Question

A stretch of highway that is 12\dfrac1412 4
1
​ 12, start fraction, 1, divided by, 4, end fraction kilometers long has speed limit signs every \dfrac78 8
7
​ start fraction, 7, divided by, 8, end fraction of a kilometer.
How many speed limit signs are on this stretch of highway?

Answer 1

Given:

• The length of the stretch of the highway is $12\frac{1}{4}$ km.

• Speed limit signs appear every $\frac{7}{8}$ fraction of a kilometer.

To find:

• How many speed limit signs are on the stretch of the highway?

Step-by-step explanation:

Speed limit appear every $\frac{7}{8}$ fraction of a kilometer.

Thus the total appearance can be found by dividing the length of the total stretch by appearance of speed limit every kilometer:

• $\quad 12\frac{1}{4}\div \frac{7}{8}$

• $=\frac{49}{4}\div \frac{7}{8}$

• $=\frac{49}{4}\times \frac{8}{7}$

• $=\bold{14}$

Answer:

Total number of appearances of speed limit is 14.

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Answer 2

Answer:

14

Step-by-step explanation:

Divide the length of the highway by the distance between signs, then rewrite 12 1/4= 12x4= 48+1= 49/4, switch the division side to multiplication then switch 7/8 to 8/7 then do 49/4 x 8/7, 49 x8= 392 then 4x7= 28 divided 392 by 28 to get 14.

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