0.9vB * tB (tB is how much from the meeting to the end for B)
D-d = 2vA * (tB+6)
Therefore:
vB*36=2vA*(tB+6)
0.9vB*tB = vA*36
divide them ->
36 / (0.9*tB) = (tB+6)/18
tB(tB+6)=720
tB^2+6tB-720=0
tB=24
tA=24+6=30min (from the meeting to B for A)
30min + 36min from the first meeting is 1h and 6min
In 36 minutes, A goes to d, and B goes to D-d.
d = vA * (36 min)
D-d = vB * (36 min).
Furthermore: d = 0.9vB * tB (tB is how much from the meeting to the end for B)
D-d = 2vA * (tB+6)
Therefore:
vB*36=2vA*(tB+6)
0.9vB*tB = vA*36
divide them ->
36 / (0.9*tB) = (tB+6)/18
tB(tB+6)=720
tB^2+6tB-720=0
tB=24
tA=24+6=30min (from the meeting to B for A)
30min + 36min from the first meeting is 1h and 6min
