2014IMAS First Round Junior Division Full Solutions

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Solution to

Fourth International Mathematics Assessment for Schools

Round 1 of Junior Division

1. What is the value of 2014 1 204  ( 2014)2 ?

（A）1 （B）1 （C） 2087

（D）4027 （E）4029

【Suggested Solution】

Original Expression 2014 1 2014   4027. Hence, we select option (D).

Answer: (D)

2. A compass costs 15.40 dollars and a ruler costs 8.65 dollars. How many more dollars does the compass cost than the ruler?

（A）7.25 dollars （B）7.75 dollars （C）24.05 dollars （D）6.25 dollars （E）6.75 dollars

The cost of a compass is more expensive than the cost of a ruler by 15.4－8.65=6.75 dollars. Hence, we select option (E).

Answer:（E）

3. The two stars in the diagram represent the same number. The sum of the three numbers in the second row is equal to twice the sum of the three numbers in the first row. What number does each star represent?

5 6 ☆

☆ 19 20

（A）7 （B）8 （C）13 （D）17 （E）18

From the given information, it shows the sum of the three numbers in the second row is equal to twice the sum of the three numbers in the first row, it follows that the difference of the sum of three numbers in the second row and the sum of three numbers in the first row equal the sum of three numbers in the first row, that is; the difference of these two rows is 19 + 20－5－6 = 28, then we have ☆ = 28－5－6 = 17. Thus, we select option (D).

Answer:（D）

4. In a restaurant, one cup of tea and two cups of coffee cost 78 dollars, while three cups of tea and one cup of coffee cost 94 dollars. How many more dollars does a cup of coffee cost than a cup of tea?

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From the given information, we know that 6 cups of tea and 2 cups of coffee cost 94 2 188  dollars, so 5 cups of tea cost 188 78 110  dollars, it follows that the cost of 1 cup of tea is 22 dollars, and cost of 1 cup of coffee is 94 22 3  28dollars, thus 1 cup of coffee is more expensive than 1 cup of tea by 28 22 6 dollars.

Hence, we select option (C).

Answer:（C）

5. When two numbers are divided by 5, the respective remainders are 4 and 2. What is the remainder when the sum of the two numbers is divided by 5?

（A）0 （B）1 （C）2 （D）3 （E）4

The sum of the remainders of two numbers is 2 + 4 = 6, continue dividing this remainder by 5, we have the final remainder 1. Hence, we select option (B).

Answer:（B）

6. A circular spinner is divided into eight equal sectors. Two of them are painted red, two black, three white and one yellow. What is the probability for the pointer of the spinner to be pointing at a black sector?

（A）3 4 （B） 1 4 （C） 3 8 （D） 1 8 （E） 1 2 【Suggested Solution】

In eight equal sectors of the circular spinner, 2 parts are black, so the probability for the pointer of the spinner to be pointing at a black sector is 2 1

8  4. Hence, we select option (B). Answer:（B）

7. To visit a friend, Rod must take the bus to the nearest Metro station, and this takes 15 minutes. He has to ride the Metro train for 20 stops, each taking 2.5 minutes. He also has to change trains twice, and it takes 3 minutes each time. Finally, after exiting the Metro, he still has to walk another 12 minutes before reaching his friend’s place. How many minutes does Rod have to spend traveling to his friend’s house?

（A）55 （B）67 （C）80 （D）83 （E）90 Black Red White Yellow White White Red Black

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From the given information, Rod must spend a total of 15 + 2.5 × 20 + 3 × 2 + 12 = 83 minutes. Hence, we select option (D) .

Answer:（D）

8. On a table there is a ring, there are 12 equally spaced beads on the ring, 3 of which are black, as shown in the diagram. Which of the following five figures cannot be obtained from the given figure by rotating the ring on the table?

（A）（B）（C）（D）（E）

Observe the arrangement as regards the number of small white circles between two black circles of the given diagram in counterclockwise direction: four white small circles, two small white circles and then three small white circles. In all the five options, only option (E) does not comply with this kind of arrangement. Hence, we select (E). Answer:（E）

9. If a, x and y are real numbers such that 2y12  ax y 0, what is the value of the product axy?

（A）0 （B）6 （C）12

（D）36 （E）impossible to determine

The two terms at the left side of the given expressions are both non-negative, it

follows the values of both equal to 0, then 2y120 and ax y 0, so we have

ax = y = 6, this implies axy = 36. Hence, we select option (D).

