JIPMAT HCF & LCM PYQs

2 solved HCF & LCM previous year questions (PYQs) from JIPMAT past year papers — attempt each and check the answer.

Q1:jipmat 2025QA › HCF & LCMEasyQA · MCQ

The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec, respectively. They all change simultaneously at 08:20:00 hours, when will they again change simultaneously?

A9:27:12 hours
B8:27:12 hours
C7:27:13 hours
D11:00:12 hours

Pick an option to attempt

The Setup: To find when cyclic events sync up perfectly IRL, we need the Least Common Multiple (LCM) of their intervals. Step 1: Find the prime factorization of each number to build the LCM.

48=16 \times 3=2^4 \times 3^1

72=8 \times 9=2^3 \times 3^2

108=4 \times 27=2^2 \times 3^3

Step 2: Calculate the LCM by taking the highest power of each prime factor present.

\text{LCM}=2^4 \times 3^3=16 \times 27

\text{LCM}=432\text{ seconds}

Step 3: Convert the seconds into minutes and seconds.

432 \div 60=7

with a remainder of

12

. So, it takes

7\text{ minutes and }12\text{ seconds}

for them to sync up again. Step 4: Add this time gap to the original starting time. Starting time: 08:20:00. Adding

0\text{ hours, }7\text{ mins, }12\text{ secs}

gives us 08:27:12. Final Answer: 8:27:12 hours

Q2:jipmat 2025QA › HCF & LCMMediumQA · MCQ

Let x be the least number which when divided by 8, 12, 20, 28, 35 leaves a remainder of 5 in each case, then the sum of digits of x is:

Pick an option to attempt

The Setup: This is an LCM boss battle. The number we're looking for (

x

) is going to be exactly

5

steps larger than the Least Common Multiple (LCM) of all those divisors. Step 1: Find the prime factorization of each divisor.

8=2^3

12=2^2 \times 3

20=2^2 \times 5

28=2^2 \times 7

35=5 \times 7

Step 2: Calculate the LCM by grabbing the highest power of each prime number present in the squads. Highest power of

2

2^3=8

. Highest power of

3

3^1=3

. Highest power of

5

5^1=5

. Highest power of

7

7^1=7

\text{LCM}=8 \times 3 \times 5 \times 7

\text{LCM}=24 \times 35=840

Step 3: Form the number

x

. Since it leaves a remainder of

5

, we just add

5

to the LCM.

x=840+5=845

Step 4: Find the sum of the digits of

x

\text{Sum}=8+4+5=17

Final Answer: 17

JIPMAT HCF & LCM — PYPs

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