Puzzle:

A spider has 8 legs. A dragonfly has 6 legs and 2 pairs of wings. A cicada has 6 legs and one pair of wings. Now we have all 3 kinds and a total of 18 insects in a cage. We have a total of 118 legs and 20 pairs of wings. How many insects do we have of each kind?

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Solution:

There are Five Spiders, Seven Dragonflies & Six Cicada.

8X+6Y+6Z = 118

2Y+Z = 20

Solving this, X=5, Y=7, Z=6

There are Five Spiders, Seven Dragonflies & Six Cicada.

Let X be the number of spiders, Y be number of Dragonflies and Z be number of cicada. Now from the puzzle,

X+Y+Z = 188X+6Y+6Z = 118

2Y+Z = 20

Solving this, X=5, Y=7, Z=6

## 8 comments:

Let spider is S, Dragonfly is D & cicada is 6 Then

8S+6D+6C = 118 ------------(1)

2D+C = 20 -------------(2)

From (2) ; C = 20 - 2D ----(3)

(3) in (1) Then

S = (6D - 2)/8

From (2) We can assume that D should be less than 10 ...Hence

min value of D is 7

means S = 5 ; D = 7 ; C = 6..

Guys....Is it correct?

Sorry ...Cicads is C

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5 spiders,7 dragonflies and 6 cicada's.

The solution is...

S=spider, D=dragon fly, C=cicada

S+D+C=18 ---- No.of Insects

8*S+6*D+6*C=118 ----- No.of legs

0*S+2*D+1*C=20 ----- No. of Wings

Now slove for S, D, C

We got>>>>

S=5, D=7 , C=6

This is the ans.....

Why cant it be

S = 2

D = 3

C = 14

spiders - 5

dragonflies - 7

cicada - 6

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