Question

15 pastries and 12 biscuit packets have been donated for a school fair. These are to be packets in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets in each. How many biscuit packets and how many pastries respectively will each box contain?
A) $3, 3$
B) $3, 4$
C) $4, 5$
D) $5, 4$

Answer 1

Hint:
We are given the number of donated pastries and biscuit packets and are asked to find the number of pastries and biscuit packets that each equal smaller unit will contain. Thus, we will take the common amount that each of the individual amounts can reach simultaneously. In other words, we will evaluate the least common multiple (L.C.M.) first and then finally calculate the amount of pastry and biscuit packets needed to reach that amount.

Complete step by step solution:
Here,
We are given that,
Number of pasties: ${n_1} = 15$
Number of biscuit packets: ${n_2} = 12$
Now,
The L.C.M. of two numbers is defined as the lowest possible number that the two individual numbers can reach. In other words, we have to simply get the least possible number which both the numbers are a factor of.
Now,
If we go through the tables of ${n_1}$ and ${n_2}$, we get
L.C.M. of ${n_1}$ and ${n_2}$ as $60$.
This means that the possible number of packets possible for equal amounts of pastries and biscuit packets is $60$.
Now,
Number of pastries in each smaller packet, ${N_1} = \dfrac{{60}}{{{n_1}}} = \dfrac{{60}}{{15}} = 4$
Number of biscuit packets in each smaller packet, ${N_2} = \dfrac{{60}}{{{n_2}}} = \dfrac{{60}}{{12}} = 5$
Thus,
The required answer is ${N_1},{N_2}$ which further turns out to be$4,5$.

Hence, the correct option is (C).

Note:
Here, we calculated the L.C.M. as the packets will finally sum up to contribute to the total amount of each item. But if we needed to find the reverse of this question, then we have to calculate the H.C.F. of the numbers.