In a class, the number of students who do Mathematics only is 14 more than those doing both Mathematics and ICT. The number doing ICT only is 5 less than the number doing both. If the total number of

From the problem statement:

1. "The number of students who do Mathematics only is 14 more than those doing both Mathematics and ICT."

   $$M_{only} = B + 14 \quad (*)$$

2. "The number of students doing ICT only is 5 less than the number doing Mathematics and ICT."

   $$I_{only} = B - 5 \quad (**)$$

3. "The number of students doing Mathematics is twice those doing ICT." Let $M_{total}$ be the total number of students doing Mathematics, and $I_{total}$ be the total number of students doing ICT.

   $$M_{total} = 2 \times I_{total} \quad (***)$$

The total number of students doing Mathematics is the sum of those doing Mathematics only and those doing both:

$$M_{total} = M_{only} + B$$

The total number of students doing ICT is the sum of those doing ICT only and those doing both:

$$I_{total} = I_{only} + B$$

Substitute $M_{only} = B + 14$ into the equation for $M_{total}$:

$$M_{total} = (B + 14) + B = 2B + 14$$

Substitute $I_{only} = B - 5$ into the equation for $I_{total}$:

$$I_{total} = (B - 5) + B = 2B - 5$$

Using $M_{total} = 2 \times I_{total}$:

$$2B + 14 = 2 \times (2B - 5)$$

Distribute the 2 on the right side:

$$2B + 14 = 4B - 10$$

Subtract $2B$ from both sides:

$$14 = 2B - 10$$

Add 10 to both sides:

$$14 + 10 = 2B$$ $$24 = 2B$$

Divide by 2:

$$B = \frac{24}{2}$$ $$B = 12$$

The number of students who do both Mathematics and ICT is $B$.

The final answer is $\text{12}$.