1. Mathematics Quiz Ideas
Level III

2. The Quiz format (5 minutes)
Quiz will have two parts
The buzzing part
The doing part
Doing part – At the beginning of the quiz, each
group is handed cards with problems to be
solved. All groups have the same problems. The
team that solves announces the solution.
This round is also open to the teachers!!! You may
pick up the cards and try your hand at the quiz as
well.

3. Warm-up!! Round:1
6 Questions
Each Q = 1 minute
Total time = 6 minutes

4. Warm –up!! Question # 1
 In a triangle the centroid is the meeting
point of the
 Altitudes
 Angle bisectors
 Medians
 Perpendicular bisectors

5. Answer:
 Medians

6. Warm -up !! Question #2
For a right angled triangle the ortho-centre
is on the:
hypotenuse
inside the triangle
on the right angle
outside the triangle

7. Answer:
 On the right angle itself

8. Warm-up!! Question # 3
 The average of 5 numbers A, B, C, D &
E is K. When we subtract L from each of
the above 5 numbers, the new average
will be:
 A – L
 E – L
 K – L
 C + L

9. Answer:
K-L

10. Warm –up!! Question #4
Use five 5s to get a target number 26
Can you plug in either addition(+),
subtraction(-), multiplication(x), division(/) and
parenthesis among five number 5s to make a
target result number 26?

11. Answer:
(5 x 5 x 5) + 5) / 5 = 26
or
(5 / 5 / 5 + 5) x 5 = 26

12. Warm-up !! Question #5
Fill in numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9
into boxes to make the equation work.

13. Answer:

14. Warm –up!! Question #6
There is a number whose double is
greater than its half by 45. Can you find
this number?

15. Answer:
30 (its double 60 is 45 more than half of it
60= 45+15)

16. Real Math!! Round : 2
6 Questions
Each Q = 2 minutes
Total time = 12 minutes

17. Real Math!! Question #1
A shopkeeper marks the list price of an
article at 400% above its cost. He gives a
discount of 80% on the list price. He
makes:
a profit of 320%
no profit no loss
loss of 10%
loss of 20%.

18. Answer:
No profit no loss

19. Real Math!! Question #2
Next door to me live four brothers. Their
average height is 74 inches and the
difference in height among the first three
men is 2 inches. The difference between
the third and the fourth man is 6 inches.
What are their heights?

20. Answer:
70, 72, 74, 80 inches tall

21. Real Math!! Question #3
 A two digit number is increased by 20%, when
its digits are reversed. Then the sum of the
digits of the number is:
 2
 7
 9
 8

22. Answers
9 (the number is 45, when reversed it is
54. 54 is 20% more than 45 )

23. Real Math!! Question #4
If (x,y) is a solution set of the following 2
equations namely
xy = 8 and
x2y + xy2 + x +y = 54, then
x2 + y2 =
62
46
20
100

24. Answer:
20

25. Real Math!! Question #5
 In a convex polygon there are two right
angles and all the remaining angles are
150° each. The number of sides of the
polygon is:
 6
 10
 8
 14

26. Answer:
8

27. Real Math!! Question # 6
The sides of a triangle are integers. The
perimeter is 8. The area is:
8
5
12.
√8

28. Answer:
√8

29. Flex your brain cells!! Round: 3
Round 1
6 Questions
Each Q = 3 minutes
Total time = 18 minutes

30. Logic!! Question #1
An artist wanted to paint a picture on a
canvas which would allow for a margin of
4 inches on top and bottom and two
inches on each side. He wanted the
picture itself to occupy 72 square inches.
What would be the smallest dimensions,
the canvas should possess?

31. Answer:
The canvas must be 10 inches wide and
20 inches high. The picture itself must be
6 inches in width and 12 inches in height.

32. Logic!! Question #2
Eight eggs look identical except one is
lighter. How can you weigh only 2 times on a
balance scale to find out which one is
lighter?

33. Solution:
If we give a number for each egg, said form 1 to 8.
We put egg 1, 2, and 3 on the left and egg 4, 5 and 6
on the right and weight them.
If they are balanced then we know egg 1 to 6 are OK.
We just need to put egg 7 on one side and egg 8 on
the other side and weight them.
Otherwise, one side will be lighter. We just assume
the left side is lighter we can just put egg 1 on the
left, egg 2 on the right, weight them one more time. If
they are balanced then egg 3 is lighter. Otherwise if
left side is lighter, ball 1 is lighter. If right side is
lighter, ball 2 is lighter.

