Solution to
Seventh International Mathematics Assessment for Schools
                Round 1 of Middle Primary Division
1. What is the value of 19 × 1 + 19 × 3 + 19 × 5 + 19 × 7 + " + 19 × 19 ?
   (A)1900            (B)1919            (C)2900          (D)2919            (E)3800
【Solution】
      19 × 1 + 19 × 3 + 19 × 5 + 19 × 7 + " + 19 × 19 = 19 × (1 + 3 + 5 + 7 + " + 19)
                                                           (1 + 19) × 10
                                                    = 19 ×
                                                                 2
                                                    = 19 × 100
                                                    = 1900
Hence (A).
                                                                           Answer:(A)
2. If (Δ × 2 − 1) × 2 = 2018 , then what is the value of Δ ?
    (A)502            (B)503            (C)504           (D)505            (E)506
【Solution】
                            (Δ × 2 − 1) × 2 = 2018
                                  Δ × 2 − 1 = 2018 ÷ 2 = 1009
                                     Δ × 2 = 1009 + 1 = 1010
                                        Δ = 1010 ÷ 2 = 505
Hence (D).
                                                                        Answer:(D)
3. Five students sit along a circle and starts to call out some numbers one-by-one.
    Student A calls out “1”, B calls out “2”, C calls out “3”, D calls out “4”, E calls
    out “5” and then it returns back to Student A who calls out “6” and so on, where
    each student increases the previously called number by one and calls it out.
    Which student calls out the number “99”?
     (A)A            (B)B             (C)C            (D)D              (E)E
【Solution】
Taking division with remainders, we get 99 = 5 × 19 + 4 . It is student D who calls 99.
Hence (D).
                                                                        Answer:(D)
4. In the figure shown below, four rectangles of the same A                           B
    size, denoted by I, II, III and IV, are placed together,            I
    where ABCD and EFGH are both squares. If rectangle
    I has a perimeter of 20 cm, then what is the perimeter,             E     F II
    in cm, of ABCD ?
     (A)40           (B)60            (C)80                         IV
                                                                        H    G
     (D)100          (E)120                                                  III
                                                                 D                      C
【Solution】
The side length of square ABCD is exactly the same as the sum of the length and
width of rectangle I. So, the perimeter of ABCD is (20 ÷ 2) × 4 = 40 cm. Hence (A).
                                                                       Answer:(A)
5. How many triangles in total are there in the figure below?
      (A)4            (B)8             (C)12             (D)14           (E)16
【Solution】
Refer to the figure below, there are four identical triangles for each case. Thus, the
total number of triangles is 4 × 4 = 16 . Hence (E).
                                                                    Answer:(E)
6. The six faces of a cube are colored using six different colors namely red, blue,
   yellow, green, black and white in some order. Turn the cube arbitrarily and it
   shows the two possible scenarios below. What is the color opposite to the face of
   green?
                                White                       Red
                                        Yellow                    Blue
                                 Red                     Blac
    (A)Red            (B)Yellow (C)Blue                 (D)Black (E)White
【Solution】
The red face is adjacent to the yellow, blue, black and white faces. Therefore, its
opposite face is colored green. Hence (A).
                                                                        Answer:(A)
7. Consider every positive integers whose digits do not include 2 and that sum of its
    digits is equal to 3 and arrange all such integers in increasing order. What is the
    sum of the three smallest integers that satisfy the conditions?
    (A)36             (B)63           (C)144            (D)206         (E)414
【Solution】
Since 3 = 1 + 1 + 1 , we can arrange such numbers in increasing order as 3, 30, 111,
300, …, etc. Therefore, the sum of the three smallest numbers is 3 + 30 + 111 = 144 .
Hence (C).
                                                                       Answer:(C)
8. In the 4 × 4 square table shown below, a △ is placed on the second row and
    second column. What is the total number of squares with sides falling on the grid
    lines and containing △?
    (A)8               (B)10           (C)11           (D)12         (E)14
【Solution】
Enumerating the squares according to the different sizes containing △. There are 1
such 1× 1 square; 4 such 2 × 2 squares; 4 such 3 × 3 squares and 1 such 4 × 4
square. In total, there are 10 such squares containing △. Hence (B).
    △                 △                △                 △                △
    △                 △                △                 △                △
                                                                      Answer:(B)
9. In the figure below, a frog jumps between the three circles. In each jump, it goes
   from one circle into another circle. It is known that the frog starts from A and
   ends at A after 4 jumps. How many different paths can the frog have?
                       A                   B                 C
    (A)4            (B)5            (C)6             (D)7            (E)8
【Solution 1】
One can enumerate the following 6 ways: ABABA, ABACA, ABCBA, ACABA,
ACACA, ACBCA. Hence (C).
                                      A       B       A
                                                   C          A
                               B         C         A ×
                        A                          B          A
                                         A         B          A
                               C
                                                   C          A
                                         B         A ×
                                                         C         A
【Solution 2】
Denote three circles between the start circle and end circle as X, Y, Z:
                                    A →X → Y → Z → A
Since X and Z are neither A, one distinguishes two cases as below:
(i) if Y is A, then X or Z each has two choices B or C, so there are 2 × 1× 2 = 4
     ways of jump.
