Mode, Mean and Median Practice

The Mode, Mean and Median, are types of averages.

The mean is the average calculated by adding the numbers and dividing by the number of items in the data set. The median is the middle value in a data set. To calculate the median, put the numbers in order, and the median will be the middle number. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers.

See our example below. The mode is the most frequently occurring number. If no number is repeated, then there is no mode.

Examples
Find the median, mode and mean of the following list:
6, 7, 8, 12, 14, 6, 7, 10

Find the mean
First add the numbers
6 + 7 + 8 + 12 + 14 + 6 + 7 + 10 = 70
There are 8 numbers in the list, so divide by 9
70/8 = 8.75 = mean

Find the median
First put the numbers in order
6, 6, 7, 7, 8, 10, 12, 14,
The data set has an even number of numbers, so the median is the average of 7 and  8. (7 + 8)/2 = 7.5

Find the mode
The mode is the most frequently occurring number. Here
6 and 7 both occur twice, so they are both considered the
mode.

 

1. Find the mean of these set of numbers – 200,000, 10,020, 30,000, 15,000 1080

a. 1080
b. 15,000
c. 256,100
d. 51,220

2. Find the mean of these set of numbers – 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

a. 55
b. 5.5
c. 11
d. 10

3. Find the mean of these set of numbers – 2.5, 10.2, 4.5, 1.25, 7.05, 20.8

a. 7.6
b. 45.6
c. 7
d. 1.25

Median Questions

4. Find the median of the set of numbers – 3, 1, 9, 7, 13, 11, 15, 21, 15, 9, 7

a. 11
b. 1
c. 9
d. 90

5. Find the median of these set of test scores taken from a class of students – 90, 80, 77, 86, 50, 91, 73, 66, 69, 45, 43, 65, 75

a. 13
b. 73
c. 9
d. 706

6. Find the median of 32, 64, 109, 67 and 3

a. 64
b. 3
c. 275
d. 5

Mode Questions

7. Find the mode from these numbers – 7,2,3,9,6,5,1,4,8

a. 1
b. 5
c. 9
d. None of the above

8. Find the mode from these numbers – 190, 280, 177, 186, 180, 291, 177, 166, 169, 165, 243, 165, 177, 243, 190

a. 190
b. 177
c. 165
d. 243

9. Find the mode from these test results – 2, 4, 2, 6, 4, 9, 6, 7, 2, 9, 7, 6, 4, 10, 10, 2, 6, 7, 9

a. 2 and 9
b. 2
c. 2 and 6
d. 2 and 7

 

 

Answer Key

 

1. D
First add all the numbers 200,000 + 10,020 + 30,000 + 15,000 + 1080 = 256,100. Then divide by 5 (the number of data provided) = 256,100/5 = 51,220

2. C
First add all the numbers 1 + 2 + 3 + 4 + 5 +6 + 7 +8 + 9 + 10 = 55. Then divide by 10 (the number of data provided) = 55/5 = 11

3. A
First add all the numbers 2.5 + 9.5 + 4.5 + 1.25 + 7.05 + 20.8 = 45.6. Then divide by 6 (the number of data provided) = 45.6/6 = 7.6

4. C
First arrange the numbers in a numerical sequence – 1, 3, 7, 7, 9, 9, 11, 13, 15, 15, 21. Next find the middle number. The median = 9

5. B
First arrange the numbers in a numerical sequence – 43, 45, 50, 65, 66, 69, 73, 75, 77, 80, 86, 90, 91. Next find the middle number. The median = 73

6. A
First arrange the numbers in a numerical sequence – 3, 32, 64, 67, 109.  Next find the middle number. The median = 64

7. D
Simply find the most recurring number. All the numbers in the series appeared only once. The answer is No Mode

8. D
Simply find the most recurring number. The most occurring number in the series is 177

9. C
Simply find the most recurring number. The most occurring numbers in the series is 2 and 6

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Geometry Practice Questions

Basic Geometry Practice

 

pythagorean3

Note: Figure not drawn to scale
1. Calculate the length of side x.

