Question 1204475: On a wall 78 2/5 cm wide, Peter is going to hang two pictures beside each other, each measuring 12 1/2 cm wide. He will leave 2 3/5 cm in between the pictures. He plans to use two nails to hang each picture, and will centre these nails leaving 4 1/2 cm between each pair. How far, in cm, from the end of the wall (on either side) will the first nail be placed?
Found 3 solutions by MathLover1, greenestamps, math_tutor2020:
Answer by MathLover1(20849)   (Show Source):
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
The arrangement is symmetrical, so we can do our calculations on one half of the wall.
(1) The width of the wall is 78.4 cm, so start at the center, at 39.2 cm.
(2) There are to be 2.6 cm between the two pictures, so move left half of that distance from the center to find the right edge of the left picture. 39.2-1.3 = 37.9.
(3) The width of each picture is 12.5 cm. Move left 12.5 cm from 37.9 to find the left edge of the left picture at 25.4 cm.
(4) There are to be 4.5 inches between the two nails for hanging each picture; the width of the picture is 12.5 cm. For the picture to hang straight, you want to have the two nails the same distance from the edges of the picture. So the distance from each nail to the edge of the picture is half of 12.5-4.5 cm, which is 4 cm.
(5) Move right those 4 cm from the left edge of the left picture at 25.4 cm to find the first nail at 25.4+4 = 29.4 cm.
ANSWER: 29.4 cm
A rough diagram (not to scale!)....
0 39.2
|-----------------------------------------|
| 1.3 |
37.9 (R edge of L picture)
| 12.5 |
25.4 (L edge of L picture)
| 4 | 4.5 | 4 |
29.4 (first nail)
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Answer: 29.4 cm
This decimal value converts to the mixed number 29 & 2/5 which can be written as 
Explanation
I'll convert each of the given mixed numbers into decimal form
78 & 2/5 = 78.4
12 & 1/2 = 12.5
2 & 3/5 = 2.6
4 & 1/2 = 4.5
Each result is exact
We start with 78.4 cm to work with. Two paintings take up 2*12.5 = 25 cm of horizontal space.
Then we also have the 2.6 cm gap between those paintings to have a total of 25+2.6 = 27.6 cm so far.
The remaining horizontal space is 78.4 - 27.6 = 50.8 cm
Split this in half to get 25.4 cm buffer gaps on either side.
Refer to this diagram
The red segments are the two paintings.
Let point A be located at 0 on the number line.
Point B is at 25.4 on the number line.
C is at 25.4+12.5 = 37.9 on the number line.
The midpoint of B and C is (B+C)/2 = (25.4+37.9)/2 = 31.65 which is the horizontal center location of the first painting.
Or we could start at B = 25.4 and move 12.5/2 = 6.25 cm to the right to arrive at 25.4+6.25 = 31.65
From 31.65 on the number line, move (4.5)/2 = 2.25 cm to the left to arrive at 31.65-2.25 = 29.4 cm which is the location of the first nail.
A similar question is found here
https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1204434.html
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