Last month a woman had a body mass of 53 kg. She reduced this by x kg so that she is now below 50 kg. Assuming that x is less than 6, find the range of the value.
Understand the Problem
The question is asking us to determine the possible values for x, the amount of weight that a woman reduced, in order to ensure her current weight is below 50 kg. She started with a weight of 53 kg and we know x is less than 6 kg.
Answer
The possible values for $x$ are in the range $3 < x < 6$.
Answer for screen readers
The possible values for $x$ are in the range $3 < x < 6$.
Steps to Solve
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Set up the inequality To find the values of $x$ that make the woman's current weight less than 50 kg, we can express her current weight mathematically. Her starting weight is 53 kg, and after reducing weight by $x$, we have: $$ 53 - x < 50 $$
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Simplify the inequality We can simplify the inequality to find the values of $x$. Start by isolating $x$: $$ 53 - x < 50 $$
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Rearrange the terms Next, we can subtract 53 from both sides of the inequality: $$ -x < 50 - 53 $$
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Calculate the right side Now, we compute the right side to get: $$ -x < -3 $$
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Multiply by -1 When multiplying or dividing an inequality by a negative number, we must reverse the inequality sign: $$ x > 3 $$
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Combine with the constraint on x We know that $x$ must also be less than 6 kg: $$ 3 < x < 6 $$
This means that the possible values for $x$ are in the range greater than 3 and less than 6.
The possible values for $x$ are in the range $3 < x < 6$.
More Information
This result indicates that the woman must reduce her weight by more than 3 kg but less than 6 kg to ensure her current weight stays below 50 kg. By controlling her weight loss within this range, she maintains her health and achieves her weight goal.
Tips
- Misinterpreting the inequality direction: Remember that when multiplying or dividing by a negative number, the inequality sign must be flipped.
- Forgetting to combine constraints: It's essential to consider the additional condition that $x$ is less than 6 throughout the solution.
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