if at least one of p, q, r is an integer, is p+q+r even?

Question

GMATBusters’ Quant Quiz Question -10 
 For questions from previous quizzes click   here

if at least one of p, q, r is an integer, is p+q+r even?
(1) p−q−r is even.
(2) (p-r)/q is odd

gvij2017 · Answer

is p+q+r+s even?

Both statements don't tell about S. 
E is answer.

numb007 · Answer

Ans: D
i) p-q-r = even
 p=even+q+r
 p+q+r=(even+q+r)+q+r=even+2(q+r)=even
 
ii)p-r/q=odd 
 p=odd*q+r
 p+q+r=odd*q+r+q+r = 2*r + q*odd +q =2*r+q(odd+1)=even+even

pk123 · Answer

if at least one of p, q, r is an integer, is p+q+r even?
(1) p−q−r is even.
(2) (p-r)/q is odd

Given at least one of p, q and r is an integer
which means minimum 1 can be an integer or all 3 can also be integers
 
Question: p+q+r is even??

Option A: p-q-r = even
p = even + q + r
2p= even + (p+q+r)
so p+q+r = 2p - even
p can be integer or real number 
in case of integer 2p will be even and entire expression p+q+r will be even
in case of real p, 2p will be real and entire expression will be real number...Even/odd only exists in case of Integers not for real numbers
Insufficient hence

option B: (p-r)/ q is odd
(p-r)/q-1 = odd -1= even
(p - r - q)/q = even
p - r - q = even * q
p = even *q +r+q
2p= even *q + (p+q+r)
p+q+r = 2p- even * q

Combining both we get q as Integer ( how from option A and option B, even *q = even)
but still p can be real number or particular Integer hence can't be said on nature of entire expression p+q+r

Hence E is the answer

willacethis · Answer

For statement 1,

We know that even-even=even and odd-odd = even.

Now, since we know that 1 of the numbers is an integer, in order for p-q-r to be even, the sum of the other 2 numbers must be an integer of the same classification (odd or even). Ex. if p is assumed to be an even integer, q+r also has to be even integer for p-(q+r) to be even.

also, if -(q+r) is even integer, +(q+r) is also supposed to be an even integer. Therefore, p+(q+r) has to be even. The same is applicable for all other combinations of sum of 2 numbers, and the third number. Therefore, p+q+r will always be even.

Hence, statement 1 is sufficient.

For statement 2,

if we take p-r = 4.5 and q = 1.5,
case 1- assuming p =5, r= 0.5
p+q+r= 5+0.5+1.5= 7= odd
case 2- assuming p=4.5, r=0, q=1.5
p+q+r= 4.5+1.5+0= 6= even

Hence statement 2 is not sufficient.

Therefore, answer is option A.

QuantMadeEasy · Answer

(1) p−q−r is even
p - (q+r) is even
when p is odd, q+r is odd
when p is even, q+r is even
sufficient

(2) (p-r)/q is odd
p-r and q may or may not be an integer 
Insufficient

IMO A

Archit3110 · Answer

#1
p−q−r is even
possible ( even-even-even ) or ( odd-odd-even)
so for all integer we get yes
and in case fraction ( 3,1/2,1/2) again 3-1/2-1/2 ; yes
sufficient
#2
(p-r)/q is odd
(4-1)/1/3 ; 9 ; but sum of P+q+r is not even
and ( 8-2)/2 ; 3 sum is even
insufficient
IMO A

if at least one of p, q, r is an integer, is p+q+r even?
(1) p−q−r is even.
(2) (p-r)/q is odd

lacktutor · Answer

if at least one of p, q, r is an integer, is p+q+r even?

( Statement1 ): p−q−r is even.
5-0.5-0.5= 4 (even) --> 5+0.5+0.5= 6 (EVEN)- YES
6.5-0.5-4 = 2 (even) --> 6.5+0.5+4= 11 (Odd) -NO
Insufficient

( Statement2 ):  \frac{(p-r)}{q }  is odd

\frac{(3.5-0.5)}{3} = 1  (odd) --> 3.5+0.5+3= 7 (odd) -NO
 \frac{(6-2)}{4} = 1  (odd) --> 6+2+4= 10 (even ) -YES
Insufficient

Taken together 1&2 , 
--> p−q−r = 5.5- 1.5-4 = 0(even)
-->  \frac{(p-r )}{q}= \frac{(5.5-4)}{(1.5)} = 1  (odd)
 p+q+r= 5.5+1.5+4= 11 (odd) --NO

--> p−q−r= 6-2-4 = 0 (even)
-->  \frac{(p-r )}{q}= \frac{(6-4)}{2}= 1  (odd)
p+q+r= 6+2+4= 12 (EVEN) -YES

Insufficient

The answer is E.

