The Secrets to Math Exam Success, Summaries of Mathematics

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THE SECRETS TO

MATH EXAM SUCCESS

BY

DANIEL O. AMAIZE

With

[Standard Math quiz questions with solutions]

INTRODUCTION

Mathematics is not a subject you can pass just by reading through the texts. You must practice problem solving on a regular basis. And you must be acquainted with the requisite formulas to solve problems in Mathematics.

Each time I write concerning Math studying tips, I remember my high school mate and friend John Makun. John hated Math and found Math problems as the toughest challenge he will ever face in life. Laughable right? Yeah! John later studied Math in School and he graduated with an upper grade. Because I told him he must fall in love with Math to excel in it, which he did. That is one of the problems that has led many students to fail Mathematics. The lack of interest in the subject kills every zeal to learn Math. And once the willingness to learn is not there, learning cannot take place. So in this write up, am going to highlight some of the problems hindering students from comprehending mathematics. Am also going to outline some guidelines on how to answer Math exam questions without stress.

Follow the guidelines in this booklet if you want to to become outstanding in your next Mathematics exams.

REASONS WHY STUDENTS FIND IT HARD TO COMPREHEND MATH

In the introductory section, I highlighted a few of the problems/hindrances to comprehending Math. Here, we are going to look at the problems in details and also proffer solutions to them.

1. A Lack of Interest And motivation: most times students find it difficult to comprehend subjects in general, the underlying problem could be a lack of interest on the part of the student. Many a time, I have seen students lose interest in a subject because the subject teacher failed to motivate the students. When a serious student finds him/herself in such a situation, the only way to succeed in such scenario is to find a way to motivate self. When I was preparing for my high school final exams, we didn’t have a good Mathematics teacher. But we (me and my Math crew) did not allow that to hinder our successes. We found a way to motivate ourselves. There were others (our peers) who used that as an excuse not to prepare very well for the final exams, because they had a lack of interest in the subject. So before you can succeed in Mathematics, your interest in the subject must be stronger than your excuses. You must find a way to motivate yourself when your school tutors fail to do their jobs.

To develop your interest and motivation in Math, you need to spend more of your leisure time in solving Math problems. Because the more problems you solve in Mathematics, the more expertise you acquire, and the more interest you will develop in the subject. Also, join the Math club in your school and make sure to participate in all Math activities. Finally, try to watch Math quiz Tv shows and familiarize yourself with the Math quiz apps.

2. Laziness on the part of students: the lazy minds always want to do only the easy or expedient things. They always want to shy away from the hard things. The truth is that there is no hiding place for lazy students in Mathematics. If you must succed in Math, you must be ready to work hard. Working hard means solving many Math problems. Because the more problems you solve, the more experience you have in solving difficult problems in Mathematics. So never you approach Mathematics with a lazy attitude. A student once asked me “how many questions do I need to solve to become good in Mathematics?.” Well my reply was simple; “You must solve at least five Math problems everyday.” Because solving five problems everyday means solving 35 problems in a week and at least 140 problems in a month. With this, you will soon become acquainted with most patterns in Math problems. This is how you can become an expert in solving Math problems.
3. Distractions: distraction is another hindrance to understanding Math. Learning Math requires focus and a lot of attention. One of the reasons students fail Mathematics is because they allow other trivial activities to distort their focus. So to become an expert in solving Math problems, you must avoid every form of distractions. Always look for convenient serene environments to study and solve problems. Avoid doing two things at the same time; solving Math and eating or solving Math and watching TV. Just avoid the distractions and you will be fine. You will be focused and you will be able to give Math all the attention she deserves.
4. Peer Influence: another reason why students fail to comprehend Math is because their peers or friends simply don’t like the subject. Most students want to do whatever their friends love and also detest whatever they hate. Many students have ended up choosing the wrong careers because of peer influence. It is not totally wrong to allow your peers to influence your academic decisions, but let such influences be for good. So what can you do when your closest friend hates to solve Math? First, you must understand that your academic/career destiny is not tied to your best friends. Yes!.... two friends can succeed in entirely different careers and many have so succeeded. So don’t allow your friend to be a source of discouragement when you actually need to understand Mathematics. Make them understand that you need Math to go further in building your career and your future. If your friends so love you, they will motivate and

encourage you to excel in your academics whether they love your choice of subjects/career or not. Secondly, you can be a source of encouragement and motivate your friends in order not to get discouraged by them. So if you don’t want your peers to discourage you, start by motivating them. Set up Math quizes with rewards for the best scorers. Also, you can assist your peer with some difficult problems and they can do same for you. Yes, the saying “When the two hands wash one another, both hands will become clean” is very true. Take a good advantage of it.

