Trenchard Boulevard begins at Ocean Street and runs directly east for

Hi All,

This is a wordy example of what's called a 'fence-post' problem. If you draw a picture, you'll find it's rather easy to deal with.

I'm going to give you some hints to get you started....

1) Draw a line segment (left to right)
2) Label the left point "Ocean" and the right point "Bay"
3) We're told that this line segment is 3 miles long and that every 1/10 of a mile there is a street that intersects this line segment. Starting next to "Ocean", draw an intersecting line and call it "1st Street", the next line is "2nd Street, the next is "3rd Street" and so on.

Without drawing every single intersection, can you figure out how many there are? (Hint: Remember that there's a beginning and an end to the 3 mile line).

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Total Perpendicular lines will be 31 (visualize by dividing a line in 3 part you need four lines,similarly dividing 3 miles in 30 parts need 31 lines
removing the first Ocean Street & Bay Street will give 29.

Ans: B

VERITAS PREP OFFICIAL SOLUTION:

Because there are intersections every tenth of a mile, and because Trenchard Boulevard is 3 miles long, there are 30 increments of 0.1 miles that divide Trenchard Boulevard. This means that there are 31 total intersections (perhaps difficult to visualize with 30 such increments, but look at a similar map with 2 increments:

I-----I-----I

there are three intersections to divide two sections with borders on each side, so you should see that you need to add one to the number of increments).

But since Bay and Ocean already have non-numbered names, 2 of the 31 intersections do not bear numerical names, so only 29 streets, 1st through 29th, fit the criteria. Therefore the highest-numbered street is 29th.

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