4 . F Look closely at the line plot of Al in Figure 1. At a frequency of 10.0 × 10^14 Hz, the Kmax is
0.1 eV, and at a frequency of 20.0 × 10^14 Hz, the Kmax is 4.2 eV. If this trend continues, as
frequency increases by 2.0 × 10^14 Hz, Kmax increases by approximately 0.8 eV. Thus, the
correct answer is (F) since at a frequency of 26 × 10^14 Hz, Kmax for Al will be considerably
greater than 5.0 eV.
5 . D We can use the given data for Metal Q and apply it to Figure 1. If Metal Q’s Kmax is 0.5 eV
and 3.0 eV at frequencies of 12.0 × 10^14 Hz and 18.0 × 10^14 Hz, respectively, we can see that
these points lie between zinc and silver. Table 1 shows the work functions of Zn and Ag as
4.30 eV and 4.73 eV, respectively. Therefore, Metal Q’s work function should be between
4.30 eV and 4.73 eV, as in (D).
Passage III
1 . C Look at Figure 1. As the concentration of Drug X increases, the percent of bacterial cells of
Strain C also increases. Therefore, (C) is the correct answer.
2 . H From Table 1, compare the values of ED 50 . Strain D has the maximum value, so (H) is the
correct answer.
3 . D In Figure 1, the percent of bacterial cells killed increases as the concentration of Drug X
increases. Thus, according to the data, a concentration of 200 mg/L would result in a greater
percentage of bacterial cells killed. Since the percentage killed at a concentration of 100
mg/L is approximately 83%, the correct answer is (D).
4 . F From the data in Table 1, Strain E has an ED 50 of 62.6. Look for a strain with an ED 50 that
when multiplied by 4 is about 62.6. Strain A has an ED 50 of 15.3, which when multiplied by
4 is approximately 60. Four times the ED 50 for Strains B, C, and D are about 120, 90, and
360, respectively. Thus, the correct answer is (F).
5 . C Examine the bars for Strain E in Figure 1. The percentages for 2 mg/L and 20 mg/L are
similar, so (B) is incorrect. The greatest difference between two concentrations is an
increase from approximately 25% to 45% from a concentration of 20 mg/L to 60 mg/L.
Passage IV
1 . B Table 1 shows that as voltage increases, power increases. For V = 0.03 V, the voltage is
between the values for Trials 1 and 2, so the expected value for P should lie between the
results for Trial 1 (0.08 W) and Trial 2 (0.32W). Therefore, (B) is the correct answer.
