help with math

[Jul18]

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1) The length of a picture with its border is 7 inches less than twice its width. If the border is 1 inch wide and its area is 62 square inches, what are the dimensions of the picture alone?

2) Angie bought some golf balls for $14. If each ball had cost $.25 less ,she could have purchased one more ball for the same amount of money. How many golf balls did Angie buy?

3) A jogger who can run an 8-minute mile starts a half mile ahead of a jogger who can run a 6-minute mi. How long will it take the faster jogger to catch the slower jogger?

Thanks a bunch!!

 

[DentalP87]

1) The length of a picture with its border is 7 inches less than twice its width. If the border is 1 inch wide and its area is 62 square inches, what are the dimensions of the picture alone?

2) Angie bought some golf balls for $14. If each ball had cost $.25 less ,she could have purchased one more ball for the same amount of money. How many golf balls did Angie buy?

3) A jogger who can run an 8-minute mile starts a half mile ahead of a jogger who can run a 6-minute mi. How long will it take the faster jogger to catch the slower jogger?

Thanks a bunch!!

hi

1) I don't understand the question

2) QP = 14

(Q + 1) (P - 0.25) = 14

solve for P

then put P into QP = 14

solv for Q

you should get Q = 7

3) (1/8)t + 1/2 = (1/6)t

solve for t

you should get t = 12min

good luck!

 

[Streetwolf]

1) The length of a picture with its border is 7 inches less than twice its width. If the border is 1 inch wide and its area is 62 square inches, what are the dimensions of the picture alone?

L = 2W - 7 (with border)

Border is 1 inch wide which means length = (L - 2) and width = (W - 2). With the previous equation the length is 2W - 9.

I'm going to assume that 62 square inches represents the area between the picture and the edge of the border. This is calculated by taking the area of the entire thing (including the border) and subtracting the area of JUST the picture (no border).

Area of everything = LW
Area of just picture = (L-2)(W-2)

So fill in the length and plug into the equation LW - (L-2)(W-2) = 62:

(2W-7)W - (2W-9)(W-2) = 62
2W^2 - 7W - 2W^2 + 4W + 9W - 18 = 62
6W - 18 = 62
6W = 80

W = 13 and 1/3.
L = 19 and 2/3.

I hate getting non-integers for answers in these problems but it works here. Those are the dimensions of the BORDER. The dimensions of the picture alone are:

W = 11 and 1/3.
L = 17 and 2/3.