AMC 10 · 2010 · #14

Grade 8 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

mean-median-mode-rangesequences-arithmeticlinear-equations-one-var convert-to-algebra ↑ Prerequisites: mean-median-mode-range

📏 Medium solution 💡 2 insights

Problem

The average of the numbers $1, 2, 3,\cdots, 98, 99,$ and $x$ is $100x$ . What is $x$ ?

Pick an answer.

(A)

$\dfrac{49}{101}$

(B)

$\dfrac{50}{101}$

(C)

$\dfrac{1}{2}$

(D)

$\dfrac{51}{101}$

(E)

$\dfrac{50}{99}$

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: The list holds the whole numbers 1 through 99 plus one extra number x, so 100 numbers in all. The average of these 100 numbers equals 100 times x. Find x.

Givens: The numbers are 1, 2, 3, ..., 98, 99, and x.; That makes 100 numbers total.; The average of all 100 numbers is 100x.

Unknowns: The value of x.

Understand

Restated: The list holds the whole numbers 1 through 99 plus one extra number x, so 100 numbers in all. The average of these 100 numbers equals 100 times x. Find x.

Givens: The numbers are 1, 2, 3, ..., 98, 99, and x.; That makes 100 numbers total.; The average of all 100 numbers is 100x.

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #5 Look for a Pattern

The word 'average' hides an equation. Turn the sentence into algebra: (sum of all 100 numbers) / 100 = 100x. First handle the fixed part as a subproblem (add 1 through 99), then solve the equation for x. A pattern in the numbers 9999 and 4950 makes the final fraction easy to reduce.

Execute — Answer: B

#7 Identify Subproblems 6.EE.A.2 Step 1

Add 1 through 99

The numbers 1 through 99 are fixed, so add them first.
Pair them from the ends: 1+99, 2+98, 3+97, and so on.
Each pair makes 100, and 50 sits alone in the middle.
That is 49 pairs of 100 plus 50, which is 4900 + 50 = 4950.
This matches the formula n(n+1)/2 with n = 99: 99 times 100 divided by 2 = 4950.

$$1 + 2 + \cdots + 99 = \dfrac{99 \cdot 100}{2} = 4950$$

💡 Pairing the ends turns a long sum into a quick multiplication.

#4 Introduce a Variable 6.EE.B.6 Step 2

Write the average as an equation

There are 100 numbers: the 99 fixed ones plus x.
Their sum is 4950 + x, so the average is (4950 + x) / 100.
The problem says this average equals 100x, so set the two expressions equal.

$$\dfrac{4950 + x}{100} = 100x$$

💡 'Average is 100x' is just an equation waiting to be written down.

#13 Convert to Algebra 8.EE.C.7 Step 3

Solve for x

Multiply both sides by 100 to clear the fraction: 4950 + x = 10000x.
Move the x from the left to the right by subtracting x from both sides: 4950 = 9999x.
Then divide both sides by 9999.

$$4950 + x = 10000x \implies 4950 = 9999x \implies x = \dfrac{4950}{9999}$$

💡 Clearing the fraction and gathering the x-terms leaves one simple division.

#5 Look for a Pattern 6.NS.B.4 Step 4

Reduce the fraction

Both numbers share the factor 99.
Notice 9999 = 100^2 - 1 = (100-1)(100+1) = 99 times 101, and 4950 = 99 times 50.
Dividing top and bottom by 99 gives 50/101.
So x = 50/101, which is choice (B).

$$x = \dfrac{4950}{9999} = \dfrac{99 \cdot 50}{99 \cdot 101} = \dfrac{50}{101}$$

💡 Spotting the shared factor 99 (via difference of squares) collapses the fraction instantly.

[1] #7 6.EE.A.2 The numbers 1 through 99 are fixed, so add them first. Pair them from the ends:

[2] #4 6.EE.B.6 There are 100 numbers: the 99 fixed ones plus x. Their sum is 4950 + x, so the a

[3] #13 8.EE.C.7 Multiply both sides by 100 to clear the fraction: 4950 + x = 10000x. Move the x

[4] #5 6.NS.B.4 Both numbers share the factor 99. Notice 9999 = 100^2 - 1 = (100-1)(100+1) = 99

Review

Reasonableness: x = 50/101 is a little less than 1/2, so 100x is a little less than 50. The 100 numbers run from 1 up to 99 plus a tiny x, so their average should be a bit under 50 as well. Both sides land just under 50, which is consistent, and 50/101 matches choice (B).

Alternative: Instead of the formula, count the fixed sum by grouping: forty-nine pairs each summing to 100, plus the middle 50. Or plug each answer choice into (4950 + x)/100 = 100x and eliminate until only (B) works.

CCSS standards used (min grade 8)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Evaluating the sum-formula n(n+1)/2 at n = 99 to get 4950.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Writing the average (4950 + x)/100 and setting it equal to 100x.)
8.EE.C.7 Solve linear equations in one variable (Clearing the fraction and collecting x-terms to solve 4950 + x = 10000x.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Dividing 4950/9999 by the common factor 99 to reduce it to 50/101.)

⭐ The word 'average' is really an equation: add everything up, divide by how many, and set it equal to what you were told.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Add 1 through 99

Write the average as an equation

Solve for x

Reduce the fraction

Review

CCSS standards used (min grade 8)

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