Since Lotte ordered 300 computers, he must get a 25% discount
Price for 1 computer with speed rating \(s\) = \(1.25^sa\)
Price for 1 computer with speed rating \(s+2\) = \(1.25^{s+2}a\)

Price for 300 computers with speed rating \(s\)= \(0.75(300P) = (225)1.25^sa\)

Now, the question is - how many computers (let it be \(n\)) could she have purchased for the same price and with a speed rating 2 higher
Note here, all the option choices are above 100 - so he must get a discount of at-least 10%

i.e. \((n)(0.90)1.25^{s+2}a = (225)1.25^sa\)

\(n = \frac{(225)1.25^sa}{(0.90)1.25^{s+2}a} = \frac{(225)(1.25^s)(a)}{(0.90)(1.25^s)(1.25^2)(a)} = \frac{225}{(0.90)1.25^2} = \frac{225}{1.40625} = 160\)

We have a match, Hence option A