Simple interest on a certain sum of money for 3 years at $18\%$ per annum is half the compound interest on Rs. 9000 for 2 years at $10\%$ per annum. The sum placed on simple interest is:

Step 1: Find the compound interest (CI) on Rs. 9000 for 2 years at $10\%$ p.a.
$$\text{Amount (A)}=P{\left(1+\frac{R}{100}\right)}^t$$
$$A=9000{\left(1+\frac{10}{100}\right)}^2=9000\times {\left(\frac{11}{10}\right)}^2$$
$$A=9000\times \frac{121}{100}=90\times 121=\text{Rs. }10890$$
$$\text{CI}=A-P=10890-9000=\text{Rs. }1890$$
Step 2: According to the problem, Simple Interest (SI) is half of this CI.
$$\text{SI}=\frac{1890}{2}=\text{Rs. }945$$
Step 3: Find the principal sum ($P_{si}$) placed on simple interest for 3 years at $18\%$ p.a.
$$\text{SI}=\frac{P_{si}\times R\times T}{100}$$
$$945=\frac{P_{si}\times 18\times 3}{100}$$
$$945=\frac{54P_{si}}{100}$$
$$P_{si}=\frac{945\times 100}{54}=\text{Rs. }1750$$
Therefore, the sum placed on simple interest is Rs 1750.