求教TT数学2中的两题

marco_polo

[attachimg]63742[/attachimg]
The figure above shows the dimensions of a semicircular cross section of a one-way tunnel.  The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel.  If vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane, what should be the limit on the height of vehicles that are allowed to use the tunnel?

A. 5½ ft
B. 7½ ft
C. 8 ½ ft
D. 9½ ft
E. 10 ft
Answer: B

请问 vehicles must clear the top of the tunnel by at least ½ foot 和隧道半径10ft不就能直接算出height limit吗？没看明白怎么用题中的其他数据

Q18:
On a certain transatlantic crossing, 20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship.  If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held round-trip tickets?

A. 33 1/3%
B. 40%
C. 50%
D. 60%
E. 66 2/3%
Answer: C

如果总乘客人数为N，held round-trip tickets and also took their cars abroad the ship的人数就是0.2N，round-trip tickets did not take their cars abroad the ship的人数就是0.6N，passengers held round-trip tickets不就是0.8N吗？没看懂题目的意思，望指教。

chinamerica

1.  form the square inside the semicircle... vehicles
must clear the top of the tunnel by at least ½ foot, which means the
height limit would be the height of the square- 1/2
not sure whether the pic can be read here, see <http://picasaweb.google.com/rainpurple/Test/photo#5165768448675372786>

2.  60 percent of the passengers with round-trip tickets did not take their cars abroad the ship means 40% of
the passengers with round-trip tickets  take their cars abroad the
ship. Notive that the percent of the PASSENGERS WITH ROUNF-TRIP
TICKETS, rather than the ship's passengers ...
while the number passengers with round-trip tickets  take their cars abroad the ship = 
    20 % of the ship’s passengers = 40% of the passengers with round-trip tickets
which implies that  the passengers with round-trip tickets = 50% of the ship’s passengers

marco_polo

Thanks chinamerica

in question 1, why should i see the dimension of the vehicle in the tunnel as a square?

chinamerica

I should say rectangle. You have to guarantee that the height of the vehile should be 0.5 ft lower than the height of the rectangle since the vehile might be a rectangular shape(actually trucks do have this shape) and  might be wide enough (takes the whole lane)  

square root of (10^2- 6^2 ) - 1/2 = 8 - 1/2 = 7.5

[此贴子已经被作者于2008-2-12 3:39:17编辑过]

mmscute

The figure above shows the dimensions of a semicircular cross section of a one-way tunnel.  The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel.  If vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane, what should be the limit on the height of vehicles that are allowed to use the tunnel?

A. 5½ ft
B. 7½ ft
C. 8 ½ ft
D. 9½ ft
E. 10 ft
Answer: B

My understanding is as follows:

(1) since the lane is 12feet wide, the width of the vehicle must be less or equal to 12feet.

(2) to make the vehicle 1/2 feet clear the top of the tunnel, then the top and the two

up-corners (right up-corner and left up-corner) of the vehicle must at least 1/2feet away from the wall of the tunnel.

(3)if we take half of the cross section of the car as a rectangular, than its width is 6ft, hight is "h"ft and the diagonal 9.5ft. We can get "h" by h^2=(9.5)^2 - 6^2

 (I don't know how to draw picture here,hope my expanation makes sense to you)