(a) Simplify the expression $3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}$ without using calculators. (b) Given two sets {2, 3, 4} and {1, 3, 5}, find the probability that the sum of a number chosen randomly from each set 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 calculators.

(b) Given two sets {2, 3, 4} and {1, 3, 5}, find the probability that the sum of a number chosen randomly from each set is greater than 3 and less than

7.

2. Solution Steps

(a) Simplifying the expression:

First, convert 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 values into the expression:

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

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

1

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

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

$\frac{64}{12} - \frac{33}{12} = \frac{64-33}{12} = \frac{31}{12}$

So the expression becomes:

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

To divide fractions, we multiply by the reciprocal:

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

The expression is now:

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

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

3

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

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

$\frac{40}{30} + \frac{177}{30} = \frac{40+177}{30} = \frac{217}{30}$

(b) Finding the probability:

The possible sums of numbers from the two sets 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 that are greater than 3 and less than 7 are 4, 5, 5,

6. There are a total of 3 x 3 = 9 possible sums.

The sums that satisfy the condition are 4, 5, 5, and

6. Thus, there are 4 such sums.

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

3. Final Answer

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

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