SAT Math: Rounding Error & Finding E − S (30-Decimal Problem)

The short answer: the correct answer is B — I and II only. Both −16 and 6 are achievable values of E − S; 10 is not. The reason it feels hard is that the problem invites you to construct an example, and constructing examples for 30 decimals at once is miserable. The right move is to bound the contribution of a single decimal to E − S and then add the bounds across all 30. That bounded-sum technique turns a five-minute brute-force scramble into a sixty-second algebra problem.

The problem, restated cleanly

List T contains 30 positive decimals, none of which is an integer, and the sum of the 30 decimals is S. The estimated sum E is built as follows: each decimal whose tenths digit is even is rounded up to the nearest integer, and each decimal whose tenths digit is odd is rounded down to the nearest integer. Exactly one-third of the decimals — so 10 of them — have an even tenths digit, which leaves 20 with an odd tenths digit. Which of the following is a possible value of E − S?

I. −16    II. 6    III. 10

If you try to pick 30 specific decimals and add them up, you will spend the whole testing block on it. The trick is to ignore the list entirely and reason about one decimal at a time.

Set up the algebra: every decimal is I + d

Write any positive non-integer decimal as x = I + d, where I is its integer part (the floor) and d is its fractional part. Because x is not an integer, we have 0 < d < 1. The tenths digit of x is the tenths digit of d.

Now consider what happens to that single decimal under the rounding rule.

Case 1: the tenths digit is even (round up)

Rounding x up to the nearest integer gives I + 1. The contribution of this single decimal to E − S is therefore

(I + 1) − (I + d) = 1 − d.

The fractional part d is in (0, 1), so 1 − d is in (0, 1) as well. But because the tenths digit is even — that is, the tenths digit is 0, 2, 4, 6, or 8 — we have d in the interval [0.0, 0.1) ∪ [0.2, 0.3) ∪ ... ∪ [0.8, 0.9). The largest d can be (excluding 0, since the decimal is non-integer) is just under 0.9, and the smallest is just above 0. So 1 − d lands in roughly (0.1, 1). Each even-tenths decimal contributes strictly less than 1 to E − S, and the contribution is strictly positive.

Case 2: the tenths digit is odd (round down)

Rounding x down to the nearest integer gives I. The contribution of this single decimal to E − S is

I − (I + d) = −d.

Because the tenths digit is odd (1, 3, 5, 7, or 9), we have d in [0.1, 0.2) ∪ [0.3, 0.4) ∪ ... ∪ [0.9, 1.0). So d is bounded between roughly 0.1 and (just under) 1.0, and the contribution −d sits in (−1, −0.1]. Each odd-tenths decimal subtracts somewhere between (just over) 0 and (just under) 1 from E − S.

Bound the whole sum

Now sum the per-decimal bounds across the full list. There are 10 even-tenths decimals and 20 odd-tenths decimals. Let U denote the total contribution from the 10 even-tenths terms and let D denote the total contribution from the 20 odd-tenths terms. Then E − S = U + D, and

• U is the sum of 10 numbers each in (0.1, 1), so U is in (1, 10).
• D is the sum of 20 numbers each in (−1, −0.1], so D is in (−20, −2].

Add the bounds: E − S = U + D ∈ (−19, −1 + 8) = (−19, 8). (The upper bound is generous because U can approach 10 only while D approaches 0, but the two are constrained together; the actual upper bound is even tighter, and we'll show it cannot reach 10.)

Now answer the three Roman numerals

With the achievable range pinned to roughly (−19, <10), we can check each candidate directly.

I. Is E − S = −16 possible?

Yes. Push the 10 even-tenths decimals to contribute very little (each just above 0.1, total just above 1) and push the 20 odd-tenths decimals to contribute as much negative as possible (each near −1, total near −20). The sum lands near 1 + (−20) = −19, and you can dial back from there to hit exactly −16. Concretely, pick 10 decimals with tenths digit 0 and fractional part just above 0.9 of an interval (so the contribution is small but positive) and 20 decimals with tenths digit 9 and fractional part just under 1.0. The sum E − S sweeps continuously through (−19, −1 + ε), and −16 is firmly inside that interval. −16 is achievable.

II. Is E − S = 6 possible?

Yes. Now push the even-tenths group hard the other way: pick d just above 0 (tenths digit 0, fractional just slightly above 0), so each contribution 1 − d is close to 1. The 10 even-tenths decimals contribute close to 10. Then pick the 20 odd-tenths decimals with tenths digit 1 and a small d (just above 0.1), so each contribution −d is close to −0.1. The 20 odd-tenths decimals contribute close to −2. Together that's close to 10 − 2 = 8, well above 6. Dial the parameters back to hit exactly 6. 6 is achievable.

III. Is E − S = 10 possible?

No — and this is the whole point of the bounding argument. The maximum possible value of U is strictly less than 10, because each of the 10 contributions 1 − d is strictly less than 1 (we have d > 0). The maximum possible value of D is at most −0.1 per term times 20 terms = −2 (actually even more negative because d > 0.1 strictly is not required, but the equality is at d = 0.1 and contributions are at most −0.1). So

max(E − S) < 10 + (−2) = 8.

Since the achievable maximum is strictly less than 8 — let alone 10 — the value 10 is impossible. 10 is not achievable.

The answer

Only I and II are possible values of E − S. The correct choice is B. I and II only.

Why this technique matters past this one problem

SAT, SHSAT, and AMC test writers love this exact pattern because it punishes brute force and rewards algebraic thinking. The recipe is:

1. Write each item in the list in a clean parametric form (here, x = I + d).
2. Compute the contribution of one item to the quantity of interest (here, E − S) in each case.
3. Bound that single contribution above and below using whatever constraint the problem gives you.
4. Sum the bounds across all items to get an interval for the whole quantity.
5. Compare each multiple-choice value against that interval.

It's the same move that solves SAT problems about averages of bounded numbers, SHSAT problems about minimum and maximum values of expressions, and AP Statistics problems about ranges of sample means. Once a student internalizes it, problems that used to take five minutes of trial and error take sixty seconds of clean algebra.

How we teach this at SOMATH

In our small-group SAT and SHSAT classes on West 79th Street, this is the kind of problem we walk through twice — once to show the technique, once to have a student narrate it back. The goal is not to memorize the move for this particular problem; it is to recognize the shape of a bounded-sum problem the next time it appears. That recognition is what separates a 700-tier SAT math score from a 780-tier one.

If your child is preparing for SAT, SHSAT, or AMC and gets stuck on problems like this — wanting to construct examples when they should be bounding — that is a teachable pattern. Our free 30-minute evaluation diagnoses exactly where the gap is and gives you a written summary within 48 hours, whether or not you enroll.