a. 6.46
b. 8.48
c. 3.6
d. 6.4

pythagorean

Note: figure not drawn to scale
2. What is the length of each side of the indicated square
above? Assume the 3 shapes around the center triangle
are square.

a. 10
b. 15
c. 20
d. 5

3. Reflect the parallelogram ABCD with the given mirror
line m.

reflection1

4. In a class, there are 8 students who take only dancing lessons. 5 students take both dancing and singing lessons. The number of students taking singing lessons is 1 less than 2 times the number of students taking dancing lessons only. If 4 students take neither of these courses,
how many students are in the class?

a. 19
b. 25
c. 27
d. 42

5. The interior angles of a triangle are given as 2x + 5, 6x
and 3x – 23. Find the supplementary of the largest angle.

a. 640
b. 720
c. 1000
d. 1080

ph17

6. In the right triangle above, |AB| = 2|BC| and |AC| =
15 cm. Find the length of |AB|.

a. 5 cm
b. 10 cm
c. 5√5 cm
d. 6√5 cm

 

perimeterarea3

Note: figure not drawn to scale
7. What is the perimeter of the equilateral ΔABC above?

a. 18 cm
b. 12 cm
c. 27 cm
d. 15 cm

perimeterarea4

Note: figure not drawn to scale
8. What is the perimeter of the above shape, assuming
the bottom portion is square?

a. 22.85 cm
b. 20 π cm
c. 15 π cm
d. 25 π cm

9. What year was Euclidian geometry disproven, and by
whom?

a. Thales – BC 500s
b. Pythagor BC 500s
c. Pierre De Fermat 1600s
d. Nikolai Lobachevsky 1830s

10. Which postulate below disproves Euclidean
geometry?

a. Through any two points, there is exactly one line.
b. If equals are added to equals, the wholes are equal.
c. Parallel postulate
d. Things which coincide with one another are equal to
one another (Reflexive property).

Answer Key 

1. B
In the question, we have a right triangle formed inside the circle. We are asked to find the length of the hypotenuse of this triangle. We can find the other two sides of the triangle by using circle properties:

The diameter of the circle is equal to 12 cm. The legs of the
right triangle are the radii of the circle; so they are 6 cm
long.

Using the Pythagorean Theorem:
(Hypotenuse)2 = (Adjacent Side)2 + (Opposite Side)2
x2 = r2 + r2
x2 = 62 + 62
x2 = 72
x = √72
x = 8.48

2. B
We see that there are three squares forming a right triangle in the middle. Two of the squares have the areas 81 m2 and 144 m2. If we denote their sides a and b respectively:

a2 = 81 and b2 = 144. The length, which is asked, is the hypotenuse; a and b are the opposite and adjacent sides of the right angle. By using the Pythagorean Theorem, we can find the value of the asked side:
Pythagorean Theorem
:
(Hypotenuse)2 = (Opposite Side)2 + (Adjacent Side)2
h2 = a2 + b2
a2 = 81 and b2 = 144 are given. So,
h2 = 81 + 144
h2 = 225
h = 15 m

3.
We reflect points A, B, C and D against the mirror line m at
right angle and we connect the new points A’, B’,C’ and D’.

reflection1a

4. C
The class can be shown by set E. Let us say, set D represents dancing lessons, and set S represents singing lessons. 8 students take only dancing lessons: s(DS) = 8
5 students take both dancing and singing lessons: s(D∩S) = 5

The number of students taking singing lessons:
s(S) = 2 * 8 – 1 = 15
s(S) = s(SD) + s(D∩S) and also s(D) = s(DS) + s(D∩S)
4 students take neither of these courses: s(E) – s(D ∪ S) = 4
s(D ∪ S) can be found by s(DS) + s(S) or s(SD) + s(D)
We are asked to find s(E):
s(E) = 4 + s(DS) + s(S) = 4 + 8 + 15 = 27