Answer:（D）

10. How many integers a satisfies 2a 7 2a 1 8?

（A）9 （B）8 （C）5 （D）4 （E）infinite

Represent the numbers on the left side of the given equation in a number line and interpret them, since the distance of two points whose coordinate is 7 and 1 is exactly 8, that means the coordinate 2a must between the two coordinates 7 and 1, so we have  7 2a1, or it is equivalent as 7 1

2 a 2

   . Since a must be integers, then the possible values of a are 3, 2,1 and 0. Therefore, there are 4 values of

a that satisfy the given equation. Hence, we select option (D).

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11. If a and b are prime numbers such that a2 7b 4 0, what is the value of a+b?

（A）5 （B）8 （C）9 （D）10 （E）13

【Suggested Solution 1】

It is obvious that a = 3 has not met the condition of the problem, then a cannot be a multiple of 3. Thus, when a is divided by 3 will give a remainder of 1. Because 2

when 4 is divided by 3 will also give a remainder of 1, so that when 7b is divisible by 3, it follows b is also multiple of 3, that is; b = 3. Therefore, a = 5 and a + b = 8. Hence, we select option (B).

【Suggested Solution 2】

Rewrite the given equation as (a2)(a2)7b, obviously a = 2, 3 does not meet the condition of the given problem, so both factors of the left side must be greater than 1 and since the two factors in the right side are prime numbers, then the two factors of the left side are also prime numbers.

When a 2 b and a 2 7, we have a = 5, b = 3 which meet the condition of the problem! So, a + b = 8.

When a 2 7 and a 2 b, we have a = 9, b = 11, does not meet the condition of the problem!

Therefore, a + b = 8. Hence, we select option (B).

Answer:（B）

12. The diagram shows a strip of paper folded along the segment AB. If ACB

60 and the area of triangle ABC is 3 cm2, what is the width, in cm, of this strip? （A）1 （B） 3 （C） 3 2 （D） 2 3 3 （E）impossible to determine 【Suggested Solution】

By method of folding paper and Alternate Interior Angles Theorem, we

have 180 60 60

2

CAB CBA   

     , so ABC is an equilateral triangle.

Let AC = x, then the altitude of ABC is 3

2 x and the area of ABC equals to

2 3 4 x cm 2 , so that 3 2 3 4 x  , this implies x = 2. Therefore, the width of the paper is 2 3 3

2

  cm.

Hence, we select option (B).

Answer:（B）

A

B C

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13. ABCD is a square of side length 10 cm. The segment BC is fixed. The segment AD moves in the plane to the segment A′D′ so that the lengths

AB, DC and AD do not change. What is the area,

in cm2, of the shaded region in the diagram when the segment A′D′ intersects the segment CD at its midpoint G? （A）50 （B）50 3  （C）60 （D）100 （E）100 3  【Suggested Solution】

Connect AA′ and D′D, Apply Cut and Paste method we know the area of the shaded region equals to the area of parallelogram AA′D′D, in parallelogram

AA′D′D we have base AD = 10 cm, altitude GD = 5

cm and so, the area is 10 × 5 = 50 cm2. Hence, we select option (A).

Answer:（A）

14. We start with an equilateral triangle ABC of area 80 cm2. We construct a right isosceles triangle BCD using BC as the hypotenuse. Then we construct an equilateral triangle using BD as a side. This continued alternately, as shown in the diagram. What is the area, in cm2, of the fourth equilateral triangle?

（A）1.25 （B）5 （C）6.4 （D）10 （E）40

Using Pythagoras’ Theorem, each side length of an equilateral triangle is 2

2 times of the previous one, so the area of the recent equilateral triangle is 1

2 that of the previous equilateral triangle. Thus, area of the fourth equilateral triangle in the given diagram is 3 1 80 10 2   _{ }    cm 2

. Hence, we select option (D).

Answer:（D）

15. We wish to spend 100 dollars to buy 18 stamps, each costing 4 dollars, 8 dollars or 10 dollars. We must buy at least 1 stamp of each of the three kinds. How many different ways can the buying of stamps be possible?

