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Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is
- a)23
- b)24
- c)23.5
- d)22.5
Correct answer is option 'D'. Can you explain this answer?
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Consider six distinct natural numbers such that the average of the two...
**Solution**
Let the six numbers be $a$, $b$, $c$, $d$, $e$, and $f$ in ascending order. Then we have:
$$\frac{a+b}{2}=14$$
$$\frac{e+f}{2}=28$$
We want to maximize the average of all six numbers, which is:
$$\frac{a+b+c+d+e+f}{6}$$
We know that $a+b=28-2(f+e)/2=28-(f+e)$. Similarly, $e+f=56-(a+b)$. So we can rewrite the average as:
$$\frac{(28-(f+e))+c+d+(56-(a+b))}{6}$$
$$=\frac{84-(a+b+c+d+e+f)}{6}$$
$$=14-\frac{(a+b+c+d+e+f)}{6}$$
So we want to minimize the sum of the six numbers. To do this, we want to make $a$ and $b$ as small as possible and $e$ and $f$ as large as possible while still satisfying the given conditions. We know that $a+b=28-(f+e)$, so to make $a$ and $b$ as small as possible, we want $f+e$ to be as large as possible. Similarly, to make $e$ and $f$ as large as possible, we want $a+b$ to be as small as possible. Therefore, we want $a$, $b$, $e$, and $f$ to be as close to each other as possible, and $c$ and $d$ to be as far away from the other four as possible. We can achieve this by letting $a=1$, $b=2$, $e=57$, and $f=58$. Then we have:
$$c+d=84-(a+b+e+f)=84-118=-34$$
So we need to choose $c$ and $d$ such that their sum is $-34$. The maximum value of their average is achieved when they are as far apart as possible, so we choose $c=0$ and $d=-34$. Then the sum of the six numbers is $1+2+0+(-34)+57+58=84$ and their average is $84/6=14$.
Therefore, the maximum possible value of the average of these six numbers is $\boxed{\textbf{(D)}\ 22.5}$.
Let the six numbers be $a$, $b$, $c$, $d$, $e$, and $f$ in ascending order. Then we have:
$$\frac{a+b}{2}=14$$
$$\frac{e+f}{2}=28$$
We want to maximize the average of all six numbers, which is:
$$\frac{a+b+c+d+e+f}{6}$$
We know that $a+b=28-2(f+e)/2=28-(f+e)$. Similarly, $e+f=56-(a+b)$. So we can rewrite the average as:
$$\frac{(28-(f+e))+c+d+(56-(a+b))}{6}$$
$$=\frac{84-(a+b+c+d+e+f)}{6}$$
$$=14-\frac{(a+b+c+d+e+f)}{6}$$
So we want to minimize the sum of the six numbers. To do this, we want to make $a$ and $b$ as small as possible and $e$ and $f$ as large as possible while still satisfying the given conditions. We know that $a+b=28-(f+e)$, so to make $a$ and $b$ as small as possible, we want $f+e$ to be as large as possible. Similarly, to make $e$ and $f$ as large as possible, we want $a+b$ to be as small as possible. Therefore, we want $a$, $b$, $e$, and $f$ to be as close to each other as possible, and $c$ and $d$ to be as far away from the other four as possible. We can achieve this by letting $a=1$, $b=2$, $e=57$, and $f=58$. Then we have:
$$c+d=84-(a+b+e+f)=84-118=-34$$
So we need to choose $c$ and $d$ such that their sum is $-34$. The maximum value of their average is achieved when they are as far apart as possible, so we choose $c=0$ and $d=-34$. Then the sum of the six numbers is $1+2+0+(-34)+57+58=84$ and their average is $84/6=14$.
Therefore, the maximum possible value of the average of these six numbers is $\boxed{\textbf{(D)}\ 22.5}$.
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Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers isa)23b)24c)23.5d)22.5Correct answer is option 'D'. Can you explain this answer?
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