AMC 10 Step by Step

2016 AMC 10A Problem 7

P1-10~2 minPrint

The mean, median, and mode of the 77 data values 60,100,x,40,50,200,9060, 100, x, 40, 50, 200, 90 are all equal to xx . What is the value of xx ?

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Key insight

Only the mean gives an equation: 540 + x must equal 7x, so x = 90. Then check that 90 is also the median and mode.

Solution

Of mean, median and mode, only the mean gives an equation directly, because the median and mode depend on where xx lands. So start with the mean.

1. Add the six known values: 60+100+40+50+200+90=54060 + 100 + 40 + 50 + 200 + 90 = 540.
2. With xx included there are 77 values with total 540+x540 + x. The mean is xx, so the total must be 77 times xx:
540+x=7x.540 + x = 7x.
3. Take xx from both sides: 540=6x540 = 6x, so x=90x = 90.
4. Check the median. Put the seven values in order: 40,50,60,90,90,100,20040, 50, 60, 90, 90, 100, 200. The middle (4th) value is 9090. Good.
5. Check the mode. The value 9090 appears twice and every other value once, so the mode is 9090. Good.

The answer is (D) 90\boxed{\textbf{(D)}\ 90}.

Why this works

When several conditions must all hold, use the one that gives a clean equation first, then test the answer against the rest. Here the mean fixes xx, and the median and mode are quick checks. Remember that xx is part of the data, so it belongs in the total.

Alternative approach

Start from the mode instead. Every listed value appears once, so for xx to be the mode it must equal one of them: 40,50,60,90,10040, 50, 60, 90, 100 or 200200. The median must also be xx, so xx has to land in the middle of the ordered list, which only 6060 and 9090 can do. Now test the mean: with x=60x = 60 the total is 600600, not 7⋅60=4207 \cdot 60 = 420; with x=90x = 90 the total is 630=7⋅90630 = 7 \cdot 90. So x=90x = 90.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The six known values sum to 60+100+40+50+200+90=54060 + 100 + 40 + 50 + 200 + 90 = 540. The mean of all seven values is xx, so
540+x7=x  ⟹  540+x=7x  ⟹  6x=540  ⟹  x=90.\frac{540 + x}{7} = x \implies 540 + x = 7x \implies 6x = 540 \implies x = 90.

Verify the other two conditions with x=90x = 90. Sorted, the data are 40,50,60,90,90,100,20040, 50, 60, 90, 90, 100, 200: the fourth value (median) is 9090, and 9090 is the only value appearing twice, so it is the mode. Everything is consistent.

The answer is (D) 90\boxed{\textbf{(D)}\ 90}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.