131. If a certain number is multiplied by 5, subtract its third from the product, divide the remainder by 10 and add to this sequentially $1 / 3, 1 / 2$ and $1 / 4$ of the original number, the result is 68. What is the number?
131. Bhaskara's Solution. If the required number were 3, then according to the condition $3 \cdot 5=15$, one third of 15 is 5. Since $15-5=10$, if we divide 10 by 10, we get one. If we add to it $\frac{1}{3}, \frac{1}{2}$ and $\frac{1}{4}$ of 3, we will have $1+1+\frac{3}{2}+\frac{3}{4}$, i.e., $\frac{17}{4}$, but it should be 68, i.e., in $68: \frac{17}{4}=$ $=16$ times more. therefore, the required number $3 \cdot 16=48$.