5. B
The interior angles of a triangle sum up to 1800:
(2x + 5) + (6x) + (3x – 23) = 180
82 NYSTCE® Mathematics Skill Practice!
11x – 18 = 180
11x = 198
x = 180
The largest angle is 6x = 6 * 18 = 1080
The supplementary of an angle is the angle which plus the
angle gives 1800. Then, the supplementary of 1080 is:
180 – 108 = 720

6. D
In the right triangle above, AB and BC are the legs and AC is the hypotenuse. For side lengths, Pythagorean Theorem is applied:
|AB|2 + |BC|2 = |AC|2
Let us say that |BC| = x. Then, |AB| = 2x:
(2x)2 + x2 = 152
5x2 = 225
x2 = 45
x = √45 = 3√5 cm
|AB| = 2x → |AB| = 6√5 cm

7. C
The perimeter of an equilateral triangle with 9 cm. sides will be
= 9 + 9 + 9 = 27 cm.

8. A
The question is to find the perimeter of a shape made by merging a square and a semi circle. Perimeter = 3 sides of the square + ½ circumference of the circle.
= (3 x 5) + ½(5 π)
= 15 + 2.5 π
Perimeter = 22.85 cm

9. D
Euclidian geometry was disproven by Nikolai Lobachevsky in 1830s.

10. C
Euclidian geometry supports parallel geometry. On the contrary, Non-Euclidian geometry is the study of geometry with curved spaces that is elliptic and hyperbolic geometry. In elliptic geometry; the inner angles of a triangle do not sum up to 1800; the sum is equal to 1800 plus the area of
the triangle. In hyperbolic geometry; the sum is equal to 1800 minus the area of the triangle. non-Euclidian geometry inspires from the shape of the world: If two meridians are selected; both intersect with the equator by 900. There is also a vertex angle in the pole. So, the inner angles sum up to 90 + 90 + vertex pole which is higher than 1800.  Another example; a person walks 10 m south, 10 m west and then 10 m north. He sees that he is where he started moving. Normally, we would say that he would be 10 m east from the starting point. Think that he is on the North Pole.
If he goes 10 m down, 10 m west and then 10 m north, he is
again on the pole.

 

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Operations with Polynomials

Adding Polynomials

1. (x3 + 5x + 3x2 +2) + (4x3 + 3x2+ 14)

a. 5x+ 9x+ 16
b. 4x3+6x2+x+14
c. 5x3+6x2+5x+16

 

2. (2x3+ 5x4 + 3x2 +12) + (7x3 + 4x2+ 3)

a. 7x3+5x2+15
b. 5x3+5x4+12+x2
c. 5x4+9x3+7x2+15

 

Subtracting Polynomials

3. (-4x2 – 5y2 – 8) – (2x2 + 3y2 + 8)

a. -4x2-5y2+8
b. -4x2-4y2-16
c. -6x2-8y2-16

 

4. (x2 + 4x) + (x2 + 8x) – (3x – 7)

a. 2x2+9x+7
b. X2-3x+7
c. 4x2-12x-7

 

5. Simplify (2x 2 + 3x)(5x – 7)

a. 10x3-14x2-21x
b. 10x3+x2-21x
c. 10x2+15x-21

 

6. Simplify (- 3x 3 +3)(- 4x 2 – 6)

a. -12x5+12x2-18
b. 12x5+18x3-12x2-18
c. 12x5+18x4-x2-18

 

Answer Key

 

  1. C
  2. C
  3. C
  4. A
  5. B
  6. B

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Study Guides and Practice Tests

Teacher Certification Study Guides and Practice Tests

 

Arizona Teacher Certification (AEPA)
California Basic Educational Skills Test (CBEST)California Teacher Of English Learners (CTEL)
Florida Teacher Certification Exam (FTCE) 
Georgia Assessment for Educators (GACE)
Illinois Certification Testing System (ICTS)
Massachusetts Test for Educator Licensure (MTEL)
Oklahoma Educators (CEOE/OSAT)