Correct Answer - B
`{:("Class Interval",f),(" "0-9,2),(" "10-19,4 f_(1)),(" "20-29,7f),(" "30-39,5 f_(2)),(" "40-49,3):}`
Here, `f = 7, f_(1) = 4, f_(2) = 5 "and" L = 19.5` (lower boundary of the highest frequency class).
Mode `=L + (Delta_(1))/(Delta_(1) + Delta_(2))` xx c
`Delta_(1) = f - f_(1) = 3`
`Delta_(2) = f - f_(2) = 2`
`bar(x) = 19.5 + (3)/(5) xx 10`
`19.5 + 6 = 25.5`.