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# A, B, and C are consecutive odd integers such that A < B < C.

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A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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01 Jun 2017, 12:28
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A, B, and C are consecutive odd integers such that A < B < C.

If A + B + C = 81, then A + C =

A) 52

B) 54

C) 56

D) 58

E) 60
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Re: A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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01 Jun 2017, 13:22
1
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Mo2men wrote:
A, B, and C are consecutive odd integers such that A < B < C.

If A + B + C = 81, then A + C =

A) 52

B) 54

C) 56

D) 58

E) 60

Say A = x - 2, B = x and C = x + 2 for some integer x. Then A + C = (x - 2) + (x + 2) = 2x.

A + B + C = 81;

(x - 2) + x + (x + 2) = 81;

3x = 81.

Then 2x = 81/1.5 = 81*2/3 = 54.

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Re: A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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01 Jun 2017, 15:15
1
B.
I took another route.
Since the sum is 81, the numbers will end in 5,7,9 respectively.
Hence the last digit is 4. Thus, B.

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A, B, and C are consecutive odd integers such that A < B < C  [#permalink]

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31 Aug 2017, 08:11
3
Mo2men wrote:
A, B, and C are consecutive odd integers such that A < B < C.

If A + B + C = 81, then A + C =

A) 52
B) 54
C) 56
D) 58
E) 60

as the numbers are consecutive so $$B$$= Average of the three numbers
Hence $$B = \frac{81}{3} = 27$$
So $$A+C = 81-27 = 54$$

Option B
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Re: A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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31 Aug 2017, 08:22
2n+1+2n+3+2n+5=81, using general form of odd numbers (2n+1,2n+3...and so on.
So n=12 after calculating. Hence numbers are 25,27,29 (after equating value of n in general form).
So A+C = 25+29=54 and answer is B

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A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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31 Aug 2017, 08:47
Mo2men wrote:
A, B, and C are consecutive odd integers such that A < B < C.

If A + B + C = 81, then A + C =

A) 52
B) 54
C) 56
D) 58
E) 60

Odd numbers are represented by $$2n+1$$

$$A$$ $$,$$ $$B$$ and $$C$$ are consecutive odd integers and $$A < B < C$$ .

Therefore $$A$$ $$,$$ $$B$$ and $$C$$ can be represented as $$= (2n+1), (2n+3)$$ and $$(2n+5)$$ respectively.

$$A + B + C = 81$$

$$(2n+1) + (2n+3) + (2n+5) = 81$$

$$2n + 1 + 2n+3+2n+5 = 81$$

$$6n + 9 = 81$$

$$6n = 81-9$$

$$6n = 72$$

$$n = \frac{72}{6} = 12$$

$$A = 2n+1 = 2(12) + 1 = 24 + 1 = 25$$

$$B = 2n+3 = 2(12) + 3 = 24 + 3 = 27$$

$$C = 2n+5 = 2(12) + 5 = 24 + 5 = 29$$

Therefore $$A + C = 25 + 29 = 54$$

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Re: A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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31 Aug 2017, 09:22
If A, B, C is consecutive odd integers, then A + B + C = 3A + 6 = 81 => 3A = 75 => A = 25, B = 27, C = 29
=> A + C = 54
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Re: A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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31 Aug 2017, 09:50
Mo2men wrote:
A, B, and C are consecutive odd integers such that A < B < C.

If A + B + C = 81, then A + C =

A) 52

B) 54

C) 56

D) 58

E) 60

Consecutive odd means
b= a+2
c=a+4
substitute the value of b and c in the given equation
a+a+2+a+4 =81
=> 3a+6=81
=> 3a= 81-6
=>a=75/3
=>a= 25
now the question is a+c
=>a+a+4
= 2a+4
=2*25+4
=50+4
=54
Hence B
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Re: A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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02 Sep 2017, 12:06
A, B, and C are consecutive, let alone that they're odd consecutive, B must be the mean; therefore B = 81/3 = 27.

A + C = 81 - B
A + C = 81 - 27 = 54

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A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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22 Dec 2017, 23:57
Bunuel chetan2u niks18

Quote:
Say A = x - 2, B = x and C = x + 2 for some integer x. Then A + C = (x - 2) + (x + 2) = 2x.

I did understand that you took this approach since it is much faster simplification and 2s get
cancelled out.

However the Q stem mentions to play with ODD nos , hence should we not take 2x+ 1, 2x+ 3...
Do we take later approach only when we have odd and even no in same Q?

In your approach if I take x = 2, or x=0 then I do not get consecutive odd numbers.
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Re: A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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23 Dec 2017, 00:54
Bunuel chetan2u niks18

Quote:
Say A = x - 2, B = x and C = x + 2 for some integer x. Then A + C = (x - 2) + (x + 2) = 2x.

I did understand that you took this approach since it is much faster simplification and 2s get
cancelled out.

However the Q stem mentions to play with ODD nos , hence should we not take 2x+ 1, 2x+ 3...
Do we take later approach only when we have odd and even no in same Q?

In your approach if I take x = 2, or x=0 then I do not get consecutive odd numbers.

x itself there denotes an odd integer. You could represent three consecutive integers as 2k - 3, 2k - 1, and 2k + 1 but it's not necessary here.
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A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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23 Dec 2017, 01:54
Bunuel chetan2u niks18

Quote:
Say A = x - 2, B = x and C = x + 2 for some integer x. Then A + C = (x - 2) + (x + 2) = 2x.

I did understand that you took this approach since it is much faster simplification and 2s get
cancelled out.

However the Q stem mentions to play with ODD nos , hence should we not take 2x+ 1, 2x+ 3...
Do we take later approach only when we have odd and even no in same Q?

In your approach if I take x = 2, or x=0 then I do not get consecutive odd numbers.

In algebra once you have defined a variable as ODD then it will never be even. So once x is defined as ODD you cannot assume x=2 or 0.

Just like in your example 2x+1, 2x+3.... etc. the implicit assumption is that x is even, so you cannot assume x=1 or any odd number because in that case 2x+1 will be even which is not possible.

So you can solve the problem using any kind of variable but once the variable is defined it stays as is.
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Re: A, B, and C are consecutive odd integers such that A < B < C.  [#permalink]

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24 Dec 2018, 12:28
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Re: A, B, and C are consecutive odd integers such that A < B < C.   [#permalink] 24 Dec 2018, 12:28
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