. ns - M1. ddle Pru.
nary Division
Solutio
1. (Also Jl) ,...
2 + 0 + 1 + r = IO, hence (A).
2. 2017- 9 = 2008, hence (C).
value 200,
3. The 2 is in the 100s column, so I·t has hence (E).
. K to row L is one square
4. From column 1 to column 3 is two squares ri ght. From row
down,
hence (A).
5. 4 x 30 = 120 therefore the apples coSt $1.20,
hence (D).
http://anzexams.com
6. (Also UP3)
Three out of the five dice showed a two on top,
hence (E).
7. 23 - 15 = 8, so it is 8 km from Marytown to Smithton,
hence (A).
8. With 4 buses, only 200 people can fit, whereas with 5 buse
s, up to 250 fit,
hence (C).
9. (Also UPS)
Only shapes 1 and 3 have five sides and the
refore are pentagons,
hence (D).
10. Before giving half of his apples to
Sally, he must have had 2
he must have had 4 apples, apples, so at the start
hence (D).
11. Counting squares, A= 8, B = 5½, C = 7, D
= 6 and E = 7.!
2'
hence (A).
ons
2017 AMC - Middle Primary 5 0 luti
1 2 (Also UP7)
Subtract the 73 and then add 37 to get
the correct answer: 157
_ 73 + 37 = 121
hence (B).
l3. (Also UPS)
The squares and triangles are worth 4 x 6 +
hexagon is worth 50 - 42 == 8 cents,
3x6 = 24 + 18 = 42 cents. Then the
hence (B).
14. For every 3 biscuits in the normal packet,
the special one contains 4.
The special packet contains 24 == 6 x 4, so the nor
mal packet must contain 6 x 3 == 18
biscuits,
hence (C).
15. (Also UPll)
Reflect the clock face in a vertical axis so that it is the right way around.
http://anzexams.com
Then the time is 10 minutes past 7,
hence (E) .
16. The shape has one SI. d e O f length 6 cm one side of length 2 cm and 10 sides of length
1cm. The perimeter is 6 + 2 + 10 = 18cm,
hence (C).
17. (Also UP12)
Alternative 1
*
Adding both left- hand SI. des, 4 hearts and 4 stars are equa1 to 32 ' so
Then in the first line,.+., + 8 - 12 so that., +., = 4 and., - 2 '
+., = 8.
hence (B).
Alternative 2
.,., .,., *.,
We can make this pattern.
.,., + + + - □12
.,., .,* ** **
+ + +
+ + +
+ + + - 20
* + * * *
+ +
.�QH AMC - Middle Primary Solutions
z1. (Also UP19)
In the net we can find the opposite faces.
• is opposite+ which rules out (D).
is opposite* which rules out (A).
•+
•
is opposite 8
which rules out (B).
We can also rule out ( C) by looking at what happens if+ is the top face of the cube
.-!
and • and ◄ are visible. This puts • on the left and ► ◄ on the right.!
The cube in (E) is possible, since with • on top, and with on the left, 8 is on
the right,
hence (E).
22. The possibilities are 223, 232,233,322,323 and 332,
hence (C).
23. (Also UP21)
cut.
The folded paper has 8 layers when it is
http://anzexams.com
original
One of these cut layers is shown on the
shown on the
square on the left, and all 8 are
right,
hence (D).
block ar e :
3, so t he p ossib le dimensions of the rectangular
24. 12 = 1x2x2x
1 X 1 X 12
1 X 2 X 6
1 X 3 X 4
2 X 2 X 3
ks, hence (C).
th ere a re
� 4 po s sible rectangular bloc
So
3 == 26. . .
kfac e a d d to 78, and 78-=- ·
All the nu mb e rs on th e cloc
. . p r
a1· s of 13 in e ach region.
25.
e do11;e by
mclud mg 2 c
Checking, this can b
hence (D).
sp lit a
·bTty that one or both lines be 3
t e ould 2
1 t·1on doesn't investigate hberps. l111 �Iiese cases the total w
o ss
. sou
Note·· Thi s i·g1·t nu m
- dig it n u m ber • to. two one- d
m
two
ary Solutions
2017 AMC: - Middle'Prim
http://anzexams.com
. two munbers . are split. However, in each case it
or 20 dependmg on wI1e tl1e1. o ne or 1
.· t 1iese totals m all three reg 0ns.
is no� possible to find lines that give
" . . o tl, 1e hundreds and tens digits are. bo th 3 or
26. The smallest any d1g1t cou Id be is 3. S
were equal to 3, then the ones d"1g1t must be
more. If both hundreds and tens crig-its
7, and the number is 337,
hence (337).
. • t wa3rs of climbing the steps, organised by what the first and
27. Here are the 13 differen
second steps are.
I 2nd - 2 II 2nd 3 // 2nd 4 // 2nd
5]
1st 1,2,3,4,5 1,2,4,5 1,2,3,5 1,2,5 1,3, 4,5 1,3,5 1,4,5
1st 2,3,4,5 2,3,5 2,4,5 2,5
1st 3 3,4,5 3,5
hence (13).
28. ,i\Tith two squares, there is only one configuration. The third s�u�re must join onto
only one of the other squares, which it can do in exactly four d1stmct ways:
hence (4).
29. The only combinations of ages that add to 28 with the difference between the largest
and smallest being 10 are 13 + 12 + 3 = 28, 14 + 10 + 4 = 28, and 15 + 8 + 5 = 28.
The products of these are 3 x 12 x 13 = 468, 14 x 10 x 4 = 560 , and 1 5 x 8 x 5 = 600.
Of these, 468 is the smallest,
hence (468).
30. (Also UP28, J26, 119, S18)
The numbers must have different first
digits, and so the smallest possible difference
wiJI be when the first digits differ by
1. Furthermore, the number formed by the
remaining digits of the smaller number
formed by the remaining digits will be as large as possible and the number
of the larger number will
The largest and smallest possible number be as small as possible.
s formed from 4 digits are 9876 and 0123, so
the two 5-digit numbers are 50123 and 49
876 and their difference is 50123 49876
247, - =
hence (247).
32 .
2017 AMC - Middle Primary. S 0 lution
s
- - --- --
