15. The sum of five unequal whole numbers is 90. The second largest of these five numbers can be at most
A.19
B.41
C.43
D.89
please explain and show work.
Let's assume the five unequal whole numbers are: a, b, c, d, and e.
According to the given information, we have the equation:
a + b + c + d + e = 90
To find the maximum value for the second-largest number, we need to consider the other four numbers as small as possible.
Since the numbers are stated to be unequal, we can assume a minimum value of 1 for one of the numbers without loss of generality.
So, let's assume a = 1.
Now, our equation becomes:
1 + b + c + d + e = 90
Considering the other four numbers as small as possible, we can assign the minimum possible value of 2 to each of the other four numbers.
So, b = 2, c = 2, d = 2, and e = 2.
Plugging in these values into the equation:
1 + 2 + 2 + 2 + 2 = 9
Therefore, the second-largest number, b, can be at most 9.
Among the options provided, the closest value to 9 is 9 itself.
So, the answer is A. 9.
To find the answer, we need to solve the problem step by step. Let's break down the problem:
We have five whole numbers, and their sum is 90. Let's represent these numbers as a, b, c, d, and e, where:
a is the largest number,
b is the second largest number,
c is the third largest number,
d is the fourth largest number,
e is the smallest number.
We know that these five numbers are unequal, so they are all distinct.
Now, we need to find the maximum value for the second largest number (b). To do so, we need to minimize the values of the other four numbers.
The smallest number (e) should be as small as possible to make room for the larger numbers. Since we need whole numbers, the smallest possible whole number is 1.
Next, we focus on the fourth largest number (d) and minimize it. To minimize d, we should make c as small as possible. Again, c should be a whole number, so the smallest possible whole number is 2.
Now, let's consider the third largest number (c). To minimize c, we make b as small as possible. We've already assigned the smallest possible value of 1 to e, the next smallest possible whole number of 2 to d, so the smallest whole number available for c is 3.
Finally, we can calculate the remaining value. We know the sum of the five numbers is 90, and we already assigned values of 1, 2, and 3 to e, d, and c. So, the remaining sum is 90 - (1 + 2 + 3) = 84.
This remaining sum (84) should be split between a and b. However, we want to maximize the value of b, so we assign the smallest possible value to a (which is 4). Now, we can determine the value of b.
a + b + c + d + e = 4 + b + 3 + 2 + 1 = 90
10 + b = 90
b = 90 - 10
b = 80
Therefore, the second largest of these five numbers can be at most 80.
The given answer options are:
A. 19
B. 41
C. 43
D. 89
So, the correct answer is not listed among the given options.