"If K Is the Sum of the Reciprocals of the Consecutive Integers From 43 to 48 Inclusive" — GMAT® Worked Solution

From Episode 41 of Real GMAT® Problems (The GMAT® Strategy Podcast). For the full strategy behind sequences problems on the GMAT®, read: GMAT® Sequences: Write Every Term, Use Fractions, and Solve What's Actually Asked.

The Problem

Source: Official Guide for GMAT® Review, 11th Edition

If K is the sum of the reciprocals of the consecutive integers from 43 to 48 inclusive, then K is closest in value to which of the following?

(A) ¹⁄₁₂

(B) ¹⁄₁₀

(C) ¹⁄₈

(D) ¹⁄₆

(E) ¹⁄₄

The Setup

Before doing any math, write what's being asked.

K is the sum of the reciprocals of the consecutive integers from 43 to 48 inclusive. Reciprocal of n is 1/n. So we want:

K = ¹⁄₄₃ + ¹⁄₄₄ + ¹⁄₄₅ + ¹⁄₄₆ + ¹⁄₄₇ + ¹⁄₄₈

Six terms. Awkward denominators. No common denominator that's reasonable to find by hand.

Notice the phrasing: "closest in value to which of the following." That phrasing is the signal — the GMAT® is not expecting an exact computation. None of the answer choices will equal the actual sum. The test is asking us to estimate.

The reliable approach: bracket the answer with a low estimate and a high estimate. The actual sum sits between them.

Step 1: Build a Low Estimate

To get a low estimate, take the SMALLEST term in the sum and pretend every term equals it.

The smallest of the six fractions is the one with the largest denominator: ¹⁄₄₈.

If all six terms were ¹⁄₄₈:

6 × ¹⁄₄₈ = ⁶⁄₄₈ = ¹⁄₈

So K is GREATER than ¹⁄₈. (We replaced the actual five larger terms with the smallest one, so the result is smaller than the real sum.)

Low estimate: K > ¹⁄₈

Step 2: Build a High Estimate

To get a high estimate, take the LARGEST term and pretend every term equals it.

The largest of the six fractions is the one with the smallest denominator: ¹⁄₄₃.

If all six terms were ¹⁄₄₃:

6 × ¹⁄₄₃ = ⁶⁄₄₃

That's hard to compare directly to the answer choices. Round 43 down to 42 to make the comparison easier — that nudges the high estimate even higher, which is fine for a bound:

⁶⁄₄₃ < ⁶⁄₄₂ = ¹⁄₇

So K is LESS than ¹⁄₇.

High estimate: K < ¹⁄₇

Step 3: Compare to the Answer Choices

We have:

¹⁄₈ < K < ¹⁄₇

Now look at the answer choices:

(A) ¹⁄₁₂ — way below the range (B) ¹⁄₁₀ — below the range (C) ¹⁄₈ — equal to the low bound (D) ¹⁄₆ — above the high bound (E) ¹⁄₄ — far above the high bound

Only (C) sits at or inside the range. Even though K is slightly larger than ¹⁄₈, it cannot reach ¹⁄₆ — that's further from the actual sum than ¹⁄₈ is.

The answer is (C).

Why This Problem Matters

About 30% of test takers miss this one. By the numbers, it's roughly 50% harder than the previous problem in this episode.

Most of the difficulty isn't the math. It's the discomfort.

Picking an answer that's "close but not exact" cuts against most of what school math conditioned us to do. We were trained to find THE answer, not the closest one. The first time you see a problem like this, the technique can feel sketchy — like maybe there's a real computation path and we're cheating by estimating.

There isn't. The exact sum of those six fractions cannot reasonably be computed by hand. The test knows that. The "closest in value to" phrasing is the test telling you that estimation is the intended path.

Three things to take away:

Watch for "closest in value to" phrasing. It is a signal that the question is built for estimation, not exact computation. Recognizing the signal early saves the panic of trying to find a common denominator for awkward fractions under time pressure.

Bracket with a low and high estimate. Take the smallest term and pretend every term equals it (low bound). Take the largest term and do the same (high bound). The actual sum sits between. Usually only one answer choice fits in the range or sits close to one of the bounds.

Trust the bracketing. It feels uneasy the first few times. With practice, it stops feeling uneasy. There are problems where this technique is the only viable path in under two minutes, and being comfortable with it pays off when one shows up on test day.

This estimation skill becomes even more important on problems with more terms — sometimes the same setup with 100 reciprocals instead of six. The technique scales. The exact computation does not.

Previous problem: A Quiz of 10 Questions Where Each Is Worth 4 Points More Than the Last — GMAT® Worked Solution

Back to the strategy article: GMAT® Sequences: Write Every Term, Use Fractions, and Solve What's Actually Asked