SHORT ANSWER PROBLEMS
1. E is the midpoint of the side BC of a triangle ABC. F is on AC, so that AC = 3 FC. The ratio of the area of quadrilateral ABEF to the area of triangle ABC is . . . : . . . .
Answer: 5:6
Since E midpoint BC then area AEB = areaAEC = 1/2 areaABC F C = 1/3AC then area CEF = 1/3 area AEC = 1/6 areaABC Area of quadrilateral ABEF is 5/6 area ABC
Ratio area ABEF toCEF would be 5 to 1
2. A rectangle is divided into four rectangles, as shown in the figure below. The areas of three of them are given in cm2
. The area of the original rectangle is . . .cm2
.
Answer: 84
21 15 =
28
A,
7 5 =
28
A, 7A= 140, A= 20
The area of the original rectangle is 15 + 21 + 28 + 20 = 84
3. A shape on the right is constructed by putting
two half circles with diameter 2 cm on the top and on the bottom of a square with 2 cm sides. A circle is inscribed in the
square as shown in the figure. The area of the shaded region is . . . cm2
Answer: 4
Area A = Area B ; Area C= Area D
Area of the shaded region is equal to area of the square.
4. The least positive integer a so that 490×a is a perfect cube number is. . . .
Answer: 700
In order to get a Perfect Cube number, since 490 = 2×5×72
, then it has to be multiple by 22
×52
×7 = 700
5. A number of cubes are stacked as shown seen in the figure below. The highest level consists of one cube, the second highest level consists of 3×3 cubes, the third highest level consists of 5×5 cubes, and so on.
If there are 2010 cubes to be stacked in this way, such that all levels are complete, some cubes may not be used. There can be as many as . . . levels.
Answer: 11
11 Number of cube needed for the 1st
level (from the top) is 12 giving the number of cube needed at the lowest level equal to 212
ratio of the area of the square to the area of the circle is . . . : . . . . [Useπ = 7.]
Answer: 7:11
See the figure below
Diameter of the circle is 2 unit length or equal to diagonal square. Since the square transfered to rhombus, the area of the rhombus would be :
L = 1
2× diagonal 1×diagonal 2 = 1
2×2×2 = 2
Area of the circle with radius 1 unit length is π, hence the area ratio between square and circle is 2 : π.
Answer: 34
see figure
8. The areas of three faces of a rectangular box are 35,55 and 77 cm2
.If the length, width and height of the box are integers, then the volume of the box is . . .cm3
.
Answer: 385
Assume q = length, h=height, w=width q×w= 55
q×h= 77 w×h= 35
w/h= 55/77 = 5/7 q/w = 77/35 = 11/5
The volume of the box is length × width × height = 11×5×7 = 385 cm3
9. In a certain year, the month of January has exactly 4 Mondays and 4 Thursdays. The day on January 1st of that year is . . . .
Answer: Friday
10. In an isosceles triangle, the measure of one of its angles is four times the other angle. The measure of the largest possible angle of the triangle, in degree, is . . . .
Answer: 120o
For ease those two angles are x and 4x.
If the third angle is x, the sum of the internal angle is x+x+ 4x= 6x or x= 30o
and 4x= 120o
.
If the third angle is 4x, the sum of the internal angle is x+ 4x+ 4x = 9x or x = 20o
and 4x= 80o
.
. . . .
Answer: 119
Number of integers which are less than 1123.5 are 2010 : 2 = 1005 numbers. Since those numbers are consecutive, then the smallest integer is 1123−(1005−1) = 119.
12. If the length of each side of a cube is decreased by 10%, then the volume of the cube is decreased by . . .%.
Answer: 27.1 %
Suppose the side of original square is 10 cm, hence the original volume would be 103 = 1000 cm3
.
Shorten the side to 10%, hence the volume become 93 = 729 cm3
The decrease of the volume is (1000−729)/1000×100% = 271/1000×100% = 27.1% 13. The operations ◦ and are defined by the following tables.
For example, 2◦3 = 2 and 23 = 3 . The value of (12)◦3 is . . . .
