Cement and sand

Rasmus mixes cement with sand for some repair work. He used 3 5/6 kg of cement and 1/2 kg more sand than cement. In all, he needs 10 kg of the mixture. Does he have enough mixture? If yes, then how much more does it have? If not, then how much more does he need?

Result

x=

Yes

No

y =  1 ⁵⁄₆ = 11/6 kg

Step-by-step explanation:

$$\begin{gathered} a=3\frac{5}{6}=3+\frac{5}{6}=\frac{3\cdot 6+5}{6}=\frac{18+5}{6}=\frac{23}{6}=3\frac{5}{6}\doteq 3.8333\text{kg} \\ b=a+\frac{1}{2}=\frac{23}{6}+\frac{1}{2}=\frac{23}{6}+\frac{3}{6}=\frac{23+3}{6}=\frac{26}{6}=\frac{13}{3}=4\frac{1}{3}\doteq 4.3333\text{kg} \\ m=a+b=\frac{23}{6}+\frac{13}{3}=\frac{23}{6}+\frac{26}{6}=\frac{23+26}{6}=\frac{49}{6}=8\frac{1}{6}\doteq 8.1667\text{kg} \\ m<10\text{kg} \\ x=1 \end{gathered}$$

$y=10-m=10-\frac{49}{6}=\frac{10\cdot 6}{6}-\frac{49}{6}=\frac{60}{6}-\frac{49}{6}=\frac{60-49}{6}=\frac{11}{6}=\frac{11}{6}\text{kg}=1.8333\text{kg}$

Try another example