Volume of the figures below?
1 answer:
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Let lizard be 'l'
Let beetle be 'b'
Let worm be 'w'
We know that the number of worms is more than the number of lizards and beetles put together. We can write this Mathematically w>b+l
We also know that we have in total 12 heads. Since each animal has one head, we can deduct that we have more than 6 worms. The possible combinations of the number of worms and beetles+lizards are
Worms = 7, Beetles+Lizards= 5
Worms = 8, Beetles+Lizards=4
Worms = 9, Beetles+Lizards=3
Worms = 10, Beetles+Lizards=2
Worms = 11, Beetles+Lizards=1 ⇒ this condition is unlikely because we can't have half beetle and half lizard and we were told that we have beetle, lizard AND worm, so none of them is 0
We will need the last constraint to work out the number of lizard Ross has. Altogether she has 26 legs. A worm has 0 legs, a beetle has 6 legs, and a lizard has 4 legs. We can form an equation
We will use the simultaneous equation to solve this. Pick one equation in term of beetle and lizard from the combination above. Let choose beetle+lizard=5
Then we have
⇒ Rearrange to get 
Substitute into
, this mean, we have 3 beetles
We can try as well using another combination
which we can rearrange to get 
and substitute into
, Mathematically, the number negative one does exist but in realization we can't have -1 lizard
We can keep trying with the other combinations to get the most sensible answer.
Trying
and substitute into , solving algebraically will give the value of 
which again is not realistic.
By following the pattern, if we try the last combination
, we will get an even more negative answer.
So it leaves us with one realistic arrangement that meets all three conditions
4l+6b+0w=26
Number of worms = 7
Number of lizards = 2
Number of beetles = 3
Let beetle be 'b'
Let worm be 'w'
We know that the number of worms is more than the number of lizards and beetles put together. We can write this Mathematically w>b+l
We also know that we have in total 12 heads. Since each animal has one head, we can deduct that we have more than 6 worms. The possible combinations of the number of worms and beetles+lizards are
Worms = 7, Beetles+Lizards= 5
Worms = 8, Beetles+Lizards=4
Worms = 9, Beetles+Lizards=3
Worms = 10, Beetles+Lizards=2
Worms = 11, Beetles+Lizards=1 ⇒ this condition is unlikely because we can't have half beetle and half lizard and we were told that we have beetle, lizard AND worm, so none of them is 0
We will need the last constraint to work out the number of lizard Ross has. Altogether she has 26 legs. A worm has 0 legs, a beetle has 6 legs, and a lizard has 4 legs. We can form an equation
We will use the simultaneous equation to solve this. Pick one equation in term of beetle and lizard from the combination above. Let choose beetle+lizard=5
Then we have
Substitute
We can try as well using another combination
We can keep trying with the other combinations to get the most sensible answer.
Trying
By following the pattern, if we try the last combination
So it leaves us with one realistic arrangement that meets all three conditions
4l+6b+0w=26
Number of worms = 7
Number of lizards = 2
Number of beetles = 3