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NandishSS
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The figure above represents a system of roads along which traffic flow Updated on: 30 May 2022, 01:18
Originally posted by NandishSS on 26 Jul 2016, 07:54.
Last edited by Bunuel on 30 May 2022, 01:18, edited 2 times in total.
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76% (01:44) correct 24% (01:45) wrong based on 1211 sessions

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The figure above represents a system of roads along which traffic flows in the direction indicated by the arrows. If the traffic were to split evenly at each fork of the road system, what percent of the traffic entering fork Z would exit at Y?

A. 87.5%
B. 75%
C. 62.5%
D. 50%
E. 12.5%

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The figure above represents a system of roads along which traffic flow 26 Jul 2016, 10:17
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NandishSS
The figure above represents a system of roads along which traffic flows in the direction indicated by the arrows. If the traffic were to split evenly at each fork of
the road system, what percent of the traffic entering fork C would exit at Y?

A. 87.5%
B. 75%
C. 62.5%
D. 50%
E. 12.5%

How to solve this question?
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The questions says traffic flow is spread evenly. That means, we have to divide the traffic along the two branches equally.

Lets say we have 100% traffic, we first slipt it into two as we have two branches at Z.

The lower branch will take the complete traffic towards Y. So, 50% of the traffic is coming towards Y via lower branch.

Now, the upper branch is further divided into two. So half of what it was carrying(50%) = 25 % would be be long each of those branches of upper branch. Now , the out of these two there is a further split in which one branch is towards Y and the other towards X. Thus, X will carry only 12.5% of the traffic while Y will carry another 12.5 from X along with the earlier 50% and 25%.

Thus, total traffic Y will carry is 87.5%. Hence , answer is A.
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The figure above represents a system of roads along which traffic flow 03 Aug 2016, 09:05
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NandishSS
The figure above represents a system of roads along which traffic flows in the direction indicated by the arrows. If the traffic were to split evenly at each fork of
the road system, what percent of the traffic entering fork C would exit at Y?

A. 87.5%
B. 75%
C. 62.5%
D. 50%
E. 12.5%

How to solve this question?
Show more

Roads divide into two after leaving from Z.

50 % move downwards and 50% move upwards

If we notice all vehicles moving downwards will end up taking exit Y.

The upward moving vehicles divide at first junction in 25% (moving upwards ) and 25 % (moving downwards). Again if we notice all vehicles moving downwards will end up taking exit Y.

Vehicles moving upwards from this point again divide, leading to 12.5% moving upwards and 12.5% moving downwards, and these will take exit Y.

Total vehicles taking exit Y= 50% +25% +12.5%= 87.5%

A is the answer
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The figure above represents a system of roads along which traffic flow 14 Jan 2022, 11:06
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Much easier to do this problem by looking at the splits in the route to X and subtracting from 1.

Counting the splits on the way to X shows only 3.

So (1/2)^3= 1/8 gets to X, meaning 7/8 gets to Y

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The figure above represents a system of roads along which traffic flow 24 Jul 2026, 04:54
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Just would like to share my method:

Split each initial fork by 50%:

1st road: split by 50%, and split by 50% until we reach X exit.

2nd road: Since 2nd road ended up on Y exit we sum 50% plus the other forks that took the 1st road but ended up in Y: 50% + 25% + 12.5% = 87.5%.

I hope it helps!
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The figure above represents a system of roads along which traffic flow 04 Aug 2026, 11:53
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FASTEST APPROACH:

There are only 2 possible exits: X or Y.
Instead of calculating all the traffic reaching Y, calculate the traffic reaching X.
To reach X, traffic must take the correct branch at 3 consecutive forks.
At each fork, traffic splits equally:
X = (1/2) × (1/2) × (1/2)
= 1/8
= 12.5%

Since all traffic must exit through either X or Y:
Y = 100% - 12.5%
= 87.5%
Answer: A

IMPORTANT:
Do NOT calculate Y by adding 50% + 25% + 12.5%.
Since there are only 2 exits, calculate the easier exit and subtract it from 100%.

TRAFFIC-FLOW / BRANCHING / SPLITTING DIAGRAMS
Default Method:
1. Assume 100 enters the system.
2. At every junction, split the ACTUAL amount arriving at that junction.
3. Follow each relevant branch.
4. Add all branches that reach the requested destination.

Example of successive 50-50 splits:
100 → 50 → 25 → 12.5
IMPORTANT:
Never keep halving the original 100.
You halve the amount arriving at THAT particular junction:
100 → 50
50 → 25
25 → 12.5
COMPLEMENT SHORTCUT:
You do NOT need to recognize the complement trick.

If it strikes you that there are only two possible exits, you can calculate the easier exit and subtract it from 100%.

But the safe default method is always:
Assume 100 enters → split actual amounts → trace branches → add amounts reaching the required destination.
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