This is a great case to think of mixtures as a "tug of war" between 2 quantities.
Imagine a number line with P standing on 60% trying to "pull" the final mixture towards itself.
Q stands on 10% trying to "pull" the final mixture towards itself.
And the final mixture actually lands on R, which stands at 25%. Q has "pulled" the final mixture 35 away from Q, but P has "pulled" the final mixture only 15 away from R. So the ratio of Q/P = 35/15 = 7/3.

Statement (1) gives us the value of Q = 455. So 455/P = 7/3, so we could solve for "P". SUFFICIENT

Statement (2) gives us R = 650. Since Q+P = 650 and Q/P = 7/3, we'd now have 3 distinct equations with 3 variables, so we'd be able to solve for any variable, including "P". SUFFICIENT.

Each statement is SUFFICIENT -- answer choice (D).

P+Q=R .....................eqn i
\(\left(60\%\ of\ P\right)+\left(10\%ofQ\right)=25\%of\ R\)
\(\left(0.6P\right)+\left(0.1Q\right)=0.25R..............equn\ ii\)
\(What\ is\ the\ value\ of\ P?\)
Substitute R in eqn ii with R in eqn i
\(0.6P+0.1Q=0.25\left(P+Q\right)...................eqn\ iii\)
From eqn iii
0.6P+0.1Q=0.25(P+Q) where Q=(455)
0.6P+0.1(455)=0.25(P+455)
0.6P+45.5=0.25P+113.75
0.6P-0.25P=113.75-45.5
\(\frac{0.35P}{0.35}=\frac{68.2}{0.35}=195ml\)
\(Statement\ 1\ is\ SUFFICIENT\)

Statement 2 => R=650ml
From eqn ii
0.6P+0.1Q=0.25R
From eqn i P+Q=R and Q=R-P
Substitute Q=R-P in equation ii
0.6P+0.1(R-P)=0.25R where R=650ml
0.6P+0.1(650-P)=0.25(650)
0.6P+65-0.1P=162.5
0.6P-0.1P=162.5-65=97.5
$$\frac{0.5P}{0.5}=\frac{97.5}{0.5}$$ $$p=195$$
Statement 2 is SUFFICIENT.
Since each statement alone is SUFFICIENT.

$$answer\ is\ Option\ D$$