$(?+\sqrt{678})\%\text{ of }1202+375÷75\times \sqrt[3]{66}=501$
What approximate value should come in place of the question mark (?) in the following questions?

The correct answer is 14.

To find the approximate value of the question mark (?), let us simplify each term of the given equation step-by-step:

The given equation is:

$(?+\sqrt{678})\%\text{of}1202+375÷75\times \sqrt[3]{66}=501$

Step 1: Approximate the square root and cube root terms
We know that ${26}^2=676$, so we can approximate the square root term as:
$\sqrt{678}\approx \sqrt{676}=26$
Similarly, we know that $4^3=64$, so we can approximate the cube root term as:
$\sqrt[3]{66}\approx \sqrt[3]{64}=4$

Step 2: Approximate the other numbers and perform division
Let's approximate 1202 to 1200 for easier calculation:
$1202\approx 1200$
Now, calculate the division term:
$375÷75=5$

Step 3: Substitute the approximated values back into the equation
Substituting these values, we get:
$(?+26)\%\text{of}1200+5\times 4\approx 501$
Simplify the multiplication:
$(?+26)\%\text{of}1200+20\approx 501$

Step 4: Solve for the unknown value (?)
Subtract 20 from both sides:
$(?+26)\%\text{of}1200\approx 501-20$
$(?+26)\%\text{of}1200\approx 481$
Now, express the percentage calculation:
$\frac{?+26}{100}\times 1200\approx 481$
$(?+26)\times 12\approx 481$
Divide both sides by 12:
$?+26\approx \frac{481}{12}$
Since $481÷12\approx 40$:
$?+26\approx 40$
$?\approx 40-26$
$?\approx 14$

Thus, the approximate value that should come in place of the question mark is 14.