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
Let spider is S, Dragonfly is D & cicada is 6 Then
ReplyDelete8S+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
ReplyDelete--
ReplyDelete5 spiders,7 dragonflies and 6 cicada's.
ReplyDeleteThe solution is...
ReplyDeleteS=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
ReplyDeleteS = 2
D = 3
C = 14
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ReplyDeletespiders - 5
ReplyDeletedragonflies - 7
cicada - 6
How can it be solve
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