5. Method of Teaching: so far, we have discussed reasons partaining to the students. Now let us dive in to a reason why students find it hard to comprehend Math, not necessarily because of the students’ direct fault but due to the method of teaching used by the teacher. Modern teaching methods are directed towards learner-centred pedagogy. The learner should be the centre of attraction and not the teacher. This ensures that every student is carried along. Involving the students in the learning process also helps the teacher to easily identify the fast and slow learners. Because, both fast and slow learners do not learn at the same pace and may not require the use of the same method of teaching. So an inappropriate method of teaching can lead to students not comprehending Math. When such slow learners are identified, the teacher is adviced to apply a different method of teaching. This will foster their understanding and comprehension.
6. Learning Dificulties: another obvious reason why students don’t comprehend Math is because of their learning difficulties. This is applicable to special students or students with impaired learning. Such students require more attention, patience and perseverance on the part of the teacher. A good teacher should be able to identify such special students in a class after spending some good time with his students. Once the special students are identified, they should be isolated from the class, and accorded more attention with better suited teaching methods.
7. Lack of Patience: solving problems generally requires patience in order to find a good solution. Solving problems in Math is not so different from the general patterns. So why not exercise some reasonable level of patience when solving problems in Math? Some Math problems may require an instant straight-forward solution, but there are others which may quite take some time. Students who are not patient enough will be put off easily by such technical problems. So if you think you can sometimes be very impatient, don’t be when solving Math. Impatience can even make you to use the wrong values or apply the wrong formula to solve an otherwise simple problem in Math. So you must learn to be very patient and diligent when solving problems in Mathematics.

HOW TO PREPARE FOR A MATH TEST/EXAM

You must be prepared before taking a test/quiz or writing an exam. There is a saying “He who fails to plan, plans to fail.” If you must pass your next Math quiz/exam, then preparation (planning) is key. The following are some preparation tips which will aid you to pass your next Math test/exam:

1. Be Very Attentive in Class: one of the reasons students fail Math tests/exams is that they are not always attentive in Math class. Math requires much focus and attention which most students don’t often realize. So you’ve got to be very attentive in Math classes in order to grab most of the basic principles which will aid you to pass your tests and exams. Being attentive in Math classes is the first preparation you need to make.
2. Have Your Private Study Regularly: as a preparation for your next Math exam, make sure to study your Math textbooks regularly. Do your Math assignments and try to solve other exercises from the textbooks on your own. Just make sure you solve a few Math problems on a daily basis. Also, try to increase the number of problems you can solve as a measure of your progress. For example, you could improve from solving three problems a day to solving five a day the next week and so on. 3. Make good summary sheets from your study and problems solved: during your study, do well to make a list of important formulas, theorems, rules and laws in Math. Also, make a list of the course objectives for each topic and do well to review the list on a daily basis.
3. Practice make perfect: do well to practice the problems you encounter on daily basis until you master the ability to solve them, and do well to verify the solutions. This will give you a good understanding of the question patterns and mode of solution to each of the problems.
4. Know Your Strengths and Weaknesses: do not be afraid of your limits, but know your limits and avoid them. If you must pass your next Math exams, you must know your areas of strength and weaknesses. Develop your areas of strength and put more efforts in improving your areas of weakness.
5. Give more preference to Math: if you must become flawless in solving Math problems, then this should become your priority. Give more attention to Math by spending more time in solving problems in Math compared to other subjects.

