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01 Sep 2022, 03:16
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
68% (01:24) correct
32%
(01:44)
wrong
based on 69
sessions
History
Date
Time
Result
Not Attempted Yet
Is m a multiple of 588?
(1) m is a multiple of 21
(2) m is a multiple of 84
(1) m is a multiple of 21
(2) m is a multiple of 84
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01 Sep 2022, 18:57
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carcass
To determine: Is m a multiple of 588?
Let us do some pre-work before the statements in order to rephrase the question into something simpler.
\(588 = 2^2*3*7^2\)
So, basically the question is asking whether m has a minimum of two 2's, one 3, and two 7's in its prime factorization because then it will be a multiple of 588, else it will not. Let us now examine the statements
(1) m is a multiple of 21
\(21 = 3*7\)
m is a multiple of 21 => m has at least one 3 and one 7 in its prime factorization
But, it could have the remaining minimum two 2's and another 7, or it could not, it is not known to us via this statement
NOT SUFFICIENT
(2) m is a multiple of 84
\(84 = 2^2*3*7\)
m is a multiple of 84 => m has at least two 2's, one 3, and one 7 in its prime factorization
But, it could have the remaining minimum one 7, or it could not, it is not known to us via this statement
NOT SUFFICIENT
Combined Together
m is a multiple of 21 and 84
When we are given information that a number is a multiple of two numbers x and y, then we know that the number is also a multiple of the LCM of x and y
Similarly m is also a multiple of LCM (21,84) => m is a multiple of 84
Alas, this brings us back to statement 2, which was not sufficient
NOT SUFFICIENT
Answer - E
02 Sep 2022, 00:32
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Question
Is m a multiple of 588?
Inference:
m = 588 * n ;in which n is an integer
m = \(2^3 * 3^1 * 7^2 \)* n
Statement 1
m is a multiple of 21
m = 7 * 3 * p
We know that m contains one 7 and one 3, but we are not sure of the composition of p.
If p contains the missing powers, then m is a multiple otherwise its not a multiple.
Hence, we cannot conclude if m a multiple of 588
Statement 2
m is a multiple of 84
m = \(2^3 \)* 3 * 7 * q
We know that m contains \(2^3 \)* 3 * 7 , but we are not sure of the composition of q.
If q contains the missing powers, then m is a multiple otherwise its not a multiple.
Hence, we cannot conclude if m a multiple of 588
Combined
From statement 1 we know that m has one three and one seven in it. From statement 2, we know that m has three twos , one seven and one three in it.
However we dont have any information if m also contains another seven in it (we need at least two seven, three twos and one three to conclude that m is a multiple of 588). Hence the combination does not help as well.
Option E
Is m a multiple of 588?
Inference:
m = 588 * n ;in which n is an integer
m = \(2^3 * 3^1 * 7^2 \)* n
Statement 1
m is a multiple of 21
m = 7 * 3 * p
We know that m contains one 7 and one 3, but we are not sure of the composition of p.
If p contains the missing powers, then m is a multiple otherwise its not a multiple.
Hence, we cannot conclude if m a multiple of 588
Statement 2
m is a multiple of 84
m = \(2^3 \)* 3 * 7 * q
We know that m contains \(2^3 \)* 3 * 7 , but we are not sure of the composition of q.
If q contains the missing powers, then m is a multiple otherwise its not a multiple.
Hence, we cannot conclude if m a multiple of 588
Combined
From statement 1 we know that m has one three and one seven in it. From statement 2, we know that m has three twos , one seven and one three in it.
However we dont have any information if m also contains another seven in it (we need at least two seven, three twos and one three to conclude that m is a multiple of 588). Hence the combination does not help as well.
Option E
