Correct Answer - B
Let `f_1 and f_2` be the missing frequencies of class intervals 10-20 and 40-50 respectively . Then
`10+f_1+25+30+f_2+10=100 rArr f_1+f_2=25` Mediann is 32 , which lies in 30-40. So the median class is 30-40
`therefore l=30, h=10, f=30, N=100 and cf=10+f_1+25=f_1+35`
Now , median `M_e=1+{hxx((N/2-cf))/f}`
`rArr 30+[10xx({50-(f_1+35)}}/(30)]=32`
`rArr 30+((15-f_1))/3=32 rArr (15-f_1)=b rarr f_1=9`
`therefore f_1=9 and f_2=25-9=16`
Hence `f_1=9 and f_2=16`
