Question:

Which two of the following numbers have a product that is between -1 and 0?
Indicate {both
of the numbers.}

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Remember that a negative exponent does not make a number negative. For any non-zero number \(a\) and positive integer \(n\), \(a^{-n} = \frac{1}{a^n}\). This is a common trap in number property questions.
Updated On: Oct 6, 2025
  • -20
  • -10
  • \(2^{-4}\)
  • \(3^{-2}\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
We need to find a pair of numbers from the given options whose product is a negative number greater than -1. The condition "between -1 and 0" can be written as \(-1<\text{product}<0\).
Step 2: Key Approach:
1. For the product to be negative, one number must be positive and the other must be negative.
2. Evaluate the positive numbers (those with negative exponents).
3. Test the possible pairs to see which product falls within the required range.
Step 3: Detailed Explanation:
First, identify the signs of the numbers.
- Options (A) -20 and (B) -10 are negative.
- Options (C) and (D) involve negative exponents, which represent reciprocals, so they are positive.
- (C) \(2^{-4} = \frac{1}{2^4} = \frac{1}{16}\)
- (D) \(3^{-2} = \frac{1}{3^2} = \frac{1}{9}\)
Now, we must form pairs with one negative and one positive number. There are four possible pairs:
- Pair 1: (A) and (C) \(\implies (-20) \times \frac{1}{16} = -\frac{20}{16} = -\frac{5}{4} = -1.25\). This is less than -1, so it is not in the range.
- Pair 2: (A) and (D) \(\implies (-20) \times \frac{1}{9} = -\frac{20}{9} \approx -2.22\). This is less than -1, so it is not in the range.
- Pair 3: (B) and (C) \(\implies (-10) \times \frac{1}{16} = -\frac{10}{16} = -\frac{5}{8} = -0.625\). This is between -1 and 0. This is a correct pair.
- Pair 4: (B) and (D) \(\implies (-10) \times \frac{1}{9} = -\frac{10}{9} \approx -1.11\). This is less than -1, so it is not in the range.
Step 4: Final Answer:
The only pair whose product, -0.625, is between -1 and 0 is -10 and \(2^{-4}\). The correct choices are (B) and (C).
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