Pretty Hard Math Questions…some Algebraic?
I have some pretty hard Algebraic questions due tomorrow ;[ And I’ve already done the ones that I could do…does anybody think they can think of an equation that would solve these questions?
1. A bag of 3 apples, 7 oranges, and 11 pears costs .04, while a bag of 2 apples, 5 oranges, and 8 pears costs .31. What is the cost of a bag of fruit consisting of 1 apple, 1 orange, and 1 pear?
2.The king took a cup filled with pure juice and drank 1/5 of its contents. When the king looked away, the court jester refilled the cup by adding water and then stirring. The king drank 1/4 of that mixture. When the king looked away again, the jester refilled the cup with more water and then stirred. Then the king drank 1.3 of the mixture. When the king looked away again, the jester refilled the cup with more water. What percentage of the final mixture was water?
3. Five counterfeit coins are mixed with nine authentic coins. If 2 coins are drawn at random, what is the probability that one coin is authentic and one is counterfeit?
4. What is the maximum number of points of intersection can be obtained when two circles and five lines intersect each other?
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1.) use a = apples, o = oranges and p= pears
now
3a + 7o + 11p= 6.04
2a+5o+ 8p=4.31
now you have to solve for each variable
so
lets get rid of a.
multiply….
-2 (3a + 7o + 11p)=-2( 6.04)
and
3(2a+5o+ 8p)=3(4.31)
then the two equations you will have are
-6a-14o-22p= -12.08
6a+15o+24p= 12.93
the a variables will cancel out when you add the equations so you have
o+2p=0.85
isolate o
so
o=0.85-2p
then you use this in
2a+5o+ 8p=4.31
2a+5(0.85-2p)+8p=4.31
2a+4.25-10p+8p=4.31
2a-2p=0.06<- use this as your FIRST equation
now plug o=0.85-2p into a different equation
3a + 7o + 11p= 6.04
3a+ 7(0.85-2p ) +11p= 6.04
3a+5.95-14p+11p= 6.04
3a-3p=0.09<- this is your second equation
now your first equation
2a-2p=0.06
and
3a-3p=0.09
now get rid of p
so
3(2a-2p)=3(0.06)
2(3a-3p)=2(0.09)
then you multiply and get
the p’s to cancel out and then
you hav the equations
6a=.18
6a=.18
or 12a=.36
so solve for a
hten you get
a + .03
so apples cost 3 cents then plug in to solve for the other variables.
2. is 78%
3. is 5/45 or 1/9 reduced
1. # of apples = a
# of oranges = o
# of pears = p
6.04 = 3a + 7o + 11p (Solve for a)
Then write the equation over and solve for o.
Then for p.
Then add a, o, and p.
I don’t know how to do the others, sorry