7. Take Study Breaks: you should not study Math without taking a break. You cannot afford to overwork your brain without giving it time for a little rest. “All work and no play they say makes Jack dull.” Always give the brain some time off to recharge and assimilate what has been committed to it. During your study breaks, take a walk and some refreshments. You can go back to the difficult problems you could not solve previously after the break.
8. Try Using Available Campus Resources: there are many Math resources available on campus. You can access free online Math tutoring on the Academic support centres. Also, do well to leverage on the Stem Learning Centres and other study groups.
9. Study in a Convenient Environment: distraction is an arc enemy of Math study. If you want your study to be efficient and effective, do well to study in a serene and convenient environment. Avoid noise and every form of distractions during Math study.
10. Test Yourself with peers and online quiz: do not be a master of yourelf but try to compare and contrast your knowledge with your peers. Compete in math quizes in school and in online quiz platforms like Typeform, Wirisquizes and others.

REASONS WHY STUDENTS MAY STILL FAIL MATH EXAMS AFTER PREPARING

WELL FOR THE EXAMS

First things first; before we look into the tips (the dos) on how to pass Math quizzes and exams, let us look at some practices to avoid (the don’ts). Please avoid these pitfalls because they will make you to lose marks and also reduce your chances of passing your next Math test/exams.

1. Failure to Read the Instructions: one of the main reasons many students fail Math quizes/exams is because they don’t read the instructions. It doesn’t matter whether this is intentional or just an oversight, you may fail your next Math quiz or exam if you do not pay attention to the instructions given before attempting to answer the questions. The instructions are the guidelines (more like a manual). Imagine you bought an entirely new device without looking at the manual. You might damage the device if you attempt to operate it with reading through the manual. So for no reason should you fail to read the instructions before answering the questions in your next Math quiz/exam.
2. Failure to Understand the Language of Math: more like other subjects, there is mode of communication in Mathematics. Be sure to understand mathematical statements and sentences. This will help you to understand the questions asked and the required answers tto the questions.

3. Illegible Handwriting: when your handwriting is not clear enough to be read, then the examiner is in trouble. He cannot mark what he cannot read. And it is very easy for an examiner to mark your work wrong if your handwriting is not clear enough to be read. So make sure to work on your handwriting before your next Math quiz/exam. Also, it is good practice to always enclose your final answer in a box to make it obvious and very visible.
4. Time Consciousness: another reason why students fail Math tests/exams is that they are not always conscious of the time allocated for the exam. The purpose of exams is to test the student’s speed and accuracy in applying the principles learnt. So if you must pass your next Math exams, then you have to be very conscious of the time allowed for the exams. So in preparation, test your speed by answering a number of questions in a given period of time.

STANDARD MATH QUESTIONS AND ANSWERS

[UNIVERSITY OF BENIN-NIGERIA]

1. Solve for a if 27a^ x 92(a + 3)^ = (3a)^4

Solution

This is a question in indices

Start by expressing each term the smallest possible base (3):

That is; (3^3 )a^ x (3^2 )2(a + 3)^ = (3a)^4

Simplify all brackets;

3 3a^ x 34a + 12^ = 34a

Apply the multiplication law of indices on the left hand side

3 (3a + 4a + 12)^ = 3(4a)

Since both bases are equal, the indice are equal. Thus ;

3a + 4a + 12 = 4a

3a + 4a – 4a = -

3a = -

a = -12/

a = -

2. Solve for x if log(x + 3) – 2log(x - 3) = log

Solution

Apply the power law: log(x + 3) – log(x – 3)^2 = log

Apply the addition law: log(

𝑥+ (𝑥−3)^2 )^ = log Since both bases are equal; (

𝑥+ (𝑥−3)^2 )^ = 2

x + 3 = 2(x – 3)^2

open the bracket x + 3 = 2(x^2 – 6x + 9)

x + 3 = 2x^2 – 12x + 18

2x^2 - 12x – x + 18 – 3 = 0

2x^2 - 13x + 15 = 0

By factorization method;

2x^2 – 10x – 3x + 15 = 0

By grouping terms; (2x^2 – 10x) – (3x + 15) = 0

Factorize each bracket 2x (x – 5) – 3(x - 5) = 0

(2x – 3)(x – 5) = 0

Either 2x – 3 = 0 or x – 5 = 0

Therefore x = 3/2 or 5

3. Calculate the rate that will generate a compound interest of $350 from a

principal of $2800 for a period of 3 years. [assume compound interest is

calculated once; at the end of the year.]

