Question

There are 156, 208, 260 students in groups A, B, C respectively. Buses are to be hired to take them for a field trip. Find the minimum number of buses to be hired if the same number of students should be accompanied in each bus.

Answer 1

Hint: We find the maximum number of students per bus by calculating HCF of 156, 208 & 260 & then divide the total numbers of students by it to find minimum numbers of buses.

Complete step by step solution: It is given that:
No of students in group A = 156
No of students in group B = 208
No of students in group C =260
Total no of students $=156+208+260=624$
First, we will find the maximum numbers of students in each bus by taking HCF of 156, 208 & 260.
Using prime factorization method:
To find the HCF of three numbers, we go this way:
\[\begin{align}
  & \left. 2 \right|\left| \!{\underline {\,
  156 \,}} \right. \\
 & \left. 2 \right|\left| \!{\underline {\,
  78 \,}} \right. \\
 & \left. 3 \right|\left| \!{\underline {\,
  39 \,}} \right. \\
 & \ \ \ 13 \\
\end{align}\] \[\begin{align}
  & \left. 2 \right|\left| \!{\underline {\,
  208 \,}} \right. \\
 & \left. 2 \right|\left| \!{\underline {\,
  104 \,}} \right. \\
 & \left. 2 \right|\left| \!{\underline {\,
  52 \,}} \right. \\
 & \left. 2 \right|\left| \!{\underline {\,
  26 \,}} \right. \\
 & \ \ \ 13 \\
\end{align}\] \[\begin{align}
  & \left. 2 \right|\left| \!{\underline {\,
  260 \,}} \right. \\
 & \left. 2 \right|\left| \!{\underline {\,
  130 \,}} \right. \\
 & \left. 5 \right|\left| \!{\underline {\,
  65 \,}} \right. \\
 & \ \ \ 13 \\
\end{align}\]
$\therefore 156=2\times 2\times 3\times 13$ $\therefore 208=2\times 2\times 2\times 2\times 13$ $\therefore 260=2\times 2\times 5\times 13$
Therefore;
\[\begin{align}
  & 156=2\times 3\times 2\times 13\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{2}^{2}}\times 3\times 13 \\
 & 208=2\times 2\times 2\times 2\times 13\ \ \ \ \ \ \ \ \ \ \ \ ={{2}^{2}}\times {{2}^{2}}\times 13 \\
 & 260=2\times 2\times 5\times 13\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{2}^{2}}\times 5\times 13 \\
\end{align}\]
HCF of 156, 208, 260$={{2}^{2}}\times 13=52$
 Maximum number of students per bus=52
 Minimum number of buses so that same numbers of students should be accompanied in each bus is given by
$=\dfrac{\text{Total}\ \text{numbers}\ \text{of}\ \text{student}}{\text{Maximum}\ \text{student}\ \text{per}\ \text{bus}}$
$=\dfrac{624}{52}=12$

Minimum number of buses to be hired, if the same number of students should be in each bus is 12.

Note: In the given problem, we have to calculate the minimum number of buses hired. So, we have found the maximum number of students in each bus. You must be very careful in finding the prime factorization of the numbers and then their HCF and you will end up making calculation mistakes.