Answer
95.
Lower limit: 90.
Upper limit: 99.
Work Step by Step
The 105 in the horizontal axis does not represent any class. 95 is the midpoint of the last class.
The class width is 10 and the midpoint of the last class is 95:
$Lower~limit~of~the~last~class=midpoint~of~the~last~class-\frac{class ~width}{2}=95-\frac{10}{2}=95-5=90$
$Upper~limit~of~the~last~class=midpoint~of~the~last~class+\frac{class ~width}{2}-1=95+\frac{10}{2}-1=95+5-1=99$
Suppose there is a 11th class. Its lower limit would be 100 (105-10/2). So, the upper limit of the last class must be 99. Otherwise the classes would overlap. Also, notice that the 10th class covers 10 different ages: 90, 91, 92, 93, 94, 95, 96, 97, 98 and 99.
