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If $a$ and $b$ are integers and $a-b$ is even, which of the following must always be even?

  1. $ab$
  2. $a^{2}+b^{2}+1$
  3. $a^{2}+b+1$
  4. $ab-b$
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Migrated from GO Mechanical 3 years ago by Arjun

3 Answers

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Best answer

Given that $:a,b\in \mathbb{Z}$

$a-b = \text{even}$

  • $\text{even} -  \text{even}= \text{even}\:\color{Green}\checkmark{}$
  • $\text{even} -  \text{odd}= \text{odd}$
  • $\text{odd} -  \text{even}= \text{odd}$
  • $\text{odd} -  \text{odd}= \text{even}\:\color{Green}\checkmark{}$

Case$1:a=\text{even}, b = \text{even}\implies a-b=\text{even}$ 

  • $\text{even} \times \text{even}= \text{even}$
  • $\text{even} \times \text{odd}= \text{even}$
  • $\text{odd} \times \text{even}= \text{even}$
  • $\text{odd} \times \text{odd}= \text{odd}$
  • $\text{even} + \text{even}= \text{even}$
  • $\text{even} + \text{odd}= \text{odd}$
  • $\text{odd} + \text{even}= \text{odd}$
  • $\text{odd} + \text{odd}= \text{even}$
  1. $ab\:\color{Green}\checkmark{}$
  2. $a^{2}+b^{2}+1$
  3. $a^{2}+b+1$
  4. $ab−b\:\color{Green}\checkmark{}$

Case$2:a=\text{odd}, b = \text{odd}\implies a-b=\text{even}$ 

  1. $ab$
  2. $a^{2}+b^{2}+1$
  3. $a^{2}+b+1$
  4. $ab−b\:\color{Green}\checkmark{}$

So, the correct answer is $(D).$ 

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2 votes
2 votes

Since $a-b$ is even, this means

  • both of them must be even or, ($a = 2n+k$ and $b=2n$ and where $k$ is even and $n$ might be even or odd so that $a-b=k$)
  • both of them must be odd ($a=n$, and $b=n+k$ where $p$ is odd and $k$ is even so that $a-b=k$)

 

Case I: Both are even, say 4 and 6

A. $ab$ = 24

B. $a^2+b^2+1$ = 53

C. $a^2+b+1$ = 23

D. $ab−b$ = 18


Case II: Both are odd, say 3 and 7

Since options 2 and 3 are not true we will look at only options 1 and 4

A. $ab$ = 21

B. NA

C. NA

D. $ab−b$ = 14


So, $ab-b$, option $D$ is the answer.

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The possible cases of a-b even is when:- 

  1. a and b is odd
  2. a and b is even so, substituting values of a=5 and b=3 or a=8 and b=4 and checking options we get answer as (d) option.  
Answer:

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