
At a certain time in a deer park, the number of heads and the number of legs of deer and human visitors was counted and it was found there are 39 heads & 132 legs. If the number of deer and human visitors in the park are ‘d’ and ‘h’ respectively, Then, find $ \left| \dfrac{d-h}{3} \right | $.
Answer
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Hint: We should know that a human will have one head and two legs and the deer will have one head and four legs. Now let us assume the total number of human visitors is equal to ‘h’ and the total number of deer is equal to d. We were given that there are 39 heads and 132 legs. By using this concept, we should find the number of deer and the number of human visitors.
Complete step-by-step solution:
Before solving the question, we should know that a human will have one head and two legs and the deer will have one head and four legs. So, the number of heads for both human and deer are equal to 2. And it is clear that the number of legs for both humans and deer is equal to 6.
Now let us assume the total number of human visitors is equal to ‘h’ and the total number of deer is equal to d.
From the question, we were given that the number of heads and the number of legs of deer and human visitors was counted and it was found there are 39 heads and 132 legs.
So, we get
\[\begin{align}
& h+d=39...(1) \\
& 2h+4d=132....(2) \\
\end{align}\]
Now we will simplify equation (2) as below,
\[\Rightarrow h+2d=66....(3)\]
Now we will subtract equation (1) and equation (3).
\[\begin{align}
& \Rightarrow (h+2d)-(h+d)=66-39 \\
& \Rightarrow d=27.....(4) \\
\end{align}\]
Now we will substitute equation (4) in equation (1).
\[\begin{align}
& \Rightarrow h+27=39 \\
& \Rightarrow h=12....(5) \\
\end{align}\]
So, we get the number of human visitors is 12 and the number of deer is equal to 27.
So, the value of
\[\left| \dfrac{d-h}{3} \right|=\left| \dfrac{27-12}{3} \right|=\left| \dfrac{15}{3} \right|=5\].
Then it is clear that the value of \[\left| \dfrac{d-h}{3} \right|\] is equal to 5.
Note: The mistakes that students can commit is while forming the equations. They must know that the number of head and legs for both human and deer and form equations in terms of d and h properly. If they form an equation for the condition for legs as $4h+2d=132$, then the whole solution will go wrong. After getting the equations 1 and 2, we can also substitute $h=39-d$ from equation 1 in equation 2 and then find d and h.
Complete step-by-step solution:
Before solving the question, we should know that a human will have one head and two legs and the deer will have one head and four legs. So, the number of heads for both human and deer are equal to 2. And it is clear that the number of legs for both humans and deer is equal to 6.
Now let us assume the total number of human visitors is equal to ‘h’ and the total number of deer is equal to d.
From the question, we were given that the number of heads and the number of legs of deer and human visitors was counted and it was found there are 39 heads and 132 legs.
So, we get
\[\begin{align}
& h+d=39...(1) \\
& 2h+4d=132....(2) \\
\end{align}\]
Now we will simplify equation (2) as below,
\[\Rightarrow h+2d=66....(3)\]
Now we will subtract equation (1) and equation (3).
\[\begin{align}
& \Rightarrow (h+2d)-(h+d)=66-39 \\
& \Rightarrow d=27.....(4) \\
\end{align}\]
Now we will substitute equation (4) in equation (1).
\[\begin{align}
& \Rightarrow h+27=39 \\
& \Rightarrow h=12....(5) \\
\end{align}\]
So, we get the number of human visitors is 12 and the number of deer is equal to 27.
So, the value of
\[\left| \dfrac{d-h}{3} \right|=\left| \dfrac{27-12}{3} \right|=\left| \dfrac{15}{3} \right|=5\].
Then it is clear that the value of \[\left| \dfrac{d-h}{3} \right|\] is equal to 5.
Note: The mistakes that students can commit is while forming the equations. They must know that the number of head and legs for both human and deer and form equations in terms of d and h properly. If they form an equation for the condition for legs as $4h+2d=132$, then the whole solution will go wrong. After getting the equations 1 and 2, we can also substitute $h=39-d$ from equation 1 in equation 2 and then find d and h.
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