If a certain culture of bacteria increases by a constant

[BTGmoderatorLU]

Source: Manhattan Prep

If a certain culture of bacteria increases by a constant factor of \(x\) every \(y\) minute, how long will it take for the culture to increase to ten-thousand times its original size?

1) \(x=10^y\)
2) In two minutes, the culture will increase to one hundred times its original size.

The OA is D

[deloitte247]

Let the number of bacteria = n
'n' increases by a factor of 'x' every 'y' minutes.
$$y\ \minutes=nx$$
$$2y\ \minutes=nx^2$$
Statement 1=>
$$x=10^y$$
$$Therefore,\ y\ \minutes=n10^y$$
$$For\ n10^4;\ y=4\ \min utes.\ Hence,\ statement\ 1\ is\ sufficient.$$

Statement 2=>
In 2 minutes, the culture will increase to one hundred times its original size.
Therefore, y=2 minutes, x=100
y=nx $$2\ \min utes=n100$$ $$2\ =n10^2$$
$$In\ 4\ \min utes=n10^4.\ Hence,\ statement\ 2\ is\ sufficient.\ $$
Conclusively, each statement alone is sufficient. The correct answer is option D.

[swerve]

We need to find \(y\), when \(xy=10^4\).

1) When \(x=10^y\), \(y\cdot 10^y =10^4\)
or could also be written as \(y = 10^{4-y}\)
By using substitution, we could say \(y\) could be between 3 and 4 \((\approx 3.4)\). Hence statement 1 is Sufficient \(\color{green}{\checkmark}\)

2) \(2\cdot x=100\) i.e. \(x=50\). By substituting the value of \(x\) in \(xy=10^4\), we could find the time taken. Hence statement 2 is Sufficient \(\color{green}{\checkmark}\)

Therefore, answer __D__ \(\color{green}{\checkmark}\)