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 mathematical tables or calculators. (b) A number is selected at random from each of the sets $\{2, 3, 4\}$ and $\{1, 3, 5\}$. Find the probability that the sum of the two numbers is greater than 3 and less than 7.

1. Problem Description

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 mathematical tables or calculators.

(b) A number is selected at random from each of the sets $\{2, 3, 4\}$ and $\{1, 3, 5\}$. Find the probability that the sum of the two numbers is greater than 3 and less than

7.

2. Solution Steps

(a) Simplify the expression $3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}$.

First, convert all mixed fractions to improper fractions:

$3\frac{4}{9} = \frac{3 \times 9 + 4}{9} = \frac{27+4}{9} = \frac{31}{9}$

$5\frac{1}{3} = \frac{5 \times 3 + 1}{3} = \frac{15+1}{3} = \frac{16}{3}$

$2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8+3}{4} = \frac{11}{4}$

$5\frac{9}{10} = \frac{5 \times 10 + 9}{10} = \frac{50+9}{10} = \frac{59}{10}$

Now, substitute these improper fractions into the expression:

$\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4}) + \frac{59}{10}$

Find a common denominator for the fractions in the parentheses:

$\frac{16}{3} - \frac{11}{4} = \frac{16 \times 4}{3 \times 4} - \frac{11 \times 3}{4 \times 3} = \frac{64}{12} - \frac{33}{12} = \frac{64-33}{12} = \frac{31}{12}$

Now, substitute back into the expression:

$\frac{31}{9} \div \frac{31}{12} + \frac{59}{10}$

Dividing by a fraction is the same as multiplying by its reciprocal:

$\frac{31}{9} \div \frac{31}{12} = \frac{31}{9} \times \frac{12}{31} = \frac{31 \times 12}{9 \times 31} = \frac{12}{9} = \frac{4}{3}$

Now, substitute back into the expression:

$\frac{4}{3} + \frac{59}{10}$

Find a common denominator:

$\frac{4}{3} + \frac{59}{10} = \frac{4 \times 10}{3 \times 10} + \frac{59 \times 3}{10 \times 3} = \frac{40}{30} + \frac{177}{30} = \frac{40+177}{30} = \frac{217}{30}$

(b) Find the probability that the sum of the two numbers is greater than 3 and less than

7. The sets are $\{2, 3, 4\}$ and $\{1, 3, 5\}$.

The possible sums are:

$2+1=3$

$2+3=5$

$2+5=7$

$3+1=4$

$3+3=6$

$3+5=8$

$4+1=5$

$4+3=7$

$4+5=9$

The sums greater than 3 and less than 7 are: 4, 5, 5,

6. The total number of possible sums is $3 \times 3 = 9$.

The number of favorable outcomes is

4. The probability is $\frac{4}{9}$.

3. Final Answer

(a) $\frac{217}{30}$

(b) $\frac{4}{9}$