34. Logic !! Question #3
 Use the given relationships between the figures and
words to find two solutions:
= L A G
= L E B
= _ _ _
= R A B
= R E G
R E B R A G = ??

35. Answer:

= L E G
 R E B R A G =
 Notice that horizontal figures both contain an L, the two vertical
figures contain an R.
 The equation with two figures both contain a B and the equation with
three figures both have a G. The circles have an A and the
diamonds an E. Therefore,
 L = horizontal R= vertical G = 3 B =2 A =
E =

36. Logic!! Question #4
We paint each face of the 4 feet cube with
the color red, then cut the cube into 64 one
foot small cubes. How many small cubes
will have no red color on any side of the
cube? How many small cubes have 1 face with red
color? How many small cubes have 3 faces with red
color?

37. Answer:
If you cut out 2 feet from each side (one foot
from each end), the remaining cubes will
have 2 x 2 x2 = 8. Only 8 cubes have no
color. For each side, the middle 4 cubes have
one face with color.
There are 6 sides for a cube. Therefore there
are 4 x 6 = 24 cubes with one red face. Only
corner cube will have 3 red face.
Since there are 8 corners, 8 cubes have 3 red
face.

38. Logic !! Question #5
I have a horse. Do you know what color it is?
A said, "I guess it is not black".
B said, "It is either brown, or gray". C said "I
know it is gray".
I said, "At least one of you is right and at least
one of you is wrong." What is the color of my
horse if the color is one of the above?

39. Answer:
If the horse is brown. then every one is
right. This is not the answer.
If the horse is black, then every one is
wrong. This is not the answer either.
Therefore, the horse is gray. To verify
the answer,
A was right, B was right, but C was
wrong.

40. Logic !! Question #6
Solve the following:
 ABCD x 4= DCBA
 ON x 4 = GO

41. Answer:
2178
X 4
8712
23
X 4
92

42. Problem solving Round :4
6 Questions
Each Q = 4 minutes
Total time = 24 minutes

43. Problem solving #1
 A, B and C can walk at the rate of 3, 4 and 5
km per hour respectively. They start from the
same place at 07.00 AM, 08.00 AM and
09.00 AM respectively. When B catches up
with A, B sends back A with a message to C.
At what time does C get the message?
 11.15 AM
 10.30 AM
 13.15 PM
 12.45 PM

44. Answer
11.15 AM

45. Problem solving #2
 In the adjoining diagram,
DABC is right-angled at A.
AB = 12 cm, AC = 16 cm.
 Then, the altitude AD is equal to:
 (a) 5.8 cm (b) 16.4 cm (c) 2.56 cm
(d) 9.6 cm
A
B
C

46. Answer:
9.6 cm

47. Problem solving #3
The sum of the infinite series ½ + ¼ +
1/8 + 1/16 ..........= 1. Then what is the
sum of the infinite series ¼ + 1/16 +
1/64 + 1/256..........?

48. Answer:
 1/3 (the diff between the denominator
and numerator of the first fraction is 3)

49. Problem solving #4
 78% of all people are gum chewers and
35 % of all people are under the age of
fifteen. Given that a person has been
selected at random , what is the
probability that the person is not a gum
chewer and above age 15?

50. Answer:
 14.3 %
22% x 65% = 14.3% are not gum
chewers or above the age of fifteen.

51. Problem solving #5
The units digit of the number
( 1 + 9 + 92 + 93 + ………+ 92010 ) is:
1
0
9
8

52. Answer:
9

53. Problem solving #6
One cement block balances evenly on the
scales with three quarters of a pound and three
quarters of a block. What is the weight of the
block?

54. Answer:
The block weighs 3 pounds
x = ¾ x + ¾ lb

55. Math Model Making!! Round :5
15 minutes for demo

56. Model #1
Elvis the Elf decided to create a landing of
marble tiles on a rectangular area 13 x 5
sq.ft. He determined he needed 65 tiles,
measuring 1 sq. foot. He purchased 65
tiles but discovered that one was
damaged. He needed to cover the area
using only 64 tiles. How did he do this?

57. Model 1: Answer
Draw a 8 x 8 square and a 13 x 5
rectangle. Divide the square into 2
congruent triangles and 2 congruent
trapezia. And put these pieces together in
the rectangle.

