Math:

Math: solve each problem.
1. for the school's intramurals, a group of students prepared 23 1/2 liters of lemonade to sell. They had 3 5/8 liters left over. How many liters of lemonade were sold?
2. My mother bought 3 3/4 kg of beef, 23/5 kg of pork, and 5 1/2 kg of chicken. How many kilograms of meat did she buy?
3. Mr. De Guzman bought 7 1/2 kg of meat. He used 2 3/4 kilograms of afritada, 3 1/8 kg for menudo, and the rest for pochero. How many kilograms of meat did he use for pochero?
4. How much greater is the difference of 8 3/4 and 4 1/2 than the sum of 1 2/3 and 1 5/8?
5. Karen's father has 18 3/8 m of nylon strings. He gave 10 5/9 m of them to his friend's fishing rod. How long is the nylon string that is left?

Final Answer:

a =  19 ⁷⁄₈ = 159/8 l
b =  13 ¹⁷⁄₂₀ = 277/20 kg
c =  1 ⁵⁄₈ = 13/8 kg
d =  23/24
e =  7 ⁵⁹⁄₇₂ = 563/72 m

Step-by-step explanation:

$$\begin{gathered} V_1=23\frac{1}{2}=23+\frac{1}{2}=\frac{23\cdot 2+1}{2}=\frac{46+1}{2}=\frac{47}{2}=23.5\text{l} \\ {V}_2=3\frac{5}{8}=3+\frac{5}{8}=\frac{3\cdot 8+5}{8}=\frac{24+5}{8}=\frac{29}{8}=3.625\text{l} \\ a=V_1-V_2=\frac{47}{2}-\frac{29}{8}=\frac{188}{8}-\frac{29}{8}=\frac{188-29}{8}=\frac{159}{8}=\frac{159}{8}\text{l}=19\frac{7}{8}\text{l}=19.875\text{l} \end{gathered}$$

$$\begin{gathered} m_1=3\frac{3}{4}=3+\frac{3}{4}=\frac{3\cdot 4+3}{4}=\frac{12+3}{4}=\frac{15}{4}=3.75\text{kg} \\ {m}_2=23/5=\frac{23}{5}=4.6\text{kg} \\ {m}_3=5\frac{1}{2}=5+\frac{1}{2}=\frac{5\cdot 2+1}{2}=\frac{10+1}{2}=\frac{11}{2}=5.5\text{kg} \\ b=m_1+m_2+m_3=\frac{15}{4}+\frac{23}{5}+\frac{11}{2}=\frac{75}{20}+\frac{92}{20}+\frac{110}{20}=\frac{75+92+110}{20}=\frac{277}{20}=\frac{277}{20}\text{kg}=13\frac{17}{20}\text{kg}=13.85\text{kg} \end{gathered}$$

$c=7\frac{1}{2}-2\frac{3}{4}-3\frac{1}{8}=\frac{13}{8}\text{kg}=1\frac{5}{8}\text{kg}=1.625\text{kg}$

$$\begin{gathered} d_1=8\frac{3}{4}-4\frac{1}{2}=\frac{35}{4}-\frac{9}{2}=\frac{35}{4}-\frac{18}{4}=\frac{35-18}{4}=\frac{17}{4}=4\frac{1}{4}=4.25 \\ {s}_1=1\frac{2}{3}+1\frac{5}{8}=\frac{5}{3}+\frac{13}{8}=\frac{40}{24}+\frac{39}{24}=\frac{40+39}{24}=\frac{79}{24}=3\frac{7}{24}\doteq 3.2917 \\ d=d_1-s_1=\frac{17}{4}-\frac{79}{24}=\frac{102}{24}-\frac{79}{24}=\frac{102-79}{24}=\frac{23}{24}=0.9583 \end{gathered}$$

$$\begin{gathered} s=18\frac{3}{8}=18+\frac{3}{8}=\frac{18\cdot 8+3}{8}=\frac{144+3}{8}=\frac{147}{8}=18.375\text{m} \\ r=10\frac{5}{9}=10+\frac{5}{9}=\frac{10\cdot 9+5}{9}=\frac{90+5}{9}=\frac{95}{9}\doteq 10.5556\text{m} \\ e=s-r=\frac{147}{8}-\frac{95}{9}=\frac{1323}{72}-\frac{760}{72}=\frac{1323-760}{72}=\frac{563}{72}=\frac{563}{72}\text{m}=7\frac{59}{72}\text{m}=7.8194\text{m} \end{gathered}$$

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