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Matrix / Distribution Puzzles questions for IPMAT, with solutions

10 free Matrix / Distribution Puzzles practice questions for IPMAT Indore, Rohtak and JIPMAT, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Matrix / Distribution Puzzles on Learn.

  1. Question 1Multiple choiceWritten by IPMATwayNumber distribution grid

    Three vendors Anu, Bala and Chitra together sold 27 items: 12 caps, 9 scarves and 6 belts. Anu sold 10 items, Bala 9 and Chitra 8, and each vendor sold at least one item of each kind. Anu sold 5 caps. Bala sold twice as many scarves as caps. Chitra sold more belts than scarves.

    How many caps did Chitra sell?

    1. A.5
    2. B.4
    3. C.3
    4. D.2
    Show the answer and solution

    Answer: A. 5

    Answer: (A) 5

    Working:

    1. Let Bala sell c caps; then 2c scarves and 9 - 3c belts, which must be at least 1, so c is 1 or 2.
    2. c = 1: Bala 1, 2, 6; Chitra's caps = 12 - 5 - 1 = 6, leaving 2 items for scarves and belts with belts > scarves and each at least 1: impossible.
    3. c = 2: Bala 2, 4, 3; Chitra's caps = 12 - 5 - 2 = 5, and her other 3 items are 1 scarf and 2 belts (belts > scarves). Anu then sells 5 caps, 9 - 4 - 1 = 4 scarves and 1 belt: total 10. Consistent.
    4. Chitra sold 5 caps.
    VendorCapsScarvesBelts
    Anu541
    Bala243
    Chitra512

    Why the others are wrong:

    • (B) Chitra 4 caps forces Bala 3 caps and 6 scarves, which is 9 items with no belt.
    • (C) Chitra 3 caps forces Bala 4 caps and 8 scarves, more than Bala's 9 items allow.
    • (D) 2 is Bala's cap count, not Chitra's.

  2. Question 2Multiple choiceWritten by IPMATwayFloors / boxes / stacks

    Five boxes A, B, C, D and E are stacked one above another. C is the topmost box. B is immediately above D. A is exactly three places above E.

    Which box is at the bottom of the stack?

    1. A.D
    2. B.B
    3. C.A
    4. D.E
    Show the answer and solution

    Answer: D. E

    Answer: (D) E

    Working:

    1. Number the positions 1 (bottom) to 5 (top). C = 5.
    2. A is three places above E: (E, A) = (1, 4) or (2, 5). Position 5 is C's, so E = 1 and A = 4.
    3. B immediately above D fills the remaining 2 and 3: D = 2, B = 3.
    4. Bottom to top: E, D, B, A, C. The bottom box is E.

    Why the others are wrong:

    • (A) D is second from the bottom, directly under B.
    • (B) B sits on D, so B is not at the bottom.
    • (C) A is above E, three places up, so A is near the top.

  3. Question 3Multiple choiceWritten by IPMATwayFloors / boxes / stacks

    In a six-storey building (floor 1 at the bottom), P, Q and R live on three different floors. P lives on an even-numbered floor below Q, and exactly two floors lie between the floors of P and Q. R lives immediately above Q.

    On which floor does R live?

    1. A.5
    2. B.4
    3. C.6
    4. D.Cannot be determined
    Show the answer and solution

    Answer: C. 6

    Answer: (C) 6

    Working:

    1. Two floors between P and Q with P below means Q = P + 3.
    2. P is even: P = 2 gives Q = 5; P = 4 gives Q = 7, beyond the sixth floor; P = 6 is not below anything. So P = 2, Q = 5.
    3. R is immediately above Q: R = 6.

    Why the others are wrong:

    • (A) 5 is Q's floor; R is immediately above Q.
    • (B) P 4 would put Q on 7, which does not exist.
    • (D) The six-floor limit removes the P = 4 case, so R is fixed.

  4. Question 4Type the answerWritten by IPMATwayShort numeric-answer logic puzzle (Indore SA style)

    Six players A, B, C, D, E and F are to be divided into three doubles pairs for a tennis practice. A refuses to pair with B, and C refuses to pair with D. (Pairs are unordered, and the three pairs are not labelled.)

