Grades 3-4 Math OTT Question 3
Training Student
?
Line
Caide
Temmy counted the number of cans of soup on each of four shelves at a supermarket. The table below shows the numbers
of cans of soup
Madc
Cans of Soup
Number
Shelf
of Cans
1 22
2 18
3 14
4 29
Tammy added all the numbers of cans of soup on these shelves and then rounded the total to the nearest 10. What is the
number of cans of soup on these shelves rounded to the nearest 10?
Practice Hint 1: Use the Sticky Note tool to make notes or calculations before selecting your answer.)
Practice Hint 2 Use the Line Guide tool to keep your place. Drag the blue handle on the right as you read.)
60
70
80

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.

step-by-step explanation:

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answer: the proof is mentioned below.

step-by-step explanation:

since, given:   lines bc = ef

prove:   lines ac > ef

          statements                                           reason

1. bc=ef                                           1. given

2. ac= ab+bc                                 2. because b is the point between the  

                                                            line segment ac

3.   ac > bc                                       3. because if a line segment is equal to

                                                          the sum of two line segments then it is

                                                          always greater then each of these

                                                          individual line segments

4. ac > ef                                         4. by equality property.