• 2021 December(E23) Math Question 32

    Word Problems

    - 1 Variable Equations

    Algebraic Problems
    Choice J
    Cost  of  3  apples  and  4  oranges=$3.47Cost\; of\; 3\; apples\; and\; 4\; oranges=\$ 3.47
    Cost  of  12  oranges=$6.36Cost\; of\; 12\; oranges=\$ 6.36
    Total  amount=$10Total\; amount=\$10
    Cost  of  1  orange={6.36}{12}=$0.53Cost\; of\; 1\; orange=\frac\{6.36\}\{12\}=\$0.53
    Cost  of  4  orange=4×0.53=$2.12Cost\; of\; 4\; orange=4\times 0.53=\$2.12
    Cost  of  3  apples=$3.47−$2.12=$1.35Cost\; of\; 3\; apples=\$3.47-\$2.12=\$1.35
    Cost  of  1  apples={1.35}{3}=$0.45Cost\; of\; 1\; apples=\frac\{1.35\}\{3\}=\$0.45

    So the number of apples she can buy today is

    Apples={10}{0.45}=22.22≈22Apples=\frac\{10\}\{0.45\}=22.22\approx 22

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