iikarthik wrote:Hi,

Need help
Pls solve these two problems and post the Explanations.

Thanks
karthik

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%

OA IS C

Q6:The figure BELOW 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
[/img]

OA IS B

1st question :
lets take 100
20 people have car and roundtrip ticket

60% have round trip ticket and no car. so 40% definetely has car n roundtrip tickets

now 40%Roundtrip = 20
= 50

2 :
when you draw a straight line from the top of the tunnel to the base which is 20 it divides in the geometric mean
i.e bd^2 = 4*16 Hence the height is 8.
Therefore 8-0.5 = 7.5
I do not know how to draw a figure else i wud have explained it

:The figure BELOW 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

Sorry , It would have been better if I illustarted my reasoning by using a drawing.
The diameter is 20 feet, but the lane is 12 feet wide, leaving a space of 4 feet on both sides of the lane.
So the length from center to the end of the lane is 12/2 = 6 feet. Draw a perpendicular line from the point at 6 feet to the circumference of the circle. We will have three sides ( a triangle ): the base (6 feet ) , the height (h) and the hypotenuse connecting base and the perpendicular line which will be the same as the raduis ( 10 feet ) . Using pythagorean, height = sqrt (raduis ^2 - base ^2) = sqrt (10^2 - 6^2 )= 8. Give a clearance of 1/2 feet . 8 - 1/2 = 7 1/2. [/img]

iikarthik wrote:
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%

We can let the number of passengers = 100.

Thus, 20 passengers held round trip tickets and took their cars aboard the ship. Since we are given that 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, the 20 passengers who held round trip tickets and took their cars aboard the ship represent 40 percent of the passengers with round-trip tickets. If n = the number of passengers with round-trip tickets, we have:

20 = 0.4n

n = 20/0.4 = 200/4 = 50

Thus, 50/100 = 50% of the ship's passengers held round-trip tickets.

Answer: C