CBSE Class 6 Mathematics Playing with Numbers Assignments Set 06

Question. Find the L.C.M of \(24, 36, 40\)
Answer: To find the L.C.M. of \(24, 36, 40\):
Using prime factorization:
\(24 = 2^3 \times 3\)
\(36 = 2^2 \times 3^2\)
\(40 = 2^3 \times 5\)
\(\text{L.C.M.} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\)

Question. Find the L.C.M of \(15, 20, 27, 81\)
Answer: To find the L.C.M. of \(15, 20, 27, 81\):
Using prime factorization:
\(15 = 3 \times 5\)
\(20 = 2^2 \times 5\)
\(27 = 3^3\)
\(81 = 3^4\)
\(\text{L.C.M.} = 2^2 \times 3^4 \times 5 = 4 \times 81 \times 5 = 1620\)

Question. Find the L.C.M of \(22, 54, 108, 135, 198\).
Answer: To find the L.C.M. of \(22, 54, 108, 135, 198\):
Using prime factorization:
\(22 = 2 \times 11\)
\(54 = 2 \times 3^3\)
\(108 = 2^2 \times 3^3\)
\(135 = 3^3 \times 5\)
\(198 = 2 \times 3^2 \times 11\)
\(\text{L.C.M.} = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940\)

Question. Find the smallest number which when diminished by \(3\) is divisible by \(21, 28, 36\) and \(45\).
Answer: Since the number when diminished by \(3\) is divisible by \(21, 28, 36\), and \(45\), this diminished number must be the L.C.M. of \(21, 28, 36\), and \(45\).
Finding the L.C.M. of \(21, 28, 36, 45\):
\(21 = 3 \times 7\)
\(28 = 2^2 \times 7\)
\(36 = 2^2 \times 3^2\)
\(45 = 3^2 \times 5\)
\(\text{L.C.M.} = 2^2 \times 3^2 \times 5 \times 7 = 4 \times 9 \times 5 \times 7 = 1260\)
Since the required number minus \(3\) is equal to the L.C.M., we have:
\(\text{Required Number} = 1260 + 3 = 1263\)

Question. Four bells ring at intervals of \(8, 12, 18\) and \(20\) minutes respectively. At what time will they ring simultaneously if they start ringing together at \(12\) Noon?
Answer: To find when they will ring simultaneously again, we need to calculate the L.C.M. of \(8, 12, 18\), and \(20\) minutes:
\(8 = 2^3\)
\(12 = 2^2 \times 3\)
\(18 = 2 \times 3^2\)
\(20 = 2^2 \times 5\)
\(\text{L.C.M.} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\text{ minutes}\)
Converting minutes to hours:
\(\frac{360}{60} = 6\text{ hours}\)
Since they start ringing together at \(12\text{ Noon}\), they will ring together again at:
\(12\text{ Noon} + 6\text{ hours} = 6\text{ P.M.}\)

Question. In the morning three persons step off together. Their steps measure \(80\text{ cm}\), \(90\text{ cm}\) and \(100\text{ cm}\) respectively. What is the minimum distance each should walk so that all can cover the distance in complete steps?
Answer: The minimum distance each should walk to cover the distance in complete steps is the L.C.M. of the step measures \(80\text{ cm}\), \(90\text{ cm}\), and \(100\text{ cm}\):
\(80 = 2^4 \times 5\)
\(90 = 2 \times 3^2 \times 5\)
\(100 = 2^2 \times 5^2\)
\(\text{L.C.M.} = 2^4 \times 3^2 \times 5^2 = 16 \times 9 \times 25 = 3600\text{ cm}\)
Therefore, the minimum distance each should walk is \(3600\text{ cm}\) (or \(36\text{ m}\)).

Question. Find the smallest 3-digit number which is exactly divisible by \(8, 12, 16\).
Answer: First, we find the L.C.M. of \(8, 12, 16\):
\(8 = 2^3\)
\(12 = 2^2 \times 3\)
\(16 = 2^4\)
\(\text{L.C.M.} = 2^4 \times 3 = 16 \times 3 = 48\)
The smallest \(3\text{-digit}\) number is \(100\).
Dividing \(100\) by \(48\):
\(100 \div 48 = 2\text{ with a remainder of } 4\)
The next multiple of \(48\) is:
\(48 \times 3 = 144\)
Therefore, the smallest \(3\text{-digit}\) number exactly divisible by \(8, 12, 16\) is \(144\).

Question. Find the greatest 3-digit number which is exactly divisible by \(4, 12, 15\).
Answer: First, we find the L.C.M. of \(4, 12, 15\):
\(4 = 2^2\)
\(12 = 2^2 \times 3\)
\(15 = 3 \times 5\)
\(\text{L.C.M.} = 2^2 \times 3 \times 5 = 60\)
The greatest \(3\text{-digit}\) number is \(999\).
Dividing \(999\) by \(60\):
\(999 \div 60 = 16\text{ with a remainder of } 39\)
Subtracting the remainder from \(999\):
\(999 - 39 = 960\)
Therefore, the greatest \(3\text{-digit}\) number exactly divisible by \(4, 12, 15\) is \(960\).

Question. Find the smallest 4 digit number which is divisible by \(12, 15, 18\).
Answer: First, we find the L.C.M. of \(12, 15, 18\):
\(12 = 2^2 \times 3\)
\(15 = 3 \times 5\)
\(18 = 2 \times 3^2\)
\(\text{L.C.M.} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\)
The smallest \(4\text{-digit}\) number is \(1000\).
Dividing \(1000\) by \(180\):
\(1000 \div 180 = 5\text{ with a remainder of } 100\)
The smallest \(4\text{-digit}\) multiple of \(180\) is:
\(180 \times 6 = 1080\)
Therefore, the smallest \(4\text{-digit}\) number divisible by \(12, 15, 18\) is \(1080\).

Question. The H.C.F of two numbers is \(24\) and their L.C.M is \(180\). of one of the number is \(36\). Find the other
Answer: We use the mathematical property:
\(\text{H.C.F.} \times \text{L.C.M.} = \text{Product of the two numbers}\)
Let the other number be \(x\).
\(24 \times 180 = 36 \times x\)
\(x = \frac{24 \times 180}{36}\)
\(x = 24 \times 5 = 120\)
Therefore, the other number is \(120\).

Question. Verify the property for the two number \(180\) and \(256\) that \(\text{HCF} \times \text{LCM} = 1\text{st No.} \times 2\text{nd No.}\)
Answer: Given numbers: \(180\) and \(256\).
Finding the H.C.F. and L.C.M. using prime factorization:
\(180 = 2^2 \times 3^2 \times 5\)
\(256 = 2^8\)
\(\text{H.C.F.} = 2^2 = 4\)
\(\text{L.C.M.} = 2^8 \times 3^2 \times 5 = 256 \times 9 \times 5 = 11520\)
Now, let's verify:
\(\text{L.H.S.} = \text{H.C.F.} \times \text{L.C.M.} = 4 \times 11520 = 46080\)
\(\text{R.H.S.} = 1\text{st No.} \times 2\text{nd No.} = 180 \times 256 = 46080\)
Since \(\text{L.H.S.} = \text{R.H.S.}\), the property is verified.

Question. The product of two numbers is \(1260\) and their H.C.F is \(12\). Find L.C.M.
Answer: We use the mathematical property:
\(\text{L.C.M.} \times \text{H.C.F.} = \text{Product of the two numbers}\)
\(\text{L.C.M.} \times 12 = 1260\)
\(\text{L.C.M.} = \frac{1260}{12} = 105\)
Therefore, the L.C.M. is \(105\).