Question

A trader has three bundles of string 777m, 315m & 588m long.What is the greatest length of string that the bundles can be cut up into without any left over string?​

Answer 1

Step-by-step explanation:

We have been given three bundles of strings having length 329m, 308m and 490m and we have to find the greatest length of string that the bundles should be cut so that no string is left. So, this greatest length will be the H.C.F of the lengths of the given 3 strings.

Let us find the H.C.F of 392, 308 and 490. Breaking these numbers into the product of their prime factors, we get,

392=2×2×2×7×7308=2×2×7×11490=2×5×7×7

Now, we know that H.C.F of a given set of numbers are the product of common factors present in all of them. So, the H.C.F of 392, 308 and 490 will be the product of all the common factors present in these three numbers.

Clearly, we can see that the factors that are common in all these three numbers are 2 and 7.

So, the required H.C.F = 2×7

= 14

Hence, the required greatest length of string that the bundles can be cut up into without any left over string is 14m.

I hope it helps you ☺️

Answer 2

Given,

The length of the first bundle$\[ = 777m\]$

The length of the second bundle$\[ = 315m\]$

The length of the third bundle$\[ = 588m\]$

To find,

The greatest length of string that the bundles can be cut up into without any leftover string.

Solution,

Know that the greatest length of string that the bundles can be cut up into without any leftover string is the HCF of the length of the given bundles.

The prime factors of $\[777{\rm{m}},315{\rm{m}}\,{\rm{and}}\,{\rm{588m}}\]$ is given as

$\[\begin{array}{l} \Rightarrow 777 = {3^1} \times {7^1} \times {37^1}\\ \Rightarrow 315 = {3^2} \times {5^1} \times {7^1}\\ \Rightarrow 588 = {2^2} \times {3^1} \times {7^2}\end{array}\]$

Therefore,

$\[\begin{array}{l} \Rightarrow 3 \times 7\\ \Rightarrow 21\end{array}\]$

Hence, the greatest length of string that the bundles can be cut up into without any leftover string is $\[21m.\]$