Solution

Future value, FV = Compound Interest + Principal (Initial value)

FV = 2800 + 350 = $

FV = P(1+R/N) ^ (NT)

Where: T = time = 3 years, P = principal = $ 2800 , R = rate =?, N = number of times

compound interest is calculated = 1

3150 = 2800(1+ 𝑅 ⁄ 1 ) ^ (1 * 3)

3150 = 2800(1 + R) 3

⁄ 2800 = (1 + R)

3

√^3150 ⁄ 2800

3

= 1 + R

3 √1.125 = 1 + R

1.04 = 1 + R

R = 1.04 -1 = 0.

Therefore R = 4%

4. Given that 134y = 1011100two, find the value of y.

Solution

This is a number base problem.

134 y = 1011100two

Convert both sides to base 10

1 * y^2 + 3 * y^1 + 4 * y^0 = 1 * 2^6 + 0 * 2^5 + 1 * 2^4 + 1* 2^3 + 1 * 2^2 + 0 * 2^1 + 0

* 2^0

y^2 + 3y + 4 = 64 + 0 + 16 + 8 + 4 + 0 + 0

y^2 + 3y + 4 = 92

y^2 + 3y + 4 – 92 = 0

y^2 + 3y - 88 = 0

by factorization; y^2 - 8y + 11y – 88 = 0

by grouping like terms; (y^2 - 8y) + (11y – 88) = 0

factorize each brackets ; y(y – 8) + 11(y – 8) = 0

(y - 8)(y + 11) = 0

Either y – 8 = 0 or y + 11 = 0

Thus; y = 8 or -

Therefore, y = 8 (since a number base cannot be negative)

5. Find the quadratic equation whose roots are such that one of the roots is 13

while the other is two-third the value of the first root.

Solution

Let the two roots be α and β. Then, β = 23 α (because one root is two-thirds the

other root)

So if α = 13 , then β = 23 * 13 = 29

The general format of a quadratic equation is;

x^2 – (sum of roots)x + product or roots = 0

x^2 – ( 13 + 29 )x + (^13 * 29 ) = 0

x^2 –(

3 + 2

9 ) + (^

2

27 ) = 0

x^2 – (

5

9 )x + (^

2

27 ) = 0

27x^2 - 15x + 2 = 0

6. Find the roots of the polynomial x^3 – 8x^2 + 19x – 12 = 0.

Solution

Test for factors of x^3 – 8x^2 + 19x – 12.

Let f(x) = x^3 – 8x^2 + 19x – 12

F(1) = (1)^3 – 8(1)^2 + 19(1) – 12 = 1 – 8 + 19 + 12 = 0

Thus, x – 1 is a factor of x^3 – 8x^2 + 19x – 12

We can get the other factors by dividing x^3 – 8x^2 + 19x – 12 by x – 1 as

follows: x^2 - 7x + 12

x – 1 x^3 – 8x^2 + 19x – 12 -(x^3 – x^2

• 7x^2 + 19x – 12

• (7x^2 + 7x
  • (12x – 12

-(12x – 12

Thus; x^3 – 8x^2 + 19x – 12 = (x – 1) (x^2 - 7x + 12 )

x^2 - 7x + 12 = x^2 - 3x – 4x + 12 = (x^2 - 3x) – (4x + 12) = x(x - 3) – 4(x - 3) = (x - 3) (x - 4) Thus; x^3 – 8x^2 + 19x – 12 = (x – 1) (x - 3) (x - 4) If (x – 1) (x - 3) (x - 4) = 0, Either x – 1 = 0, x – 3 = 0 or x – 4 = 0 Therefore, x = 1, 3 or 4

7. Find the equation of the straightline passing through point P(3, 4) and

perpendicular to the line 4x + 3y + 6 = 0.

