Correct Answer – B
Let x and y be the upper and lower class limit in a frequency distribution.
Now, mid value of a class `=(x+y)/2=10` (given)
`rArr x+y=20`………..(i)
Also, given that, width of class =6
`rArr x-y=6`………….(ii)
On adding Eqs. (i) and (ii), we get
`2x=20+6`
`rArr 2x=26 rArr x=13`
On putting x=13 in Eq. (i), we get
`13+y=20 rArr y=7`
Hence, the lower limit of the class is 7

Correct Answer – B
Let the lower limit be a. Then, upper limit=a+3.
Mid value of `=(a+a+6)/(2)=(2a+6)/(2)=a+3`
`therefore a+3=10 Rightarrow a=7`
`therefore` lower limit of the class is 7.