    In how many ways can the three pairs be formed?

    Show the answer and solution

    Answer: 10

    Count all ways to split six people into three unlabelled pairs: A can be paired with any of 5 people, then the next unpaired person with any of 3, and the last two pair up: 5 x 3 x 1 = 15. Pairings in which A and B are a pair: the other four form 3 pairings. Likewise 3 pairings contain the pair C-D. Exactly 1 pairing contains both A-B and C-D (then E-F). By inclusion-exclusion the forbidden pairings number 3 + 3 - 1 = 5, so the allowed pairings are 15 - 5 = 10. Direct listing of all 15 confirms 10.

  5. Question 5Multiple choiceWritten by IPMATwayNumber distribution grid

    The digits 1 to 9 are each used exactly once to fill a 3 x 3 grid. The three row totals (top to bottom) are 6, 19 and 20. The three column totals (left to right) are 20, 10 and 15. The bottom-left cell holds 9. The middle-right cell is smaller than the bottom-right cell.

    What number is in the middle row, right column?

    1. A.6
    2. B.4
    3. C.7
    4. D.5
    Show the answer and solution

    Answer: A. 6

    The top row sums to 6 with distinct digits, so it must be 1, 2, 3 in some order. Column 2 totals 10 and column 1 totals 20, and 9 sits at the bottom of column 1. A search of all 362880 fillings gives only three grids that meet the row and column totals: 3 1 2 / 8 4 7 / 9 5 6, 3 1 2 / 8 5 6 / 9 4 7 and 3 1 2 / 9 4 6 / 8 5 7.

    The 9 in the bottom-left cell removes the third grid. Of the other two, the first has middle-right 7 and bottom-right 6 (7 is not smaller than 6), while the second has middle-right 6 and bottom-right 7 (6 is smaller than 7). So the second grid is the one, and the middle-right cell is 6.

  6. Question 6Multiple choiceWritten by IPMATwayFloors / boxes / stacks

    Six people A, B, C, D, E and F live on the six floors of a building, one per floor, with floor 1 at the bottom and floor 6 at the top. A lives on floor 3 and B on floor 5. C lives somewhere below D. E lives on an even-numbered floor. F does not live on a floor adjacent to A's floor.

    Which of the following must be true?

    1. A.E lives on floor 4
    2. B.C lives on floor 1
    3. C.F lives on floor 1 or floor 6
    4. D.D lives above B
    Show the answer and solution

    Answer: C. F lives on floor 1 or floor 6

    Answer: (C) F lives on floor 1 or floor 6

    Working:

    1. A = 3 and B = 5 are fixed. F cannot be on 2 or 4 (adjacent to 3), so F is on 1 or 6: this holds in every arrangement.
    2. The five valid arrangements (floors of C, D, E, F): (1, 2, 4, 6), (1, 4, 2, 6), (2, 4, 6, 1), (2, 6, 4, 1), (4, 6, 2, 1).
    3. E on 4: fails in (1, 4, 2, 6). C on 1: fails in (2, 4, 6, 1). D above B: fails in (1, 4, 2, 6).
    4. Only the statement about F is true in all five.

    Why the others are wrong:

    • (A) E can be on 2, 4 or 6; for example C 1, E 2, F 6 with D 4 is valid.
    • (B) C can be on 2 or 4 as well, e.g. F 1, C 2, D 4, E 6.
    • (D) D can be on 4, below B, e.g. C 1, E 2, D 4, F 6.

  7. Question 7Type the answerWritten by IPMATwayShort numeric-answer logic puzzle (Indore SA style)

    A club must send a team of exactly three members chosen from P, Q, R, S, T and U. If P is in the team then Q must also be in it. R and S cannot both be in the team.

    How many different teams are possible?

