Skip to content
IPMATway
Sign in

LR / DI · free practice

Quant-flavoured reasoning (money, bounds, shifts, counting) questions for IPMAT, with solutions

10 free Quant-flavoured reasoning (money, bounds, shifts, counting) practice questions for IPMAT Indore, Rohtak and JIPMAT, 2 from past papers, 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 Quant-flavoured reasoning (money, bounds, shifts, counting) on Learn.

  1. Question 1Multiple choiceAsked in JIPMAT 2024Money and notes

    A cashier has ₹ 620 in the denominations of two-rupee notes, ten-rupee notes and fifty-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he/she has ?

    1. A.25
    2. B.35
    3. C.12
    4. D.30
    Show the answer and solution

    Answer: D. 30

    Answer: (D) 30

    Working:

    1. Let each denomination have n notes: 2n + 10n + 50n = 62n = 620.
    2. n = 10, so the total number of notes is 3 × 10 = 30.

    Why the others are wrong:

    • (A) 25 is not a multiple of 3, so three equal counts cannot add to it.
    • (B) 35 is not a multiple of 3 either.
    • (C) 12 notes is 4 sets, worth only 248.

  2. Question 2Multiple choiceWritten by IPMATwayCounting logic

    The pages of a book are numbered 1 to 250. How many digits are used in all to print the page numbers?

    1. A.650
    2. B.642
    3. C.600
    4. D.750
    Show the answer and solution

    Answer: B. 642

    Answer: (B) 642

    Working:

    1. Pages 1-9: 9 digits. Pages 10-99: 90 × 2 = 180. Pages 100-250: 151 × 3 = 453.
    2. Total 642.

    Why the others are wrong:

    • (A) 650 is a rough guess.
    • (C) 600 treats pages 100-250 as 150 three-digit pages and rounds.
    • (D) 750 is 250 × 3.

  3. Question 3Multiple choiceWritten by IPMATwayCounting logic

    In how many ways can 5 people stand in a row if two particular people, A and B, must stand next to each other?

    1. A.120
    2. B.48
    3. C.24
    4. D.72
    Show the answer and solution

    Answer: B. 48

    Answer: (B) 48

    Working:

    1. Treat AB as one unit: 4 units can be arranged in 4! = 24 ways.
    2. A and B can swap inside the block: × 2 = 48.

    Why the others are wrong:

    • (A) 120 ignores the adjacency condition.
    • (C) 24 forgets the internal order.
    • (D) 72 is 3 × 24.

  4. Question 4Multiple choiceAsked in IPMAT Rohtak 2020Integer bounds

    Nurse Ranji has worked more night shifts in a row than Nurse Lalita, who has worked five. Nurse Sunita has worked fifteen night shifts in a row, more than Nurses Rajni and Lalita combined. Nurse Anita has worked eight night shifts in a row, less than Nurse Rajni. How many night shifts in a row has Nurse Rajni Worked?

    1. A.8
    2. B.9
    3. C.10
    4. D.11
    Show the answer and solution

    Answer: B. 9

    Answer: (B) 9

    Working:

    1. Lalita worked 5. Sunita (15) worked more than Rajni + Lalita: Rajni + 5 < 15, so Rajni < 10.
    2. Anita (8) worked less than Rajni: Rajni > 8.
    3. The only whole number with 8 < Rajni < 10 is 9.

    Why the others are wrong:

    • (A) 8 is Anita's count; Rajni has worked more.
    • (C) 10 would make Rajni + Lalita equal 15, not less than 15.
    • (D) 11 breaks the Sunita condition.

  5. Question 5Multiple choiceWritten by IPMATwayInteger bounds

    The product of two consecutive positive integers exceeds their sum by 41. What is the smaller integer?

    1. A.7
    2. B.6
    3. C.8
    4. D.9
    Show the answer and solution

    Answer: A. 7

    Answer: (A) 7

    Working:

    1. n² + n - 2n - 1 = 41, so n² - n = 42.
    2. n = 7 (7 × 6 = 42).

    Why the others are wrong:

    • (B) 6 gives a difference of 29.
    • (C) 8 gives 55.
    • (D) 9 gives 71.

