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09 Jun 2016, 03:49
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B
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Difficulty:
35%
(medium)
Question Stats:
72% (01:57) correct
28%
(02:03)
wrong
based on 411
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What is the smallest of six consecutive odd integers whose average (arithmetic mean) is x + 2?
A. x - 5
B. x - 3
C. x - 1
D. x
E. x + 1
A. x - 5
B. x - 3
C. x - 1
D. x
E. x + 1
09 Oct 2017, 16:52
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Bunuel
We can let the first (or smallest) odd integer = n, and thus we have:
n, n + 2, n + 4, n + 6, n + 8, and n + 10 as the six consecutive odd integers.
Since the average of terms in an evenly spaced set is (first number + last number)/2:
(n + n + 10)/2 = x + 2
2n + 10 = 2x + 4
2n = 2x - 6
n = x - 3
Answer: B
01 Mar 2020, 14:31
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Bunuel
A student asked me to solve this question so.....
I'd typically use @ScottTargetTestPrep's approach, but here's another route.
Let k = the smallest odd integer
So, k + 2 = the next odd integer (since each consecutive odd integer is always 2 greater than the odd integer before it)
And k + 4 = the next odd integer
And k + 6 = the next odd integer
And k + 8 = the next odd integer
And k + 10 = the last (6th) odd integer
Since the average of the six integers is \(x+2\), we can write: \(\frac{k + (k+2)+(k+4)+(k+6)+(k+8)+(k+10)}{6}=x+2\)
Simplify: \(\frac{6k + 30}{6}=x+2\)
Multiply both sides of the equation by 6 to get: \(6k + 30=6x+12\)
What is the smallest of six consecutive odd integers?
Since k is the smallest odd integer, we need to solve this equation for k.
Take: \(6k + 30=6x+12\)
Subtract 30 from both sides: \(6k=6x-18\)
Divide both sides by 6 to get: \(k = x-3\)
Answer: B
Cheers,
Brent
