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21 Oct 2021, 03:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (02:02) correct
26%
(01:51)
wrong
based on 362
sessions
History
Date
Time
Result
Not Attempted Yet
If \(\frac{x−8}{6}\) is a positive integer, which of the following must be true?
(A) \(x\) is a multiple of 14.
(B) \(x ≥ 20\)
(C) \(x + 10\) is a multiple of 6.
(D) \(\frac{x+8}{6}\) is an integer.
(E) \(\frac{x−8}{12}\) is an integer.
(A) \(x\) is a multiple of 14.
(B) \(x ≥ 20\)
(C) \(x + 10\) is a multiple of 6.
(D) \(\frac{x+8}{6}\) is an integer.
(E) \(\frac{x−8}{12}\) is an integer.
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21 Oct 2021, 05:27
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Bunuel
Understand the question and try to infer as much as information possible
\(\frac{x−8}{6}\) is a positive integer
To be positive
x> 8
to be an integer
x- 8 must be multiple of 6
so we need to find a number greater than 8
Let us choose two values for x, which satisfy the given equation
14 & 20
and put in the options
(A) \(x\) is a multiple of 14.
x= 20 is not a multiple of 14
incorrect
(B) \(x ≥ 20\)
x =14 is less than 20
incorrect
(C) \(x + 10\) is a multiple of 6.
It satisfies both x = 14 & 20
(D) \(\frac{x+8}{6}\) is an integer.
if we put x = 14 then OD is not true
incorrect
(E) \(\frac{x−8}{12}\) is an integer.[/quote]
if we put x = 14 then OE is not true
incorrect
Answer C
26 Jan 2022, 07:58
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Let the positive integer be = I
Given,
x-8/6 = I
x-8 = 6I
x = 6I+8
(A) x is a multiple of 14.
If I = 1 then x = 14
If I = 2 then x = 20
Therefore, NOT always True.
(B) x≥20
If I = 1 then x = 14
If I = 2 then x = 20
Therefore, NOT always True.
(C) x+10 is a multiple of 6.
The last equation we got was:
x = 6I + 8
Adding 10 to both sides:
x+10 = 6I + 18 ---------- Dividing this equation by 6
\(\frac{x+10}{6}\) = \(\frac{6I}{6} + \frac{18}{6}\)
Both 6I and 18 are completely divisible by 6. Hence, x+10 is a multiple of 6.
Always TRUE (C)
(D) x+8/6 is an integer.
The last equation we got was:
x = 6I + 8
Adding 8 to both sides:
x+8 = 6I + 16
Divide both sides by 6:
\(\frac{x+8}{6} = \frac{6I}{6} + \frac{16}{6}\)
16/6 is NOT an integer.
Therefore, FALSE.
(E) x−8/12 is an integer.
The last equation we got was:
x = 6I + 8
Subtracting 8 to both sides:
x-8 = 6I + 8 - 8
Divide both sides by 12:
\(\frac{x-8}{12} = \frac{6I}{12}\)
If I=1 then 6I/12 is NOT an integer.
If I=2 then 6I/12 is an integer.
Therefore, NOT always True.
Given,
x-8/6 = I
x-8 = 6I
x = 6I+8
(A) x is a multiple of 14.
If I = 1 then x = 14
If I = 2 then x = 20
Therefore, NOT always True.
(B) x≥20
If I = 1 then x = 14
If I = 2 then x = 20
Therefore, NOT always True.
(C) x+10 is a multiple of 6.
The last equation we got was:
x = 6I + 8
Adding 10 to both sides:
x+10 = 6I + 18 ---------- Dividing this equation by 6
\(\frac{x+10}{6}\) = \(\frac{6I}{6} + \frac{18}{6}\)
Both 6I and 18 are completely divisible by 6. Hence, x+10 is a multiple of 6.
Always TRUE (C)
(D) x+8/6 is an integer.
The last equation we got was:
x = 6I + 8
Adding 8 to both sides:
x+8 = 6I + 16
Divide both sides by 6:
\(\frac{x+8}{6} = \frac{6I}{6} + \frac{16}{6}\)
16/6 is NOT an integer.
Therefore, FALSE.
(E) x−8/12 is an integer.
The last equation we got was:
x = 6I + 8
Subtracting 8 to both sides:
x-8 = 6I + 8 - 8
Divide both sides by 12:
\(\frac{x-8}{12} = \frac{6I}{12}\)
If I=1 then 6I/12 is NOT an integer.
If I=2 then 6I/12 is an integer.
Therefore, NOT always True.
