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If x and y are consecutive integers such that x < y and x

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
Expert replies
by DivyaD » Fri Jan 18, 2019 4:14 am
If x and y are consecutive integers such that x < y and x is odd, which of the following must be true?
I. xy - y is odd.
II. x(x + y) is even.
III. x2 + y2 is odd.

a) None
b) III only
c) I and II
d) I and III
e) II and III
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Source: — Quantitative Reasoning |

DivyaD wrote:If x and y are consecutive integers such that x < y and x is odd, which of the following must be true?
I. xy - y is odd.
II. x(x + y) is even.
III. x² + y² is odd.

a) None
b) III only
c) I and II
d) I and III
e) II and III
GIVEN: x and y are consecutive integers such that x < y
So, if x is ODD, then y is EVEN

One option here is to replace the x's and y's with ODD and EVEN and apply the rules for ODDs and EVENs.
Or we can just replace x and y with odd and even numbers. Let's do that.

Replace x with 1 and replace y with 2 to get:
I. xy - y is odd.
xy - y = (1)(2) - 2 = 0, which is EVEN
Statement I is not true
ELIMINATE C and D

II. x(x + y) is even.
x(x + y) = (1)(1 + 2) = 3, which is ODD
Statement II is not true.
ELIMINATE E

III. x² + y² is odd.
x² + y² = 1² + 2² = 5, which is ODD
Statement III is true

Answer: B

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Fri Jan 18, 2019 5:12 pm, edited 1 time in total.
Brent Hanneson - Creator of GMATPrepNow.com
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by fskilnik@GMATH » Fri Jan 18, 2019 4:49 pm
DivyaD wrote:If x and y are consecutive integers such that x < y and x is odd, which of the following must be true?
I. xy - y is odd.
II. x(x + y) is even.
III. x^2 + y^2 is odd.

a) None
b) III only
c) I and II
d) I and III
e) II and III
$$\left. \matrix{
x\,\,{\rm{odd}} \hfill \cr
y = x + 1\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,?\,\,:\,\,{\rm{true}}$$

$$\left( {\rm{I}} \right)\,\,y\left( {x - 1} \right) = \underbrace {\left( {x + 1} \right)}_{{\rm{even}}}\underbrace {\left( {x - 1} \right)}_{{\rm{even}}}\,\,\mathop {\rm{ = }}\limits^? \,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {\rm{C}} \right),\left( {\rm{D}} \right)\,\,{\rm{refuted}}$$
$$\left( {{\rm{II}}} \right)\,\,x\left( {x + y} \right) = \underbrace {\,\,x\,\,}_{{\rm{odd}}}\underbrace {\left( {2x + 1} \right)}_{{\rm{odd}}}\,\,\mathop {\rm{ = }}\limits^? \,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {\rm{E}} \right)\,\,{\rm{refuted}}\,\,\,\left( {{\rm{also}}} \right)$$
$$\left( {{\rm{III}}} \right)\,\,{x^2} + {y^2} = \underbrace {\,\,{x^2}\,\,}_{{\rm{odd}}} + \underbrace {{{\left( {x + 1} \right)}^2}}_{{\rm{even}}}\,\,\mathop {\rm{ = }}\limits^? \,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {\rm{B}} \right)\,\,$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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Oops - pressed the wrong button.

As you were!
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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