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

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

Let first cyclist and second cyclist's speed be a and b km/min respectively.

So total distance of track = 36 (a+b) .........(1)

Let x be the time in which B reaches point A after these 36 mins. So A reaches point A in x+6 mins.

Distance travelled by A = 36a + (x+6)a .........(2)
Distance travelled by B = 36b + 0.9xb .........(3)

Solve these three equations to get x = 24.

Therefore time taken by first cyclist = 36 + 30 mins = 66 mins

IMO C.

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


Now:
Cyclist 1: speed is 'a' and then after 36 mins it is '2a'. thus distance travelled: '36a' and '36b'
Cyclist 2: speed is 'b' and then after 36 mins it is '0.9b'. thus distance travelled: '36b' and '36a'

say cyclist 1 took 1 hr 6 mins.

thus cyclist 2 took 1 hr.
thus cyclist 2 travelled '36a' in 24 mins at speed of '0.9b'

(36a/0.9b) = 24 mins
5a = 3b

now distance '36b' = '60a' (as 5a = 3b).
thus cyclist 1 travelled '60a' at speed of '2a' taking (60a/2a) = 30 mins.
Hence finishing 6mins after cyclist 2

Let the distance between A and B be d km
Original speed of cyclist 1 be v1 km/hr
Original speed of cyclist 2 be v2 km/hr
They met after 36 minutes

So, v1 * 36/60 + v2 * 36/60 = d
(v1 + v2) * 36/60 = d
v1 + v2 = d * 60/36 = 5d/3 km/hr
After meeting, cyclist 1 doubled his speed to 2v1 km/hr
Cyclist 2 reduced his speed by 10% to 0.9v2 km/hr
Time taken by cyclist 2 to reach A after meeting = d/(0.9v2) minutes
Time taken by cyclist 1 to reach B after meeting = d/(2v1) minutes
It is given: d/(2v1) - d/(0.9v2) = 6
Solving this: v1 = 3 km/hr, v2 = 6 km/hr
Original time taken by cyclist 1 to reach B = d/v1 = 60 minutes

So the total time is 60 + 36 = 96 minutes = 1 hour 36 minutes

Therefore, the answer is E.

Is C the answer?

1 hr 6 mins

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Let starting point of A be A and B be B. Let them meet at a point C
Let A start cycling with speed X metres/minute. B at Y metres/minute. Therefore, post their meeting at C, speed of A is 2X and speed of B is 0.9Y

Let t1 be time taken by cyclistA to reach C from A (and thus the time taken by cyclistB to reach C from B). its given t1 = 36 minutes
Let t2 be time taken by cyclistA to reach B from C i.e., rest of journey. Therefore time taken by cyclistB to reach A is t2-6. We need to find t1+t2. since we have t1, we have to find t2.

Now, length of segment CB = (2X)*(t2)= (Y)*(t1). This implies, X*t2 = 18*Y ... (1)
length of segment AB = 36X = 0.9Y(t2-6) . using equation (1) and eliminating Y (since we are working towards t2), we get
40X = ((X*(t2^2))/18) - (X*t2)/3

We know X is not 0. So cancelling X, multiplying LHS and RHS by 18 and then rearranging terms gives us a quadratic equation in t2 as follows: (t2^2)-(6*t2)-720=0

Solving it gives values as t2 = -24 or t2 = 30. Since t2 is time in minutes, it cant be negative. therefore t2 = 30 minutes. Total time is t1+t2 = 66 minutes = 1 hour 6 minutes

Great question!

we know from the text that it takes b 64 minutes and so a does it in 70 minutes

so its c

Let us assume speed of first cyclist be u and speed of the second cyclist be 10v.
Now, let us assume that when both met for the first time, the first cyclist covered a distance of y and second covered a distance of x.
Hence. y=36*u and x=36*10v
Now considering the journey after they met (assume t is the time required by the second cyclist to travel from the point of meet to point A):
9v*t=y and 2u*(6+t)=x
solving the above: 180/t=6+t/4
720=(6+t)(t)
hence t=24
The time required= 36+24+6=106 min or 1 hour 6 min.

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 ?

Let speed of first cyclist be Sa and speed of second cyclist be Sb
36 minutes = 6/10 = 3/5 hrs
Distance traveled by first cyclist in 3/5 hrs = Sa*3/5......1
Distance traveled by second cyclist in 3/5 hrs = Sb*3/5......2
After they cross each other, second cyclist speed is 0.9Sb and he takes say, t hrs to cover the balance distance;
and the first cyclist speed is 2Sa and he takes (t+1/10) hrs to cover the balance distance
Distance traveled by first cyclist in (t+1/10) is 2Sa*(t+1/10).....3
Distance traveled by second cyclist in t is 0.9Sb*t .....4
Since, distance in 1 is equal to distance in 4 and distance in 2 is equal to distance in 3, we can solve the equations to get t = 24/60 = 2/5 hrs
therefore, the first cyclist takes 3/5 + 2/5 + 1/10 hrs = 11/10 = (1 + 1/10) hrs = 1hr 6 minutes to travel from point A to B

Answer C

Taking the help of answers C 1hr 6 min
Let the distance covered by A to reach the meeting point be X
Distance covered by A in next 30 mins at double speed = (5/3)X (Using T1/T2=D1/D2 as well as S1/S2=T2/T1)
Now time taken by B to cover (5/3)X at 0.9 times its original speed should be 40 mins
Time take by B to cover X distance is (40/T2=(5/3X)/X)
T2=24 which satisfies our given condition hence C is our answer

Q: 36 mins + (t+6)mins) = t + 42 mins =?

