Answer the following question based on the information provided.

Liam deposited $x$ dollars in account A for three years at a rate of 12% p.a. simple interest, whereas Sophia deposited $(x+4000)$ dollars in account B for two years at 20% p.a. compound interest. Find the amount deposited by Sophia if the total interest accumulated from both accounts is 3360 dollars.

The correct option is 6000.

Step 1: Understand the given details for both accounts

Account A (Liam's deposit):
Principal ($P_{\mathrm{A}}$) = $x$
Rate of interest ($R_{\mathrm{A}}$) = 12% p.a. (Simple Interest)
Time ($T_{\mathrm{A}}$) = 3 years

Account B (Sophia's deposit):
Principal ($P_{\mathrm{B}}$) = $x+4000$
Rate of interest ($R_{\mathrm{B}}$) = 20% p.a. (Compound Interest)
Time ($T_{\mathrm{B}}$) = 2 years

Total interest accumulated from both accounts = 3360 dollars

Step 2: Calculate simple interest from Account A ($I_{\mathrm{A}}$)

Using the simple interest formula:

$I_{\mathrm{A}}=\frac{P\times R\times T}{100}$

$I_{\mathrm{A}}=\frac{x\times 12\times 3}{100}=0.36x$

Step 3: Calculate compound interest from Account B ($I_{\mathrm{B}}$)

Using the compound interest formula:

$I_{\mathrm{B}}=P\times \left[{\left(1+\frac{R}{100}\right)}^T-1\right]$

$I_{\mathrm{B}}=(x+4000)\times \left[{\left(1+\frac{20}{100}\right)}^2-1\right]$

$I_{\mathrm{B}}=(x+4000)\times \left[{\left(1.2\right)}^2-1\right]$

$I_{\mathrm{B}}=(x+4000)\times (1.44-1)$

$I_{\mathrm{B}}=0.44(x+4000)=0.44x+1760$

Step 4: Formulate the equation using total interest

$I_{\mathrm{A}}+I_{\mathrm{B}}=3360$

$0.36x+0.44x+1760=3360$

$0.80x+1760=3360$

$0.80x=3360-1760$

$0.80x=1600$

$x=\frac{1600}{0.80}=2000$

Step 5: Compute Sophia's deposit

Sophia deposited $(x+4000)$ dollars.

$\text{Sophia's deposit}=2000+4000=6000$

Hence, the amount deposited by Sophia is 6000 dollars.