Simple mathematics question

[Kibishi]

Given CLASS MARK and SIZE OF CLASS INTERVALS are 5 and 9 respectively.Find
(i)UPPER LIMIT
(ii)LOWER LIMIT
N.B:Upper boundary=upper limit+0.5
and,
Lower boundary=lower limit-0.5

 

[itakiamo]

let x=upper limit and y=lower limit
classmark=(x+y)/2=5 then x+y=10(first equation)
Size of class interval=(x+0.5)-(y-0.5)=9 then x-y+1=9
x-y=8(second equation) by using simultenous equation
x=9 and y=8, therefore upper limit is 9 and lower limit is 8

 

[Kibishi]

let x=upper limit and y=lower limit
classmark=(x+y)/2=5 then x+y=10(first equation)
Size of class interval=(x+0.5)-(y-0.5)=9 then x-y+1=9
x-y=8(second equation) by using simultenous equation
x=9 and y=8, therefore upper limit is 9 and lower limit is 8

Hatua zote upo sahihi mkuu itakatikiamo.
Hebu solve vizuri hiyo simultaneous equation.

 

[Cndoh Jr]

Upo vzur mkuu ila umejsahau kdogo kwny kusolv simultaneous eqn... ..upper ni 9 na lower ni 1