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01 Nov 2021, 04:45
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A
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Difficulty:
65%
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Question Stats:
60% (02:11) correct
40%
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wrong
based on 382
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\(\frac{(3.999999)}{(2.001)} - \frac{(3.999996)}{(2.002)} = ?\)
(A) \(\frac{(1)}{(1000)}\)
(B) \(\frac{(1)}{(10000)}\)
(C) \(\frac{(1)}{(100000)}\)
(D) \(\frac{(1)}{(1000000)}\)
(E) \(\frac{(1)}{(10000000)}\)
(A) \(\frac{(1)}{(1000)}\)
(B) \(\frac{(1)}{(10000)}\)
(C) \(\frac{(1)}{(100000)}\)
(D) \(\frac{(1)}{(1000000)}\)
(E) \(\frac{(1)}{(10000000)}\)
01 Nov 2021, 05:13
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Bunuel
\(\frac{(3.999999)}{(2.001)} - \frac{(3.999996)}{(2.002)} \)
\(\frac{(4-0.000001)}{(2.001)} - \frac{(4-0.000004)}{(2.002)} \)
\(\frac{(2^2-(0.001)^2)}{(2.001)} - \frac{(2^2-(0.002)^2)}{(2.002)} \)
Now we know that \(a^2-b^2=(a-b)(a+b)\).
\(\frac{(2+0.001)(2-0.001)}{2.001}-\frac{(2-0.002)(2+0.002)}{2.002}\)
\(\frac{2.001*1.999}{2.001}-\frac{2.002*1.998}{2.002}\)
\(1.999-1.998=0.001=\frac{1}{1000}\)
A
Updated on: 01 Nov 2021, 06:07
Originally posted by GMATVirtuoso on 01 Nov 2021, 05:09.
Last edited by GMATVirtuoso on 01 Nov 2021, 06:07, edited 8 times in total.
Last edited by GMATVirtuoso on 01 Nov 2021, 06:07, edited 8 times in total.
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On taking LCM of the denominators:-
(6 decimals*3 decimals - 6 decimals*3 decimals)/6 decimals
Least significant digits of the two numbers in the numerator are going to be 9*2 = 8 and 6*1 = 6. Since the least significant digit cannot be 0 on subtraction of these two numbers, and each number has 9 decimals, there will be 9 decimals post subtraction.
=> 9 decimals/6 decimals = 3 decimals
So answer is (A) 1/1000
Posted from my mobile device
(6 decimals*3 decimals - 6 decimals*3 decimals)/6 decimals
Least significant digits of the two numbers in the numerator are going to be 9*2 = 8 and 6*1 = 6. Since the least significant digit cannot be 0 on subtraction of these two numbers, and each number has 9 decimals, there will be 9 decimals post subtraction.
=> 9 decimals/6 decimals = 3 decimals
So answer is (A) 1/1000
Posted from my mobile device
