DS problem

[abhasjha]

If m, n, and p are three-digit integers and m + n = p, is the sum of the units digits of m and n at least 2 more than the sum of the tens digits of m and n?

(1) The tens digit of p is greater than the sum of the tens digits of m and n.

(2) The tens and units digits of p are equal.

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Hi abhasjha,

This DS question is a bit layered; it requires a bit of "playing around", and understanding of how basic arithmetic "works" and TESTING Values.

We're told that M, N and P are three-digit numbers and M + N = P. We can write this information in this way....

M = _ _ _
N = _ _ _
-----------
P = _ _ _

We're asked if the sum of the UNIT'S DIGITS of M and N is at least 2 more than the sum of the TENS DIGITS of M and N. This is a YES/NO question.

Fact 1: The TENS DIGIT of P is > SUM of the TENS DIGITS of M and N

Here's an example to illustrate what Fact 1 "means":

M = 105
N = 105
-------
P = 210

Here, the ten's digit of P (1) is > the sum of the ten's digits of M and N (0).

Mathematically, the ONLY way for the the ten's digit of P to be greater than the sum of the ten's digits of M and N is IF the Units digits sum to 10 or greater; this would "carry over" and make the ten's digit of P "1 greater."

The arithmetic rules built into this Fact are:
1) The Units digits of M and N must sum to 10 or greater
2) The Tens digits of M and N must sum to 8 or less (since the ten's digit of P is GREATER than the sum of the ten's digits of M and N. The ten's digit of P could be 9, but the sum of the ten's digits of M and N must be LESS than that).

Thus, the sum of the unit's digits of M and N will ALWAYS be at least 2 greater than the sum of the ten's digits of M and N.
Fact 1 is SUFFICIENT.

Fact 2: The ten's and unit's digits of P are equal.

While Fact 1 took a lot of work to figure out, Fact 2 is rather straight-forward. Let's TEST Values:

M = 100
N = 100
---------
P = 200
The answer to the question is NO

M = 105
N = 106
---------
P = 211
The answer to the question is YES
Fact 2 is INSUFFICIENT

Final Answer: A

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

[GMATinsight]

If m, n, and p are three-digit integers and m + n = p, is the sum of the units digits of m and n at least 2 more than the sum of the tens digits of m and n?

(1) The tens digit of p is greater than the sum of the tens digits of m and n.

(2) The tens and units digits of p are equal.

If m, n, and p are three-digit integers

Let,
m = abc = 100a + 10b + c
m = def = 100d + 10e + f
p = pqr = 100p + 10q + r

Since, and m + n = p

Then, (100a + 10b + c) + (100d + 10e + f) = (100p + 10q + r)
i.e. 100(a+d) + 10(b+e) + (c+f) = 100p + 10q + r

Question : Is (c+f) > 2 + (b+e) ?

Statement 1) The tens digit of p is greater than the sum of the tens digits of m and n
q > (b+e)
q will be greater than (b+e) only if there is one carry forward from the sum of Unit digits and the sum of tens digits with carry forward is not becoming a two digit number.
therefore, the sum of unit digits must be greater than or equal to 10 and sum of tens digits can't be greater than 8 as atleast one will be carry forward.

Therefore, Sum of Unit digits of m and n will certainly be greater than Sum of tens digits of m and n by atleast 2
SUFFICIENT

Statement 2) The tens and units digits of p are equal

Here sum of the Unit digit can be 12 and sum of tens digits of m and n can be 1 leading to the answer YES for the question

alternatively, sum of the Unit digit can be 2 and sum of tens digits of m and n can be 2 as well leading to the answer NO for the question
INSUFFICIENT

Answer: Option A

[GMATGuruNY]

If m, n, and p are three-digit integers and m + n = p, is the sum of the units digits of m and n at least 2 more than the sum of the tens digits of m and n?

(1) The tens digit of p is greater than the sum of the tens digits of m and n.

(2) The tens and units digits of p are equal.

Let m = ABC, n = DEF, and p = GHI, where the letters A through I represent digits.
Since m + n = p, we get the following sum:

ABC
DEF
GHI

Question stem, rephrased: Is (C+F) - (B+E) ≥ 2?

Statement 1: The tens digit of p is greater than the sum of the tens digits of m and n.
Since H > B+E, the MAXIMUM possible value of B+E is 8, in which case H=9:
A3C
D5F
G9I

In order that H > B+E, the LEAST possible value for C+F is 10, so that a "1" is carried over from the units place to the tens place.
To illustrate:
A38
D52
G90
In this case, (C+F) - (B+E) = (8+2) - (3+5) = 2.

If the value of B+E decreases, the difference between C+F and B+E will INCREASE.
Thus, it must be true that (C+F) - (B+E) ≥ 2.
SUFFICIENT.

Statement 2: The tens and units digits of p are equal.
The sum could look like this:
111
111
222
In this case, (C+F) - (B+E) = (1+1) - (1+1) = 0.

The sum could look like this:
179
109
288
In this case, (C+F) - (B+E) = (9+9) - (7+0) = 11.

Since (C+F) - (B+E) is LESS THAN 2 in the first case but GREATER THAN 2 in the second case, INSUFFICIENT.

The correct answer is A.