Math Problem Statement

(99-8-11) Tenglamani yeching. (\frac{x - 12}{0.3 \cdot 3 \frac{1}{3} + 7}) \div \frac{3}{8} = 1 Options: A) 25, B) 14, C) 15, D) 16, E) 18

Solution

Let's solve this problem step by step.

The equation is:

(x12)0.3313+7÷38=1\frac{(x - 12)}{0.3 \cdot 3 \frac{1}{3} + 7} \div \frac{3}{8} = 1

Step 1: Simplify the Denominator

  1. The denominator is 0.3313+70.3 \cdot 3 \frac{1}{3} + 7.

    Converting 3133 \frac{1}{3} to an improper fraction: 313=1033 \frac{1}{3} = \frac{10}{3}

  2. Now substitute this back: 0.3103+70.3 \cdot \frac{10}{3} + 7

  3. Calculate 0.31030.3 \cdot \frac{10}{3}: 0.3103=310103=10.3 \cdot \frac{10}{3} = \frac{3}{10} \cdot \frac{10}{3} = 1

  4. So the denominator becomes: 1+7=81 + 7 = 8

Step 2: Rewrite the Equation

The equation now is: (x12)8÷38=1\frac{(x - 12)}{8} \div \frac{3}{8} = 1

Step 3: Simplify the Division

Dividing by 38\frac{3}{8} is equivalent to multiplying by 83\frac{8}{3}: (x12)883=1\frac{(x - 12)}{8} \cdot \frac{8}{3} = 1

The 88 in the numerator and denominator cancels out: x123=1\frac{x - 12}{3} = 1

Step 4: Solve for xx

Multiply both sides by 3: x12=3x - 12 = 3

Now, add 12 to both sides: x=15x = 15

Final Answer

The answer is: 15\boxed{15}

Would you like further details on any step, or do you have any questions?


Further Questions

  1. How would this problem change if the initial equation had different fractions?
  2. Can you explain the impact of multiplying and dividing by fractions on equations?
  3. Why is it important to simplify fractions early in multi-step problems?
  4. What alternative methods could solve similar equations?
  5. How would you check the answer to ensure it's correct?

Tip

Always simplify complex fractions and mixed numbers in steps to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Algebra
Fractional Equations
Mixed Numbers

Formulas

Fraction division formula: a / b ÷ c / d = a / b * d / c
Mixed number to improper fraction conversion: a b/c = (ac + b) / c

Theorems

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Suitable Grade Level

Grades 9-10