Assuming total distance = d
Cyclist 1 - x distance in 36 mins
Cyclist 2 - (d-x) distance in 36 mins

After 36 mins,
Cyclist 1 - rate doubled (2x/36 = x/18)
Cyclist 2 - speed reduced by 10% [90% of (d-x)/36 = (d-x)/40]

Cyclist 1 - Rate*time = (d-x)
=> (x/18)*(t+6) = (d-x) ----- (1)
Cyclist 2 - Rate*time = x (remaining distance)
=> [t(d-x)]/40 = x
Substituting (d-x) from (1)
=> t{(x/18)*(t+6)} = 40x
=> t(t+6) = 720
=> t^2 +6t - 720 = 0
=> (t+30)(t-24) = 0
=> t = 24 (cannot be -ve)

Q: t + 42 mins =?
= 66 mins or 1 hour 6 mins [C]

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

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
If we go by reverse calculation, it means B taken 54 Minutes hence not correct

B. 1 hour and 4 minutes
If we go by reverse calculation, it means B taken 58 Minutes hence not correct

C. 1 hour and 6 minutes
If we go by reverse calculation, it means B taken 60 Minutes hence correct option

D. 1 hour and 10 minutes
If we go by reverse calculation, it means B taken 64 Minutes hence not correct

E. 1 hour and 36 minutes
If we go by reverse calculation, it means B taken 90 Minutes hence not correct

Option C is correct

let total distance between A and B be d
let the speeds of the cyclists be a,b respectively

using relative speed concept, you know

d=36(a+b) ----->1

also,
d= 36a+(t+6)0.9a------>2
d= 36a+(t) 2b------------>3

from 1 and 2, we get at + 6a = 40b -----> 4
from 1 and 3, we get b= (18a)/t ---------> 5

substitute 5 in 4 and solve for t. You get t as 24 mins

So total time taken by cyclist 1 is 24+36+6 = 1 hr and 6 mins

Before the meeting, A's distance = 36x and B's distance = 36y
After the meeting, A's distance = (time + 6) * 2x and B's distance = time * 0.9y

1) A's before meeting distance = B's after meeting distance
i.e. \(36x = time * 0.9y\)
therefore, \(time = \frac{36x}{0.9y}\)
2) B's before meeting distance = A's after meeting distance
i.e. \(36y = (time + 6) * 2x\)

Substituting Equation 1 into 2: \(\frac{36y}{2x} = \frac{36x}{0.9y }+ 6\)

Let \(a = \frac{y}{x}\)

Solving, \(18a = \frac{40}{a} + 6\)
\(18a^2 = 40 + 6a\)
\(18a^2 - 6a - 40 = 0\)
\(9a^2 - 3a - 20 = 0\)
\(9a^2 - 15a + 12a - 20 = 0\)
\(3a(3a - 5) + 4(3a - 5) = 0\)
\((3a-5) (3a+4) = 0\)
\(a = +\frac{5}{3} or a = -\frac{4}{3}\)

Time = \(36 + \frac{36y}{2x}\)
= \(36 + 18 * \frac{5}{3} \)
= \(36+30\)
= \(66 \)

Therefore, option C is the correct answer.

From the diagram we can see that, first and second cyclist met at point C after 36 mins. After that second cyclist took "t" mins to reach "A" and the first cyclist took "t+6" mins to reach point "B".

Let SI is speed of first cyclist from "A" to "C", then 2*SI is its speed from "C" to "B". Similarly, if SII is the speed of second cyclist from "B" to "C", then 0.9*SII is its speed from "C" to "A".

From point "A" to "C", for the same distance, first and second cyclist took different time to travel. So their speeds are inversely proportional to their time of travel.

Therefore, SI/0.9*SII = t/36.

Similarly from "C" to "B" we get:- 2*SI/SII = 36/(t+6)

Eliminating SIand SII from the above equations we get a quadratic equation:- t^2 + 6t - 720 = 0.
So we get t = 24 mins. Hence the time required for the first cyclist to go from "A" to "B" is 36 + 24 + 6 = 66 mins, that is 1 hour 6 mins

Given,
2 cyclists (suppose them A and B) met at a point in 36 minutes while both were moving at constant speeds.
Let the speeds of A and B be u and v respectively.
So at the meeting point the distance covered by:

Using Distance = Speed * Time

Dis. A = 36 * u
Dis. B = 36 * v

Now after crossing each other A travelled the distance that B travelled and vice versa. But the speeds and time were different.
Considering the time taken for cyclist B as tWe can make the equations from the data given on the questins out to be:

Dis. A (Remaining distance travelled by Cyclist B after crossing) = 0.9v * t
Dis. B (Remaining distance travelled by Cyclist A after crossing) = 2u * (t + 6)

Replacing the value of distances from the above equations, we get:

\(36u = 0.9vt\)
\(36v = 2u(t + 6)\)

Substituting value of either u or v from the 1st equation in the second and simplifying gives us the equation:

\(t^2 + 6t -720 = 0\)

Solving for t gives us t = 24 or -30
Since, t cannot be negative t = 24.
So total time taken by A = 36 + (t +6) = 36 + 24 + 6 = 66 minutes or 1 hour and 6 minutes.