Word Problems: HCF & LCM - Class 5 Mathematics | Quick Revision Study Guide

Q1: Three friends, Nina, Ravi, and Vikas, want to start a reading group and meet every few days. If Nina can meet every 8 days, Ravi every 12 days, and Vikas every 16 days, how often will all three meet together?

Sol: To find when all three friends can meet together, we need to find the LCM of 8, 12 and 16.
Prime factors:

• 8 = 2 × 2 × 2
• 12 = 2 × 2 × 3
• 16 = 2 × 2 × 2 × 2

LCM: Take the highest power of each prime that appears.

• Highest power of 2 = 2 × 2 × 2 × 2 (from 16)
• Highest power of 3 = 3 (from 12)

LCM calculation:

• LCM = 2 × 2 × 2 × 2 × 3 = 16 × 3 = 48

All three friends will meet together every 48 days.

Q2: A gardener wants to plant rows with an equal number of two types of flowers. If one flower comes in bunches of 15 and the other in bunches of 25, what is the smallest number of each type of flower he can plant in each row?

Sol: We need the smallest number that is a multiple of both 15 and 25, so find the LCM of 15 and 25.
Prime factors:

• 15 = 3 × 5
• 25 = 5 × 5

LCM: Take the highest power of each prime.

• Highest power of 5 = 5 × 5 (from 25)
• Highest power of 3 = 3 (from 15)

LCM calculation:

• LCM = 3 × 5 × 5 = 3 × 25 = 75

The gardener can plant 75 of each type of flower in each row.

Q3: Two cyclists start cycling on a circular track from the same point but in opposite directions. One completes a round in 40 seconds and the other in 30 seconds. After how many seconds will they meet at the starting point?

Sol: To find when both cyclists will be at the starting point together, find the LCM of 40 and 30.
Prime factors:

• 40 = 2 × 2 × 2 × 5
• 30 = 2 × 3 × 5

LCM: Take the highest power of each prime.

• Highest power of 2 = 2 × 2 × 2 (from 40)
• Highest power of 3 = 3 (from 30)
• Highest power of 5 = 5 (from both)

LCM calculation:

• LCM = 2 × 2 × 2 × 3 × 5 = 8 × 3 × 5 = 24 × 5 = 120

They will meet at the starting point every 120 seconds.

Q4: Two bells ring at intervals of 18 minutes and 24 minutes respectively. If they ring together at 8:00 AM, at what time will they next ring together?

Sol: Find the LCM of 18 and 24.
Prime factors:

• 18 = 2 × 3 × 3
• 24 = 2 × 2 × 2 × 3

LCM: Take the highest power of each prime.

• Highest power of 2 = 2 × 2 × 2 (from 24)
• Highest power of 3 = 3 × 3 (from 18)

LCM calculation:

• LCM = 2 × 2 × 2 × 3 × 3 = 8 × 9 = 72

72 minutes after 8:00 AM is 9:12 AM. So the bells will next ring together at 9:12 AM.

Q5: Anna, Ben, and Carl go to a gym every 6, 8, and 12 days, respectively. If they all met at the gym today, after how many days will all three meet again at the gym?

Sol: Find the LCM of 6, 8 and 12.
Prime factors:

• 6 = 2 × 3
• 8 = 2 × 2 × 2
• 12 = 2 × 2 × 3

LCM: Take the highest power of each prime.

• Highest power of 2 = 2 × 2 × 2 (from 8)
• Highest power of 3 = 3 (from 6 or 12)

LCM calculation:

• LCM = 2 × 2 × 2 × 3 = 8 × 3 = 24

All three will meet again after 24 days.

Q6: Pawan wants to plant 48 onion plants and 32 cabbage plants in his vegetable garden. What is the greatest number of rows possible if each row has the same number of onion plants and the same number of cabbage plants?

Sol: We need the highest common factor (HCF) of 48 and 32 so that each row has the same number of each vegetable.
Factors:

• Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
• Factors of 32: 1, 2, 4, 8, 16, 32

The highest common factor is 16.

So Pawan can make 16 rows.
Each row will have 48 ÷ 16 = 3 onion plants and 32 ÷ 16 = 2 cabbage plants.

Q7: Sanjay has 15 toffees, Raj has 20, and Sanjana has 25 toffees. Each of them wants to make packets of toffees so that there is an equal number of toffees in each packet with no toffee left behind. How many maximum toffees should be there in each packet?

Sol: We must find the HCF of 15, 20 and 25 so that each packet has the same number and no toffee is left over.
Factors:

• Factors of 15: 1, 3, 5, 15
• Factors of 20: 1, 2, 4, 5, 10, 20
• Factors of 25: 1, 5, 25

The highest common factor is 5.

Therefore, the largest possible number of toffees in each packet is 5.

Q8: A gardener wants to plant 72 rose bushes and 90 tulip bulbs in a garden. If he wants each row to contain the same number of rose bushes and the same number of tulip bulbs, what is the greatest number of rows he can plant?

Sol: Find the HCF of 72 and 90.
Factors:

• Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
• Factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

The highest common factor is 18.

So the gardener can make 18 rows.
Each row will have 72 ÷ 18 = 4 rose bushes and 90 ÷ 18 = 5 tulip bulbs.

Q9: Three friends have 24 apples, 36 oranges, and 48 bananas. They want to distribute these fruits into identical baskets without any fruit left over. What is the largest number of fruits that can go into each basket?

Sol: We need the HCF of 24, 36 and 48.
Factors:

• Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
• Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
• Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

The highest common factor is 12.

Therefore, the largest number of fruits that can go into each basket is 12.

Q10: Rajesh has 56 pencils, 84 erasers, and 112 sharpeners. He wants to distribute these items equally among several students such that each student gets the same number of pencils, erasers, and sharpeners. What is the maximum number of students he can distribute these items to?

Sol: Find the HCF of 56, 84 and 112.
Factors:

• Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56
• Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
• Factors of 112: 1, 2, 4, 7, 8, 14, 16, 28, 56, 112

The highest common factor is 28.

Rajesh can distribute the items to a maximum of 28 students.
Each student will get 56 ÷ 28 = 2 pencils, 84 ÷ 28 = 3 erasers and 112 ÷ 28 = 4 sharpeners.