Solution

If 4x + 3y + 6 = 0

y = − 4𝑥 ⁄ 3 – 2

the slope of this line is − 4 ⁄ 3. So any line perpendicular to this should have a slope of 3 ⁄ 4. Therefore the equation of the line passing through point (3, 4) and perpendicular to this line should be: (y – 4) = 3 ⁄ 4 (𝑥 − 3) 4(y – 4) =3(x – 3) 4y – 16 = 3x – 9 Therefore, the equation is 4y – 3x – 7 = 0

8. Given that f(t) = 2t^4 – 10t-3^ + 8y^2 + 12, find f^1 (-2).

Solution

First differentiate f(t)

f(t) = 2t^4 – 10t-3^ + 8y^2 + 12

f^1 (t) =

𝑑𝑦

𝑑𝑡 = 8t

3 + 30t-4 + 16t = 8t 3 + 30

𝑡^4 + 16t

substitute t = -

f^1 (-2) = 8(-2)^3 + (−2)^304 + 16(-2)

f^1 (-2) = 8(-8) +

30

16 + 16(-2)

f^1 (-2) = -64 +

15

8 + -

therefore f^1 (-2) = -

753

8 or - 94

1 8

9. A piece of land is in the shape of the curved surface of a cone. Calculate the curved surface area of the land if the cone has a base diameter of 24units and a vertical height of 5units. [π = 3.142] Solution

r = d/2 = 24/2 = 12

h = 5

l = √𝑟^2 + ℎ^2 = √12^2 + 5^2 = √144 + 25 = √169 = 13

curved surface area of a cone = πrl = 3.142 * 12 * 13

Therefore, the curved surface area = 490.088squ. units

10. If 8 workmen could do a job in 40 days and they all work at the same rate, how

many workmen are required to do a job thrice larger than the first in 10 days?

Solution

If 8 workmen could do a job in 40 days

If the job is three times large, 1 workmen will be able to do the job in:

8 * 40 days = 320 days

Number of workmen required to do the job in 10 days = 32010 = 32

If the job is thrice larger than the first, then the number of workmen required

= 32 x 3

Therefore, the number of workmen required = 96 workmen

11. The annual percentage increase in population of a town is 5.5% of her initial

population. Calculate the population density of the town four years after her initial

population was 1,200,000 people. [The land area measures 100 by 64 acres]

Solution

Area of land = 100 * 64 = 6400 acres.sq.

Population after the first year =

100 * 1,200,000 = 1,266,000 people

Population after the second year =

100 * 1,266,000 = 1,335,630 people

Population after the third year =

100 * 1,335,630 = 1,409,090 people

Population after the fourth year = 105.5 100 * 1,409,090 = 1,486,590 people

Population density =

𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛

𝑎𝑟𝑒𝑎 =^

1,486, 6400

Therefore, the population density = 232 people per sq. acres

12. Given that 3:x = 2:4 = 5:y, find the value of 3𝑥

(^2) +𝑦 𝑦

Solution

3:x = 2:4 = 5:y

from the above equations, 3:x = 2:4 or 3 𝑥 = 24

that is; y =

3 ∗ 4

2 =^6

similarly, 2:4 = 5:y or

2

4 =^

5 𝑦

y =

4 ∗ 5 2 =^10

3𝑥^2 +𝑦

𝑦 =^

3(6^2 )+

10 =^

3(36)+

10 =^

108 + 10

10 =^

128 10

Therefore

3𝑥^2 +𝑦

𝑦 =^ 12.

13. given that μ = {x: x is an integer and 0 < x ≤ 15}

A = {x: x is a prime number 0 < x ≤ 15}

B = {x: x is an odd number 0 < x ≤ 15} C = {x: x is a even number 0 < x ≤ 15}

Determine ((AnB) U (AnC))^1

Solution

A = {2, 3, 5, 7, 11, 13}

B = {1, 3, 5, 7, 9, 11, 13, 15}

C = {2, 4, 6, 8, 10, 12, 14}

AnB = {3, 5, 7, 11, 13}

AnB = {2}

(AnB) U (AnC) = {2, 3, 5, 7, 11, 13}

Therefore ((AnB) U (AnC))^1 = {1, 4, 6, 8, 9, 10, 12, 14, 16}

14. In a village of 120 people, the three most read News papers are the Dailys, Guardian and the Tribune. In a survey carried out from in 2022, it was observed that 40 people read the Daily, 50

10

read the Guardian, 30 read the Tribune while 10 people read all three news papers. If 15 people read the Guardian and the Daily news papers, 12 read Tribune and the Guardian while 18 people read the Dailys and Tribune, how many people read at least two news papers?

