Around the World in 80 Questions (Day 1): The first cyclist from point

Question

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

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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

Flower_Child · Accepted Answer

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

turbojuly · Answer

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

Kinshook · Answer

Given: 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.

Asked: How much time did the first cyclist take to travel from point A to point B ?

Attachment:

Screenshot 2023-07-17 at 9.00.44 PM.png [ 28.67 KiB | Viewed 8178 times ]

Let the speed of cyclist A and cyclist B be Va & Vb respectively

Total distance from point A to point B = 36/60 (Va + Vb) = .6 (Va + Vb)

Time taken by cyclist A to reach point B = .6 + .6Vb/2Va = .6 + .3Vb/Va
Time taken by cyclist B to reach point A = .6 + .6Va/.9Vb = .6 + 2Va/3Vb

Time taken by cyclist A to reach point B - Time taken by cyclist B to reach point A = 6/60 hours = .1 hours
.3Vb/Va - 2Va/3Vb = .1
Let Va/Vb = x

$.3/x - 2x/3 = .1$

$9/x - 20x = 3$

$20x^2 + 3x - 9 = 0$

x = Va/Vb = .6

Time taken by cyclist A to reach point B = .6 + .6Vb/2Va = .6 + .3Vb/Va = .6 + .3/.6 = .6 + .5 = 1.1 hours = 1 hour 6 minutes

IMO C

Ray007 · Answer

Let speed of first cyclist be x and speed of second cyclist by y. Let T be time take by second cyclist to reach after from the meeting point.
Distance = speed x time

36x=0.9yT
36y = 2x(T+6) 
y=2x(T+6)/36

36x = 0.9(2x(T+6)/36)T
Simplifying this, equation becomes : T^2 + 6T - 720

Solve for T : 24 min

From A to B, first cyclist took, 36+24+6 mins --> 1 hour 6 mins

missionmba2025 · Answer

Speed of first cyclist = a units / min
Speed of second cyclist = b units / min

Until A and B meet -

Distance traveled by A in 36 mins = 36a
Distance traveled by B in 36 mins = 36b

After A and B meet -

Time taken by A = 36b/2a = 18b/a
Time taken by B = 36a/0.9b = 40a/b

18b/a - 40a/b = 6 ---(1)

Divide by 2

9b/a - 20a/b = 3

Let b/a = x

9x - 20/x = 3

9x^2 - 20 - 3x = 0

x = 5/3

Total time = 36 + 36 * 5/6 = 36 + 30 = 66

IMO C

sujoykrdatta · Answer

Let the meeting point between A and B is P
Also, let the cyclist from A be named X and the one from B be named Y
X covered AP in 36 min, Y covered BP in 36 min
Let A would have covered PB in x min, but at twice his speed, he will take x/2 min
Let B would have covered PA in y min, but at 90% of his speed, he will take y/0.9 = 10y/9 min
Thus: x/2 = 6 + 10y/9 ... (i)
Also, X covered AP in 36 min and BP in x min (usual speed)
Y covered BP in 36 min and AP in y min (usual speed)
=> speed ratio = 36/y = x/36 => y = 1296/x

From (i): x/2 = 6 + 1440/x
=> 1440/x = 6 + x/2

Solving, we get: x = 60
Thus, actual time taken = x/2 = 30 mins to cover PB
Thus, total time = 36 + 30 = 66 min

Answer C

Wadree · Answer

let, they meet at Y which is between A and B. let, after meeting, second cyclist, with 90% of his speed, takes X time to reach A. so after meet, first cyclist takes X+6 minutes to reach B with double his speed. so, with his original speed, first cyclist would take double time to reach B from Y which is 2X+12. and second cyclist went from B to Y in 36 minutes. 
again, second cyclist, with 90% of his speed takes X time to go from Y to A. so he will take 9X/10 time to go from Y to A with his original speed. and first cyclist went from A to Y in 36 minutes. so......

first cyclist(original speed):
A---------Y = 36 minutes
Y---------B = 2X+12

second cyclist(original speed):
B---------Y = 36 minutes
Y---------A = 9X/10

so (2X+12)/36=36/(9X/10)
 X^2 + 6X - 720 = 0
 X = 24 , -30
 X = 24
 first cyclist from A to B takes 36 + X + 6 = 1 hour 6 minute

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