SHSAT Sunset

Vicente and Carla each ran 8 laps

2023 Handbook

2023 Test A Math · Question 52 of 57

Question

Vicente and Carla each ran 8 laps around a track. They started at the same time and place. If Vicente ran 1.5 times as fast as Carla, how many laps did Carla have left to finish when Vicente finished his 8th lap?

  • A2122\frac{1}{2}
  • B2232\frac{2}{3}
  • C3133\frac{1}{3}
  • D5135\frac{1}{3}
Show the answer

Correct answer: B

Vicente ran 1.5 times as fast as Carla, so that means for every lap Carla finished, Vicente finished 1.5 laps. Set up a proportion to calculate c, the number of laps Carla had completed when Vicente finished 8 laps:

c1=81.5\frac{c}{1} = \frac{8}{1.5} Cross multiply.
1.5c=8(1)1.5c = 8(1) Apply the multiplicative inverse property; divide both sides of the equation by 1.5.
c=513c = 5\frac{1}{3}

So, the number of laps Carla still had to finish is 8−513=2238 - 5\frac{1}{3} = 2\frac{2}{3}.

Question analysis

This ratio word problem says Vicente and Carla each ran 8 laps, starting at the same time and place, and Vicente ran 1.5 times as fast. It asks how many laps Carla still had to run when Vicente finished. Students reason with a speed ratio and mixed numbers, and giving the laps run rather than the laps remaining is an easy slip.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 rate problems usually give a speed and a time. Here only the ratio of two speeds is given, so students must reason that distances run in the same time share that ratio, then work with a fractional number of laps.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 4, 44 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

No time is given, so students must reason that in equal time distance scales with speed, which turns the race into one division and one subtraction. Both steps hide a trap: the division by 1.5 leaves a remainder that is easy to misread, and subtracting a mixed number from 8 needs regrouping. The result of the first step, Carla's laps run, is itself a choice, so the question also rewards rereading what is asked.

  1. Turning words into math moderate Understanding the question

    The phrase “1.5 times as fast” has to become distances in the same time: while Vicente runs 8 laps, Carla runs 8 ÷ 1.5, not 8 − 1.5 or 8 × 1.5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic moderate Working it out

    Dividing 8 by 1.5 gives 16/3, or 5 1/3, a division by a decimal with a remainder, and 8 − 5 1/3 then needs a whole regrouped into thirds. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer minor Choosing the answer

    The choice 2 1/2 is only a sixth of a lap from 2 2/3, so a rough estimate of how far Carla trails cannot separate them; only the exact quotient does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: In the same amount of time the faster runner covers 1.5 times the distance, so divide Vicente's laps by 1.5 to get Carla's, then subtract that from 8.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, rate and time
d=r×td = r \times t
d is the distance, r is the rate or speed, and t is the time. When two runners run for the same time, the faster one goes farther by the same factor as the speed. So 1.5 times as fast means 1.5 times as far.
Dividing by a fraction
a÷mn=a×nma \div \frac{m}{n} = a \times \frac{n}{m}
a is any number and m over n is a fraction. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down.

  1. Sort out the question

    Vicente and Carla each run 8 laps, starting together. Vicente runs 1.5 times as fast as Carla. You need how many laps Carla still has to run when Vicente finishes. First find how far Carla has run at that moment.

  2. Compare their distances

    d=r×td = r \times t

    They have run for the same time. So Vicente’s distance is 1.5 times Carla’s. Let c be the laps Carla has run.

    8=1.5×c8 = 1.5 \times c

    To find c, divide 8 by 1.5.

  3. Divide 8 by 1.5

    Write 1.5 as the fraction 32\frac{3}{2}.

    a÷mn=a×nma \div \frac{m}{n} = a \times \frac{n}{m}

    c=8÷32=8×23=163=513c = 8 \div \frac{3}{2} = 8 \times \frac{2}{3} = \frac{16}{3} = 5\frac{1}{3}

    Carla has run 5135\frac{1}{3} laps. Watch out: that is how far she has run, not what is left. Stopping here gives choice D.

  4. Find the laps she has left

    8−513=243−163=83=2238 - 5\frac{1}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} = 2\frac{2}{3}

    Watch out: 8 − 5135\frac{1}{3} is not 3133\frac{1}{3}. You must break 1 whole into thirds first. Answering 3133\frac{1}{3} is choice C.

    So the answer is B, 2232\frac{2}{3} laps.

Check your answer

Multiply back: 1.5×513=32×163=81.5 \times 5\frac{1}{3} = \frac{3}{2} \times \frac{16}{3} = 8. So when Carla has run 5135\frac{1}{3} laps, Vicente has run 8. And 513+223=85\frac{1}{3} + 2\frac{2}{3} = 8, her full race.

More questions like this

The original DOE question

Question 109 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 109. Open page 89 of the handbook (PDF)