I need some help with some consectuive number problems...?

Question

I'm doing my hw and i cannot solve a few problems about consectuive numbers and equations, but i cannot figure them out. I am not looking for answers to the questions; im looking for how to solve them. I know how to use vairables and such, and to solve for the variables, but i cannot figure out the proper equation. Can someone please tell me what i should do? Here are the questions:

1. The sum of four consecutive numbers exceeds the difference of five times the fourth term and three times the third term by twenty-seven.

2. A shipping company told its customer that the average weight of five boxes is thirty-three pounds. It knows that the weights of four of the boxes are 25 pounds, 30 pounds, 30 pounds, and 35 pounds. The shipper lost the information on the fifth box. Help him find out the weight of the fifth box.

3.The sum of four consecutive even numbers is twelve more than three times the third term. What are the four numbers?

Four number are added together. The second n

umber is three more thn the first number. The third number is two times the sum of the first two numbers. And the fourth number is three times two less than the second number. If the sum of all the four numbers is equal to one more than seven times the second number, what are all the four numbers?

MOST HELPFUL ANSWER IS GUARANTEED 10 POINTS

I already knew the x + (x + 1) + (x + 2) + (x + 3)...its just in the middle somewhere i keep messing up, because i always get lik e 24 or 27

Puggy- Great Answer...very helpful..but i got the same thing, and its wrong.

tom4bucs- ty very much, very helpful indeed

disposable, you made a mistake

never mind dispoasable
im in 7th grade

Dispoasable, you are amazing!! THx, you made also figure out what i was doing wrong! You win the 10 points..just wait untill for a few hours...i had made up this rule that when u had taken things out of parentheses, the signs go to opposites
it was close though

Answer 1

n+(n+1)+(n+2)+(n+3) -27=5(n+3) - 3(n+2)
4n+6-27=5n+15-3n-6
4n-21=2n+9
2n=30
n=15
15, 16, 17, 18

check:15+16+17+18=66
5(18)-3(17)=90-51=39, which is 27 less than 66, so it works

2. 5 times 33=165
subtract 25+30+30+35 (which is 120)
so the last package must weigh 45 lbs

3. n+(n+2)+(n+4)+(n+6)-12=3(n+4)
4n+12-12=3n+12
n=12

so, 12, 14, 16, 18 are the numbers
check: 12+14+16+18=60
16x3=48 , which is 12 less than the sum of the 4 numbers, so it works.

4. x=first number
x+3=second number
2(2x+3)=third number
3(x+1)=fourth number

x+(x+3)+2(2x+3)+3(x+1)-1=7(X+3)
2X+3+4X+6+3X+3-1=7X+21
9x+11=7x+21
2x=10
x=5

so the numbers are:
5, 8, 26, 18

check 5+8+26+18=57
7x8=56, which is one less than the sum, so it works

Answer 2

(1) If you don't know one number, use a variable. Call it (n). The second number is one greater than the first, because they're consecutive. It's (n+1). The third number is one more than the second. It's (n+2). The fourth number is (n+3).

Now read the problem and substitute the values to write your equation. solving after that should be a piece of cake. :)

"The sum of four consecutive numbers" (add them up):
n + (n+1) + (n+2) + (n+3)
"exceeds" (is greater than by a certain number... in this case, it's the "by twenty-seven" at the end. Since the sum exceeds the rest, it's 27 more than the rest, or subtract 27 to make it equal):
n + (n+1) + (n+2) + (n+3) - 27 =
"the difference of" (the answer to a subtraction problem:
n + (n+1) + (n+2) + (n+3) - 27 = ___ - ___
"five times the fourth term" (we defined this above)
n + (n+1) + (n+2) + (n+3) - 27 = 5(n+3) - ___
"and three times the third term" (again, defined above)
n + (n+1) + (n+2) + (n+3) - 27 = 5(n+3) - 3(n+2)

Combine like terms on the left, distribute and combine on the right, then solve.

(2) You know the average of five weights is their sum divided by five. You know four of the five... for the unknown one, use a variable. I'll use (x).

Average = Sum ÷ 5
33 = (25 + 30 + 30 + 35 + x) ÷ 5

I'd multiply both sides by five to get rid of any possible fractions.

(3) Again, the furst even number is unknown... call it (n). Because we're using consecutive evens (they go up by two's), the next numbers would be (n+2), (n+4), and (n+6).

"The sum of four consecutive even numbers" (add 'em up)
n + (n+2) + (n+4) + (n+6)
"is" (equals)
n + (n+2) + (n+4) + (n+6) =
"twelve more than" (something plus twelve)
n + (n+2) + (n+4) + (n+6) = ____ + 12
"three times the third term"
n + (n+2) + (n+4) + (n+6) = 3(n+4) + 12

Solve the same way as the first one.

