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

(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)

First convert the mixed fractions to improper fractions:

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

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

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

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

Then, evaluate the expression inside the parenthesis:

$5\frac{1}{3} - 2\frac{3}{4} = \frac{16}{3} - \frac{11}{4} = \frac{16 \times 4}{3 \times 4} - \frac{11 \times 3}{4 \times 3} = \frac{64}{12} - \frac{33}{12} = \frac{64-33}{12} = \frac{31}{12}$

Now, evaluate the division:

$3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) = \frac{31}{9} \div \frac{31}{12} = \frac{31}{9} \times \frac{12}{31} = \frac{31 \times 12}{9 \times 31} = \frac{12}{9} = \frac{4}{3}$

Finally, evaluate the addition:

$\frac{4}{3} + 5\frac{9}{10} = \frac{4}{3} + \frac{59}{10} = \frac{4 \times 10}{3 \times 10} + \frac{59 \times 3}{10 \times 3} = \frac{40}{30} + \frac{177}{30} = \frac{40+177}{30} = \frac{217}{30} = 7\frac{7}{30}$

(b)

The set A = {2, 3, 4} and the set B = {1, 3, 5}.

All possible outcomes when choosing one number from each set are:

(2,1), (2,3), (2,5)

(3,1), (3,3), (3,5)

(4,1), (4,3), (4,5)

Total number of outcomes = 3 x 3 = 9

The 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 corresponding outcomes are: (3,1), (2,3), (4,1), (3,3).

There are 4 favorable outcomes.

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

3. Final Answer

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

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