Question

A forester wants to plant 66 apple trees, 88 banana trees, and 110 mango trees in equal rows. Find the number of minimum rows required.

Answer 1

Hint: As a minimum number of rows are asked, so we need to put the maximum number of trees possible in each row. Therefore, the number of trees in each row is equal to the HCF of the number of the three trees. Once we get the number of trees in each row, just divide the total number of trees by the number of trees per row to get the answer.

Complete step-by-step solution:
A minimum number of rows are asked, so we need to put the maximum number of trees possible in each row. Also, it is mentioned that the number of trees in each row is the same and no row consists of more than one type of tree. So, the number of trees in each row must be the factor of the number of trees of each type. So, we can say that the number of trees in each row must be equal to the HCF of the number of trees of each type, i.e., the highest common factor of 66, 88, and 110.
The prime factorization of 66 is,
$ \Rightarrow 66 = 2 \times 3 \times 11$
The prime factorization of 88 is,
$ \Rightarrow 88 = 2 \times 2 \times 2 \times 11$
The prime factorization of 110 is,
$ \Rightarrow 110 = 2 \times 5 \times 11$
Since $2 \times 11$ is common in the prime factorization of all numbers. So,
$ \Rightarrow HCF = 2 \times 11 = 22$
Thus, the number of trees in each row will be 22 trees.
Now, find the total number of trees,
$ \Rightarrow $ Total trees $ = 66 + 88 + 110$
Add the terms,
$ \Rightarrow $ Total trees $ = 264$
Now, divide the total number of trees by trees in each row to find the number of rows required,
$ \Rightarrow $ Total number of rows $ = \dfrac{{264}}{{22}}$
Cancel out the common factors,
$\therefore $ Total number of rows $ = 12$

Hence, the number of minimum rows required is 12 rows.

Note: The idea of projecting the whole question into a single term, the highest common factor is very important. You must learn the application of the highest common factor, least common multiple to real-life incidents. While taking common if you get two 3’s as common, as two pairs of 3’s then 3 must be multiplied twice and do not even combine them.