4 numbers are placed in a bag. two are removed and total 2.4. The numbers are returned to the bag and the process is repeated 4 more times. The totals are 2.4 2.5 2.6 2.7 and 2.8. What are the 4 numbers in the bag? please let me know how you arrived at the answer. I was good at grade 5 math 34 yrs ago!
1.15, 1.25, 1.35 and 1.45
I used a series of "what if" steps and got lucky. The two lowest numbers have to add up to 2.4 and the two highest have to add up to 2.8. I could not make it work if the numbers were divisible by 0.1, so I tried numbers divisible by 0.05 and ending in 5.
Which of the following shows the correct order of the numbers below?
()2, , 0.45
To find the four numbers in the bag, we can work backward from the given totals and the number of times the process was repeated.
Let's start with the first round: two numbers are removed, and the total is 2.4. Since there were four numbers in the bag initially and two were removed, we can use algebra to represent the unknown numbers as variables:
Let's say the first number removed is X1 and the second number removed is X2. Therefore, the sum of X1 and X2 is 2.4:
X1 + X2 = 2.4 ---(Equation 1)
Since the numbers are returned to the bag, the total count of numbers in the bag remains the same. In this case, there are still four numbers. Therefore, we can represent the remaining two numbers as Y1 and Y2:
X1 + X2 + Y1 + Y2 = 4 ---(Equation 2)
Now, let's move on to the second round: two numbers are removed, and the total is 2.5. As in the first round, we can represent the unknown numbers with variables:
X3 + X4 = 2.5 ---(Equation 3)
And the count of numbers in the bag remains four:
X3 + X4 + Y1 + Y2 = 4 ---(Equation 4)
Continuing this pattern for the remaining rounds, we have:
X5 + X6 = 2.6 ---(Equation 5)
X5 + X6 + Y1 + Y2 = 4 ---(Equation 6)
X7 + X8 = 2.7 ---(Equation 7)
X7 + X8 + Y1 + Y2 = 4 ---(Equation 8)
X9 + X10 = 2.8 ---(Equation 9)
X9 + X10 + Y1 + Y2 = 4 ---(Equation 10)
Now we have a system of equations with ten variables. We can solve these equations simultaneously using substitution or elimination methods.
By simplifying Equation 2, 4, 6, 8, and 10, we can deduce that:
Y1 + Y2 = 4 - (X1 + X2) ---(Equation 11)
Y1 + Y2 = 4 - (X3 + X4) ---(Equation 12)
Y1 + Y2 = 4 - (X5 + X6) ---(Equation 13)
Y1 + Y2 = 4 - (X7 + X8) ---(Equation 14)
Y1 + Y2 = 4 - (X9 + X10) ---(Equation 15)
From these equations, we observe that Y1 + Y2 remains the same throughout all the rounds. We can conclude that the values of Y1 and Y2 are constant. Therefore, we can rewrite Equation 11 to Equation 15 as:
Y1 + Y2 = C ---(Equation 16)
Now, we only have six equations left:
X1 + X2 = 2.4 ---(Equation 1)
X3 + X4 = 2.5 ---(Equation 3)
X5 + X6 = 2.6 ---(Equation 5)
X7 + X8 = 2.7 ---(Equation 7)
X9 + X10 = 2.8 ---(Equation 9)
Y1 + Y2 = C ---(Equation 16)
Since the values of Y1 and Y2 do not change in any round, we can focus on finding the values of X1, X2, X3, X4, X5, X6, X7, X8, X9, and X10 that satisfy Equations 1, 3, 5, 7, 9, and 16.
To find these values, we can substitute values for X1, X3, X5, X7, X9, and C, and solve the system of equations iteratively. There might be multiple solutions, so we need to continue checking for consistency among all the equations until we find a valid set of values.
Alternatively, one could solve this system of equations numerically using specialized software or programming languages like Python to find the exact values for X1, X2, X3, X4, X5, X6, X7, X8, X9, and X10.