Math Problem Statement

There was 11/13 of a bottle of shampoo left. Georgio used 2/7 of what was left on a customer. How much of the original bottle was left?

Solution

Let's break down the problem:

  1. Amount of Shampoo Initially Left:
    Originally, 1113\frac{11}{13} of the bottle was left.

  2. Amount Georgio Used:
    Georgio used 27\frac{2}{7} of this remaining amount, so we multiply: 27×1113=2×117×13=2291\frac{2}{7} \times \frac{11}{13} = \frac{2 \times 11}{7 \times 13} = \frac{22}{91}

  3. Remaining Shampoo After Georgio Used Some:
    To find out how much of the original bottle is left, we subtract the amount used from the initial amount left: 11132291\frac{11}{13} - \frac{22}{91}

    To perform this subtraction, we need a common denominator. The least common multiple of 13 and 91 is 91, so we convert 1113\frac{11}{13} to a fraction with 91 as the denominator: 1113=11×713×7=7791\frac{11}{13} = \frac{11 \times 7}{13 \times 7} = \frac{77}{91}

    Now we subtract: 77912291=772291=5591\frac{77}{91} - \frac{22}{91} = \frac{77 - 22}{91} = \frac{55}{91}

So, 5591\frac{55}{91} of the original bottle of shampoo was left.

Would you like further clarification on any of these steps?


  1. If Georgio used 3/7 of the remaining shampoo, how much of the original bottle would be left?
  2. What fraction of the entire bottle did Georgio use?
  3. If the original bottle was 500 ml, how many ml were left after Georgio’s use?
  4. How much more would Georgio need to use to reach half of the original bottle?
  5. If Georgio used another 1/5 of what was left, how much of the original bottle would remain?

Tip: When working with fractions, finding a common denominator helps simplify addition and subtraction.

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Math Problem Analysis

Mathematical Concepts

Fractions
Multiplication of Fractions
Subtraction of Fractions

Formulas

Multiplication of fractions: (a/b) * (c/d) = (a * c) / (b * d)
Subtraction with common denominators: (a/b) - (c/b) = (a - c) / b

Theorems

Basic operations with fractions

Suitable Grade Level

Grades 5-7