(4) The first number is unknown: (n)
The second number is three more than the first: (n+3)
The third is two times the sum of the first two. Adding those gives you (2n+3), then double it: (4n+6)
"The fourth is three times two less than the second." The second is (n+3), two less than that is (n+3-2) or (n+1). Triple that: (3n+3)

For the equation, "The sum of all the four numbers"
(n) + (n+3) + (4n+6) + (3n+3)
"is equal to"
(n) + (n+3) + (4n+6) + (3n+3) =
"one more than"
(n) + (n+3) + (4n+6) + (3n+3) = ____ + 1
"seven times the second number"
(n) + (n+3) + (4n+6) + (3n+3) = 7(n+3) + 1

One last piece of advice... don't forget to answer the question, not just solve the equation. For example, in problem number 3, you can solve for (n), but that only gives the first of four numbers your teacher or the text's author wants. Be sure to answer with all four numbers, if that's what they're looking for.

You've got it from here, big guy... good luck! :)

Answer 3

for # 1:

let x = the first number
consecutive means the numbers are one right after the other (ex. 4,5,6,...)
this means that the second number will be x+1 because the number will be one more than x
the third number will be x+2 in order to make the numbers consecutive
the fourth number is x+3
so now you have:
x = first number
x+1 = second number
x+2 = third number
x+3 = fourth number
sum means to add the numbers together so you add
x+x+1+x+2+x+3 which equals 4x+6
five times the fourth term would be 5(x+3) or 5x+15 because the x+3 = the fourth number
three times the third term would be 3(x+2) or 3x+6
the word "difference" in the equation tells you to subtract these two so it would be (5x+15) - (3x+6) or 2x + 9
it says the sum of four consectutive numbers exceeds the 2x+9 by 27, so you must add 27 to 2x + 9 to make them equal
this gives you 2x + 36
now you set this equal to 4x + 6 and solve for x
4x + 6 = 2x + 36 subtract 6
4x = 2x + 30 subtract 2x
2x = 30 divide by 2
x = 15
so the first number is 15
the second number (x+1) is 16
the third number (x+2) is 17
and the fourth number (x+3) is 18


for # 2:
to find the average of a set of numbers, you must add up all the numbers and divide by the amount of numbers you had
to do this you would add 25 + 30 + 30 + 35 + x (x represents the fifth number)
when you do this you get 120 + x
then you must divide this by 5 because there is 5 total boxes
since you cannot divide x by 5 you leave it as a fraction and put it over 5 and set it equal to 33 because the average weight of boxes is 33 pounds
so you have 120 + x/5 = 33
you multiply by 5 to get rid of the fraction
now you have 120 + x = 165
subtract 120
you get x = 45 pounds


to do the third one you do the same as number 1
x = first number
x+2= second number (you add 2 because they are consecutive even numbers not just consecutive and there is 2 numbers between each even number)
x+4 = third number
x+6 = fourth number
add the numbers together and get
x + x+2 + x+4 + x + 6 which is 4x + 12
it is 12 more than 3 times the third term
this means to multiply (x+4) by 3 and add 12
you get 3x +12 + 12 or 3x + 24
you then set this equal to 4x+12
so you have 4x+12 = 3x + 24 subtract 12
4x = 3x + 12 subtract 3x
x= 12
so the second number (x+2) is 14
the third number (x+4) is 16
and the fourth number (x+6) is 18


hope this helped
please choose as best answer if it did :)

Answer 4

NOTE ABOUT #1: Since the sum exceeds the difference of the two numbers by 27, in order to make the equation balanced, you must add the 27 to the difference, not the sum. For example, if our two numbers are 10 and 5, and we say a (10) exceeds b (5) by 5, we say a = b + 5 (10 = 5 + 5). Tell me where I made my mistake and I will try to correct it... I've gone over my answers and cannot find the mistake you reference)

#1)

1st = n
2nd = n+1
3rd = n+2
4th = n+3

the sum of the 4 is 27 greater than 5*4th - 3*3rd
n+(n+1)+(n+2)+(n+3) = 5(n+3)-3(n+2) + 27
4n + 6 = 5n + 15 - 3n - 6 + 27
4n + 6 = 2n + 36
2n = 30
n = 15
(15, 16, 17, 18)

#2)

(x1 + x2 + x3 + x4 + x5) / 5 = 33
(25 + 30 + 30 + 35 + x)/5 = 33
120 + x = 33 * 5
120 + x = 165
x = 45 lbs

