Quants Questions for CLAT | QB Set 12

A school is organising a “Math Relay Challenge” for four houses — Red, Blue, Green and Yellow. Each house must send students in groups at regular intervals for different activities. The Red House sends a group every 12 minutes, the Blue House every 18 minutes, and the Green House every 24 minutes. All three groups start together at 9:00 AM from the school auditorium. The organisers want to know after how much time all three houses will again send their groups together.

Meanwhile, the Yellow House is preparing gift packets for participants. They have 72 chocolates, 108 pencils, and 144 stickers. The teacher wants to divide these items into identical packets in such a way that each packet contains the same number of chocolates, pencils, and stickers without leaving any item unused. The maximum number of packets possible is to be found.

The principal also notices that understanding HCF and LCM is important in real-life planning. While HCF helps in dividing things into equal groups efficiently, LCM helps in synchronising repeated events happening at different intervals. The students are therefore asked to solve several practical problems based on these concepts during the competition.

Questions

Q1. After how many minutes will the Red, Blue and Green Houses again send their groups together?

A. 48 minutes
B. 60 minutes
C. 66 minutes
D. 72 minutes

Q2. What is the maximum number of identical packets that can be made using 72 chocolates, 108 pencils and 144 stickers?

A. 36
B. 24
C. 18
D. 12

Q3. If another house sends students every 30 minutes along with the three houses mentioned, after how many minutes will all four houses meet together again?

A. 240 minutes
B. 300 minutes
C. 180 minutes
D. 360 minutes

Q4. What will be the number of chocolates in each packet if the maximum number of packets is made?

A. 4
B. 2
C. 6
D. 8

Q5. Two bells ring at intervals of 45 minutes and 60 minutes respectively. If they ring together at 8:00 AM, when will they next ring together?

A. 9:30 AM
B. 10:00 AM
C. 11:00 AM
D. 10:30 AM

Answers with Detailed Explanations

Q1. Correct Answer: D. 72 minutes

To find when all houses will send groups together again, the LCM of 12, 18 and 24 must be calculated.

Prime factorisation:

12 = 2² × 3
18 = 2 × 3²
24 = 2³ × 3

Taking highest powers:

2³ × 3²
= 8 × 9
= 72

Therefore, all houses will meet together again after 72 minutes.

Q2. Correct Answer: A. 36

To make the maximum number of identical packets, the HCF of 72, 108 and 144 is required.

Prime factorisation:

72 = 2³ × 3²
108 = 2² × 3³
144 = 2⁴ × 3²

Taking common lowest powers:

2² × 3²
= 4 × 9
= 36

Hence, the maximum number of packets possible is 36.

Q3. Correct Answer: D. 360 minutes

The intervals are 12, 18, 24 and 30 minutes.

Find the LCM:

12 = 2² × 3
18 = 2 × 3²
24 = 2³ × 3
30 = 2 × 3 × 5

Taking highest powers:

2³ × 3² × 5
= 8 × 9 × 5
= 360

Therefore, all four houses will meet together again after 360 minutes.

Q4. Correct Answer: B. 2

From Question 2, the maximum number of packets is 36.

Number of chocolates in each packet:

72 ÷ 36 = 2

Therefore, each packet will contain 2 chocolates.

Q5. Correct Answer: C. 11:00 AM

Intervals are 45 minutes and 60 minutes.

Find the LCM:

45 = 3² × 5
60 = 2² × 3 × 5

LCM = 2² × 3² × 5
= 4 × 9 × 5
= 180 minutes

180 minutes = 3 hours.

Starting from 8:00 AM:

8:00 AM + 3 hours = 11:00 AM

Hence, the correct answer is 11:00 AM.