jjjinapinch wrote:In cross section, a tunnel that carries one lane of one-way traffic is a semicircle with radius 4.2 m. Is the tunnel large enough to accommodate the truck that is approaching the entrance to the tunnel?
(1) The maximum width of the truck is 2.4 m
(2) The maximum height of the truck is 4 m
Official Guide question
Answer: C
We are given that the entrance of the shape of the tunnel is a semi-circle that has a radius of 4.2 meters.
We need to ensure whether the tunnel is large enough to accommodate the truck.
To answer this question we need to ensure that the truck's cross section that is of rectangular shape can be accommodated if the truck follows the middle-most path inside the tunnel.
Since the radius of the tunnel is 4.2 m, the diameter of the tunnel = 2*4.2 = 8.4 m. For the truck to pass through the tunnel, the width of the truck < 8.4 m.
Since the radius of the tunnel is 4.2 m, the height of the tunnel at the center = 4.2 m (Maximum). As we move from the center of the semi-circular cross section of the tunnel to the wall of the tunnel, the height of the tunnel keeps decreasing. For the truck to pass through the tunnel, the maximum height of the truck < the height of the tunnel at the vertical edge of the truck.
Statement 1: The maximum width of the truck is 2.4 m.
Clearly insufficient as we do not know the height of the truck. If the height of the truck is too less, it will pass; however, if it is too high, it will not. Insufficient.
Statement 2: The maximum height of the truck is 4 m.
Clearly insufficient as we do not know the width of the truck. If the width of the truck is too less, it will pass; however, if it is too high, it will not. Insufficient.
Statement 1 & 2 combined:
Since the maximum width of the truck is 2.4 m, which is less than the diameter (8.4 m) of the tunnel, it is not a problem.
We must now calculate the height of the tunnel where the vertical edge of the truck is positioned; say that position is V.
Assume that the truck follows the middle-most path.
Height of the tunnel at position V = √[(Radius)^2 – (half of width of the truck)^2]
= √[(4.2)^2 – (1.2)^2]
Case 1: If √[(4.2)^2 – (1.2)^2] < 4 (Max. height of the truck); the answer is YES.
Case 2: If √[(4.2)^2 – (1.2)^2] = 4 (Max. height of the truck); the answer is YES.
Case 3: If √[(4.2)^2 – (1.2)^2] > 4 (Max. height of the truck); the answer is NO.
We need not calculate the value of √[(4.2)^2 – (1.2)^2]; whatever be the value of √[(4.2)^2 – (1.2)^2], we are sure that either the TRUCK would PASS or the TRUCK would NOT PASS. The answer is unique. Sufficient.
The correct answer:
C
For your curiosity to know whether the truck would pass through or not (Not need though in a DS question), let's calculate the value of √[(4.2)^2 – (1.2)^2 = √[(4.2 + 1.2)*(4.2 – 1.2)] = √[5.4 * 3] = √16.2
Since √16.2 < 4 (Max. height of the truck), the truck would pass.
Hope this helps!
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