#3)

n + (n+1) + (n+2) + (n+3) = 3(n+2) + 12
4n + 6 = 3n + 6 + 12
n = 12
(12, 13, 14, 15)

#4)

x1, x2, x3, and x4

x2 = x1 + 3
x3 = 2 (x1 + x2) = 2(x1 + (x1 + 3))
x4 = 3(x2 - 2) = 3(x1 + 3 - 2)

now we have all 4 numbers using a single variable (x1), so let's change it to just x to simplify things...

we know the sum of all 4 is 1 more than 7 time x2 (or x+3) so

x + (x+3) + 2(x + x +3) + 3(x + 3 -2) = 7(x+3) + 1
2x + 3 + 4x + 6 + 3x + 3 = 7x + 21 + 1
9x +12 = 7x + 22
2x = 10
x = 5

so...

x1 = 5
x2 = x1 + 3 = 8
x3 = 2 (2x1 + 3) = 4x1 + 6 = 26
x4 = 3 (x1 + 1) = 3(6) = 18


Hope this helps :)

Answer 5

#2 is extremely easy if you follow this method, you wont find it in a book.

so its 25+30+30+35+x

first you want the average to be 33 so make it that way, think of it like this.

25 + 8=33
30 + 3=33
30 + 3=33
35 -2=33

then you add up the 8+3+3-2= 12 and the average number you wanted it to be which was 33.

so 12+33= 45
25+30+30+35+45= 165/5=33

method number 2.

you know that the average you want is 33 and there are going to be 5 numbers. so think of it as x/5 = 33.

Then multiply both sides by 5 so its 5 * x/5= 33 *5.

so you get x = 165

then 25+30+30+35= 120
so 165-120= 45

This is just how my brain works, Im a math genious but I never read the textbook. This is just how I worked them out in my head. If you reason through your math instead of copy and pasting the formulas in your head itll eventually much easier.

Answer 6

Remember the definition of consecutive numbers; they are numbers that are one after the other. So 5,6,7,8 are all consecutive numbers.

Algebraically, if you're given a consecutive number x, it would be easy to calculate the consecutive numbers after that (x+1, x+2, x=3, and so forth). That's the whole idea of consecutive numbers and algebra.

1. The sum of four consecutive numbers exceeds the difference of five times the fourth term and three times the third term by twenty-seven.

Our first step is to assign our consecutive numbers.
Let x = the first consecutive number. Then
x+1 = the second
x+2 = the third
x+3 = the fourth.

To represent the sum of these four consecutive numbers, it would be:

x + (x+1) + (x+2) + (x+3) = S

Let's call this S for now, for readability purposes.

We're told that this exceeds the difference of five times the fourth term and three times the third term by 27.

So if this sum exceeds the difference by 27, we know right away we're going to add 27 to the sum to see if it equals the description.

S + 27 =

"The difference of five times the fourth term and three times the third term".

Note that "difference" means subtraction. the fourth term is equal to (x+3) [please see the beginning of this problem to be reminded], and the third term is (x+2). 5 times the fourth term and three times the third term means:

S + 27 = 5(x+3) - 3(x+2)

Key things: "difference", and noting what each term actually is.

Let's put S back.

x + (x+1) + (x+2) + (x+3) + 27 = 5(x+3) - 3(x+2)

Now, we combine like terms.

4x + 33 = 5(x+3) - 3(x+2)

Then we expand the right hand side.

4x + 33 = 5x + 15 - 3x - 6
4x + 33 = 2x + 9
2x = -24
x = -12

Therefore, since x represents your first number, you know that your consecutive numbers are x = -12, -11, -10, -9

Answer 7

Let n be the lowest number in the consecutive set - then the next number is (n + 1), the next (n + 2), and so on. Then you can start reasoning with these, e.g. in question 1 the numbers add up to
n + (n + 1) + (n + 2) + (n + 3)

Should be enough to get you on your way!

Answer 8

The sum of four consecutive numbers is n+(n+1)+(n+2)+(n+3)

question 3
n+(n+1)+(n+2)+(n+3)=12+3*(n+2)

Answer 9

1) you could figure it out with an equation: (x)+(x+1)+(x+2) < 86 the three terms on the left represent 3 consecutive integers once you figure out x, substitute the value into the equation to see if it works. i got 27, 28, 29 2) you could figure it out with an equation: (x)+(x+1)+(x+2) = 2(x+2) + 20 so... 3x+3 = 2x+24 after you combine like terms on each side subtract 2x... x+3=24 subtract 3... x=21

Answer 10

here is the set up for #2
25 + 30 +30 +35 +x = 5 times 33

go for it