1. **Stating the problem:** We have six numbers with the following information: - Lowest number = 37 - Range = 24 - Mode = 43 - Median = 46 - One number is a multiple of 11 We need to find the mean of these six numbers. 2. **Understanding the given data:** - Range is the difference between the highest and lowest numbers: $$\text{Range} = \text{Highest} - \text{Lowest}$$ - Mode is the number that appears most frequently. - Median is the middle value when numbers are arranged in order. For six numbers, median is the average of the 3rd and 4th numbers. 3. **Find the highest number:** Given lowest = 37 and range = 24, $$\text{Highest} = 37 + 24 = 61$$ 4. **Arrange the six numbers in ascending order:** Let the numbers be $$a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5 \leq a_6$$ We know: - $$a_1 = 37$$ - $$a_6 = 61$$ - Median $$= \frac{a_3 + a_4}{2} = 46 \Rightarrow a_3 + a_4 = 92$$ - Mode = 43 means 43 appears at least twice. 5. **Include the mode 43 twice:** Set $$a_2 = 43$$ and $$a_3 = 43$$ (since mode appears most frequently and 43 is less than median 46, placing 43 at positions 2 and 3 is logical). 6. **Find $$a_4$$ using median:** $$a_3 + a_4 = 92 \Rightarrow 43 + a_4 = 92 \Rightarrow a_4 = 49$$ 7. **Now the numbers are:** $$37, 43, 43, 49, a_5, 61$$ 8. **One number is a multiple of 11:** Multiples of 11 between 37 and 61 are 44 and 55. Since 43 and 49 are fixed, $$a_5$$ can be 44 or 55. 9. **Check if 44 or 55 fits:** - If $$a_5 = 44$$, order is $$37, 43, 43, 49, 44, 61$$ which is not ascending because 44 < 49. - If $$a_5 = 55$$, order is $$37, 43, 43, 49, 55, 61$$ which is ascending. So, $$a_5 = 55$$. 10. **Calculate the mean:** $$\text{Mean} = \frac{37 + 43 + 43 + 49 + 55 + 61}{6} = \frac{288}{6} = 48$$ **Final answer:** The mean of the six numbers is **48**.