The reciprocal of 1/4 is added

2024 Handbook

2024 Test A Math · Question 30 of 57

Question

The reciprocal of $\frac{1}{4}$ is added to the reciprocal of 3. What is the reciprocal of this sum?

A $\frac{3}{13}$
B $\frac{3}{4}$
C $\frac{4}{5}$
D $\frac{13}{4}$

Show the answer

Correct answer: A

The reciprocal of $\frac{1}{4}$ is added to the reciprocal of 3.

$\frac{4}{1} + \frac{1}{3} = \frac{12}{3} + \frac{1}{3} = \frac{13}{3}$

The reciprocal of this sum is $\frac{3}{13}$ .

Question analysis

This fractions question asks for the reciprocal of a sum: the reciprocal of one fourth is added to the reciprocal of 3, and students must flip the total. It tests knowing what a reciprocal is and adding a whole number to a fraction. Keeping the three flips in the right order, without skipping one, is what makes it tricky.

NYS standards: 6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students use reciprocals when dividing fractions, usually one at a time. This item chains three reciprocals around one addition, with positive numbers only, so the demand is following the wording in order rather than new content.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is easy, so the question is about order: flip each number, add, then flip the total. The common failures are flipping at the wrong moment, term by term instead of the whole sum, and adding a whole number to a fraction straight across. Writing the full expression 1 ÷ (4 + 1/3) before doing any arithmetic keeps the three reciprocals in their places.

Turning words into math moderate Understanding the question
The two sentences hold three reciprocals and a sum, which must become 1 ÷ (4 + 1/3) before any arithmetic; "this sum" means flipping the total, not each part. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Adding 4 and 1/3 means writing 4 as 12/3 to get 13/3, a whole number plus a fraction, where adding tops and bottoms gives 5/4 instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work in the order of the words: flip each number to get its reciprocal, add the two using a common denominator, and then flip that sum.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: weak
Multiplies the two reciprocals instead of adding them, 4 × 1/3 = 4/3, then flips that product to 3/4.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds 4/1 and 1/3 straight across, numerators and denominators, getting 5/4, then flips it to 4/5.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Flips each term of 4 + 1/3 instead of the whole sum, as if the reciprocal of a sum were the sum of reciprocals: 1/4 + 3 = 13/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal: $\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$; a is the top of a fraction and b is the bottom. The reciprocal is the fraction flipped upside down. A whole number such as 3 is the fraction 3 over 1, so its reciprocal is 1 over 3. A number times its reciprocal always makes 1.
Adding fractions with the same bottom: $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$; c is the shared bottom number, called the denominator. a and b are the tops, called numerators. When the pieces are the same size, you just add how many pieces you have. The size of the pieces, c, stays the same.

Sort out the wording
The question has three reciprocals in it. A reciprocal is a fraction flipped upside down. Do the jobs in the order of the words:
1. Flip $\frac{1}{4}$ .
2. Flip 3.
3. Add the two results to get the sum.
4. Flip that sum.
The last flip gives the answer.
Flip each number
$\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$
Flip $\frac{1}{4}$ to get $\frac{4}{1}$ , which is 4. The number 3 is the same as $\frac{3}{1}$ , so its reciprocal is $\frac{1}{3}$ .
Add the two reciprocals
You need $4 + \frac{1}{3}$ . To add, both numbers need the same bottom. So write 4 in thirds. One whole is 3 thirds, so 4 wholes are 12 thirds.
$\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$
$\frac{12}{3} + \frac{1}{3} = \frac{13}{3}$
The sum is $\frac{13}{3}$ . Watch out: don’t add the tops and the bottoms straight across. That gives $\frac{5}{4}$ , which flips to $\frac{4}{5}$ , choice C.
Flip the sum
$\text{reciprocal of } \frac{13}{3} = \frac{3}{13}$
Watch out: flip the whole sum, not each part. Flipping 4 and $\frac{1}{3}$ one at a time takes you back to $\frac{1}{4} + 3 = \frac{13}{4}$ , choice D.
So the answer is A, $\frac{3}{13}$ .

Check your answer

A number times its reciprocal is 1. Test it: $\frac{13}{3} \times \frac{3}{13} = \frac{39}{39} = 1$ . Also estimate. The sum is a bit more than 4, so its reciprocal should be a bit less than $\frac{1}{4}$ . $\frac{3}{13}$ is about 0.23, just under 0.25.

The original DOE question

Question 87 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 87. Open page 83 of the handbook (PDF)