（A）1 （B）2 （C）3 （D）4 （E）5 A B C D G A′ _D′ A B C D A B C D G A′ D′

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Let x represent the number of stamps that cost 8 dollars and y represent the number of stamps that cost 10 dollars, then 18 x y represent the number of stamps that cost 4 dollars. From the given information, we obtain 8x10y4(18 x y) 100 and after simplifying we have 2x3y14, notice that both x and y must be positive integers, then there are only two possible solution sets: x = 4, y = 2 and x = 1, y = 4. By trial and check, these two sets of solution meet the condition 18 x y is a positive integer, so there are two different ways of buying the stamps. Hence, we select option (B).

Answer:（B）

16. The sectors MAB and MCD are inside the square ABCD of side length 10 cm, as shown in the diagram. What is the total area, in cm2, of these two sectors, correct to 1 decimal place? Take π=3.14.

（A）52.3 （B）78.5 （C）104.7 （D）157.0 （E）314.0

From the given information, △BCM is an equilateral triangle, then ∠ABM =∠MCD =30°.

Total area of the two sectors is 3.14 102 30 2 52.3 360

    cm2. Hence, option (A).

Answer:（A）

17. Three different positive integers m, n and p are such that (m3)(n3)(p 3) 4. What is the value of m + n + p?

（A）5 （B）6 （C）8 （D）14 （E）15

From the given information, m3, n3, p3 are three distinct integers such that their product is 4. From parity checking, maybe all the three integers are positive or one is positive integer and two negative integers. When all the three integers are positive, then the least product is 1 × 2 × 3 = 6, does not meet the condition of the problem. When one of them is positive and the other two are negative, let us assume

3 0

m  , n 3 0, p 3 0, due to the fact that the positive integers less than 3 are 1 and 2, so that m, n must be equal to 1 and 2, and so p = 5. The sum of these three integers is 8. Hence, we select option (C).

Answer:（C）

A

B C

D M

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18. If x y 0 and x2  y2 4xy，what is the value of x y x y   ? （A） 3 （B） 3 （C）3 （D） 6 （E） 6 【Suggested Solution】

Since x y 0 so we know that both xy, x y are negative numbers, it follows that x y 0 x y  _  , and 2 ₂ ₂ 2 2 2 6 3 2 2 x y x y xy xy x y x y xy xy    _   _ _  _  _ _   . Therefore x y 3 x y  _

 。Hence, we select option (A).

Answer:（A）

19. The diagram shows two right triangles OAC and OBD. The lengths of three of the segments AB, AC, CD and DB are 12 cm, 6 cm and 3 cm. What is the number of possible lengths, in cm, of the fourth segment?

（A）2 （B）3 （C）4 （D）5 （E）6

AB is obvious the longest among the four line segments, then AB = 12 or AB = x:

(1) If AB = 12: When CD = x, 122 x2  (3 6)2, x3 7. When CD = 6, 122 62  (3 x)2, x6 33. When CD = 3. 122 32  (6 x)2, x3 156. (2) If AB = x：When CD = 12, x2 122  (6 3)2, x15. When CD = 6. x2 62 (12 3) 2, x3 29. When CD = 3. x2 32 (12 6) 2, x3 37. Therefore, there are 6 possible values of x. Hence, we select option (E).

Answer:（E）

20. For any real number x, we denote by [x] the greatest integer not greater than x. For example, [ ] 3  and [  ] 4. How many positive integers n satisfy

100 1 n n      _ _{ }       ? （A）1 （B）2 （C）3 （D）4 （E）5 A C B O D

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Since 10010, we know that when the values of n increases then the value of 100 n n      _ _       

will then decrease. Among the values of n as 11, 10, 9, 8, 7, 6 we

discovered only 10, 9 and 8 satisfy the condition of the problem. Thus, there are 3 possible values of n. Hence, we select option (C).

Answer: （C） 【Remarks】When m, n are positive integers, we have

x x m mn n         _{    }  _{  }     , then the values of 100₂ 1 n  _{ }     , which is equivalent as 2 100 1 2 n   , that is n = 8, 9, 10.

