PQ and QP represent two-digit numbers having P and Q as their digits.

Question

PQ and QP represent two-digit numbers having P and Q as their digits. RSR is a three-digit number having the digits R and S. What is the value of P + Q + R + S?

(1) PQ + QP = RSR.
(2) P, Q, R and S are distinct non-zero digits.

Luckisnoexcuse · Accepted Answer

(1)

R cannot be 2 and has to be 1 as sum of 2 two digit numbers cannot exceed 198 (99+99)

SO now since R is 1 S can take 1, 2, 3, 4, 5, 6, 7, 8, 9, 0

but since PQ and QP = RSR

P and Q cannot take 0 
and one can only be odd digit and one even 
Also one of them has to be greater than 5

92 + 29 = 121
83 + 38 = 121
74 + 47 = 121

in all cases P+Q = 11 an R+S = 3
Sufficient

(2) Not sufficient ( We cannot take same values of P and Q)

A

srikanth9502 · Answer

PQ and QP can be written in following ways

PQ = 10P+Q (Example: 21=10(2)+1)
QP = 10Q+P

PQ+QP = 10P+Q+10Q+P = 11(P+Q)
Given that PQ+QP = RSR

so, RSR = 11(P+Q)
-> RSR is a multiple of 11

Since it is given that the unit's digit and the hundred's digit are same i.e R
The need to be multiplied by either 11 (11*11 = 121), 22(11*22 = 242), 33....
But P+Q cannot exceed 18(as the max possible single digit integer is 9)

So, P+Q = 11 and RSR = 121

niks18 · Answer

it is evident that P, Q, R & S are single digit number

Statement 1 : solve the addition problem in a conventional way to get-
 PQ
+QP
------
RSR

maximum sum of two single digits can be 9+9=18

as Q+P=R so S=P+Q+1 (there has to be a carry forward of 1 as R & S are different digits)

and finally the hundreds place i.e R will be 1 (carry forward from the sum of P+Q)

so now we have R=1 & S=2 and P+Q=11 (because sum of P & Q has to yield a unit's digit 1 and maximum possible sum of any two single digit is 18)

Hence P+Q+R+S=11+1+2=14.  Sufficient

Statement 2 : no relation provided for the digits. Hence  Insufficient

Option  A

EMPOWERgmatRichC · Answer

Hi All,

While this question might look complex, it's actually based on some basic arithmetic rules. If you don't recognize the "theory" behind this question, then you could still solve it with a bit of 'brute force' arithmetic and a little logic.

We're told that QP and PQ are two 2-digit numbers that have the same digits (just in reverse-order) and that RSR is a 3-digit number. We're asked for the value of P+Q+R+S.

1) PQ + QP = RSR.

With Fact 1, notice that the sum of the two 2-digit numbers is a 3-digit number. In this situation, the 3-digit number MUST begin with a 1 (there's no other possibility since the highest sum of two 2-digit numbers is 99+99 = 198). Thus, R = 1....

PQ + QP = 1S1

By extension, Q+P must "end" in a 1 AND PQ+QP must be large enough to create a 3-digit sum. From here, you can brute-force the options and see what happens.... 
P=2, Q=9... 29+92 = 121.... so S=2 and the answer to the question is 2+9+1+2 = 14 
P=3, Q=8... 38+83 = 121..... so S=2 and the answer to the question is 3+8+1+2 = 14

Interesting that the resulting sum stayed exactly the SAME. There are only a few options left, but if you map them out, you'll end up with the exact same answer every time... the answer to the question is ALWAYS 14. 
Fact 1 is SUFFICIENT

2) P, Q, R, and S are distinct non-zero digits.

With this Fact, we know that the 4 digits are all DIFFERENT non-zero integers, but there are multiple possible answers to the given question. 
Fact 2 is INSUFFICIENT

Final Answer:  A

GMAT assassins aren't born, they're made, 
Rich

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PQ and QP represent two-digit numbers having P and Q as their digits.

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