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What is the value of 512 div 16 + 4 xx (...

What is the value of `512 div 16 + 4 xx (5)/(2) " of " ((1)/(8) +(1)/(3))`?

A

`(455)/(12)`

B

`(449)/(12)`

C

`(439)/(12)`

D

`(443)/(12)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the expression \( 512 \div 16 + 4 \times \left( \frac{5}{2} \right) \text{ of } \left( \frac{1}{8} + \frac{1}{3} \right) \), we will follow the order of operations (BODMAS/BIDMAS). ### Step-by-Step Solution: 1. **Evaluate the expression inside the brackets:** \[ \frac{1}{8} + \frac{1}{3} \] To add these fractions, we need a common denominator. The least common multiple of 8 and 3 is 24. \[ \frac{1}{8} = \frac{3}{24}, \quad \frac{1}{3} = \frac{8}{24} \] Therefore, \[ \frac{1}{8} + \frac{1}{3} = \frac{3}{24} + \frac{8}{24} = \frac{11}{24} \] **Hint:** Find a common denominator to add fractions. 2. **Substitute back into the expression:** Now, substitute \( \frac{11}{24} \) back into the expression: \[ 512 \div 16 + 4 \times \left( \frac{5}{2} \right) \times \frac{11}{24} \] 3. **Perform the division:** \[ 512 \div 16 = 32 \] **Hint:** Division is performed before addition. 4. **Calculate the multiplication:** First, calculate \( 4 \times \frac{5}{2} \): \[ 4 \times \frac{5}{2} = \frac{20}{2} = 10 \] Now multiply this result by \( \frac{11}{24} \): \[ 10 \times \frac{11}{24} = \frac{110}{24} \] **Hint:** Multiply the whole number by the fraction. 5. **Simplify \( \frac{110}{24} \):** Divide both the numerator and denominator by their greatest common divisor (GCD), which is 2: \[ \frac{110 \div 2}{24 \div 2} = \frac{55}{12} \] **Hint:** Always simplify fractions when possible. 6. **Add the results:** Now we add \( 32 \) and \( \frac{55}{12} \): Convert \( 32 \) to a fraction with a denominator of 12: \[ 32 = \frac{384}{12} \] Now add: \[ \frac{384}{12} + \frac{55}{12} = \frac{384 + 55}{12} = \frac{439}{12} \] **Hint:** Convert whole numbers to fractions with a common denominator before adding. ### Final Answer: The value of the expression is \( \frac{439}{12} \).
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