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0.6^3 or 0.6^1/3
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Updated on: 04 Mar 2022, 06:29
Updated on: 04 Mar 2022, 06:29
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Question Stats:
66% (00:42) correct
33% (00:33) wrong based on 18 sessions
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Quantity A |
Quantity B |
\(0.6^3\) |
\(0.6^{\frac{1}{3}}\) |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: 0.6^3 or 0.6^1/3
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03 Mar 2022, 11:59
03 Mar 2022, 11:59
1
Carcass wrote:
Quantity A |
Quantity B |
\(0.6^3\) |
\(0.6^{\frac{1}{3}}\) |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Part A is simple, we can just multiply 0.6 three times in calc and get 0.216
Part B is little tricky as we have a decimal, so lets park it aside for the moment and focus on 6^1/3 ----> Closest perfect cube is 8 and it is 2^3, so 6^1/3 will defiantly needs to be lesser than 2 and if we divide it by 10, we will get a decimal no. less than 0.2. Hence making option B smaller than A
Ans is A
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Re: 0.6^3 or 0.6^1/3
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04 Mar 2022, 06:18
04 Mar 2022, 06:18
2
1
AkkuJi wrote:
Carcass wrote:
Quantity A |
Quantity B |
\(0.6^3\) |
\(0.6^{\frac{1}{3}}\) |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Part A is simple, we can just multiply 0.6 three times in calc and get 0.216
Part B is little tricky as we have a decimal, so lets park it aside for the moment and focus on 6^1/3 ----> Closest perfect cube is 8 and it is 2^3, so 6^1/3 will defiantly needs to be lesser than 2 and if we divide it by 10, we will get a decimal no. less than 0.2. Hence making option B smaller than A
Ans is A
AkkuJi
The denominator has a power of \(\frac{1}{3}\) too, so it cannot be parked aside.
Col. A: \(0.6^3\)
Col. B: \(0.6^{\frac{1}{3}}\)
Cubing both sides;
Col. A: \(0.6^9\)
Col. B: \(0.6\)
Col. A: \((\frac{6}{10})^9\)
Col. B: \(\frac{6}{10}\)
Remember: Whenever 0 < x < 1, on increasing the power the value decreases
Using the same logic,
Power in Col. A is 9 whereas, power in Col. B is 1
Clearly, Col. A < Col. B
Hence, option B
Use your on-screen calculator
\(0.6^3 = 0.216\)
cube-root of \(0.6 = 0.8434\)
