Qr question

[FutureDental88]

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FROM ADA FREE SAMPLE TEST:

What is the maximum number of 3-inch
squares (squares that are three inches on
a side) that can be cut from a sheet of tin
19 x 23 inches?


I took the area of the large one: which is : 19x23 : 437
Then I found the area of small one. 3x3 : 9


Then i did 437/9. = 48.7777,


Meaning i can cut a maximum om 48 squares.


But the answer says 42. What did i do wrong?


A. 42
B. 48
C. 49
D. 145
E. 146

 

[bnakken]

You found the total areas of both of them. That means that you would have 1/2 squares to make up the leftover edges and such.

You can solve this by simply looking at each side. On the side with 19 inches you could fit 6 squares 3inches each = 18 inches (You still have an inch left but that is just scrap metal.) Then look at the other side. You can fit 7 squares in 23 inches. Again you'll have two inches of scrap metal on that side. Then you just multiply those two numbers you found for each side: 6 squares x 7 squares= 42 total squares.

 

[FutureDental88]

Thank you ! I see it now.

But i remember there being a different way of doing it... Any other suggestions?

 

[dentalWorks]

FROM ADA FREE SAMPLE TEST:

What is the maximum number of 3-inch
squares (squares that are three inches on
a side) that can be cut from a sheet of tin
19 x 23 inches?


I took the area of the large one: which is : 19x23 : 437
Then I found the area of small one. 3x3 : 9


Then i did 437/9. = 48.7777,


Meaning i can cut a maximum om 48 squares.


But the answer says 42. What did i do wrong?


A. 42
B. 48
C. 49
D. 145
E. 146

19 / 3 = 6.x <---this side is 6 perfect squares and some change
23 / 3 = 7.x <---this side is 7 perfect squares and some change

6 * 7 = 42