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 randomly selected 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 randomly selected 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 $3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}$.

First, convert the mixed numbers to improper fractions:

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

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

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

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

So the expression becomes:

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

Find a common denominator for $\frac{16}{3}$ and $\frac{11}{4}$, which is

1

2. $\frac{16}{3} = \frac{16 \times 4}{3 \times 4} = \frac{64}{12}$

$\frac{11}{4} = \frac{11 \times 3}{4 \times 3} = \frac{33}{12}$

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

So the expression becomes:

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

When dividing fractions, we multiply by the reciprocal of the second fraction:

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

So the expression becomes:

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

Find a common denominator for $\frac{4}{3}$ and $\frac{59}{10}$, which is

3

0. $\frac{4}{3} = \frac{4 \times 10}{3 \times 10} = \frac{40}{30}$

$\frac{59}{10} = \frac{59 \times 3}{10 \times 3} = \frac{177}{30}$

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

$\frac{217}{30} = 7\frac{7}{30}$

(b) Let 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 pairs are:

(2,1) sum is 3

(2,3) sum is 5

(2,5) sum is 7

(3,1) sum is 4

(3,3) sum is 6

(3,5) sum is 8

(4,1) sum is 5

(4,3) sum is 7

(4,5) sum is 9

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

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

6. These sums correspond to the pairs: (3,1), (2,3), (4,1), (3,3).

So, there are 4 such pairs.

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

3. Final Answer

(a) $\frac{217}{30} = 7\frac{7}{30}$

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