Beaker A contains sugar solution with 18 percent sugar.beaker B contains sugar solution with 12 percent sugar.How much of each must he mixed together to get solution of 16 percent sugar weighing 240 gm of it ?

Let the weight of the solution from Beaker A be x grams and the weight of the solution from Beaker B be y grams.

Since the total weight of the final solution is 240 gm, we have:

⇒ x + y = 240 ………………..(1)

The sugar content in Beaker A is 18%, and in Beaker B it is 12%. The final mixture should have 16% sugar.

⇒ $\dfrac{18}{100}x + \dfrac{12}{100}y = \dfrac{16}{100} \times 240$

⇒ 18x + 12y = 16 $\times$ 240

⇒ 18x + 12y = 3,840

⇒ 3x + 2y = 640 ………………..(2)

Multiply equation (1) by 3:

(x + y = 240) x 3

⇒ 3x + 3y = 720 …………………(3)

Subtracting equation (2) from equation (3),

$\begin{matrix} & 3x & + & 2y & = & 640 \ & 3x & + & 3y & = & 720 \ & - & &- & & - \ \hline & & & -y & = & -80 \ \end{matrix}$

⇒ y = 80

Substituting the value of y in equation (1), we get:

⇒ x + 80 = 240

⇒ x = 240 - 80

⇒ x = 160

Hence, the weight of beaker A = 160 gm and the weight of beaker B = 80 gm.