Maven Securities interview questions

How Maven Securities's process runs, as candidates and the firm describe it, and the questions candidates report, each with a worked solution. Free, no sign-up.

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How Maven Securities interviews

Sources: Glassdoor: Maven Securities trader

Questions (10)

Try each one before opening the solution. Interviewers change the numbers, so learn the method.

Graduate Trader · Fourth round · Market making · reported by a candidate

In a market-making game you sell one pen at each of 13, 16, 19 and 22. The pen's value turns out to be 22.

  1. (a) What is your position (in pens)?
    Answer and worked solution

    Answer: -4

    Key idea. Selling makes your position negative.

    1. Four sales: −4.
  2. (b) What is your profit or loss, marked at 22?
    Answer and worked solution

    Answer: -18

    Key idea. P&L = cash received − value of what you sold.

    1. Received 13 + 16 + 19 + 22 = 70.
    2. Value of 4 pens at 22: 88.
    3. P&L = −18.
  3. (c) What is the highest whole-number price at which buying all four back would still leave you with a profit?
    Answer and worked solution

    Answer: 17

    Key idea. Buying back 4 pens must cost less than the 70 received.

    1. 4p < 70, so p < 17.5.
    2. Highest whole number: 17.

    Check. The lesson: you kept selling into a rising price. Notice when you're being lifted repeatedly and move your quotes up.

Source: Glassdoor · Try it in the app →

Graduate Trader · Fourth round · Combinatorics · reported by a candidate

Count the rectangles on an 8 × 8 chessboard, squares included.

  1. How many?
    Answer and worked solution

    Answer: 1296

    Key idea. A rectangle is fixed by choosing 2 vertical and 2 horizontal grid lines.

    1. An 8 × 8 board has 9 lines each way.
    2. C(9, 2)² = 36² = 1,296.

    Check. Squares alone: 1² + 2² + … + 8² = 204.

Source: Glassdoor · Try it in the app →

Graduate Trader · Fourth round · Logic · reported by a candidate

You have a balance scale and some balls that look identical, except that one has a different weight.

  1. (a) Eight balls, and the odd one is heavier. Fewest weighings to guarantee finding it?
    Answer and worked solution

    Answer: 2

    Key idea. Each weighing has 3 outcomes, so k weighings distinguish at most 3k cases.

    1. 8 cases ≤ 3² = 9.
    2. Weigh 3 v 3. Balanced: weigh the other 2. Unbalanced: take the heavy 3 and weigh 1 v 1.
    3. 2 weighings.
  2. (b) Twelve balls, and you don't know whether the odd one is heavier or lighter. Fewest weighings to find it (and whether it's heavy or light)?
    Answer and worked solution

    Answer: 3

    Key idea. Count cases: 12 balls × heavier or lighter = 24 ≤ 3³ = 27.

    1. Weigh 4 v 4 first.
    2. Then rearrange balls between the pans so every outcome splits the remaining cases about evenly.
    3. 3 weighings suffice (and 2 can't, since 9 < 24).

Source: Glassdoor · Try it in the app →

Graduate Trader · Fourth round · Mental maths · reported by a candidate

In your head:

  1. 17/7 to ten decimal places
    Answer and worked solution

    Answer: 17/7 ≈ 2.4286

    Key idea. 17/7 = 2 + 3/7, and sevenths cycle through 142857.

    1. 3/7 = 0.428571428571…
    2. 17/7 = 2.4285714285|71…
    3. To ten places: 2.4285714286.

Source: Glassdoor · Try it in the app →

Graduate Trader · Fourth round · Expected value · reported by a candidate

n people each get a name sticker for someone in the group, assigned at random (a random permutation). Start with one person, go to the person named on their sticker, and so on until you return: that's a cycle. Every person is in exactly one cycle.

  1. For n = 4, what is the expected number of cycles?
    Answer and worked solution

    Answer: 25/12 ≈ 2.0833

    Key idea. Build the permutation one person at a time: the k-th person closes a new cycle with probability 1/k.

