Question

Three sets of English mathematics and science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. Total number of stacks will be:
A. \[14\]
B. \[21\]
C. \[22\]
D. \[48\]

Answer 1

Hint: Here we calculate the length of each stack by taking the highest common factor of the number of books of each subject. Calculate the total number of books and use a unitary method to find the number of stacks made.
* Unitary method helps us to find the value of a single by dividing the value of multiple objects by the number of objects and helps us to find the value of multiple objects by multiplying the value of single object to the number of objects.
* HCF of the numbers a, b and c is the largest number that divides all the three numbers.

Complete step-by-step answer:
We are given three sets of books of English(336) Mathematics(240) and Science(96).
We calculate the total number of books.
Total number of books \[ = \]Number of books of English \[ + \] Number of books of Mathematics \[ + \] Number of books of Science.
Substituting the number of books of each subject on RHS we get,
\[ \Rightarrow \]Total number of books \[ = 336 + 240 + 96\]
\[ \Rightarrow \]Total number of books \[ = 672\]
Now we have to make stacks of books in such a way that all books are stacked subject wise and the height of the stacks is the same. So we calculate the highest number which divides all three sets of books as that number of books will form a stack.
Now we calculate HCF of the three numbers 336, 240 and 96.
We write each number in the form of its prime factors.
\[336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7\]
\[240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5\]
\[96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3\]
Since, we know \[\underbrace {a \times a \times ..... \times a}_n = {a^n}\]. Therefore, we can write
\[336 = {2^4} \times 3 \times 7\]
\[240 = {2^4} \times 3 \times 5\]
\[96 = {2^4} \times 2 \times 3\]
HCF is the highest common divisor, so from the prime factorization above we say that \[{2^4} \times 3\]is the highest number which divides all the three numbers.
\[ \Rightarrow \]HCF \[ = {2^4} \times 3\]
\[ \Rightarrow \]HCF \[ = 2 \times 2 \times 2 \times 2 \times 3\]
\[ \Rightarrow \]HCF \[ = 48\]
 So, each stack contains 48 books.
Now using the unitary method we can find the number of stacks that can be made when we are given a total number of books i.e. 672 and length of 1 stack.
Number of stacks formed by 48 books \[ = 1\]
Number of stacks formed by 1 book \[ = \dfrac{1}{{48}}\]
Number of stacks formed by 672 books \[ = \dfrac{1}{{48}} \times 672\]
Cancel out the terms from the numerator and denominator of the fraction.
 Number of stacks formed by 672 books \[ = 14\]

So, option A is correct.

Note: Students many times make the mistake of making stacks of 1 book of each subject i.e. stack of 3 books which will contradict our condition of height as the number of books in each subject are different.