\(\text{We define }k\text{ as the common multiplier for last year's ratio "2:3:2".}\)
\(o = 2k\)
\(e = 3k\)
\(b = 2k\)
\(\text{From "15 oak trees have been removed" we translate oak now to }2k - 15\text{.}\)
\(o_{\text{now}} = 2k - 15\)
\(\text{From "90 elm trees have been planted" we translate elm now to }3k + 90\text{.}\)
\(e_{\text{now}} = 3k + 90\)
\(\text{Birch is unchanged, so }b_{\text{now}} = 2k\text{. The current total }T\text{ is the sum of the three current counts.}\)
\(T = (2k - 15) + (3k + 90) + 2k\)
\(\text{From "elm trees now account for 3/4 of the total" we write the fraction condition for elm.}\)
\(\frac{3k + 90}{T} = \frac{3}{4}\)
\(\text{Substitute }T\text{ and solve the resulting linear equation for }k\text{.}\)
\(\frac{3k + 90}{(2k - 15) + (3k + 90) + 2k} = \frac{3}{4}\)
\(\frac{3k + 90}{7k + 75} = \frac{3}{4}\)
\(4(3k + 90) = 3(7k + 75)\)
\(12k + 360 = 21k + 225\)
\(360 - 225 = 21k - 12k\)
\(135 = 9k\)
\(k = 15\)
\(\text{Compute the current total }T\text{ using }k = 15\text{.}\)
\(T = 7k + 75\)
\(= 7 \times 15 + 75\)
\(= 105 + 75\)
\(= 180\)
Answer A
Hope this helps!
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