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A quantitative comparison question states:

if $r<s<t$ and the average arithmetic mean of r , s and t is 90 . Which of the following is bigger a)The average of s and t or b)$90$.

The answer is a. However I cant quiet figure out how they got this:

I know that : $\frac{r+s+t}{3}=90$ so Average of s and t would be $\frac{270-r}{2}$. How can I precisely say that this average is definitely greater than 90 ?

Edit: After reviewing some of the pointers here is what I got. Please Let me know if this is correct

since $s+t = 270 -r$ so $\frac{270-r}{2}> 90$ thus $270-r > 180$ so $90-r >0$ Now since r cant be negative which is impossible (even if it is) the $90-r > 0$ Is this correct ?

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2 Answers 2

up vote 1 down vote accepted

Suppose that the average of $s$ and $t$ is $90$ or more; then $\frac{s+t}2\ge 90$, so $s+t\ge 180$, and therefore $r+s+t\ge r+180$. On the other hand, you know that $r+s+t=270$, so $270\ge r+180$, and therefore $90\ge r$. Is it possible to have $90\le r<s<t$ and $r+s+t=270$?

Added: You ask whether this argument is correct:

since $s+t = 270 -r$ so $\frac{270-r}{2}> 90$ thus $270-r > 180$ so $90-r >0$ Now since $r$ cant be negative which is impossible (even if it is) the $90-r > 0$

It needs a bit of work. The fact that $s+t=270-r$ does not imply that $\frac{270-r}2>90$, so your first so is inappropriate. What you want to say is this:

Suppose that $\frac{270-r}2>90$; then $270-r>180$, so $90-r>0$.

Next, it’s true that $r$ can’t be negative, but this is irrelevant. What you can conclude here is that $r<90$.

Unfortunately, none of this really helps: it just shows that there is no immediate problem with the assumption that the average of $s$ and $t$ is more than $90$. In order to show that this actually is the case, you want to suppose that it isn’t, and derive a contradiction. Thus, you really ought to start out:

Suppose that $\frac{270-r}2\le 90$; then $270-r\le 180$, so $90-r\le 0$, and therefore $90\le r$.

Now you can go on to argue as follows:

By hypothesis, $r<s<t$, so we have $90\le r<s<t$. This implies that $$r+s+t>r+r+r\ge 3\cdot 90=270$$ and hence that $$\frac{r+s+t}3>\frac{270}3=90\;.$$ This, however, contradicts the hypothesis that the arithmetic mean of $r,s$, and $t$ is $90$. Thus, the supposition that $\frac{270-r}2\le 90$ must be false, i.e., it must be true that $\frac{270-r}2>90$ and therefore that the arithmetic mean of $s$ and $t$ is more than $90$.

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Just added to the question.. could you please tell me if my method is correct ? –  Rajeshwar Jul 13 '12 at 21:33

To show that $\frac{270-r}{2}$ is bigger than $90$ is equivalent to showing that $\frac{270-r}{2}-90$ is positive. But $\frac{270-r}{2}-90$ simplifies to $\frac{90-r}{2}$. So we want to show that $90-r$ is positive. Showing that will not be hard.

Added: Note that since $r \lt s \lt $, we have $$r+r+r \lt r+s+t=270,$$ so $r \lt 90$, and therefore $90-r$ is positive.

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Just need to use inequality here.. Thanks for the great pointer –  Rajeshwar Jul 13 '12 at 21:10
    
Just added to my question . Did I solve it correct ? –  Rajeshwar Jul 13 '12 at 21:34
    
since r cant be negative which is impossible (even if it is) still 90>0 –  Rajeshwar Jul 13 '12 at 21:34
    
@Rajeshwar: I do not think the rest of what you wrote makes the logic clear enough. –  André Nicolas Jul 13 '12 at 21:37
    
@Rajeshwar: Note that the argument works perfectly well if some or all of $r$, $s$, $t$ are negative. So there is no reason to assume that $r \gt 0$. –  André Nicolas Jul 13 '12 at 22:18

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