(a) As suggested by the hint the concentrations are halved:

A_{1} = e_{1}c_{1}l_{1}
= 4200 X 1.50X10^{-4} X 1 = 0.630

A_{2} = e_{2}c_{2}l_{2}
= 4200 X 1.30X10^{-4} X 1 = 0.365

A = 0.630 + 0.365 = 0.995

(b)Alternatively you could think of the 1 cm cell as being 1/2
cm of the 1st solution and 1/2 cm of the 2nd solution.

Therefore A = 1.00/2 = 0.995 as before.

In either case:

A = 0.995 = log I_{0}/I

I_{0}/I = 9.89

%T = 100 I/I_{0} = 100/9.89 = 10.1%