There is a list of seven numbers the average of the first four numbers is 5 and the average of the last four numbers is 8. If the average of all seven number is 6 4/7, then what number is common to both sets of four numbers
Let's call the seven numbers a, b, c, d, e, f, and g (in that order).
We know that:
(a + b + c + d)/4 = 5
And:
(d + e + f + g)/4 = 8
And:
(a + b + c + d + e + f + g)/7 = 6 4/7
To find the number that is common to both sets of four numbers, we need to find which number appears in both equations.
Let's start by simplifying the third equation:
(a + b + c + d + e + f + g)/7 = 6 4/7
Multiplying both sides by 7:
a + b + c + d + e + f + g = 46
We can use this equation to eliminate some variables.
Let's start with (a + b + c + d)/4 = 5
Multiplying both sides by 4:
a + b + c + d = 20
We can substitute this into the previous equation:
20 + e + f + g = 46
Simplifying:
e + f + g = 26
Now let's look at the second equation:
(d + e + f + g)/4 = 8
Multiplying both sides by 4:
d + e + f + g = 32
We can substitute in our previous equation:
d + 26 = 32
Solving for d:
d = 6
Therefore, the number common to both sets of four numbers is d, which is 6.
How many four-digit positive integers are there such that the leftmost digit odd, the second digit is even, and all four digits are different?
We have to count the number of possibilities for the two remaining even digits and the one odd digit (we don't have to worry about the leftmost digit, as it must be odd). Once we've chosen these digits, there is only one way to arrange them.
There are 5 choices for the leftmost digit (1, 3, 5, 7, or 9).
There are 4 choices for the second digit (0, 2, 4, or 6).
There are then 7 choices for the third digit (since we've already used one digit).
And finally, there are 6 choices for the fourth digit (since we've now used two digits).
So the total number of possible numbers is:
5 × 4 × 7 × 6 = 840
Therefore, there are 840 four-digit positive integers such that the leftmost digit is odd, the second digit is even, and all four digits are different.
The last Monday of a particular month is on the 27th day of the month. What day of the week in the first day of the month?
We can approach this problem by working backwards from the given date and using the fact that there are always 7 days in a week.
Since the last Monday of the month is on the 27th, we know that there are exactly 6 days from the last Monday to the end of the month: the 28th, 29th, 30th, and 31st (if it is that kind of month).
Going back another 21 days (3 weeks), we should again arrive on a Monday. This Monday is the first Monday of the month, and it falls on the 27th - 21 = 6th day of the month.
So the first day of the month is a Monday + 6 days = $\boxed{\text{Sunday}}$.
What the ANSWER
The answer is Sunday.
Granny Smith has $63. Elberta has two dollars more than Anjou and Anjou has 1/3 as much as much as Granny Smith. How many dollars does Elberta have
Let's call the amount of money that Anjou has "x".
We know that Granny Smith has 3 times as much as Anjou, so Granny Smith has:
3x = 63
Solving for x:
x = 63/3 = 21
So Anjou has $21.
We also know that Elberta has 2 dollars more than Anjou, so Elberta has:
21 + 2 = $23
Therefore, Elberta has $23.