CATS:
If a cat can see its self; then 3.
Rap^:
a+b+c=100
5a+b+1/4c=100
so
20a+4b+c=400
subtracting
4a+4b+4c=400
you get
16a-3c=0
since 16 and 5 are relatively prime
a=3 and c=16
and
b=100-3-16=81
So there are 3 dogs, 81 cats and 16 mice
Check
3*5+1*81+16/4=15+81+4=100.
What!
3Dogs=$45
81 Cats=$81
16 Mice=$4
Total = $130
You dun only got $100! :wink:
Stormy:
Pet Shop
3 dogs.............$45
41 cats............$41
56 mice...........$14
_________________
100 animals...$100
Which all goes to show; There is only one way to skin a cat!
Let D stand for the number of dogs, C for the number of cats, and M stand for the number of mice.
Then you can write two equations:
D + C + M = 100 and
15 D + 1 C + 0.25 M = 100.00.
It wasn't going to be five dogs because five dogs would eat up $75 and then you'd only have $25 with which to buy another 95 animals, and it wasn't going to work.
And you know it wasn't zero dogs because that was one of the conditions of the problem.
So, use trial and error with dogs = 4, then 3, 2, and 1 to see which works. The answer turns out to be 3 dogs, 41 cats and 56 mice.
The following Javascript solves the problem:
for (Dogs=1; Dogs<=6; Dogs++) {
for (Cats=1; Cats<=99; Cats++) {
for (Mice=1; Mice<=399; Mice++) {
Cost=1500*Dogs+100*Cats+25*Mice;
if ((Dogs*1+Cats*1+Mice*1 == 100) && (Cost == 10000))
{alert(Dogs+' '+Cats+' '+Mice);
}}}}
A certain A2K member has 5 daughters: Emily, Jane, Betsy, Abigail, and Nancy.
Their fortunes are as follows:
The first two and the last two have $19,000;
The first four, $19,200;
The last four, $20,000;
The first and the last three, $20,500;
The first three and the last, $21,300.
They all live in CA so they are pretty rich.
What is the fortune of each