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The squares of two consecutive positive integers differ by 55. What is

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The squares of two consecutive positive integers differ by 55. What is  [#permalink]

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New post 18 Feb 2019, 00:53
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[GMAT math practice question]

The squares of two consecutive positive integers differ by 55. What is the smaller of the two integers?

A. 27
B. 29
C. 30
D. 32
E. 35

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Re: The squares of two consecutive positive integers differ by 55. What is  [#permalink]

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New post 18 Feb 2019, 01:00
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1
Let the consecutive integers be a and a+1
\((a+1)^2-a^2 = 55\)
\(a^2+2a+1-a^2 = 55\)
\(2a = 54\)
\(a = 27\)

The consecutive integers are 27 and 28, 27 being the smallest.

A is the answer.
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Re: The squares of two consecutive positive integers differ by 55. What is  [#permalink]

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New post 18 Feb 2019, 01:10
MathRevolution wrote:
[GMAT math practice question]

The squares of two consecutive positive integers differ by 55. What is the smaller of the two integers?

A. 27
B. 29
C. 30
D. 32
E. 35



given
square of two consecutive no is 55
lets solve considering answer options
27
(28^2-27^2) = 55

(28+27)*(28-27) = 55
55*1=55
LHS = RHS
sufficient
IMO A
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Re: The squares of two consecutive positive integers differ by 55. What is  [#permalink]

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New post 18 Feb 2019, 01:19
MathRevolution wrote:
[GMAT math practice question]

The squares of two consecutive positive integers differ by 55. What is the smaller of the two integers?

A. 27
B. 29
C. 30
D. 32
E. 35


Let smaller term = x
Next consecutive term = x+1

So now as per given, "The squares of two consecutive positive integers differ by 55"

\([m](x-1)^2\)[/m] - \(x^2\) = 55

just expand this

to get x = 27
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Re: The squares of two consecutive positive integers differ by 55. What is  [#permalink]

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New post 18 Feb 2019, 02:59
MathRevolution wrote:
[GMAT math practice question]

The squares of two consecutive positive integers differ by 55. What is the smaller of the two integers?

A. 27
B. 29
C. 30
D. 32
E. 35


*** consecutive positive integers.

x and x + 1

\(( x + 1)^2 - x^2 = 55\)

( x + 1 + x) ( x + 1 - x) = 55

2x + 1 =55

x = 27.


A is the correct answer.
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Re: The squares of two consecutive positive integers differ by 55. What is  [#permalink]

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New post 18 Feb 2019, 07:21
Top Contributor
MathRevolution wrote:
[GMAT math practice question]

The squares of two consecutive positive integers differ by 55. What is the smaller of the two integers?

A. 27
B. 29
C. 30
D. 32
E. 35


Let x = the smaller integer
So, x+1 = the larger integer (since the numbers are CONSECUTIVE)

The squares of two consecutive positive integers differ by 55.
We can write: (x + 1)² - x² = 55
Expand: x² + 2x + 1 - x² = 55
Simplify: 2x + 1 = 55
So: 2x = 54
Solve: x = 54/2 = 27

Answer: A

Cheers,
Brent
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Re: The squares of two consecutive positive integers differ by 55. What is  [#permalink]

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New post 20 Feb 2019, 18:36
=>

Let the two consecutive positive integers be \(n\) and \(n+1\).
Then \((n+1)^2 – n^2 = 55\), so \(2n+1 = 55.\)
It follows that \(2n = 54\) and \(n = 27\).

Therefore, the answer is A.
Answer: A
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Re: The squares of two consecutive positive integers differ by 55. What is   [#permalink] 20 Feb 2019, 18:36
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