Explanation:

- You have a frequency distribution table that provides different class intervals and their respective frequencies.

- The sum of all the frequencies is 120.

- Given: Frequency for class 0-20 is 17, for class 40-60 is 32, and for class 80-100 is 19.

- The frequencies for classes 20-40 and 60-80 are expressed as p+q and p-3q.

- Total frequency equation: $17+(p+q)+32+(p-3q)+19=120$.

- Simplify the equation: $68+2p-2q=120$.

- This gives: $2p-2q=52$, resulting in $p-q=26$. [Equation 1]

- The mean of the distribution is 50.

- To find mean: $[17\times 10]+[(p+q)\times 30]+[32\times 50]+[(p-3q)\times 70]+[19\times 90]=50\times 120$.

- Simplify mean equation: $170+30p+30q+1600+70p-210q+1710=6000$.

- Combine and simplify: $100p-180q=2520$.

- Further simplify: $5p-9q=126$. [Equation 2]

- Solve Equations 1 and 2 to find q:

- Solving simultaneously: From Equation 1, $p=q+26$.

- Substitute into Equation 2: $5(q+26)-9q=126$, simplify to: $5q+130-9q=126$.

- Solve: $-4q=-4$ results in $q=1$.

- ?? Highlight the correct option: 1, 1.

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