Let l, t, b represent the numbers of lions, tigers, bears, respectively.
2l +3t +3b = 156 . . . . . . . 156 meals per day are supplied l +t = 3b . . . . . . . . . . . . . . there are 3 times as many great cats as bears l +t +b = 68 . . . . . . . . . . . there are a total of 68 animals
_____ The last 2 equations tell you .. 4b = 68 .. b = 17 Subtracting 3 times the last equation from the first gives .. -l = -48 There are 48 lions, 3 tigers, and 17 bears.
Let L = number of lions Let T = number of tigers Let B = number of bears
If one lion eats 2 meals per day, then L number of lions eat 2L meals per day. Tigers eat 3T meals per day, and bears eat 3B meals per day.
In one day, all animals combined ate 156 meals.
That gives us our first equation.
2L + 3T + 3B = 156
The lions and tigers are considered great cats. The bears are not cats. There are 3 times as many great cats as bears. This gives us our second equation.
L + T = 3B
In all there are 68 animals in the zoo. From here we get our third equation.
L + T + B = 68
Our system of equations is:
2L + 3T + 3B = 156 L + T = 3B L + T + B = 68
Now we solve the system of equations.
Subtract the second equation from the third equation.
B = -3B + 68
4B = 68
B = 17
Substitute B = 17 into the first and second equations.
2L + 3T + 3B = 156
2L + 3T + 3(17) = 156
2L + 3T + 51 = 156
2L + 3T = 105 New Equation A
L + T = 3B
L + T = 3(17)
L + T = 51 Equation B
Equations A and B have only two variables, so we can solve them as a system of two equations in two unknowns.
2L + 3T = 105 L + T = 51
Rewrite equation A. Multiply Equation B by -2 and write underneath Equation A. Then add the equations.
What is 32% of 75?
Y is 32% of 75.
Equation: Y = P% * X.
Solving our equation for Y:
Y = P% * X.
Y = 32% * 75.
Converting percent to decimal:
p = 32%/100 = 0.32.
Y = 0.32 * 75 .
Y = 24.
Answer is 24.
