This was on my test and have a few more question that is from my test.
18.The Sum of five unequal natural numbers is 90. The second largest of these five numbers can be at the most?
A.19
B.41(41 is the right answer)
C.44
D.43
E.42
please show work.
22.In what month does the 301st day of the year occur?
The answer is October,but is there a work for this problem?
A.August
B.September
C.October
D.November
E.December
32.For how many positive integers for "n" in the following inequality?
4.2<3n<57
A.16
B.17
C.18
D.15
E.14
the answer is 17, but i'm not sure how.
42. If a=-9 and b=10, what is the value of a^2+4a(divide)2+3b?
A.93
B.-69
C.52.5
D.7.5
E.32
the answer is 93, but i got -69. Please help me with this question.
Making the first 3 as small as possible, they would have to be 1, 2, and 3
let the 4th be x and the 5th be y
then
1+2+3+x+y = 90
x+y = 84
remember, y has to be the largest
we want x to be as large as possible but still be less than y
x y
.. ..
.. ..
40 44
41 43
42 42 --->but they can't be equal
43 41 ---> x is no longer less than y
so it looks like x = 41 and y = 43
the largest that the second largest number can be is 41
22. The average number of days per month is 30
301/30 = 10.0333
and the 10th month is October
or
365-301 = 64 days left
but Dec + Nov = 31+30 = 61
so that puts you back into October
32.
4.2 < 3n < 57
divide by 3
1.4 < n < 19
So n can be 2,3,4, ... , 17,18
Count them, there are 17
42.
a^2 + 4a/2 + 3b
= 81 + 4(-9)/2 + 3(10)
= 81 - 18 + 30
= 93
Sure! Let's go through each question step by step:
18. The first step is to set up an equation based on the given information. We know that the sum of five unequal natural numbers is 90. Let's call the five numbers a, b, c, d, and e. So, we have the equation: a + b + c + d + e = 90.
Now, we want to find the second largest number, which means we need to determine the maximum value that the second largest number can take. Let's consider the extreme case where the other four numbers are minimized. In this case, if we let a = 1, b = 2, c = 3, and d = 4, then the remaining number e would be 80 (90 - 1 - 2 - 3 - 4). So, the second largest number can be at most 80.
Since the answer choices given are all less than or equal to 80, we can conclude that the correct answer is B. 41 is the right answer.
22. To determine the month when the 301st day of the year occurs, we can divide 301 by the number of days in each month. By doing so, we find that 301 divided by 30 gives us a quotient of 10 and a remainder of 1. This tells us that the 301st day occurs in October.
So, the correct answer is C. October.
32. To solve the inequality 4.2 < 3n < 57, we need to isolate n.
First, we divide the entire inequality by 3 to get: 1.4 < n < 19.
Next, we note that we are looking for positive integers for n. Since 1.4 is not an integer, we can conclude that n must be greater than 1.4.
The largest positive integer, less than 19, is 18. Therefore, there are a total of 18 positive integers for n.
So, the correct answer is C. 18.
42. The expression a^2 + 4a/2 + 3b can be simplified using the given values of a and b.
Substituting a = -9 and b = 10, we get:
(-9)^2 + 4(-9)/2 + 3(10) = 81 - 36 + 30 = 93.
Therefore, the correct answer is A. 93.
I hope this helps clarify the solutions to the questions on your test!