Trenchard Boulevard begins at Ocean Street and runs...

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AAPL: Moderator; Posts: 2632; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Wed Feb 21, 2018 2:43 pm

Trenchard Boulevard begins at Ocean Street and runs directly east for 3 miles until it ends when it meets Bay Street. Trenchard Boulevard is intersected exactly every tenth of a mile by a perpendicular street, and each of those streets other than Ocean Street and Bay Street is given a number beginning at 1st Street (one block east of Ocean Street) and continuing consecutively (2nd Street, 3rd Street, etc.) until the highest-numbered street one block west of Bay Street. What is the highest-numbered street that intersects Trenchard Boulevard?

A. 28th Street
B. 29th Street
C. 30th Street
D. 31st Street
E. 32nd Street

The OA is B.

I don't have clear this PS question. I appreciate if any expert explains it to me. Thank you so much.

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Source: Beat The GMAT — Problem Solving | Wed Feb 21, 2018 2:43 pm

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Tue Apr 03, 2018 10:01 am

The distance from ocean street to Bay street is 3 miles long.
Every 1/10 of a mile,there is a street that intersects this line segment. This is given as
$$\frac{1}{10}\cdot x=1mile$$
$$x=10\cdot1=10\ inter\cept\ or\ streets$$
$$Therefore,\ \frac{1}{10}\cdot x=3miles$$
$$x=10\cdot3=30\ intercept\ or\ streets\ between\ ocean\ street\ and\ bay\ street.$$
$$There\ will\ be\ 29\ intercepts\ or\ street.\ \left(answer\ is\ B\right)$$

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by [email protected] » Wed Apr 04, 2018 3:49 pm

Hi All,

We're told that Trenchard Boulevard begins at Ocean Street and runs directly east for 3 miles until it ends when it meets Bay Street. Trenchard Boulevard is intersected exactly every tenth of a mile by a perpendicular street, and each of those streets other than Ocean Street and Bay Street is given a number beginning at 1st Street (one block east of Ocean Street) and continuing consecutively (2nd Street, 3rd Street, etc.) until the highest-numbered street one block west of Bay Street. We're asked for the highest-numbered street that intersects Trenchard Boulevard.

This is a variation on a 'fence post' problem. Typically, in these types of questions you're asked to calculate the total number of items in a sequence (including the first AND last terms). For example, if you read pages 2 through 5 of a book (inclusive), then how many pages have you read? The answer is 4 (NOT 3) because you read pages 2, 3, 4 and 5 --> 4 pages. Here, we're asked to total up the number of streets BETWEEN Ocean Street and Bay street (since those streets are each given a number in the form 1, 2, 3, etc. all the way up to the last street BEFORE Bay Street).

The distance between Ocean Street and Bay Street is 3 miles and a new perpendicular street appears every 1/10 of a mile. Thus, there are (3)/(1/10) = 30 streets that to the east of Ocean Street - but that INCLUDES Bay Street (which we are told NOT to count). Thus, there are 30 - 1 = 29 streets and the last is labeled "29th Street."

Final Answer: B

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

Contact Rich at [email protected]

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Terry@ThePrincetonReview: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Mon Mar 12, 2018 10:26 amFollowed by: 1 member; Location: Minneapolis, Minnesota, USA

by Terry@ThePrincetonReview » Wed Apr 04, 2018 4:46 pm

I would probably take the "brute force" approach here and map it out carefully using cross-hatches. Why? Because in a question where the street count is only 30, it doesn't really take any more time to do it using the brute force method than by applying mathematical principles. I did this question in less than 40 seconds, most of it spent reading and understanding the question.

If the question involved a higher street number (say 370), I would of course rely on the mathematical approach, but I might still use a "scaled down" brute-force example to verify my approach, depending on how confident / nervous I was feeling that day and whether the question threw in any additional curve ball.

* How many units of length are involved? -- 30 (3 divided by 1/10 = 3 * 10).
* How many separators is that? -- 30 + 1 = 31.
* Each street is a separator. Thus, there are 31 separators/streets from Ocean Street to Bay Street, inclusive. Because the numbered streets 1, 2, 3 ... begin 1 block east of Ocean, let's number the separators/streets 0 to 30, making Ocean Street the equivalent of 0 (the 1st separator) and Bay Street the equivalent of 30 (the 31st separator). Neither Ocean nor Bay is actually numbered. The highest-numbered street is the one right before Bay Street, the second-to-last separator. Counting 0 to 30 where 30 is actually named "Bay Street", that makes 29 the highest-numbered street.

Terry Serres
The Princeton Review
Certified Instructor, Content Developer
Online Tutoring
[email protected]

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