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Question analysis.
p = 6.72 when rounded off to 2 digits.(hundredth is 2)
so p = 6.72A
if A={0,1,2,3,4} then hundredth digit is 2 else its 1.

Statement 1:
A = 3 which is sufficient

Statement 2:
A + hundredth digit = 5
hundredth digit min value = 1.
so A(max) = 4.
This is sufficient

Hence answer is D
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Bunuel
When rounded to the nearest hundredths digit, the number p becomes 6.72. What is the hundredths digit of p?

(1) The thousandths digit of p is 3.
(2) The sum of the hundredths and the thousandths digits of p is 5

When rounded to the nearest hundredths digit, the number p becomes 6.72. What is the hundredths digit of p.
=> Means \(6.715\leq{p}<6.725\). Thus p can be 1 or 2.

(1) The thousandths digit of p is 3.
So two possibilities..
\(6.713\), but when rounded off to nearest hundredths 6.713=6.71....NOT possible, \(p\neq{1}\)
\(6.723\), and when rounded off to nearest hundredths 6.723=6.72....Yes, possible
SO p=2.
Sufficient

(2) The sum of the hundredths and the thousandths digits of p is 5
We know p can be 1 or 2.
If p=1, then only possibility is 6.714, as 1+4=5, but when rounded off to nearest hundredths 6.714=6.71....NOT possible, \(p\neq{1}\)
If p=2, then only possibility is 6.723, as 2+3=5 and when rounded off to nearest hundredths 6.723=6.72....Yes
p=2
Suff

D
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