CAT DILR practice paper: part 1 of 2

A DILR set pays nothing until it is decoded: its table or grid has to stand on your sheet before the first question can be answered, and a set that takes 8 minutes to decode leaves 12 for everything else in a 20-minute half. Part 1 of this DILR paper gives you two such sets, with four questions each.

Sit it on your own clock, decode each set on paper, and answer. Afterwards a program proves every answer by trying every case the set allows, and an analysis sheet shows which set paid and which one held your minutes.

Two sets, eight questions, twenty minutes: the blueprint and the rules

The plan of the half, MySmartCoPilot’s own, since CAT does not publish how many sets or questions a section holds:

Three of the eight answers are typed: questions 1, 3 and 6. Score a practice paper as the CAT 2026 FAQs score CAT, in every section alike: a right answer earns 3, a wrong option costs 1, and a wrong typed answer or a blank earns 0. This half is worth 24 marks. Mark each part on its own, add the two parts for the whole section, and read the analysis sheet before the score.

Sit it on a stopwatch, pressing Lap after the scan, after decoding each set and after each answer. Decoding means building the set’s table or grid on your sheet before any question; most of a set’s minutes go there. CAT gives an on-screen calculator throughout the test, so a basic calculator is fair here, and so is one sheet for rough work.

Here is a set of the paper’s kind in miniature. Om, Pia and Ravi take seats 1, 2 and 3 of a row; Pia does not sit at an end, and Ravi sits somewhere to the right of Om. Who sits in seat 3? Pia must take seat 2, the only seat that is not an end, and Om must sit left of Ravi, so the row is Om, Pia, Ravi, and the answer is Ravi. A checker proves an answer like that by trying every order the set allows:

"""The sample set of the lesson, solved the way a checker proves any arrangement set: try every order.

Om, Pia and Ravi take seats 1, 2 and 3 of a row. Pia does not sit at an end. Ravi sits somewhere to the right of Om.
"""
from itertools import permutations

kept = []
for order in permutations(("Om", "Pia", "Ravi")):  # order[0] sits in seat 1
    seat = {name: i + 1 for i, name in enumerate(order)}
    if seat["Pia"] in (1, 3):
        verdict = "out: Pia sits at an end"
    elif seat["Ravi"] < seat["Om"]:
        verdict = "out: Ravi sits left of Om"
    else:
        verdict = "fits both clues"
        kept.append(order)
    print(f"  seats 1-3: {', '.join(order):<16} {verdict}")
print(f"{len(kept)} of 6 orders fit; seat 3: {', '.join(o[2] for o in kept)}")

  seats 1-3: Om, Pia, Ravi    fits both clues
  seats 1-3: Om, Ravi, Pia    out: Pia sits at an end
  seats 1-3: Pia, Om, Ravi    out: Pia sits at an end
  seats 1-3: Pia, Ravi, Om    out: Pia sits at an end
  seats 1-3: Ravi, Om, Pia    out: Pia sits at an end
  seats 1-3: Ravi, Pia, Om    out: Ravi sits left of Om
1 of 6 orders fit; seat 3: Ravi

One order of the six survives both clues. Every set in this paper is proved the same way, on a larger scale.

Set A and questions 1 to 4: four branches, one month’s sales Pro: opens with a Pro, Premium or Ultimate pass