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07 Aug 2015, 06:17
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Difficulty:
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Question Stats:
75% (01:43) correct
25%
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If O and P are each circular regions, what is the area of the smaller of these regions?
1) The difference between the areas of the regions O and P is 21π
2) The difference between the circumferences of regions O and P is 6π
1) The difference between the areas of the regions O and P is 21π
2) The difference between the circumferences of regions O and P is 6π
07 Aug 2015, 07:10
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noTh1ng
If O and P are each circular regions, what is the area of the smaller of these regions?
1) The difference between the areas of the regions O and P is 21π
2) The difference between the circumferences of regions O and P is 6π
1) The difference between the areas of the regions O and P is 21π
2) The difference between the circumferences of regions O and P is 6π
Statement 1:
If the difference between the areas is \(21\pi\), then smaller area could be anything. For example, if smaller area is 2π, larger area is 23π; and if smaller area is 4π, larger area is 25π. Hence insufficient.
Statement 2:
Same reasoning as in the previous statement, smaller area could have any radius. Hence, insufficient.
Combined:
Statement 1 gives \(\pi{R}^{2} - \pi{r}^{2} = 21\pi\), which implies that \({R}^{2} - {r}^{2} = 21\), i.e. (R - r)(R + r) = 21.
Statement 2 gives \({2}\pi{R} - {2}\pi{r} = 6\pi\), which implies that R - r = 3
From the above two: R - r = 3 and R + r = 7, which gives r = 2.
Hence, the area can be determined using both statements together.
10 Aug 2015, 12:27
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noTh1ng
If O and P are each circular regions, what is the area of the smaller of these regions?
1) The difference between the areas of the regions O and P is 21π
2) The difference between the circumferences of regions O and P is 6π
1) The difference between the areas of the regions O and P is 21π
2) The difference between the circumferences of regions O and P is 6π
Let the radius of Circle O be R and P be r
St 1:
π* R^2 - π*r^2 = 21π
R^2 - r^2 = 21
(R+r)(R-r) = 21
We do not know anything about R or r, hence Not Sufficient.
St 2:
2πR - 2πr = 6π => R - r = 3. Not sufficient
Combining 1 and 2,
R+r = 7 and R-r = 3
Solving we can find R = 5 and r = 2.
Hence we can area of the smaller region.
Sufficient.
Option C.
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