Find the nearest number to $12199$ which is exactly divisible by the product of the first four prime numbers.

The price of commodity $X$ increases by $40 \text{ paise}$ every year,while the price of commodity $Y$ increases by $15 \text{ paise}$ every year. If in $2001$,the price of commodity $X$ was $Rs. 4.20$ and that of $Y$ was $Rs. 6.30$,in which year will commodity $X$ cost $40 \text{ paise}$ more than commodity $Y$?

The sum of three numbers is $68$. If the ratio between the first and second is $2:3$ and that between the second and third is $5:3$,then the second number is

Arranging the following in descending order,we get $\sqrt[3]{4}, \sqrt{2}, \sqrt[6]{3}, \sqrt[4]{5}$.

The product of two numbers is $120$. The sum of their squares is $289$. The sum of the two numbers is

$(\sqrt[3]{8000} + \sqrt[3]{0.027} - \sqrt[3]{0.216})$ is equal to: