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 the numbers 8, 12, and 16.
Prime factors:
- 8 = 2 × 2 × 2
- 12 = 2 × 2 × 3
- 16 = 2 × 2 × 2 × 2
LCM: The highest power of each prime number:
- 2 × 2 × 2 × 2 (from 16)
- 3 (from 12)
LCM calculation:
- LCM = 2 × 2 × 2 × 2 × 3 = 16 × 3 = 48
All three friends will meet 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: To find the smallest number of each type of flower that can be equally divided among the rows, we need to find the LCM of 15 and 25.
Prime factors:
- 15 = 3 × 5
- 25 = 5 × 5
LCM: The highest power of each prime number:
- 5 × 5 (from 25)
- 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 meet at the starting point, we need to find the LCM of their cycling times.
LCM of 40 and 30:
- 40 = 2 × 2 × 2 × 5
- 30 = 2 × 3 × 3 × 5
LCM calculation:
- LCM = 2 × 2 × 2 × 3 × 3 × 5 = 8 × 9 × 5 = 360
They will meet at the starting point every 360 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: LCM of 18 and 24:
- 18 = 2 × 3 × 3
- 24 = 2 × 2 × 2 × 3
LCM calculation:
- LCM = 2 × 2 × 2 × 3 × 3 = 8 × 9 = 72
The bells will next ring together at 8:00 AM + 72 minutes = 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: LCM of 6, 8, and 12:
- 6 = 2 × 3
- 8 = 2 × 2 × 2
- 12 = 2 × 2 × 3
LCM calculation:
- LCM = 2 × 2 × 2 × 3 = 8 × 3 = 24
All three will meet again at the gym 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: To find the greatest number of rows, we need to determine the highest common factor (HCF) of 48 and 32.
- The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
- The factors of 32 are: 1, 2, 4, 8, 16, 32
The highest common factor (HCF) of 48 and 32 is 16 .
Thus, Pawan can plant the vegetables in 16 rows , with each row containing 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 are 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 need to find the HCF of 15, 20, and 25.
- The factors of 15 are: 1, 3, 5, 15
- The factors of 20 are: 1, 2, 4, 5, 10, 20
- The factors of 25 are: 1, 5, 25
The highest common factor (HCF) of 15, 20, and 25 is 5.
So, the maximum number of toffees in each packet should be 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.
- The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
- The factors of 90 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
The highest common factor (HCF) of 72 and 90 is 18.
Therefore, the gardener can plant the flowers in 18 rows , with each row containing 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 to find the HCF of 24, 36, and 48.
- The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
- The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36
- The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The highest common factor (HCF) of 24, 36, and 48 is 12 .
Thus, 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.
- The factors of 56 are: 1, 2, 4, 7, 8, 14, 28, 56
- The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
- The factors of 112 are: 1, 2, 4, 7, 8, 14, 16, 28, 56, 112
The highest common factor (HCF) of 56, 84, and 112 is 28.
Therefore, Rajesh can distribute the items to a maximum of 28 students , with each student getting 56/28 = 2 pencils, 84/28 = 3 erasers, and 112/28 = 4 sharpeners.
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