(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

(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) We need to simplify the expression $3 \frac{4}{9} \div (5 \frac{1}{3} - 2 \frac{3}{4}) + 5 \frac{9}{10}$.

First, convert the mixed numbers to improper fractions:

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

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

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

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

Now, we have $\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4}) + \frac{59}{10}$.

First, we simplify the expression inside the parentheses:

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

Now, we have $\frac{31}{9} \div \frac{31}{12} + \frac{59}{10}$.

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

Now, we have $\frac{4}{3} + \frac{59}{10}$.

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

(b) The sets are $A = \{2, 3, 4\}$ and $B = \{1, 3, 5\}$. We want to find the probability that the sum of a number chosen from $A$ and a number chosen from $B$ is greater than 3 and less than

7. The possible sums are:

$2 + 1 = 3$

$2 + 3 = 5$

$2 + 5 = 7$

$3 + 1 = 4$

$3 + 3 = 6$

$3 + 5 = 8$

$4 + 1 = 5$

$4 + 3 = 7$

$4 + 5 = 9$

There are $3 \cdot 3 = 9$ possible sums.

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

6. There are 4 such sums.

So, the probability is $\frac{4}{9}$.

3. Final Answer

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

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