Let hare be X, cat be Y and dog be Z then,1X+1Y+0z=10.....(i) 1X+0y+1Z=20.....(ii) 0x+1Y+1Z=24.....(iii) Solving for three unknowns(x,y and z) we get 3,7 and 17 respectively. Therefore,in the place where all three animals involved hare,cat and dog ie x+y+z=3+7+17=27
6 comments:
The answer is 27.
Let hare be X, cat be Y and dog be Z
then,1X+1Y+0z=10.....(i)
1X+0y+1Z=20.....(ii)
0x+1Y+1Z=24.....(iii)
Solving for three unknowns(x,y and z) we get 3,7 and 17 respectively.
Therefore,in the place where all three animals involved hare,cat and dog ie x+y+z=3+7+17=27
it is very interesting puzzle,as it involves some mathematics#AQU/E/10/T/0040
the answer is 27£ AQU/12/T/0017
Nice the Answers are correct
ok sir,and the marking scheme was clearly shown!#AQU/E/10/T/0040
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