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Mathematics Quiz Ideas 
Level III
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.
Warm-up!! Round:1 
6 Questions 
Each Q = 1 minute 
Total time = 6 minutes
Warm –up!! Question # 1 
 In a triangle the centroid is the meeting 
point of the 
 Altitudes 
 Angle bisectors 
 Medians 
 Perpendicular bisectors
Answer: 
 Medians
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
Answer: 
 On the right angle itself
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
Answer: 
K-L
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?
Answer: 
(5 x 5 x 5) + 5) / 5 = 26 
or 
(5 / 5 / 5 + 5) x 5 = 26
Warm-up !! Question #5 
Fill in numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 
into boxes to make the equation work.
Answer:
Warm –up!! Question #6 
There is a number whose double is 
greater than its half by 45. Can you find 
this number?
Answer: 
30 (its double 60 is 45 more than half of it 
60= 45+15)
Real Math!! Round : 2 
6 Questions 
Each Q = 2 minutes 
Total time = 12 minutes
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%.
Answer: 
No profit no loss
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?
Answer: 
70, 72, 74, 80 inches tall
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
Answers 
9 (the number is 45, when reversed it is 
54. 54 is 20% more than 45 )
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
Answer: 
20
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
Answer: 
8
Real Math!! Question # 6 
The sides of a triangle are integers. The 
perimeter is 8. The area is: 
8 
5 
12. 
√8
Answer: 
√8
Flex your brain cells!! Round: 3 
Round 1 
6 Questions 
Each Q = 3 minutes 
Total time = 18 minutes
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?
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.
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?
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.
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 = ??
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 =
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?
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.
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?
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.
Logic !! Question #6 
Solve the following: 
 ABCD x 4= DCBA 
 ON x 4 = GO
Answer: 
2178 
X 4 
8712 
23 
X 4 
92
Problem solving Round :4 
6 Questions 
Each Q = 4 minutes 
Total time = 24 minutes
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
Answer 
11.15 AM
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
Answer: 
9.6 cm
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..........?
Answer: 
 1/3 (the diff between the denominator 
and numerator of the first fraction is 3)
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?
Answer: 
 14.3 % 
22% x 65% = 14.3% are not gum 
chewers or above the age of fifteen.
Problem solving #5 
The units digit of the number 
( 1 + 9 + 92 + 93 + ………+ 92010 ) is: 
1 
0 
9 
8
Answer: 
9
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?
Answer: 
The block weighs 3 pounds 
x = ¾ x + ¾ lb
Math Model Making!! Round :5 
15 minutes for demo
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?
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.
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
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.
Model #3 
Proof of Pythagoras theorem 
Using two squares of 17 units and 4 right 
triangles of 5, 12 and 13 units
Model #3 : Answer 
b 
a 
a b 
b 
a b 
a 
a 
a 
a 
b 
b
Model #4 
(a+b)3 = a3 + b3 + 3a2b + 3ab2 
How do you form a cone from a sector of 
a circle?
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°.
Model #5 
Determine the area of a circle using the 
area formula for a rectangle. 
Determine the area of a trapezium.
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.
Model #6 
This is a cube. Put it together and 
determine what algebraic identity it 
represents.
Model #6 : Answer 
(a+b+c)3 = a3 + b3 + c3 + 3a2b + 3ab2 + 
3a2c + 3ac2+ 3b2c +3 bc2 + 6 abc
Written work!! Round :6 
Answer discussion time: 10 
minutes
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?
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.)
Question 2 
Solve the cryptarithm : 
 NOON 
 MOON 
+ SOON 
 JUNE
Solution: 
 2442 
5442 
+ 1442 
9326
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?
Answer
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?
Answer:
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?
Answer:
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 
________?
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

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Mathematics high school level quiz - Part I

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

Hinweis der Redaktion

  1. 70,72,74,80
  2. 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 = 1002). 10X + 5Y + 0.5Z = 100From 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.The answer is: 1 Bull + 9 Cows + 90 Calves.
  3. 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.