High School Maths problem Sum

Hi Guys,

I am stuck with this High school maths problem sum which is killing me !

My tutor dosen't allow me to use algebra to solve this question =(


Here goes the question :

Alex bought some fishes and dogs for $96. He bought 3 times as many dogs as fishes. He paid $30 more for the fishes than the dogs. Each fish cost $10.40 more than each dog.

a)How many Fishes did he buy?

b)What is the cost of each dog?


Any kind soul pls help !

This question is killing me !

You have 4 equations is 4 unknowns. Let F = number of fish, D = number of dogs, Pf = price of a fish, Pd = price of a dog:

Quote:

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Originally Posted by liukawa

Alex bought some fishes and dogs for $96.

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F(Pf) +D(Pd) = 96

Quote:

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Originally Posted by liukawa

He bought 3 times as many dogs as fishes.

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D = 3F

Quote:

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Originally Posted by liukawa

He paid $30 more for the fishes than the dogs.

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F x Pf = 30 +D(Pd)

Quote:

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Originally Posted by liukawa

Each fish cost $10.40 more than each dog.

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Pf=10.40 +Pd

Now to work through these four equations:

1. Put the value for D from equation 2 into equation 1. This reduces the problem to 3 equations in 3 unknowns:

F(Pf) + 3F(Pf) = 96
F(Pf) = 30 +3F(Pd)
Pf=10.4+Pd

2. Put the value for Pf from the third equation into the first two:
F(10.4+Pd) + 3F(Pd) = 96
F(10.4 + Pd) = 30 + 3F(Pd)

3. Subtract the 2nd equation from the first:
3F(Pd) = 96-(30+ 3F(Pd)); so F xPd = 11

4. Put F(Pd) =11 into the first equation:
10.40 F + 11 +3(11) = 96

Which yields F = 5

Now you can back substitue this into the previous equations to get the values for Pd, Pf, and D.

You should be able to use whatever method you want. I see solutions involving less algebra (e.g. no system of equations) but eventually you'll have to use at least a little algebra, unless you're plainly guessing and checking.