looks like C, my rationale
given class A
a1,a2,a3.........g , with range g-a1=r
also class B
b1,b2,b3........h with range h-b1=s
we need to know does a1>b1, or if a1=g-r, b1=h-s, is g+s>h+r, to the statements
(1) insuff as it gives us only range in A< range in B. r<s, it can easily be proved with simple numbers,
A:1,2,3, r=3-1=2. B: 2,2,5, s=5-2=3. r<s,(2<3) and a1=1 b1=2. a1<b1. but another series
A:2,2,3. r=3-2=1. B: 1,4,5. s=5-1=4 r<s (1<4) but a1=2>b1=1
(2) also nothing useful, g>h
A:1,1,4, B:2,2,3 here g>h (4>3) and a1=1<b1=2. another example
A:2,2,3, B:1,2,2 here g>h ( 3>2) but a1=2>b1=1
both sum (1)+(2) i mean
r<s
h<g , and come r+h<s+g, exactly what we need to prove from the start, g+s>h+r
so C

Class A:

Range=High-Low

r=g-LowA

Class B:

Range=High-Low

s=h-LowB

LowA>LowB?

stmt1,

r<s

No info abt least heights

Not suff

stmt2,

g>h

No info abt least heights

Not suff

Combining 1 and 2,

r<s or -r>-s

g>h

Add both inequalities,

g-r<h-s

LowA>LowB

Sufficient

Pick C

selango wrote:
r<s or -r>-s

g>h

Add both inequalities,

g-r<h-s

Good call, Saves time

Thanks Clock60, Selango and beat_gmat_09.