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

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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.

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

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New post 01 Jun 2017, 14:15
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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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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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New post 31 Aug 2017, 07:22
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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New post 31 Aug 2017, 07:47
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 \(A + C = 25 + 29 = 54\)

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

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New post 31 Aug 2017, 08:22
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
The answer is B
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Re: A, B, and C are consecutive odd integers such that A < B < C. [#permalink]

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New post 31 Aug 2017, 08:50
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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New post 02 Sep 2017, 11:06
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

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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New post 22 Dec 2017, 22:57
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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New post 22 Dec 2017, 23:54
adkikani wrote:
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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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 23 Dec 2017, 00:54
adkikani wrote:
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.


Hi adkikani

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.
A, B, and C are consecutive odd integers such that A < B < C.   [#permalink] 23 Dec 2017, 00:54
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