I am unable to understand christoph's explanation. However, the answer is E. Both are not sufficient.
Quick solution. Assume that t2=0 - in other words, that the second city has no sales tax. Then the answer is immediately clear. Both conditions are satisfied (as long as t1>0 and Z1>0), yet the price of the computer in the second city, Z2, can be anything.
Second approach: intuition. What if the first city has higher taxes (t1>t2)? This is not enough: the second city can have more expensive computers. What if the actual sales tax you would pay in the first city is higher (Z1*t1>Z2*t2)? This is not enough: the first city may have ridiculous taxes but cheap computers. What if both are true? Same thing. High taxes in the first city, but cheap computers.
More formally, we want to see if it is possible that Z1+Z1*t1 < Z2+Z2*t2. Let's assume that all numbers are nonzero. (If there is an example, there is also an example with nonzero numbers, because of continuity.) I will also assume that all numbers are positive. Then, without loss of generality, we can scale Z1 and Z2 by the same positive number. This is because if Z1+Z1*t1 < Z2+Z2*t2 was true, it will stay true. The same holds for the conditions t1>t2 and Z1*t1>Z2*t2.
Without loss of generality, let Z2=1. (If, for example, Z1=33 and Z2=3, change it to Z1=11 and Z2=1. This will not change anything in this problem.)
Similarly, without loss of generality, let t2=1.
Thus, we want to see if it is possible that Z1+Z1*t1=Z1*(1+t1) < 2. We are given that t1>1 and/or that Z1*t1>1.
As you can see from the plot, all three inequalities can be simultaneously satisfied. Thus, with a relatively high t1 (taxes in the first city) and low Z1 (computer price in the first city) it is still possible that the overall price of a computer with taxes included is lower in the first city. The answer is E.
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Sergey Orshanskiy, Ph.D.