RBSE Solutions Class 6 Maths Chapter 2 Relation Among Numbers Exercise 2.4

 

Question 1. Find out L.C.M. of the following.
(i) 10, 15
(ii) 14, 28
(iii) 12, 18 and 27
(iv) 48, 56 and 72
Answer: Finding L.C.M. of numbers:
(i) For 10 and 15:
We list the prime factors for each number and then multiply the highest powers of all prime factors.

Thus, the L.C.M. of 10 and 15 is \( 2 \times 3 \times 5 = 30 \).
(ii) For 14 and 28:
We find the smallest common multiple by prime factorization.Therefore, the L.C.M. of 14 and 28 is \( 2 \times 2 \times 7 = 28 \).
(iii) For 12, 18 and 27:
We list the prime factors to find their least common multiple.So, the L.C.M. of 12, 18 and 27 is \( 2 \times 2 \times 3 \times 3 \times 3 = 108 \).
(iv) For 48, 56 and 72:
We perform prime factorization to find the LCM for these three numbers.Therefore, the L.C.M. of 48, 56 and 72 is \( 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 = 1008 \). Finding the LCM helps identify the smallest number that is a multiple of all given numbers.
In simple words: To find the L.C.M., we multiply the highest powers of all prime factors found in the numbers. This gives us the smallest number that all the original numbers can divide into evenly.

🎯 Exam Tip: Always show your prime factorization steps clearly when finding the L.C.M. for multiple numbers, as it helps prevent errors and earns partial marks.

 

Question 2. Minimum how many mangoes can be divided into the group of 5 and 6 completely?
Answer: To find the minimum number of mangoes that can be divided completely into groups of 5 and 6, we need to find the Least Common Multiple (L.C.M.) of 5 and 6.
We can find the L.C.M. of 5 and 6 by listing their multiples or using prime factorization.
Multiples of 5: 5, 10, 15, 20, 25, **30**, 35 ...
Multiples of 6: 6, 12, 18, 24, **30**, 36 ...
The smallest common multiple is 30.
Alternatively, using prime factorization:
Prime factors of 5 = 5
Prime factors of 6 = \( 2 \times 3 \)
L.C.M. (5, 6) = \( 2 \times 3 \times 5 = 30 \).
So, 30 mangoes are required. This ensures every mango can be perfectly sorted without any left over.
In simple words: We need to find the smallest number that can be divided by both 5 and 6 without any remainder. This number is called the Least Common Multiple, which is 30.

🎯 Exam Tip: When a question asks for the "minimum" or "smallest" quantity that can be divided by several numbers, it's a clear signal to calculate the Least Common Multiple (L.C.M.).

 

Question 3. Sneha and Vansh go to market every 3rd and 5th day. They went to market today. After how many days will they go to market together?
Answer: Sneha goes to the market every 3rd day, and Vansh goes every 5th day. To find out when they will go to the market together again, we need to find the Least Common Multiple (L.C.M.) of 3 and 5.
Since 3 and 5 are prime numbers, their L.C.M. is simply their product.
L.C.M. of 3 and 5 = \( 3 \times 5 = 15 \).
So, they will go to the market together again after 15 days. This is a common way to find when two periodic events will happen at the same time again.
In simple words: Sneha visits every 3 days and Vansh every 5 days. We need to find the smallest number of days that is a multiple of both 3 and 5. This number is 15, so they will meet again on the 15th day.

🎯 Exam Tip: For small prime numbers, the L.C.M. is always their product. Remember that finding the L.C.M. is key for problems involving recurring events.

 

Question 4. Harish, Kareem and Rakesh take a round of the ground in 6, 8 and 12 minutes. If all of three start at 6 O'clock then after how long time they will be together?
Answer: Harish takes 6 minutes, Kareem takes 8 minutes, and Rakesh takes 12 minutes to complete one round of the ground. To find when they will all be together at the starting point again, we need to calculate the Least Common Multiple (L.C.M.) of 6, 8, and 12.
We will use prime factorization to find the L.C.M.:

L.C.M. of 6, 8 and 12 = \( 2 \times 2 \times 2 \times 3 = 24 \).
So, they will all be together at the starting point again after 24 minutes. If they started at 6 O'clock, they will meet again at 24 minutes past 6. This is a common application of LCM in real-world scenarios.
In simple words: We need to find the earliest time when all three runners will be at the start line together again. This is found by calculating the L.C.M. of their individual round times (6, 8, and 12 minutes), which is 24 minutes.

🎯 Exam Tip: When dealing with cycles or repeated events and trying to find when they will align again, the Least Common Multiple (L.C.M.) is the correct mathematical tool to use.

 

Question 5. Find out the smallest number which is divisible by 16, 20 and 24 completely.
Answer: To find the smallest number that is completely divisible by 16, 20, and 24, we need to calculate their Least Common Multiple (L.C.M.).
We can find the L.C.M. using the prime factorization method:
Prime factorization of 16 = \( 2 \times 2 \times 2 \times 2 = 2^4 \)
Prime factorization of 20 = \( 2 \times 2 \times 5 = 2^2 \times 5^1 \)
Prime factorization of 24 = \( 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1 \)
To find the L.C.M., we take the highest power of each prime factor present in any of the numbers:
Highest power of 2 is \( 2^4 \)
Highest power of 3 is \( 3^1 \)
Highest power of 5 is \( 5^1 \)
L.C.M. = \( 2^4 \times 3^1 \times 5^1 = 16 \times 3 \times 5 = 240 \).
So, the smallest number that is completely divisible by 16, 20, and 24 is 240. This number is the first common multiple of all three numbers.
In simple words: We need to find the smallest number that 16, 20, and 24 can all divide into without leaving a remainder. This is the L.C.M., which is found to be 240.

🎯 Exam Tip: When prime factorizing numbers to find the L.C.M., remember to use the *highest* power of each unique prime factor that appears in any of the numbers.

 

Question 6. A blue bulb keeps flashing on the interval of every 60 seconds and a red bulb keeps flashing on the interval of every 90 seconds. If both the bulbs are switched on at 5 “O” clock together then after how much time both will flash together?
Answer: The blue bulb flashes every 60 seconds, and the red bulb flashes every 90 seconds. To find when both bulbs will flash together again, we need to calculate the Least Common Multiple (L.C.M.) of 60 and 90.
We will use prime factorization to find the L.C.M.:

L.C.M. of 60 and 90 = \( 2 \times 2 \times 3 \times 3 \times 5 = 180 \).
This means both bulbs will flash together after 180 seconds. We convert 180 seconds to minutes by dividing by 60: \( 180 \div 60 = 3 \) minutes. So, after 3 minutes, they will flash together. If they started at 5:00 O'clock, they will flash together again at 5:03 O'clock. This method is useful for synchronizing events that happen at different time intervals.
In simple words: One bulb flashes every 60 seconds, and the other every 90 seconds. We find the smallest time they will both flash at the same moment by calculating their L.C.M., which is 180 seconds, or 3 minutes. So, they will flash together at 5:03 O'clock.

🎯 Exam Tip: Always remember to convert units if the question asks for the answer in a different unit (e.g., seconds to minutes) to ensure your final answer is correct and clear.