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12 Dec 2023, 08:08
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Official Solution:
This problem can be solved by testing the answers or it can be solved algebraically (though that method is not easy!). Both solutions are shown.
Test the Answers
When testing answers, start with answer (B) or (D). Note that is equivalent to 62.5%.
(B) D:d is in the ratio of 5:3. Let the diameters D and d be 10 and 6 units, respectively, so that the radii of the corresponding circles are 5 and 3 units. The sum of the areas of the two smaller circles is thus π(52) + π(32) = 34π square units. The diameter of the large circle is 10 + 6 = 16 units, so the large circle’s radius is 8 units and its area is π(82) = 64π square units. 34π is a bit more than half of 64π, but not more than 60%, so choice (B) is incorrect.
Try (D) next. D:d is in the ratio of 3:1. Let the diameters D and d be 6 and 2 units, respectively, so that the radii of the corresponding circles are 3 and 1 units. The sum of the areas of the two smaller circles is thus π(32) + π(12) = 10π square units. The diameter of the large circle is 6 + 2 = 8 units, so the large circle’s radius is 4 units and its area is π(42) = 16π square units. 10π / 16π = 5 / 8, so choice (D) is correct.
Strategy tip: If neither of those solutions had worked, then you would have looked at how close they were to the desired fraction of 5/8 and chosen accordingly. For example, if B were below 60% but D were above 65%, then the correct answer must be in the middle (answer C). If D had also been below 60%, then the correct answer would have been even greater (E).
Algebraic Solution
Let R be the radius of the circle whose diameter is D, and r the radius of the circle whose diameter is d. Then the ratio R:r is the same as the answer to the problem, since D:d = 2R:2r = R:r. Also, the diameter of the largest circle is 2R + 2r and its radius is half that value, or R + r.
The sum of the areas of the two smaller circles is , and the area of the large circle is . From the statement given in the problem, the former is five-eighths of the latter:
Divide out π:
Multiply by 8 to get rid of the fraction: 8R2 + 8r2 = 5R2 + 10Rr + 5r2
3R2 – 10Rr + 3r2 = 0
(3R – r)(R – 3r) = 0
3R = r or R = 3r
Since R > r, the 3R = r solution is impossible. Therefore, R = 3r, and so the desired ratio is R:r = 3:1.
This problem can be solved by testing the answers or it can be solved algebraically (though that method is not easy!). Both solutions are shown.
Test the Answers
When testing answers, start with answer (B) or (D). Note that is equivalent to 62.5%.
(B) D:d is in the ratio of 5:3. Let the diameters D and d be 10 and 6 units, respectively, so that the radii of the corresponding circles are 5 and 3 units. The sum of the areas of the two smaller circles is thus π(52) + π(32) = 34π square units. The diameter of the large circle is 10 + 6 = 16 units, so the large circle’s radius is 8 units and its area is π(82) = 64π square units. 34π is a bit more than half of 64π, but not more than 60%, so choice (B) is incorrect.
Try (D) next. D:d is in the ratio of 3:1. Let the diameters D and d be 6 and 2 units, respectively, so that the radii of the corresponding circles are 3 and 1 units. The sum of the areas of the two smaller circles is thus π(32) + π(12) = 10π square units. The diameter of the large circle is 6 + 2 = 8 units, so the large circle’s radius is 4 units and its area is π(42) = 16π square units. 10π / 16π = 5 / 8, so choice (D) is correct.
Strategy tip: If neither of those solutions had worked, then you would have looked at how close they were to the desired fraction of 5/8 and chosen accordingly. For example, if B were below 60% but D were above 65%, then the correct answer must be in the middle (answer C). If D had also been below 60%, then the correct answer would have been even greater (E).
Algebraic Solution
Let R be the radius of the circle whose diameter is D, and r the radius of the circle whose diameter is d. Then the ratio R:r is the same as the answer to the problem, since D:d = 2R:2r = R:r. Also, the diameter of the largest circle is 2R + 2r and its radius is half that value, or R + r.
The sum of the areas of the two smaller circles is , and the area of the large circle is . From the statement given in the problem, the former is five-eighths of the latter:
Divide out π:
Multiply by 8 to get rid of the fraction: 8R2 + 8r2 = 5R2 + 10Rr + 5r2
3R2 – 10Rr + 3r2 = 0
(3R – r)(R – 3r) = 0
3R = r or R = 3r
Since R > r, the 3R = r solution is impossible. Therefore, R = 3r, and so the desired ratio is R:r = 3:1.