Aviral1995 · Answer

CONSIDER ST-1
let us take values as P=3.5, Q=1,R=0.5
P-Q-R= 2= EVEN
P+Q+R IS ODD

let us take values as P=3, Q=0.5, R=0.5
P-Q-R =2=EVEN
P+Q+R IS EVEN

THEREFORE ST-1 itself is not sufficient

Consider ST-2
let us take values as P=3.5, Q=1,R=0.5
(p-r)/q =3 , which is odd
p+q+r=odd

let us take values as P=3, Q=0.5, R=0.5
(p-r)/q =5 , which is odd
p+q+r=even

therefore st-2 itself is not sufficient

using above 2 examples, even both statements together cannot help us to determine that p+q+r is even
Therefore E

Raxit85 · Answer

if at least one of p, q, r is an integer, is p+q+r even?
(1) p−q−r is even.
(2) (p-r)/q is odd

There are three cases,
i) one of p,q,r is integer
ii) two of p,q,r are integer
iii) all three are integer

1)p−q−r is even.
case iii)If, p=5, q=2, r= 1, then p−q−r=2, which is even. So, p+q+r = 8, even
case i) p=10, q=1.5, r=4.5, then p−q−r=4, which is even. So, p+q+r = 16, even
case ii) p=10.5, q=1.5, r=1,then p−q−r=8, which is even. So, p+q+r = 13, odd.
So, insufficient
 
(2) (p-r)/q is odd
case iii) If, p = 8, q=2, r=2, then (p-r)/q =6/2=3, which is odd. So, p+q+r=12, even
case ii) p=5, q=1.5, r=0.5, then (p-r)/q =4.5/1.5=3, which is odd. So, p+q+r=7, odd
case i) p=10.5, q=3 , r=1.5, then (p-r)/q =9/3=3, which is odd. So, p+q+r=15, odd
So, insufficient.

1) + 2) By combining,
p = 8, q=2, r=2 satisfies both statements condition. So, p+q+r=12, even
p=10.5, q=3 , r=1.5 satisfies both statements condition. So, p+q+r=15, odd
So, insufficient.

Hence, Ans. is E.

fauji · Answer

S1:  p−q−r is even
Multiple possibilities:
p,q, and r can be as follows to result to be even:
e,e,e
e,o,o
o,o,e 
Using values in p+q+r... no unique values are coming. .....Insufficient
 S2:  (p-r)/q is odd
Multiple possibilities:
Numerator/Denominator can be Odd/Odd. but not in all cases, for e.g. 7/3 ....Insufficient

Combine  S1 & S2: 
Similar scenario as no unique combination exists.....Insufficient

IMO Option E!

mohagar · Answer

A is the option only possile

hiranmay · Answer

if at least one of p, q, r is an integer, is p+q+r even?
(1) p−q−r is even. -->  insuff: if p = 8, q = 2, r = 2, then p+q+r=12=even, but if p = 5/2, q = 2, r= 1/2, p−q−r = 0 = even, but p+q+r = 3+2=5=odd 
(2) (p-r)/q is odd-->  insuff: if p = 8, q = 2, r = 2, then p+q+r=12=even, but if p = 5/2, q = 2, r= 1/2, p+q+r = 3+2=5=odd 
 combining (1) + (2), we can't find a unique solution
So  answer: E

AndreV · Answer

Answer: E
1) Insufficient
if p=7, q=3 and r=4 (this combination meets the requirement that at least one of the numbers is an integer), then p-q-r=0, which is an even number. Then, p+q+r=14, which is even.
if p=7.2, q=3.2 and r=4 (this combination meets the requirement that at least one of the numbers is an integer), then p-q-r=0, which is an even number. Then, p+q+r=14.4, which is clearly not an even number.
Two different answer. Insufficient
1) Insufficient
if p=7, q=3 and r=4 (this combination meets the requirement that at least one of the numbers is an integer), then (p-r)/q=1, which is an odd number. Then, p+q+r=14, which is even.
if p=7.2, q=3.2 and r=4 (this combination meets the requirement that at least one of the numbers is an integer), then (p-r)/q=1, which is an odd number. Then, p+q+r=14.4, which is clearly not an even number.
Two different answer. Insufficient
1)+2) Insufficient
If we select the same set of numbers as before, we'll obtain the same result. Insufficient.

testtakerstrategy · Answer

Is there a way to do this without plugging in?

My logic ended up at D:

1) p-q-r = even -> then p+q+r should be even too, because the negative signs are just reversed for each (each number will stay in the same "category" ex. if "a" is even, "-a" is also even).
Sufficient.

2) This is a fractional division. If the result is ODD, we know both the numerator and denominator are odd. For the numerator: if p-r is odd, that means one must be ODD and the other must be EVEN -> then we know p/q/r will be either OEO or EOO. Either way, adding these up results in an even number.

Where is this going wrong?

ShaggiShow09 · Answer

You are not considering the case where say P and Q are not integers. For e.g. P is 5.3, Q is 1.3 and R is 2. Therefore, P-Q-R = 2 which is even. But, P+Q+R = 8.6 which is  not even an integer , let alone an even no.

In my view, the rules of odd-even apply only to integers.

if at least one of p, q, r is an integer, is p+q+r even?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

if at least one of p, q, r is an integer, is p+q+r even?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards