Two IV tubings of lengths 70 cm and 100 cm are to be cut into pieces of equal length, without wastage. Find the greatest possible length of each piece.

We need to cut the IV tubings into pieces of equal length with no wastage. This means the length of each piece must be a common divisor of both 70 and 100. The greatest such divisor will be the greatest common divisor (GCD) of 70 and 100. 1. Write the two numbers: \(70\) and \(100\). 2. We find the GCD: - Factorize the numbers: - \(70 = 2 \times 5 \times 7\) - \(100 = 2^2 \times 5^2\) - The common prime factors are \(2\) and \(5\). Using the lowest powers of these common factors: - For \(2\): lowest power is \(2^1\) - For \(5\): lowest power is \(5^1\) - Multiply these: \[ \gcd(70, 100) = 2 \times 5 = 10 \] 3. The greatest possible length for each piece is therefore \(10\) cm. Thus, the pieces should each be \(10\) cm long.