    1. E[cycles] = 1 + 1/2 + … + 1/n.
    2. n = 4: 1 + ½ + ⅓ + ¼ = 25/12 ≈ 2.08.

    Check. For large n, about ln n + 0.577.

Interviewers may push further: For large n, about how many? (ln n + 0.577.)

Source: Glassdoor · Try it in the app →

Graduate Trader · Second round · Combinatorics · reported by a candidate

A bag holds 6 blue, 6 green and 6 red blocks. You draw 5 without replacement to build a tower.

  1. What is the probability the tower uses exactly two colours?
    Answer and worked solution

    Answer: 0.27311

    Key idea. Count draws using exactly two colours.

    1. Choose the two colours: 3 ways.
    2. Draws of 5 from those 12 blocks: C(12, 5) = 792, minus 2 × C(6, 5) = 12 single-colour draws: 780.
    3. All draws: C(18, 5) = 8,568.
    4. P = 3 × 780/8,568 = 65/238 ≈ 0.273.

    Trap. Forgetting to remove single-colour draws.

Source: Glassdoor · Try it in the app →

Graduate Trader · Online test · Mental maths · reported by a candidate

Examples from Maven's first-round numerical test.

Reported as 60 problems in 10 minutes without a calculator.

  1. (a) 35 × 26
    Answer and worked solution

    Answer: 910

    Key idea. Halve one factor and double the other.

    1. 35 × 26 = 70 × 13 = 910.
  2. (b) 15 × 76
    Answer and worked solution

    Answer: 1140

    Key idea. Same trick.

    1. 15 × 76 = 30 × 38 = 1,140.

Source: Glassdoor · Try it in the app →

Trader · Interview · Behavioural · reported by a candidate

Behavioural questions candidates report:

Answer one out loud, as you would in the interview.

Model answer

Use a short story: the situation in one sentence, what you did (not the team), the result with a number if you have one, and what you'd do differently. Aim for about two minutes.

On the trading mistake: own it quickly, explain the reasoning at the time, say what you'd do now. Interviewers care more about how you review errors than about the error.

Source: Glassdoor · Try it in the app →

Graduate Trader (Chicago) · Interview · Combinatorics · reported by a candidate

Count the distinct arrangements of the letters of MISSISSIPPI.

  1. How many are there?
    Answer and worked solution

    Answer: 34650

    Key idea. Arrangements of a multiset: n! divided by the factorial of each letter's count.

    1. 11 letters: M ×1, I ×4, S ×4, P ×2.
    2. 11!/(4! 4! 2!) = 39,916,800/1,152 = 34,650.

    Trap. Forgetting to divide by 2! for the two Ps.

Source: Wall Street Oasis: Maven Securities interview reports · Try it in the app →

Trading intern · Interview · Mental maths · reported by a candidate

Which is larger, 1040 or 40!? Then find the two consecutive factorials that 1040 lies between.

  1. 1040 lies between (n − 1)! and n!. What is n?
    Answer and worked solution

    Answer: 35

    Key idea. Work in powers of 10.

    1. 40! ≈ 8.2 × 10⁴⁷, far bigger than 10⁴⁰.
    2. 30! ≈ 2.65 × 10³². Multiply up: 31! ≈ 8.2 × 10³³, 32! ≈ 2.6 × 10³⁵, 33! ≈ 8.7 × 10³⁶, 34! ≈ 3.0 × 10³⁸, 35! ≈ 1.03 × 10⁴⁰.
    3. So 34! < 10⁴⁰ < 35!: n = 35.

    Check. Stirling: log₁₀ 35! ≈ 35 log₁₀(35/e) + ½ log₁₀(70π) ≈ 38.8 + 1.2 = 40.0.

Source: Wall Street Oasis: Maven Securities interview reports · Try it in the app →

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