The problem has two parts. (a) Simplify the expression $3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}$ without using a calculator. (b) Find the probability that the sum of two numbers, one selected at random from the set $\{2, 3, 4\}$ and the other from the set $\{1, 3, 5\}$, is greater than 3 and less than 7.

ArithmeticFractionsMixed NumbersArithmetic OperationsProbability
2025/4/19

1. Problem Description

The problem has two parts.
(a) Simplify the expression 349÷(513234)+59103\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10} without using a calculator.
(b) Find the probability that the sum of two numbers, one selected at random from the set {2,3,4}\{2, 3, 4\} and the other from the set {1,3,5}\{1, 3, 5\}, is greater than 3 and less than
7.

2. Solution Steps

(a) Simplify the expression: 349÷(513234)+59103\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}
First, convert the mixed numbers to improper fractions:
349=3×9+49=27+49=3193\frac{4}{9} = \frac{3 \times 9 + 4}{9} = \frac{27+4}{9} = \frac{31}{9}
513=5×3+13=15+13=1635\frac{1}{3} = \frac{5 \times 3 + 1}{3} = \frac{15+1}{3} = \frac{16}{3}
234=2×4+34=8+34=1142\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8+3}{4} = \frac{11}{4}
5910=5×10+910=50+910=59105\frac{9}{10} = \frac{5 \times 10 + 9}{10} = \frac{50+9}{10} = \frac{59}{10}
Now, substitute the improper fractions into the expression:
319÷(163114)+5910\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4}) + \frac{59}{10}
Find a common denominator for the fractions inside the parentheses:
163114=16×43×411×34×3=64123312=643312=3112\frac{16}{3} - \frac{11}{4} = \frac{16 \times 4}{3 \times 4} - \frac{11 \times 3}{4 \times 3} = \frac{64}{12} - \frac{33}{12} = \frac{64-33}{12} = \frac{31}{12}
Now, the expression becomes:
319÷3112+5910\frac{31}{9} \div \frac{31}{12} + \frac{59}{10}
Dividing by a fraction is the same as multiplying by its reciprocal:
319÷3112=319×1231=31×129×31=129=43\frac{31}{9} \div \frac{31}{12} = \frac{31}{9} \times \frac{12}{31} = \frac{31 \times 12}{9 \times 31} = \frac{12}{9} = \frac{4}{3}
Now, the expression becomes:
43+5910\frac{4}{3} + \frac{59}{10}
Find a common denominator for the fractions:
43+5910=4×103×10+59×310×3=4030+17730=40+17730=21730\frac{4}{3} + \frac{59}{10} = \frac{4 \times 10}{3 \times 10} + \frac{59 \times 3}{10 \times 3} = \frac{40}{30} + \frac{177}{30} = \frac{40+177}{30} = \frac{217}{30}
Convert the improper fraction to a mixed number:
21730=7730\frac{217}{30} = 7\frac{7}{30}
(b) Find the probability.
The possible sums are:
2+1 = 3
2+3 = 5
2+5 = 7
3+1 = 4
3+3 = 6
3+5 = 8
4+1 = 5
4+3 = 7
4+5 = 9
Total number of possible sums is 3×3=93 \times 3 = 9.
The sums that are greater than 3 and less than 7 are: 4, 5, 5,

6. There are 4 such sums.

The probability is the number of favorable outcomes divided by the total number of possible outcomes:
Probability =49= \frac{4}{9}

3. Final Answer

(a) 77307\frac{7}{30}
(b) 49\frac{4}{9}

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