Solution D = 40 G = 50 U = 120 17 5 33 8 2 10 T = 30

From the Venn Diagram above:

Daily Newspaper only = 40 – (8 + 5 + 10) = 40 – 23 = 17

Gaurdian Newspaper only = 50 – (5 + 10 + 2) = 50 – 17 = 33

Tribune Newspaper only = 30 – (8 + 10 + 2) = 30 – 20 = 10

At least two Newspapers means that: either they read all three or any two Newspapers

= DnGnT or DnG or DnT or GnT

= 10 + 8 + 5 + 2

Therefore number of people reading at least two Newspapers = 25 people

15. The number of books n, sold by a bookstore is partly constant and partly varies inversely as the price p, of a book. 80 books were sold when the price of a book was $500 and 50 books were sold when the price of a book was $700. Find the number of books sold when the price of a book is $100.

Solution

n α k + (^1) 𝑝

n = k + 𝑐𝑝 (where k and c are constants)

when n = 80 and p = $500;

80 = k + 500 𝑐

500 * 80 = 500 * k + 500 * 500 𝑐

40000 = 500k + c ------------------------------- (1)

When n = 50 and p = 700;

50 = k + 700 𝑐

50 * 700 = 700 * k + 700 * 700 𝑐

35000 = 700k + c ------------------------------- (2)

Subtracting (2) from (1) gives

5000 = - 200k

k = 5000/-

k = -

from (1):

40000 = 500 * -25 + c

40000 = -12500 + c

c = 40000 + 12500

c = 52500

n = (^52500) 𝑝 – 25 (relationship)

when p = 100

n = 52500100 – 25

n = 525 – 25

Therefore n = 500

16. Find the inverse of the Matrix below.

A =

2 −3 1

−2 1 4

Solution

det (A) = 2(12 – 5) - -3(4 - - 10) + 1(1 - - 6)

= 2(7) + 3 (14) + 1(7) = 14 + 42 + 7

det (A) = 63

AT^ =

2 1 −

−3 3 1

det (Ai) = |^35 14 | | −3 1 14 | | −3 1 35 |

| 15 −2 4 | | 21 −2 4 | | 21 15 |

| −1 3 21 | | −3^2 −2 1 | | −3^2 13 |

det (Ai) =

7 −13 18

adj (A) =

7 −13 18

*

+ − +

− + −

+ − +

adj (A) =

−14 10 −

7 −8 9

A-1^ = (^) det(𝐴)^1 * adj (A) = 631

−14 10 −

7 −8 9

Therefore, A-1^ =

1 9

13 63

2 7 − 63

10 63

− 7 1 9

− 63

1 7

17. Calculate the standard deviation of the following scores: 10, 12, 18, 14, 16, 20.

Solution

Mean (m) =

10 + 12 + 18 + 16 +14 + 20 6 = 15 S/N x (m = 15 ) x - m (x – m)^2 1 10 - 5 25 2 12 - 3 9 3 18 3 9 4 16 1 1 5 14 - 1 1 6 20 5 25 n = 6 ∑(x – m)^2 = 70

Standard deviation = √∑(𝐱 – 𝐦)𝟐𝑛 = √𝟕𝟎 6

Standard deviation = 3.

18. A bag contains 8 red balls, 6 white balls and 10 blue balls. If two balls are picked from the bag at random, one after the other without replacement, what is the probability that both balls are of different color?

Solution

The three probabilities are: (R, W) or (R, B) or (W, B)

Total no. of balls = 6 + 8 + 10 = 24

Pr (R) = 𝟖 24 =^

𝟏 3

Pr (W) =

𝟔 24 =^

𝟏 4

Pr (B) = 𝟏𝟎 24 =^

𝟓 12

Pr (different colors) = (

𝟏 3 *^

𝟔 23 ) + (^

𝟏 3 *^

𝟏𝟎 23 ) + (^

𝟏 4 *^

𝟏𝟎 23 )

= 𝟐 46 +^

𝟏𝟎 69 +^

𝟓 46

Pr (different colors) = 𝟔 + 𝟐𝟎 + 𝟏𝟓 138

Pr (different colors) =

𝟒𝟏 138