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Which is greater: 10^(-100) + 99^100 or 9^(-99) + 100^99?
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26 Oct 2017, 23:44
26 Oct 2017, 23:44
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Question Stats:
45% (02:08) correct
54% (00:57) wrong based on 33 sessions
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Quantity A |
Quantity B |
10^(-100) + 99^100 |
9^(-99) + 100^99 |
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
Kudos for correct solution.
Re: Which is greater: 10^(-100) + 99^100 or 9^(-99) + 100^99?
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18 May 2018, 01:08
18 May 2018, 01:08
Can Someone pls explain the solution.
Re: Which is greater: 10^(-100) + 99^100 or 9^(-99) + 100^99?
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01 Feb 2023, 23:26
01 Feb 2023, 23:26
1
In order to understand the problem, we first need to analyse one key trick, the trick is
Given two numbers suppose "a" and "b" such that, b > a;
then we can conclude that, a^b > b^a, exceptions for the numbers 0,1,2,3;
Now, we can see that,
99^100 > 100^99
Now looking at the first part, for each quantity, we get,
10^(-100) and 9^(-99)
= 1/10^100 and 1/9^99
The above terms are so small infront of the their second term, that we can completely neglect them.
Hence the final question becomes, to compare between 99^100 and 100^99
And clearly 99^100 > 100^99
So Answer is option A
Given two numbers suppose "a" and "b" such that, b > a;
then we can conclude that, a^b > b^a, exceptions for the numbers 0,1,2,3;
Now, we can see that,
99^100 > 100^99
Now looking at the first part, for each quantity, we get,
10^(-100) and 9^(-99)
= 1/10^100 and 1/9^99
The above terms are so small infront of the their second term, that we can completely neglect them.
Hence the final question becomes, to compare between 99^100 and 100^99
And clearly 99^100 > 100^99
So Answer is option A
