Question

The length, breadth and height of a room are 825cm, 675cm and 450cm respectively. Find the longest tapes which can measure the three dimensions of the room exactly.
(a) 60cm
(b) 65cm
(c) 70cm
(d) 75cm

Answer 1

Hint: To find the largest tapes, take the HCF of length, breadth and height. Find HCF using a prime factorization method. The HCF will give the largest tape.

Complete step-by-step answer:
Given the length of room = 825cm.
Breadth of the room = 675cm.
Height of the room = 450cm.
We need to find the longest tape that is capable of measuring the sides.
So to find the length of tape, we need the length of tape which is a factor of 825, 675 and 450cm.
So if we need to find the tape of the highest possible length that we need to find HCF of 825, 675 and 450.
By prime factorization, we can find the HCF. Prime factorization is finding which prime number multiplies together to make the original number.
A prime number is a number greater than 1 that cannot be made into a whole number. For a prime number the only factors are 1 and itself.
Now let us find the prime factorization of 825, 675 and 450.
$ 825=3\times 5\times 5\times 11=3\times {{5}^{2}}\times 11 $
$  675=3\times 3\times 3\times 5\times 5={{3}^{3}}\times {{5}^{2}} $
$ 450=2\times 3\times 5\times 3\times 5=2\times {{3}^{2}}\times {{5}^{2}} $
From the above, we can say that \[\left( 3\times {{5}^{2}} \right)\] is common for 825, 675 and 450.
\[\therefore \] HCF (825, 675, 450) = \[3\times 5\times 5=75\].
Hence, we got HCF = 75.
\[\therefore \] The longest tape which can measure the three dimensions of a room exactly will be 75cm long.
\[\therefore \] Option (d) is the correct answer.

Note: We can also find HCF using different methods.
$ \therefore 825=3\times 5\times 5\times 11 $
$  675=3\times 3\times 3\times 5\times 5 $
$ 450=2\times 3\times 3\times 5\times 5 $