Latest Elitmus Aptitude Question SOLUTION: Is (p+r)(q+r) odd? 1. p is odd, q is odd. 2. q is odd, r is odd. Which is correct? a-1 only b-2 only c-Both 1 & 2 d-Neither 1 nor 2

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Number System

Elitmus Question Papers:

(#M40164392) ELITMUS Numerical Ability question Elitmus 16/4/17 Keep an EYE

Is (p+r)(q+r) odd?
1. p is odd, q is odd.
2. q is odd, r is odd.
Which is correct?
a-1 only
b-2 only
c-Both 1 & 2
d-Neither 1 nor 2

Asked In Elitmus Harnirvair (7 years ago)

Solved sandeep Read Solution (14)

Is this Puzzle helpful?   (50)   (17) Submit Your Solution Number System

Read Solution (14) :

satish kumar  (7 years ago)

Neither 1 nor 2

Is this solution Helpfull? Yes (29) | No (4)  

Runni Kumari  (7 years ago)

ANS: B

2 only

statement 1: if we assume
CASE1: r is even
(odd[p]+even[r])(odd[q]+even[r])
(odd[p+r])(odd[q+r])
=>odd*odd= odd
CASE 2: r is odd
(odd[p]+odd[r])(odd[q]+odd[r])
=>even*even= even
Statement 2:
odd[q]+odd[r]=even
even*any number (p+r)=even
so, with only statement it can be determined that (p+r)(q+r) is not odd

and is b-2 only

Is this solution Helpfull? Yes (12) | No (6)  

shubham goswami  (7 years ago)

Ans- 1 only
p is odd, q is odd
(odd + even)(odd + even)
(odd)*(odd) = odd
so answer is p is odd, q is odd

Is this solution Helpfull? Yes (10) | No (15)  

sandeep  (7 years ago) Best Solution

st1=> p is odd , q is odd
if r is even then
P(odd)+r(even)=odd
q(odd)+r(even)=odd
result= odd*odd=odd
else
if r is odd then
P(odd)+r(odd)=even
q(odd)+r(odd)=even
result even*even= even
in stmt 1 dosn't give unique answer so stmt one is not ture
in stmt 2
q(odd)+r(odd)=even
so
(P+r) is odd or even doesn't mean the multiplication of two if one is even the result always be even
so answer found no form ans: b-2

Is this solution Helpfull? Yes (10) | No (17)  

Susmeeta Sau  (6 years ago)

I don't know you guys are thinking from which point of view
But you need Both to ans the question

Is this solution Helpfull? Yes (3) | No (1)  

yogesh  (5 years ago)

stmt 1: take p=3(odd),q=5(odd) , first put r=2(even) we get (p+r) (q+r)=5*7=35(odd),
now ,put r=9(odd) we get (p+r) (q+r)=12*14=168(even) so stmt 1 not satisfied condition
stmt 2: take q=3(odd), r=5(odd) first put p=2(even), we get (p+r) (q+r)=7*5=35(odd)
now put p=3(odd) we get (p+r) (q+r)=8*8=64(even) so stmt 2 also not satisfied
ANSWER is: d-Neither 1 nor 2

Is this solution Helpfull? Yes (3) | No   

Shubham  (6 years ago)

Answer is b-2
Sol:
First consider Statement 1: Let us suppose r is odd, then p(odd)+r(odd)--> even.
q(odd)+r(odd)--->even.
Now, let us suppose r is even, then p(odd)+r(even)--->odd.
& q(odd)+r(even)-->odd.
Hence nounique answer if we consider just statement 1.

Similarly evaluate this for statement 2, You will find out that statement 2 gives unique answer.
Hence, answer is b-2.

Is this solution Helpfull? Yes (2) | No (1)  

M.R.Hareesh  (7 years ago)

p is odd and q is odd. then sum of odd and even gives odd number and final result is product of two odd numbers is odd

Is this solution Helpfull? Yes (1) | No (2)  

Adil khan  (6 years ago)

statement 1
(odd+even)(odd+even)=odd
(odd+odd)(odd+odd) =even so we can see question is depend upon r we can not tell
statement 2
(even+odd)(odd+odd)
so whatever the value of (p+r) ,(q+r) will be always even so
it is sufficient to tell that NO
its not odd
so ans will be
only statement 2 required to tell
Note - for who dont know the concept of data sufficiency
in question asking is it odd so if so u dont have to proof its odd
u just have to tell yes or no by statement so if yes then its true if its no its allso true
only will be wrong if its ans is ambiguos means when its give yes or no both
u are not sure .

Is this solution Helpfull? Yes (1) | No (1)  

Deepak Mishra  (6 years ago)

(p+r).(q+r)= odd (acc. to given situation)
RULE:- odd*odd=Odd
Option 1- (3+2).(5+2)= odd
(Let P and Q as an odd no and R as an even no.)
5.7= odd
35= odd
Option 2- (2+5).(7+5)=odd
(Let q and r as an odd no. And P as an even no)
84 is not an odd no

Hence option 1 (1only) satisfy the condition

Is this solution Helpfull? Yes (1) | No (1)  

deepak saroha  (7 years ago)

2 only
q(odd)+r(odd)=even

Is this solution Helpfull? Yes | No (5)  

sudhansu  (6 years ago)

2 only since first have 2 answers with 2 diff types of value of r.But 2nd case
If p=odd & r is odd then then p+r=even
(Anything)×(even)=even
Since this gives answer of our question that given term is not odd.

Is this solution Helpfull? Yes | No (1)  

poonam chandra  (5 years ago)

p-odd, q-odd, r-odd

(odd+odd)(odd+odd)
(even)(even)
even
is (p+r)(q+r) odd ans d-Neither 1 nor 2

Is this solution Helpfull? Yes | No   

Akanshi Agarwal  (4 years ago)

Neither 1 nor 2

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.