  6. Question 6Multiple choiceWritten by IPMATwayMoney and notes

    A person has ₹ 330 in ₹ 1, ₹ 5 and ₹ 10 notes, with the same total value in each denomination. How many notes does he have?

    1. A.132
    2. B.143
    3. C.110
    4. D.121
    Show the answer and solution

    Answer: B. 143

    Answer: (B) 143

    Working:

    1. Equal value means ₹ 110 per denomination.
    2. Notes: 110 of ₹ 1, 22 of ₹ 5, 11 of ₹ 10 = 143.

    Why the others are wrong:

    • (A) 132 has no basis in the data.
    • (C) 110 counts only the ₹ 1 notes.
    • (D) 121 forgets the ₹ 5 notes.

  7. Question 7Multiple choiceWritten by IPMATwayWork shifts and cycles

    Buses on route A leave a depot every 18 minutes and buses on route B every 24 minutes. Both leave together at 6:00 a.m. At what time do they leave together for the FOURTH time, counting 6:00 a.m. as the first?

    1. A.10:48 a.m.
    2. B.8:24 a.m.
    3. C.1:12 p.m.
    4. D.9:36 a.m.
    Show the answer and solution

    Answer: D. 9:36 a.m.

    Both routes leave together whenever the elapsed time is a multiple of both 18 and 24, i.e. of LCM(18, 24) = 72 minutes. The joint departures are at 6:00, 6:00 + 72 min, 6:00 + 144 min, 6:00 + 216 min, and so on. Counting 6:00 a.m. as the first, the fourth is 216 minutes = 3 hours 36 minutes later, at 9:36 a.m. The usual slip is to multiply 72 by 4, which skips past the fourth meeting to the fifth.

  8. Question 8Multiple choiceWritten by IPMATwayWork shifts and cycles

    A watchman works 4 nights in a row and then takes 2 nights off, again and again. He starts his first working night on 1 March. On how many nights of March (31 days) does he work?

    1. A.21
    2. B.20
    3. C.24
    4. D.22
    Show the answer and solution

    Answer: A. 21

    His pattern repeats every 6 nights: work on nights 1 to 4 of each cycle, off on nights 5 and 6. March has 31 nights, which is 5 full cycles (30 nights) plus 1 extra night. The 5 full cycles give 5 x 4 = 20 working nights. The extra night, 31 March, is night 1 of the next cycle, which is a working night. So he works 20 + 1 = 21 nights in March.

  9. Question 9Multiple choiceWritten by IPMATwayMinimum and maximum overlap

    Of 100 students, 60 passed Maths and 70 passed Science. What is the largest possible number of students who passed EXACTLY one of the two subjects?

    1. A.100
    2. B.30
    3. C.130
    4. D.70
    Show the answer and solution

    Answer: D. 70

    Let x students pass both. Then exactly one subject = (60 - x) + (70 - x) = 130 - 2x, which is largest when x is smallest. Students passing at least one subject = 130 - x, which cannot exceed 100, so x is at least 30. With x = 30: 30 passed only Maths, 40 only Science, 30 both and nobody failed both, which adds to 100. Exactly one = 30 + 40 = 70. So the largest possible number is 70.

  10. Question 10Multiple choiceWritten by IPMATwayMinimum and maximum overlap

    In a class of 30 students, every student plays at least one of three games: 18 play cricket, 15 play football and 12 play hockey. What is the largest possible number of students who play all three games?

    1. A.7
    2. B.12
    3. C.15
    4. D.8
    Show the answer and solution

    Answer: A. 7

    Add the three game counts: 18 + 15 + 12 = 45 'plays'. Every student plays at least one game, so 45 = (exactly one) + 2 x (exactly two) + 3 x (all three), and 30 = (exactly one) + (exactly two) + (all three). Subtracting, (exactly two) + 2 x (all three) = 15. So all three is at most 7 (since 2 x 8 = 16 > 15). Seven is possible: 7 play all three, 1 plays cricket and football, and the rest play one game (cricket only 10, football only 7, hockey only 5), which gives 18, 15, 12 and a total of 30. So the largest number is 7, not the naive 12.