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# Two friends A and B simultaneously start running around a circular tra

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Math Expert
Joined: 02 Sep 2009
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Two friends A and B simultaneously start running around a circular tra  [#permalink]

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20 Apr 2020, 05:25
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95% (hard)

Question Stats:

24% (01:56) correct 76% (01:58) wrong based on 134 sessions

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Two friends A and B simultaneously start running around a circular track in the same direction, from the same point. A runs at 6 m/s and B runs at b m/s, where b is a positive integer. If they cross each other at exactly two points on the circular track, how many values can b take?

A. 1
B. 2
C. 3
D. 4
E. 5

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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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04 Jun 2020, 17:04
2
If two runners moves in same direction with simplified ratio of speed a:b where a>b then they meet a-b times in the circular track.
And if the runners moves in opposite direction then they meet a+b times on a circular track.
Now,
6:b can be 3:1, 1:3, 3:5
So b can be 2, 18 and 10 respectively.
Don't take 2:4 because it will reduced to 1:2 so only three values of b are possible.

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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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08 Jun 2020, 12:48
2
2
Bunuel wrote:
Two friends A and B simultaneously start running around a circular track in the same direction, from the same point. A runs at 6 m/s and B runs at b m/s, where b is a positive integer. If they cross each other at exactly two points on the circular track, how many values can b take?

A. 1
B. 2
C. 3
D. 4
E. 5

Solution:

If two objects are moving around a circle at different speeds, the number of distinct points on the circle at which they will meet can be determined by the difference between the (most) reduced ratios of their speeds.

Since b can be either greater than 6 or less than 6, let’s first assume that b < 6. If b < 6, we can let the reduced ratio be 6/b = r/s such r - s = 2. Notice that after being reduced, r < 6. So (r, s) can be (3, 1), (5, 3).

If (r, s) = (3, 1), we have 6/b = 3/1 and this gives us b = 2.

If (r, s) = (5, 3), we have 6/b = 5/3 and this gives us b = 18/5.

However, since b is a positive integer, b can’t be 18/5. So we only have one value for b when b < 6. Next, let’s assume now b > 6. Again we can let the reduced ratio be 6/b = r/s such s - r = 2. Notice that after being reduced, r < 6. So (r, s) can be (1, 3), (3, 5).

If (r, s) = (1, 3), we have 6/b = 1/3 and this gives us b = 18.

If (r, s) = (3, 5), we have 6/b = 3/5 and this gives us b = 10.

Since both values of b are integers, we have 2 integer values for b when b > 6. Therefore, altogether, there are 3 integer values for b.

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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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06 Jun 2020, 10:00
1
let say time taken by both A and B to meet is same
so Ta=Tb
and no of turns taken by B = n

Ta=Tb
2(pi)r/6=n2(pi)r/b

n=6/b
nb=6

so in order to do 2 meeting n=2
so, 2*b=6
b=3

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Two friends A and B simultaneously start running around a circular tra  [#permalink]

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21 Apr 2020, 05:03
Bunuel wrote:
Two friends A and B simultaneously start running around a circular track in the same direction, from the same point. A runs at 6 m/s and B runs at b m/s, where b is a positive integer. If they cross each other at exactly two points on the circular track, how many values can b take?

A. 1
B. 2
C. 3
D. 4
E. 5

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Solution:

• Speed of A = 6 m/s
• Speed of b = b m/s
o Ratio of A and B = 6: b = m : n [After eliminating the common factors]
NOTE:- Number of distinct points at which they (the bodies in circular motion) will meet can be determined by finding out the reduced ratios of their speeds (m: n).
• m -n =2 or n-m =2
• Case I: If m >n and m-n = 2
o Then, 6 > b, the possible value of b = 2.
 $$\frac{6}{2}$$ = $$\frac{3}{1}$$, and $$3-1 = 2$$
• Case II: If m <n and n-m = 2
o Then, 6 < b, the possible value of b = 10 or18.
 $$\frac{6}{18} = \frac{1}{3}$$, and $$3-1 = 2$$
$$\frac{6}{10} = \frac{3}{5}$$, and $$5-3 = 2$$
• There are 3 different possible values of b.
Hence, the correct answer is Option C.
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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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04 Jun 2020, 10:23
Hi chetan2u , any easier way to do this ?
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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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15 Jun 2020, 21:43
ScottTargetTestPrep
If two objects are moving around a circle at different speeds, the number of distinct points on the circle at which they will meet can be determined by the difference between the (most) reduced ratios of their speeds.

Sorry if it too elementary, can I check if the above statement is applicable irrespective of their direction i.e. same or opposite. If not, how does it change?
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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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15 Jun 2020, 21:54
Bunuel wrote:
Two friends A and B simultaneously start running around a circular track in the same direction, from the same point. A runs at 6 m/s and B runs at b m/s, where b is a positive integer. If they cross each other at exactly two points on the circular track, how many values can b take?

A. 1
B. 2
C. 3
D. 4
E. 5

The number of meeting points of two runner in same direction on a circular track = a-b

The number of meeting points of two runner, running in opposite direction on a circular track = a+b

la-bl = 2 where a = 6

so b = 4 or 8

IMO B
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Two friends A and B simultaneously start running around a circular tra  [#permalink]

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19 Jun 2020, 06:05
ScottTargetTestPrep wrote:
Bunuel wrote:
Two friends A and B simultaneously start running around a circular track in the same direction, from the same point. A runs at 6 m/s and B runs at b m/s, where b is a positive integer. If they cross each other at exactly two points on the circular track, how many values can b take?

