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25 Jun 2021, 02:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (02:12) correct
31%
(03:25)
wrong
based on 13
sessions
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A cylinder, a hemisphere and a cone have identical base and same height. The area of their curved surfaces will be in the ratio -
(A) 1:1:√2
(B) √2 : √2 : 2
(C) 2 : √2 : 2
(D) √2 : √2 : 1
(E) √2 : 1: √2
(A) 1:1:√2
(B) √2 : √2 : 2
(C) 2 : √2 : 2
(D) √2 : √2 : 1
(E) √2 : 1: √2
25 Jun 2021, 04:04
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Solution:
We know CSA of Hemisphere = \(2 \pi r^2\).
From the hemisphere, we get that the radius \(r\) and height \(h\) are both the same. So, \(r = h\)
cylinder cone hemisphere.png [ 21.69 KiB | Viewed 3200 times ]
CSA of Cylinder = \(2 \pi r h = 2\pi r^2\)
CSA of Cone = \(\pi r l = \pi r \sqrt{h^2+r^2} = \pi r \sqrt{r^2+r^2} = \pi r \sqrt{2r^2} = \sqrt{2}\pi r^2\)
So, CSA of Cylinder : Hemisphere : Cone = \(2\pi r^2 : 2\pi r^2 : \sqrt{2}\pi r^2\)
CSA of Cylinder : Hemisphere : Cone = \(\sqrt{2} : \sqrt{2} : 1\)
Hence the right answer is Option D.
We know CSA of Hemisphere = \(2 \pi r^2\).
From the hemisphere, we get that the radius \(r\) and height \(h\) are both the same. So, \(r = h\)
Attachment:
cylinder cone hemisphere.png [ 21.69 KiB | Viewed 3200 times ]
CSA of Cylinder = \(2 \pi r h = 2\pi r^2\)
CSA of Cone = \(\pi r l = \pi r \sqrt{h^2+r^2} = \pi r \sqrt{r^2+r^2} = \pi r \sqrt{2r^2} = \sqrt{2}\pi r^2\)
So, CSA of Cylinder : Hemisphere : Cone = \(2\pi r^2 : 2\pi r^2 : \sqrt{2}\pi r^2\)
CSA of Cylinder : Hemisphere : Cone = \(\sqrt{2} : \sqrt{2} : 1\)
Hence the right answer is Option D.
25 Jun 2021, 04:21
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This is way out of the scope of the GMAT if you actually want to solve the question properly -- you don't need to know anything about finding surface areas of cones or spheres on the GMAT.
You can answer the question in 5 seconds though: if you multiply the ratio in answer A by √2, you get answer B. So if A were correct, B would also be correct, and a question can't have two right answers. So A and B must both be wrong. Similarly if you multiply answer E by √2, you get answer C, so those must both also be wrong. D must be the answer.
It's never true though that the GMAT would ask a question where the technique I used above was the only reasonable way to get an answer. It's always possible to solve real GMAT questions using GMAT-level math, and within 2 minutes, and you can't do that with this question. But if anyone does want a full solution: I've never heard anyone talk about the "height" of a sphere or hemisphere, but I gather that's the radius (the vertical distance from circular base to the top point is the radius), which I'll call r. So if the "height" of the hemisphere is the height of the cone and of the cylinder, then the heights of the cone and cylinder are just r.
The surface area of a cylinder is the only relevant thing here to GMAT test takers. That surface area is 2πr^2 + 2πrh, but the "2πr^2" part is the area of the two circular ends. We just want the 'curved surface', which is the part that 'wraps around' the cylinder, and that has surface area 2πrh, and if h = r, that surface area is 2πr^2.
The surface area of a sphere (not tested on the GMAT) is 4πr^2. The surface area of half a sphere (a hemisphere) is thus just 2πr^2, plus the area of the circular base, but we're not meant to include that here.
The surface area of a cone (no way you'd ever need this on the GMAT - and I had to look it up) is the area of the circular base, which we ignore here, plus πr√(h^2 + r^2), and if h = r, that becomes πr√(r^2 + r^2) = πr√(2r^2) = √2 * πr^2
So their curved surface areas are 2πr^2, 2πr^2 and √2 πr^2, and dividing by πr^2, those are in a 2 to 2 to √2 ratio, which, dividing by √2, is the same as a √2 to √2 to 1 ratio.
You can answer the question in 5 seconds though: if you multiply the ratio in answer A by √2, you get answer B. So if A were correct, B would also be correct, and a question can't have two right answers. So A and B must both be wrong. Similarly if you multiply answer E by √2, you get answer C, so those must both also be wrong. D must be the answer.
It's never true though that the GMAT would ask a question where the technique I used above was the only reasonable way to get an answer. It's always possible to solve real GMAT questions using GMAT-level math, and within 2 minutes, and you can't do that with this question. But if anyone does want a full solution: I've never heard anyone talk about the "height" of a sphere or hemisphere, but I gather that's the radius (the vertical distance from circular base to the top point is the radius), which I'll call r. So if the "height" of the hemisphere is the height of the cone and of the cylinder, then the heights of the cone and cylinder are just r.
The surface area of a cylinder is the only relevant thing here to GMAT test takers. That surface area is 2πr^2 + 2πrh, but the "2πr^2" part is the area of the two circular ends. We just want the 'curved surface', which is the part that 'wraps around' the cylinder, and that has surface area 2πrh, and if h = r, that surface area is 2πr^2.
The surface area of a sphere (not tested on the GMAT) is 4πr^2. The surface area of half a sphere (a hemisphere) is thus just 2πr^2, plus the area of the circular base, but we're not meant to include that here.
The surface area of a cone (no way you'd ever need this on the GMAT - and I had to look it up) is the area of the circular base, which we ignore here, plus πr√(h^2 + r^2), and if h = r, that becomes πr√(r^2 + r^2) = πr√(2r^2) = √2 * πr^2
So their curved surface areas are 2πr^2, 2πr^2 and √2 πr^2, and dividing by πr^2, those are in a 2 to 2 to √2 ratio, which, dividing by √2, is the same as a √2 to √2 to 1 ratio.
