d Word Problem 21 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Word Problem 21

Problem 21: There are some pigeons and hares in a zoo. If heads are counted, these are 200. If legs are counted, they are 580. The number of hares in the zoo are ;

a) 50                        b) 150                c) 90                 d) 120



We are given following information;

Heads = 200

Legs = 580

Hares = ?

Pigeons = ?

We suppose that ;

Number of Hares = y

Number of Pigeons = z

Before we proceed further, keep in mind that each hare has four legs —  and each pigeon has two legs !

So, total number of heads is equal to total number of hares and pigeons which is  200


\[ z+y=200——–(1) \]

Each pigeons has 2 legs – means z pigeons will have 2z legs. Similarly each hare has 4 legs — means y hares will have 4y legs ! But total legs are 580;


\[ 2z+4y=580 ——– (2) \]

Multiplying eq (1) with 2 w get,

\[ 2z+2y=400 ——(3) \]

Subtracting eq (3) from eq(2) we get;

\[ 2y=180 \]


\[ y=\frac{180}{2}=90 \]

Since y is number of heads of hares – so number of hares is 90

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