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.
OA A
Need help how the answer is A.
PQ and QP represent two-digit numbers
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To satisfy Statement 1, P and Q must sum to 11, with the result that RSR = 121:stevecultt wrote:Need help how the answer is A.
29+92 = 121
38+83 = 121
47+74 = 121
56+65 = 121.
In every case, P+Q = 11 and R+S = 3.
Thus, P+Q+R+S = 11+3 = 14.
SUFFICIENT.
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Statement 1:stevecultt wrote: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.
OA A
Need help how the answer is A.
Since the two-digit numbers, PQ and QP add to form a three-digit number RSR, the value
of the digit in the hundreds place must be '1.'
=ƒ> R = 1
Thus, P and Q should be such digits which add up to give 1 as the unit digit (so that the unit digit of R = 1 is obtained in the sum).
Thus, possible values of P and Q are:
- P = 2, Q = 9 ƒ=> PQ + QP = 29 + 92 = 121 ƒ=> P + Q + R + S = 2 + 9 + 1 + 2 = 14
P = 3, Q = 8 ƒ=> PQ + QP = 38 + 83 = 121 ƒ=> P + Q + R + S = 3 + 8 + 1 + 2 = 14
P = 4, Q = 7 ƒ=> PQ + QP = 47 + 74 = 121 ƒ=> P + Q + R + S = 4 + 7 + 1 + 2 = 14
P = 5, Q = 6 ƒ=> PQ + QP = 56 + 65 = 121 ƒ=> P + Q + R + S = 5 + 6 + 1 + 2 = 14
Statement 2:
We know that P, Q, R, and S are distinct non-zero digits.
However, their values cannot be determined. - Insufficient
The correct answer: A
Hope this helps!
Relevant book: Manhattan Review GMAT Data Sufficiency Guide
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Hi stevecultt,
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
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 = 10P + Q
QP = 10Q + P
so
A slightly more thorough breakdown of S1::
PQ + QP = 10P + Q + 10Q + P => 11P + 11Q => 11 * (P + Q)
RSR = 100R + 10S + R = 101R + 10S
Since 11*(P+Q) is at most 11*(9+9), we know that it's less than 200.
From that 101R + 10S must also be less than 200, so R = 1. (By convention in these cryptarithms, the leading digit can't be 0.) That gives us
11*(P + Q) = 101 + 10S
or
11*(P+Q) = 1S1
The only threedigit multiple of 11 that begins and ends in 1 is 121, so S1 is sufficient.
QP = 10Q + P
so
A slightly more thorough breakdown of S1::
PQ + QP = 10P + Q + 10Q + P => 11P + 11Q => 11 * (P + Q)
RSR = 100R + 10S + R = 101R + 10S
Since 11*(P+Q) is at most 11*(9+9), we know that it's less than 200.
From that 101R + 10S must also be less than 200, so R = 1. (By convention in these cryptarithms, the leading digit can't be 0.) That gives us
11*(P + Q) = 101 + 10S
or
11*(P+Q) = 1S1
The only threedigit multiple of 11 that begins and ends in 1 is 121, so S1 is sufficient.