Does anyone know how to solve these problems using equations?

Question

I'm really stuck on the following questions:


1) The diagram shows a road from A to E.

A to B is 5km more than D to E.

C to D is twice the distance from A to B

C is midway between B and D

If the total distance from A to E is 91km, find the distance from D to E

2) The sum of four consecutive whole numbers is 98. Let the first number be x and write down the other three numbers in terms of x . Find the four numbers.

3) The sum of four consecutive odd numbers is 216. Find the numbers.


Please give working out and answer.

Thanks,
p_q

In an arithmagon, the number in a square is the sum of the two numbers in the two circles either side of it. Find x in these arithmagons.

(a) http://www.picturehost.co.uk/links/arithmagon-for-a/

(b)
http://www.picturehost.co.uk/links/arithmagon-for-b/

(c)
http://www.picturehost.co.uk/links/arithmagon-for-c/

^
4)

(Plz show working out with answer)

Answer 1

1)
91 = AB + BC + CD + DE
91 = DE + 5 + 2DE + 10 + 2DE + 10 + DE
6DE = 91 - 25
6DE = 66
DE = 11 km

2)
x + x + 1 + x + 2 + x + 3 = 98
4x = 92
x = 23
The numbers are 23, 24, 25, 26

3)
216/4 = 54
The numbers are
51, 53, 55, 57

Answer 2

2) As given, x is the first number. Since these are consecutive whole numbers, it follows that:
x + 1 is the second number
x + 2 is the third number
x + 3 is the fourth number.

We know that their sum is equal to 98; therefore

x + (x + 1) + (x + 2) + (x + 3) = 98

Now, solve for x. Combine all the x terms, and they become 4x. Combine the numbers, and they become 1+2+3 = 6, so we have

4x + 6 = 98

Subtract 6 both sides,

4x = 92

Divide both sides by 4

x = 23

That means if 23 is the first number, the next three are 24, 25, and 26.

3) Same deal.
Let x = the first number. Then
x + 1, x + 2, x + 3, are the 2nd, 3rd, and 4th numbers respectively. So

x + x + 1 + x + 2 + x + 3 = 216
4x + 6 = 216
4x = 212
x = 54

So the four numbers are 54, 55, 56, 57

Answer 3

1)
AB = DE+5
CD = 2AB
BC = CD
AE = 91

AE = AB + BC + CD + DE
AE = AB + CD + 2AB + (AB - 5)
91 = AB + 2AB + 2AB + (AB - 5)
96 = 6AB
AB = 16

therefore, DE = AB - 5 = 16 - 5 = 11km

2) x + (x+1) + (x+2) + (x+3) = 98
4x + 6 = 98
4x = 92
x = 23

therefore, the 4 numbers are 23, 24, 25, 26

3) 2a+1 is the first odd number, 2a+3, 2a+5, 2a+7
216 = (2a+1) + (2a+3) + (2a+5) +(2a+7)
216 = 8a + 16
8a = 200
a = 25
therefore the four numbers are 25, 27, 29, 31