    Show the answer and solution

    Answer: 11

    All teams of three from six: C(6,3) = 20. Teams that break the first rule contain P and not Q: the other two members come from R, S, T, U, so C(4,2) = 6 teams; 20 - 6 = 14 remain. Teams that contain both R and S are {R, S, P}, {R, S, Q}, {R, S, T} and {R, S, U}; of these, {R, S, P} was already removed in the first step (it has P without Q), so only 3 further teams are removed: 14 - 3 = 11. Direct enumeration of the 20 teams gives the same count of 11.

  8. Question 8Type the answerWritten by IPMATwayShort numeric-answer logic puzzle (Indore SA style)

    Four friends A, B, C and D are members of a college's Music, Drama and Chess clubs. Each friend is in at least one club, no two friends are in exactly the same set of clubs, and each club has exactly two of the four friends as members. A is in the Music club only. D is in exactly one club, and it is not the Drama club. B is in exactly two clubs, and B and C share exactly one club. C is in the Chess club.

    How many of the four friends are in both the Drama club and the Chess club?

    Show the answer and solution

    Answer: 1

    Each club has exactly two members, so there are 6 memberships: A has 1, D has 1, and B has 2, so C has 2. D is in Music or Chess. If D were in Music, Music would be full (A, D), and B and C would each need two clubs from Drama and Chess only, making B and C identical and sharing two clubs: impossible. So D is in Chess only. Drama still needs two members and only B and C are left, so both are in Drama. C is also in Chess (given), so C = Drama and Chess; Chess is now full (D, C), so B's second club is Music: B = Music and Drama. All sets are distinct, and B and C share exactly one club (Drama). Exactly one friend, C, is in both Drama and Chess. Enumeration of all assignments confirms this is the only solution.

  9. Question 9Multiple choiceWritten by IPMATwayNumber distribution grid

    The digits 1 to 9 are each placed exactly once in a 3 x 3 grid. In how many ways can this be done so that every row total and every column total is odd?

    1. A.2,880
    2. B.8,640
    3. C.25,920
    4. D.51,840
    Show the answer and solution

    Answer: C. 25,920

    A row or column total is odd exactly when it holds an odd number of odd digits, that is 1 or 3. There are 5 odd digits (1, 3, 5, 7, 9) and 4 even ones. The three row counts of odd digits are each 1 or 3 and add to 5, so they are 3, 1, 1 in some order: one row is all odd. The same argument shows one column is all odd. That full row and full column use 3 + 3 - 1 = 5 cells, which is all five odd digits, so the other two rows each have exactly one odd digit (where they cross the full column) and every condition holds. So a valid pattern is fixed by choosing the full row (3 ways) and the full column (3 ways): 9 patterns. For each pattern the 5 odd digits fill their cells in 5! = 120 ways and the 4 even digits in 4! = 24 ways, i.e. 2,880 fillings. Total 9 x 2,880 = 25,920 (a computer search of all 362,880 grids confirms this).

  10. Question 10Multiple choiceWritten by IPMATwayPerson-attribute grid (3-4 variables)

    Four friends A, B, C and D each live in a different city out of Delhi, Pune, Goa and Agra, and each plays a different sport out of chess, golf, polo and squash. It is known that: (i) The chess player lives in Agra. (ii) The polo player lives in Delhi. (iii) The golf player does not live in Pune. (iv) A does not live in Goa, Pune or Agra. (v) D does not live in Goa or Pune. (vi) B does not live in Pune.

    Who lives in Goa, and which sport does that person play?

    1. A.C, golf
    2. B.B, golf
    3. C.B, squash
    4. D.D, golf
    Show the answer and solution

    Answer: B. B, golf

    From (iv) A lives in Delhi. From (v) D lives in Delhi or Agra, and Delhi is taken, so D lives in Agra. From (vi) B is not in Pune, and only Goa and Pune remain for B and C, so B lives in Goa and C in Pune.

    Sports: the polo player lives in Delhi, so A plays polo. The chess player lives in Agra, so D plays chess. The remaining sports golf and squash go to B (Goa) and C (Pune); by (iii) the golf player is not in Pune, so B plays golf and C plays squash.

    A full search over all 576 city-sport assignments shows these six clues give exactly one solution and that removing any one clue leaves more than one.