Question

At a certain time in a deer park, the number of heads and the number of legs of deer and human visitor were counted and it was found that there were 41 heads and 136 legs. Find the number of deers and human visitors in the park.​

Answer 1

$\sf\large\underline\blue{Let:}$

$\tt{\implies The\: number\:_{(deer)}=x}$

$\tt{\implies The\: number\:_{(human\: visitor)}=y}$

$\sf\large\underline\blue{To\: Find:}$

$\tt{\implies The\: number\:_{(human\: visitor\:and\:deer)}=?}$

$\sf\large\underline\blue{SolutioN:}$

• As we know that , A deer and human has one head and A deer has 4 legs and human has 2 legs so, we have to assume the number of legs for deer for x and the number of human for y than setting up the equation and calculate the value of x and y]

$\tt\purple{\implies Given\:in\:the\: question:-}$

$\tt{\implies The\: number\:_{(head\:deer+human)}=41}$

$\rm{\implies x+y=41------(i)}$

$\tt{\implies The\: number\:_{(legs\:human\: visitor+deer)}=136}$

$\rm{\implies 4x+2y=136-------(ii)}$

• In eq 1st multiplying by 4 than subtracting from eq 2nd here]

$\rm{\implies 4x+4y=164}$

$\rm{\implies 4x+2y=136}$

• By solving equation we get ,here]

$\rm{\implies 2y=28\implies y=14}$

• Now putting the value of y=14 in eq 1st]

$\rm{\implies x+y=41}$

$\rm{\implies x+14=41}$

$\rm{\implies x=41-14\implies x=27}$

$\sf\large{Hence,}$

$\tt{\implies The\: number\:_{(deer)}=27}$

$\tt{\implies The\: number\:_{(human\: visitor)}=14}$

Answer 2

• Given -

• At a certain time in a deer park, the number of heads and the number of legs of deer and human visitor were counted and it was found that there were 41 heads and 136 legs

• Let -

• The no. of deer = x
• The no. of human Visitor = y

• To Find -

• Total No. of deer in park
• Total No. of human visitor in park

• Solution -

• Here, No other values are given. So, No. of legs are considered here.
• No. of legs in Deer = 4
• No. of legs in Human visitor = 2

Here, It is given that there are 41 heads. Let form eqn according to given statement,

☆ $⟹x + y = 41</strong><strong> </strong><strong>.</strong><strong>.</strong><strong>.</strong><strong> </strong><strong>(</strong><strong>1</strong><strong>)</strong><strong>$

Here, It is given that there are 136 legs. Let Form Eqn according to given Conditions,

☆ $⟹4x + 2y = 136</strong><strong> </strong><strong>.</strong><strong>.</strong><strong>.</strong><strong> </strong><strong>(</strong><strong>2</strong><strong>)</strong><strong>$

Multiplying Eqn (1) by 4 to get equal Terms,

☆ $4x + 4y = 164 ... (3)$

Solving Eqn (3) & (2) -

$⟹4x + 4y = 164$

$⟹4x + 2y = 136$

On Solving, we get

$⟹2y = 28$

$⟹y = \frac{\cancel{28}}{\cancel{2}}$

$⟹y = 14$

Now Placing this value in any of eqn, let's place this value in eqn (2)

$⟹4x + 2(14) = 136$

$⟹4x + 28 = 136$

$⟹4x = 136 - 28$

$⟹4x = 108$

$⟹x = \frac{\cancel{108}}{\cancel{4}}$

$⟹x = 27$

⛬ Total no. of deers in park is 27 & Total no. of Human Visitor in parks are 14