(ii) if Y is not A, then X is determined by Y, as well as Z is also determined. There
     are two choices of Y so there are 2 × 1× 1 = 2 ways of jump.
Totally there are 4 + 2 = 6 ways of jump. Hence (C).
                                                                          Answer:(C)
10. In 1202 A.D., Italian mathematician Fibonacci (1170~1250) wrote in his book
     《Liber Abaci》 the following interesting problem: Exactly two months after
     their birth, a pair of rabbits will give birth to a new pair (one male and one
     female) and then give birth to a pair each month after that. There is only one pair
     of new born rabbits at the beginning. If the rabbits never die, how many pair of
     rabbits are there after exactly 12 months?
      (A)144           (B)233             (C)234            (D)235       (E)377
【Solution】
By iteration, we get the following sequence:
                      1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …
where the 13th number is 233. The sequence is the so-called Fibonacci sequence,
where each term (starting from the third) is the sum of the two terms before it.
Hence (B).
                                                                          Answer:(B)
11. Define “ *” as an operation such that 4 ∗ 2 = 82 , 6 ∗ 3 = 183 , 8 ∗ 4 = 324 ,
     9 ∗ 3 = 276 and 9 ∗ 5 = 454 . What is the value of 10 ∗ 2 ?
      (A)55            (B)125             (C)202            (D)208       (E)2002
【Solution】
From the examples it shows that the result is the products followed by the difference.
As 10 × 2 = 20 , 10 − 2 = 8 , we have 10 ∗ 2 = 208 . Hence (D).
                                                                          Answer:(D)
12. Bob got a score of 94 on foreign language test, and his average score on the
     native language and math tests is 97. What is his average score on these three
     tests?
      (A)94            (B)94.5         (C)95            (D)95.5         (E)96
【Solution 1】
His total score of native language and math is 97 × 2 = 194 , and total score of three
subjects is 194 + 94 = 288 , so the average of three is 288 ÷ 3 = 96 . Hence (E).
【Solution 2】
Since the average score on native language and math is 97 and the score for foreign
language is 94, the scores on native language and math are totally (97 − 94) × 2 = 6
more than the score for foreign language. Thus the average score of the three tests is
6 ÷ 3 = 2 more than the score for foreign language, i.e. the average score of the three
tests is 94 + 2 = 96 . Hence (E).
                                                                         Answer:(E)
13. Color the surfaces of a 5 × 5 × 5 wooden cube red, then cut it into smaller
     1 × 1 ×1 cube pieces. How many of these smaller cubes have exactly two red
     faces?
      (A)36            (B)48           (C)61            (D)90           (E)98
【Solution】
Unit cubes with two red faces are exactly those adjacent to edges of the large cube
but not at the corner. There are 5 − 2 = 3 such unit cubes on each edge of the large
cube, who has 12 edges. The total number of unit cubes with two red faces is then
12 × 3 = 36 . Hence (A).
                                                                         Answer:(A)
14. A palindrome number is a positive integer that is the same when read forwards or
     backwards. The numbers 909 and 1221 are examples of palindromes. How many
     palindrome numbers are there between 10 and 1000?
      (A)90            (B)99           (C)100           (D)106          (E)108
【Solution】
A two-digit number is a palindrome number if the two digits are the same but
non-zero, so there are 9 of them. A three-digit number is a palindrome number if the
first digit is equal to the third and non-zero, there are 90 of them. 1000 is not a
palindrome number. Totally there are 9 + 90 = 99 of palindrome numbers between
10 and 1000. Hence (B).
                                                                         Answer:(B)
15. In the month of February of some year, there are more Saturdays than any other
     days in a week. What day is the last day of this month?
      (A)Wednesday                (B)Thursday               (C)Friday
                    (D)Saturday                (E)Sunday
【Solution】
February in a flat year has 28 days, which is a multiple of 7. Each day in a week
appears four times in a flat year February. For a leap year February, the first day or
the last day is the same day in a week, which appear 5 times. Hence (D).
                                                                         Answer:(D)
16. Four students namely Annie, Benny, Charlie and Deany all paid the same amount
    of money to buy some number of notebooks together. After distributing the
    notebooks, Annie, Benny and Charlie got 6, 7 and 11 notebooks more than Deany,
    respectively. As such, to be fair, Annie, Benny and Charlie gave back a total of
    $48 to Deany. What is the price of each notebook?
     (A)2            (B)6             (C)8           (D)12          (E)16
【Solution】
Totally A, B, C takes 6 + 7 + 11 = 24 notebooks more than D. D takes 24 ÷ 4 = 6
notebooks less than average. He is paid back $48. Each notebook counts for
48 ÷ 6 = 8 dollars. Hence (C).
                                                                    Answer:(C)
17. A train left town A at 8:30 AM some day and arrived at town B at 1:50 AM of the
    next day. There is no time difference between the two places. How long did the
    train travelled for the trip?
     (A)5 hours 20 minutes (B)10 hours 20 minutes (C)15 hours 20 minutes
                   (D)16 hours 20 minutes (E)17 hours 20 minutes
【Solution】
The train used 15 hours 30 minutes on the first day; and 1 hours 50 minutes on the
next day. The total time is 17 hours 20 minutes. Hence (E).
