BTGmoderatorDC wrote: ↑Mon Jan 20, 2020 5:56 pm
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.
OA
A
Source: Manhattan Prep
z
Say
• m = xyz;
• n = abc; and
• p = rst
where for m, hundreds digit is x, tens digit is y and units digit is z; similarly for n and p.
Given that p = m + n
We have to determine whether z + c ≥ y + b + 2
Let's take each statement one by one.
(1) The tens digit of p is greater than the sum of the tens digits of m and n.
=> s > y + b
This is only possible if there is a carry of "1" from the sum of units digits of m and n.
Since s ≤ 9, we have y + b ≤ 8. Since there's a carry of '1' from the addition of units digit of m and n, we have z + c ≥ 10. Thus, from y + b ≤ 8 and z + c ≥ 10, we have z + c ≥ y + b + 2. Sufficient
(2) The tens and units digits of p are equal.
=> s = t
Case 1: Say m = xyz = 123 and n = abc = 132, thus, p = rst = m + n = 123 + 132 = 255
However, we see that z + c = y + b. The answer is no.
Case 2: Say m = xyz = 102 and n = abc = 101, thus, p = rst = m + n = 102 + 101 = 211
However, we see that z + c = 11 and y + b = 0; thus, z + c ≥ y + b + 2. The answer is yes.
The correct answer:
A
Hope this helps!
-Jay
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