21. In the diagram, AH is perpendicular to BC, AB = BC < AC, and AD is the bisector ∠BAC. If ∠DAH = 21°, what is the measure, in degrees, of ∠BAC?

【Suggested Solution】 Let BAC.

Since AB = BC then   C BAC,

Because AD is the angle bisector of BC, then 1

2 2

DAC BAC      . In right-angled triangle ACH, the two acute angles are complementary, then

21 90

2



     , so that   46 .

Answer: 046

22. How many four-digit numbers are divisible by all of 2, 3, 4, 5, 6, 7 and 8?

The Least Common Multiple of 2, 3, 4, 5, 6, 7 and 8 is 3 × 5 × 7 × 8 = 840, then 840 × 2, 840 × 3, 840 × 4, … ,840 × 11

We have a total of 10 four-digit numbers that satisfy the given condition.

Answer: 010

23. Each of A, B, C and D has some apples. A has as many apples as the other three together. B has half as many apples as the other three together. C has one-sixth as many apples as the other three together. How many times D’s number of apples will be equal to total number of apples of A, B and C?

A

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From the given information, A has 1 1

1 1  2 the apples of all the four persons, B has

1 1

1 2 3 the apples of all the four persons, C has

1 1

1 6 7 the apples of all the

four persons, then D has 1 1 1 1 1

2 3 7 42

 

_   _

  the apples of all the four persons.

Therefore, the total number of apples that A, B and C have is 42 1 41  times that

D has.

Answer: 041

24. In how many ways can 31 be expressed in the form a + b + c (a b c), where

a, b and c are prime numbers? 【Suggested Solution】

From the condition, a + b + c = 31 ( a b c). Since 3a31 then a10, so the possible values of a are 2, 3, 5, 7.

If a = 2, then b + c = 29, no solutions!

If a = 3, then b + c = 28, there are 2 possible solutions: (b ,c) = (5, 23), (11, 17). If a = 5, then b + c = 26, there are 2 possible solutions: (b ,c) = (7, 19), (13, 13). If a = 7, then b + c = 24, there are 2 possible solutions: (b ,c) = (7, 17), (11, 13). In summary, there are a total of 2 + 2 + 2 = 6 ways to express 31 as sum of three distinct prime numbers.

Answer: 006

25. Each of the three dimensions of a cuboid of volume a cm3 is an integral number of cm. The cuboid is placed on a table. The total surface area of the five visible faces is a cm2. Find the minimum value of a.

Let the length, width (or breadth) and height of the cuboid represent as x, y, z cm, respectively, then from the given information we have xy + 2yz + 2zx = xyz. Divide both side of the above equation by xyz, 1 2 2 1

z   x y .

When 1 2 2 1

3

z   x y , then x = y = 6, z = 3, xyz = 108,

Suppose not all of 1

z , 2 x, 2 y is equal to 1

3, then one of them greater than 1 3.

Assume 1 1

3

z  , then z < 3, and because

1 1

z  so that z > 1, then z = 2, so the given

equation becomes 1 1 1

4

x  y .

Thus, the possible (x, y) = (5, 20), (6, 12), (8, 8), (12, 6), (20, 5). If follows the corresponding value of xyz = 200, 144, 128.

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When 2 1 3

x  （similarly for

2 1

3

y  ）, then x < 6, and because

2 1 x  then x > 2, so that x = 3, 4, 5. Suppose x = 3, we have 1 2 2 3

z  y , the solution of (y, z) = (7, 21), (8, 12), (9, 9), (12,

6), (15, 5), (24, 4). The corresponding product of xyz = 441, 288, 243, 225, 216. Suppose x = 4, we have 1 2 1

2

z  y , the solution of (y, z) = (5, 10), (6, 6), (8, 4), (12,

3). The corresponding product of xyz = 200, 144, 128. Suppose x = 5, we have 1 2 3

5

z  y , then the solution of (y, z) = (4, 10), (5, 5), (20, 2).

The corresponding value of xyz = 200, 125.

In summary, the minimum volume of a is 108 cm3, it is obtained when length is 6 cm, width is 6 cm and height is 3 cm.

Answer: 108 【Remarks】This problem is related with AM-GM Inequality.