Answer: (12)◦3 = 3◦3 = 3
14. The number of zeros in the end digits in the product of
1×5×10×15×20×25×30×35×40×45×50×55×60×65×70×75×80×85×90×95 is. . . . From the product factors above, 109
has 9 zero digits. Product of 2×2×5×5 results 2 zero digits.
15. In the addition sentence below
each letter represents a different digit. The number of all possible digits represented by ’E’ is . . . .
16. A piece of rod is 16 meters long. There is a device that can divide any piece of rod into two equal pieces. We can use this device 8 times. In the end, we will have 9 pieces of rod. The maximum possible difference between the length of the longest piece and the length of shortest piece is . . . meters.
Answer: 715 16
After the nth
cut, the length of the shortest piece is 16 2n m.
Hence, the maximum difference of the length of the longest piece and the length of the shortest piece is 8− 1
16 = 7 15 16 m.
I +II = 1
19. For any positive integer n, let d(n) be the sum of digits in n.
For example, d(123) = 1 + 2 + 3 = 6 andd(7879) = 7 + 8 + 7 + 9 = 31.
20. The numbers 1447, 1005 and 1231 have something in common. Each is a four-digit number beginning with 1 that has exactly two identical digits. There are. . .such num-bers.
Answer: 432
We cluster the set of such numbers into two subsets :
• Those numbers having two 1′s.
• Those numbers having two identical digits other than 1. Let 1xyz be the number.
The number of such numbers in subset (1):
The second digit 1 may be placed in x or y or z. (There are three possibilities). The others two digits must be distinct, so that there are 9×8 possibilities. Thus, there are 3×9×8 = 216 such numbers.
The number of such numbers in subset (2):
There are nine possibilities for the two identical digits, and there are three possibil-ities for its position (xy, xz,andyz). There are eight possibilities for the remaining digit Thus there are 9×3×8 = 216 such numbers.
21. Let n be a positive integer greater than 1. By the length of n, we mean the number of factors in the representation of n as a product of prime numbers. For example, the ’length’ of the number 90 is 4, since 90 = 2×3×3×5. The number of odd numbers between 2 and 100 having ’length’ 3 is . . . .
Answer: 5
One of the prime factor is 3, because 5×5×5>100. This allows only five numbers: 3×3×3 = 27, 3×3×5 = 45, 3×3×7 = 63, 3×5×5 = 75 and 3×3×11 = 99. 22. A rectangle intersects a circle as shown:
AB = 8cm, BC = 9cm andDE = 6cm. The length of EF is . . .cm.
Answer: 13
Draw in the perpendicular bisector of chord EF, as shown in the figure. This bisector must go through the center O of the circle. Since P Q is perpendicular to chord BC as well, it bisects BC. SinceP Q//AD, we have DQ=AP = 8 + 9
2 = 25
2 .
Thus EF = 2EQ= 2(DQ−DE) = 13.
Alternate Solution
Drop perpendiculars BR and CS as shown in the second figure, yielding rectangles BCSR and ABRD.
slipping around the circuit shown. At the start the top face is 3. At the end point, the number displayed on the top face is . . . .
Answer: 6
As the diagram in the question shows, after one rotation a 6 is on the top face. After a second rotation a 4 is on the top face. The next rotation brings the face that was on the back of the die to the top; since 2 was on the front this is 5. A further rotation brings 3 to the top. With another rotation along the path, 1 is displayed, followed by 4, and finally 6.
24. LetA, B and Cbe three distinct prime numbers. IfA×B×C is even andA×B×C > 100, then the smallest possible value of A+B +C is . . . .
Answer: 18
Since the product of the three numbers is even, then one of them is 2.
Therefore, the sum of the remaining two is as small as possible, whereas the product is at least 51. Let the two prime numbers be A and B.
If A= 3 then the possible value for B is 17. Thus, A+B = 20. If A= 5 then the possible value for B is 11. Thus, A+B = 16. If A= 7 then the possible value for B is 11. Thus, A+B = 18. If A= 11 then the possible value for B is 5. Thus,A+B = 16. If A= 13 then the possible value for B is 5. Thus,A+B = 18.
25. The product of a × b ×c× d = 2010, where a, b, c and d are positive integers and a < b < c < d. There are . . . different solutions for a, b, c and d.
Answer: 7