A. 1
B. 2
C. 3
D. 4
E. 5

Solution:

If two objects are moving around a circle at different speeds, the number of distinct points on the circle at which they will meet can be determined by the difference between the (most) reduced ratios of their speeds.

Since b can be either greater than 6 or less than 6, let’s first assume that b < 6. If b < 6, we can let the reduced ratio be 6/b = r/s such r - s = 2. Notice that after being reduced, r < 6. So (r, s) can be (3, 1), (5, 3).

If (r, s) = (3, 1), we have 6/b = 3/1 and this gives us b = 2.

If (r, s) = (5, 3), we have 6/b = 5/3 and this gives us b = 18/5.

However, since b is a positive integer, b can’t be 18/5. So we only have one value for b when b < 6. Next, let’s assume now b > 6. Again we can let the reduced ratio be 6/b = r/s such s - r = 2. Notice that after being reduced, r < 6. So (r, s) can be (1, 3), (3, 5).

If (r, s) = (1, 3), we have 6/b = 1/3 and this gives us b = 18.

If (r, s) = (3, 5), we have 6/b = 3/5 and this gives us b = 10.

Since both values of b are integers, we have 2 integer values for b when b > 6. Therefore, altogether, there are 3 integer values for b.

Hi Scott,
Why can't (r,s) take the values (4,2) and (2,4) when r>s and s>r respectively?
Is it that the values of r must be factors of 6? That would eliminate (4,2) as an option but (2,4) is still viable.
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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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19 Jun 2020, 06:15
Bunuel wrote:
Two friends A and B simultaneously start running around a circular track in the same direction, from the same point. A runs at 6 m/s and B runs at b m/s, where b is a positive integer. If they cross each other at exactly two points on the circular track, how many values can b take?

A. 1
B. 2
C. 3
D. 4
E. 5

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Bunuel what if the both start together with same speed which is 6. then B reduces his speed <6 then again increases>6 , then again decreases less than 6.

In this case b can have 4 values. . or does crossing also mean that if both are at same pace and one takes the lead from there.
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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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27 Jun 2020, 04:55
ShreyasJavahar wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
Two friends A and B simultaneously start running around a circular track in the same direction, from the same point. A runs at 6 m/s and B runs at b m/s, where b is a positive integer. If they cross each other at exactly two points on the circular track, how many values can b take?

A. 1
B. 2
C. 3
D. 4
E. 5

Solution:

If two objects are moving around a circle at different speeds, the number of distinct points on the circle at which they will meet can be determined by the difference between the (most) reduced ratios of their speeds.

Since b can be either greater than 6 or less than 6, let’s first assume that b < 6. If b < 6, we can let the reduced ratio be 6/b = r/s such r - s = 2. Notice that after being reduced, r < 6. So (r, s) can be (3, 1), (5, 3).

If (r, s) = (3, 1), we have 6/b = 3/1 and this gives us b = 2.

If (r, s) = (5, 3), we have 6/b = 5/3 and this gives us b = 18/5.

However, since b is a positive integer, b can’t be 18/5. So we only have one value for b when b < 6. Next, let’s assume now b > 6. Again we can let the reduced ratio be 6/b = r/s such s - r = 2. Notice that after being reduced, r < 6. So (r, s) can be (1, 3), (3, 5).

If (r, s) = (1, 3), we have 6/b = 1/3 and this gives us b = 18.

If (r, s) = (3, 5), we have 6/b = 3/5 and this gives us b = 10.

Since both values of b are integers, we have 2 integer values for b when b > 6. Therefore, altogether, there are 3 integer values for b.

Hi Scott,
Why can't (r,s) take the values (4,2) and (2,4) when r>s and s>r respectively?
Is it that the values of r must be factors of 6? That would eliminate (4,2) as an option but (2,4) is still viable.

Response:

According to my solution, (r, s) must satisfy two things: 1) the difference between them must equal 2, and 2) r/s must be in the most reduced terms (meaning r and s must be relatively prime). Your examples of (4, 2) and (2, 4) both meet the first requirement; however, neither 4/2 nor 2/4 is in the most reduced terms. We can reduce 4/2 to 2/1 and 2/4 to 1/2, but in that case the difference is not 2 anymore. That’s why (r, s) can’t take the values (4, 2) or (2, 4).
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Re: Two friends A and B simultaneously start running around a circular tra  [#permalink]

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29 Jun 2020, 07:53
vardaanaro wrote:
ScottTargetTestPrep
If two objects are moving around a circle at different speeds, the number of distinct points on the circle at which they will meet can be determined by the difference between the (most) reduced ratios of their speeds.

Sorry if it too elementary, can I check if the above statement is applicable irrespective of their direction i.e. same or opposite. If not, how does it change?

Response:

The statement depends on whether the two objects move in the same direction or in opposite directions. If r/s is the most reduced ratio of their speeds, then they will meet in r - s points if they move in the same direction and r + s points if they move in opposite directions.
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Re: Two friends A and B simultaneously start running around a circular tra   [#permalink] 29 Jun 2020, 07:53