Question

One-fourth of a herd of deer has gone to the forest. One-third of the total number is grazing in a field and the remaining 15 are drinking water on the bank of a river. Find the total number of deer.

Answer 1

Hint: To solve the given question, we will assume that the total number of deer is t, the total number of deer which have gone to the forest is x, the total number of which is grazing in a field is y. Now, we will find the relation between t and x and t and y. Then we will write x and y in terms of t and then we will make use of the fact that the sum of x, y, and 15 is it to get the total deer.

Complete step by step answer:
It is given in the question that there is a herd of a deer. We will assume that there are ‘t’ deer in that herd. The number of deer which are gone to the forest is a part of that herd. Let their number be x. The total number of deer which are grazing in a field is y.
Now it is given in the question that the number of deer which have gone to the forest is one-fourth of the total. Thus, we have,
\[x=\dfrac{t}{4}.....\left( i \right)\]
The number of deer which are grazing in the field is one – third of the total. Thus, we have,
\[y=\dfrac{t}{3}....\left( ii \right)\]
Now there are 15 deer which are drinking water. The total number of deer will be equal to the sum of deer which have gone to the forest, deer which are grazing and deer which are drinking water. Thus, we will get,
\[x+y+15=t....\left( iii \right)\]
Now, we will put the value of x and y from (i) and (ii) to (iii). Thus, we will get,
\[\Rightarrow \dfrac{t}{4}+\dfrac{t}{3}+15=t\]
\[\Rightarrow t-\dfrac{t}{4}-\dfrac{t}{3}=15\]
\[\Rightarrow t\left( 1-\dfrac{1}{4}-\dfrac{1}{3} \right)=15\]
\[\Rightarrow t\left( 1-\dfrac{7}{12} \right)=15\]
\[\Rightarrow t\left( \dfrac{5}{12} \right)=15\]
\[\Rightarrow t=\dfrac{12\times 15}{5}\]
\[\Rightarrow t=36\]
So, there are a total of 36 deer in that herd.

Note: The alternate method of solving the above question is shown below. There are some deer which have gone to the forest and some which are grazing in the field. Let there be x and y respectively. Their sum is equal to the total, i.e. 15. So, we can say that,
\[x+y=t-15\]
\[\Rightarrow \dfrac{t}{4}+\dfrac{t}{3}=t-15\]
\[\Rightarrow \dfrac{7t}{12}=t-15\]
\[\Rightarrow t-\dfrac{7t}{12}=15\]
\[\Rightarrow \dfrac{5t}{12}=15\]
\[\Rightarrow t=36\]