58. Model # 2
Verify the following identity
(a+b)2 = a2 + b2 + 2ab
Verify the following identity
(a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2cd

59. Model #2 : Answer
With 1 square of a2 units, 1 square of b2
units and 2 rectangles of area ab units.
Take a square of 18 units x 18 units and
divide into sections of 10 units, 6 units and
2 units. And demonstrate.

60. Model #3
Proof of Pythagoras theorem
Using two squares of 17 units and 4 right
triangles of 5, 12 and 13 units

61. Model #3 : Answer
b
a
a b
b
a b
a
a
a
a
b
b

62. Model #4
(a+b)3 = a3 + b3 + 3a2b + 3ab2
How do you form a cone from a sector of
a circle?

63. Model #4 : Answer
 Take cubes of dimension a units and b
units and 3 cuboids, using the above
dimensions. Place them together to show
the identity.
Draw a circle and cut out a sector of
degree measure 120°.

64. Model #5
Determine the area of a circle using the
area formula for a rectangle.
Determine the area of a trapezium.

65. Model #5: Answer
Draw a circle. Cut it into sectors. And
arrange the sectors to form a rectangle.
Using the formula for area of a rectangle,
arrive at the area of a circle.
Draw two congruent trapezia. Place them
in such a way as to get a parallelogram.
Now get the area of the trapezium.

66. Model #6
This is a cube. Put it together and
determine what algebraic identity it
represents.

67. Model #6 : Answer
(a+b+c)3 = a3 + b3 + c3 + 3a2b + 3ab2 +
3a2c + 3ac2+ 3b2c +3 bc2 + 6 abc

68. Written work!! Round :6
Answer discussion time: 10
minutes

69. Question 1
Long time ago, a rancher spent $100.00 and
bought 100 animals. There were 3 kind of
animals he bought: each Bull cost him $10.00,
each Cow cost him $5.00 and each Calf cost
him $0.50. How many he bought for each kind of
animal?

70. Answer:
 1 Bull + 9 Cows + 90 Calves.
(Assume the number of bulls he bought was X, the number of
cows he bought was Y, and the number of calves he bought
was Z. Thus, we have 2 equations:
1). X + Y + Z = 100
2). 10X + 5Y + 0.5Z = 100
From equation 1, we can have Z = 100 - X – Y. If we substitute Z
in equation 2, then we can come up with the equation:
19X + 9Y = 100.
Since X and Y must be whole numbers, when X = 1 (minimum),
Y will be the maximum (9). When Y = 1, X will be the
maximum(4). We can easily figure out X = 1 and Y = 9. From
equation 1, we can figure out Z = 90.)

71. Question 2
Solve the cryptarithm :
 NOON
 MOON
+ SOON
 JUNE

72. Solution:
 2442
5442
+ 1442
9326

73. Question 3
Four married couples played a tennis
tournament of “mixed doubles”. A man
and a woman always played against a
man and a woman. However, no person
ever played with or against any other
person more than once. They all played
together in two courts on three successive
days. Can you show how they could have
done it?

74. Answer

75. Question 4
Two women were selling marbles in the market place –
one at three for a paisa and the other at two for a paisa.
One day both of them were obliged to return home when
they each had thirty marbles unsold. They put together
the two lots of marbles and handing them over to a friend
asked her to sell them at five for 2 paise. According to
them, 3 for one paisa and 2 for 1 paisa was the same as
5 for 2 paisa.
Now they were expecting to get 25 paise for the marbles,
as they would have got, had they sold separately. But
much to their surprise, they got only 24 paise for the
entire lot. Now where did the one paisa go? Can you
explain the mystery?

76. Answer:

77. Question 5
All the nine digits are arranged here
so as to form four square numbers. – 9,
81, 324, 576.
How would you put them together so as
to form a single smallest possible
square number and a single largest
possible square number?

78. Answer:

79. Question # 6
 Cubes and squares can be one and the
same. But if this so happens, they need
a new name. Squbes sounds ok, but can
you now tell where the next one is at?
64 729 4096 15625
________?

80. Answer:
 The next one is 46656.
Disregarding the number 1, these
are the four consecutive lowest
numbers that are both cubes and
squares.
64 = 82 or 43 729 = 272 or 93
4096= 642 or 163 15625 = 1252
or 253 and the fifth is
46656 = 2162 or 363