math problems(: please help!*?

Question

*The cost of 12 oranges and 7 apples is $5.36. Eight oranges and 5 apples cost $3.68. Find the cost of each orange.


*Sam invested a total of $10,000, part at 8% and the rest at 3%. How much was invested at 8% if he earned a total of $600 in one year? (Use the formula , Interest = principal * time * rate)


*In an office there was a small cash box. One day Jane took half of the money plus $1 more. Then Mike took half of the remaining money plus $1 more. Rob then took the remaining $6. How many dollars were originally in the box?


*The sum of two numbers is 8 less than twice the first number. Their difference is 4 less than twice the second number. What is the larger number?

Answer 1

let the cost of oranges=x, let the cost of apples =y

12x + 7y = 5.36 and 8x + 5y = 3.68

(You know in the first case that 12 times the cost of oranges and 7 times the cost of apples is 5.36. Same logic for the second equation)

multiply both sides of the first equation by two (2) to give 24x + 14y =10.72 (this is allowed as long as you do the same thing to both sides of the equation). Multiply the second equation by three (3) both sides to yield 24x + 15y=11.04

24x + 14y = 10.72
24x + 15y = 11.04

Subtract the bottom equation from the top equation (each term) to yield

0x -1y=0.22

or y= $0.22

that is, apples are $0.22 apiece.

pick either of the original equations and insert $0.22 in place of the y value.

8x + 5*0.22 = $3.68
8x +1.10 =$3.68

Subtract a dollar ten from both sides of the equation
8x =2.58
divide both sides by 8
x=0.86
oranges are $0.86 each.

The rest of the questions use the same logic, it's called solving simultaneous equations.

Answer 2

1. This means if x are the oranges and y the apples that:
12x +7y =5.36
8x + 5y =3.68 <=>x=(3.68-5y)/8 so the first becomes:
12*(3.68-5y)/8 +7y=5.36
3*(3.68-5y)/2 +7y=5.36
(11.06-15y)2 + 7y=5.36, by multiplying with 2 we have:
11.06 -15y+14y=10.72
-y=-0.34
y=0.34
so oranges cost x=(3.68-5y)/8=
=(3.68-5*0.34)/8=1.98/8~0.25$

2. Let x be the part invested in 8% and y the one at 3%.
we have x+y=10,000
and 600=(8/100)x+(3/100)y for both parts.
which is by multiplying with 100 we have
60000=8x+3y
From the 1st we have y=10,000-x
so the 2nd becomes: 60,000=8x+3*10,000-3x
60,000-30,000=5x
x=30,000/5
x=6,000

3. Let x be the money in the box. Jane took x/2+1.
Mike took (x/2+1)/2 +1=x/4 + 1/2 +1
Then Rob took the remaining 6 $. So the money of Jane, Mike and Rob all together are x. So we have
x=x/2 +1+x/4 +1/2 +1+6
we multiply by 4 so that there are quarters and seconds:
4x=2x+4+x+2+4+24
4x-2x-x=24
x=24$

4. Let x be the 1st number and y be the 2nd
we have x+y=2x-8
and x-y=2y-4 (2)
by adding we have
(x+y)+(x-y)=(2x-8)+(2y-4)
2x=2x+2y-12
2y=12
y=6
so from (2) we have x-6=2*6-4
x=14 so the larger number is 14

Answer 3

#3. ((6 + 1) * 2) + 1) * 2 = $30
remaining money = 6, add 1 dollar then times 2 for the missing half. add another dollar and times 2 again for the first halfing

#4. 2a - 8 = a + b
a = b + 8

a - b = 2b - 4
a + 4 = 3b

a = (a + 4)/3 + 8

sorry, cant do anymore, no time.
can someone else check my work?

Answer 4

12 O +7A= 5.36
8 O + 5A = 3.68
60 O + 35A = 26.8
-56 O -35A = -25.76
4 O = 1.04
O = $.26

X = part invested at 8%
10000-X = amount invested at 3%
So .08x+.03(10000-X) = 600
.08x +300 -.03x = 600
.05x = 300
X = $6000 = amount invested at 8%

30

x= 1st number
y = 2nd number
x+y = 2x-8 --> x-y = 8
x-y = 2y-4--> -x+3y = 4
2y = 12 --> y= 6
So x= 14 = larger number

Answer 5

Multiply .07 by 13,888.60 (which I'm guessing is the cost of the car), then add the sales tax (the amount you get when you multiply those two numbers) to the original cost of the car. That's it :]