                                                                     Answer:(E)
18. Mike placed 4 identical squares, each with side length 5 cm and are
    non-overlapping, to form a new figure as shown below. Find the perimeter, in cm,
    of this new figure.
    (A)15            (B)20            (C)30          (D)45         (E)50
【Solution】
The figure can be cut into three regions as below.
The left and right part must be covered by one square each and the middle region is
covered by two. The region has the same length of circumference as the minimum
rectangles covering it, which is 5 × 3 = 15 cm wide and 5 × 2 = 10 cm high. The
circumference is (15 + 10) × 2 = 50 cm. Hence (E).
                        A                              B
                          D                                   C          Answer:(E)
19. Adam owns an old watch, which is slower than a normal watch by the same
    amount of time for each hour. At 8 o'clock in the morning, the old watch reads 8
    o'clock. At 9 o'clock in the morning, it reads 8:58. What time does it read when
    the real time is 4 o'clock in the afternoon?
     (A)3:42                      (B)3:44                     (C)3:46
                   (D)4:08                      (E)4:16
【Solution】
The watch is 2 minutes slower every hour. There are 8 hours from 8 am to 4 pm. The
watch is then 2 × 8 = 16 minutes behind. It reads 3:44. Hence (B).
                                                                         Answer:(B)
20. Identical equilateral triangles are placed together into two figures as shown below.
    One can cover Figure B using 5 pieces of Figure A. How many different ways
    can we do the covering?
                        Figure A
                                                      Figure B
    (A)3            (B)4             (C)5             (D)6             (E)7
【Solution】
The triangle X can be covered in three ways:
Case (i)                                 X
        Then another piece of A is determined as below
                                         X
            The blank triangle area can be covered in two ways:
                               X                           X
            This case contributes two cases in total
Case (ii)
                                              X
         This case is symmetric to case (i), so it contributes two cases.
Case (iii)
                                            X
            The blank area has only one covering method.
                                              X
          This case contributes 1.
In total, there are 2 + 2 + 1 = 5 ways of covering. Hence (C).
                                                                            Answer:(C)
21. Put some unit cubes into a 3D model such that in the model, each cube touches
    some other cubes in at least one point. It is known that the model looks like the
    figure below from three directions of upright front, left and top. What is the least
    number of unit cubes needed to make such a model?
【Solution】
It is obvious the model is contained in a cube of edge length 3 and contains some of
the 27 unit cubes of the large cube. The unit cube in the center has to appear in the
model. The cubes at the center of faces of the large cube must not appear. The cubes
at the middle of edges of the large cube must not appear. Moreover, one of the two
unit cubes at any adjacent corners of the large cube must appear. It is then obvious by
pigeonhole principle, one needs at least 5 unit cubes. One such construction is in the
figure below.
                                                                        Answer:005
22. How many ways can we divide 6 students into 3 groups so that each group has
     exactly 2 students?
【Solution】
Denote the six persons as A, B, C, D, E, F. There are five possibilities for the person
in the same group as A. Without loss of generality, assume A and B are in the same
group. Then there are 3 possibilities for the person in the same group as C. The last
group is also determined. Totally there are 5 × 3 = 15 ways of division.
                                                                         Answer:015
23. In the figure below, ABCD, EFGH and AJKL are squares. The area of AJKL is
    2018 cm2. If rectangles EFCI and JBFK both have an area of 1360 cm2, then
    what is the area, in cm2, of CGHI?
                             A               L       D
                                                         I   H
                                         E
                         J                       K
                             B            F          C       G
【Solution】
Since DL = AD − AL = AB − AJ = BJ and EFCI has the same area as JBFK, we
have EF = JK . Square AJKL and EFGH are of the same area. Then CGHI has area
2018 − 1360 = 658 cm2.
                                                                Answer:658
24. Three pairs of red, four pairs of yellow and five pairs of white socks are placed in
    a bag. Now, blindly take a sock out each time. How many socked are needed to
    be taken out to guarantee having six pairs of socks? (Note: Two socks of the
    same color are considered a pair)
【Solution】
Since two socks of the same color forms a pair, one needs 12 socks. Now there are
three colors, there might be two single socks, so at least one needs to take out
12 + 1 + 1 = 14 socks.
                                                                         Answer:014
25. Cut the 6 × 6 square table below into rectangles along grid lines such that no
    two rectangles are identical. What is the maximum number of rectangles one can
    get? (Note: A square is considered a rectangle.)
【Solution】
For more rectangles, the rectangles we get need to be as small in area as possible.
There one rectangle of area 1, 2 or 3; two rectangles of area 4(1 × 4 and 2 × 2 ); one of
area 5; two of area 6(1 × 6 and 2 × 3 ). These are 7 rectangles of smallest area. Since
   1 + 2 + 3 + 4 + 4 + 5 + 6 + 6 = 31 < 6 × 6 = 36 < 1 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + 7 = 38 ,
There are at most 8 different rectangles, while the figure below shows one such
example.
                                                                             Answer:008