This is the thread for NJTC mathematics tutorial. Check out for the timetable lecture starts today by 9am and ends by 3pm.
TUTORS
Dejt4u
raayah
logoDwhiz
goofyone.
Thanks. And be orderly. BEST OF LUCKS.

Ok..
Ladies and gentlemen...
On behalf of the tutor-generals here, I welcome you to this educative thread..

To start with, I shall post all the topics you are expected to know very well in mathematics before you walk in to your exam center..

The next post shall therefore contain the syllabus in sequential order..
Note that the co-ordinator of this UTME- maths thread is Goofyone
Keep following

Okay let's have the questions

SECTION I: NUMBER AND
NUMERATION.
1. Number bases:
(a) operations in different number bases from 2 to 10;
(b) conversion from one base to another including fractional parts.

2. Fractions, Decimals, Approximations
and Percentages:
(a) fractions and decimals
(b) significant figures
(c) decimal places
(d) percentage errors
(e) simple interest
(f) profit and loss per cent
(g) ratio, proportion and rate

3. Indices, Logarithms and Surds:
(a) laws of indices
(b) standard form
(c) laws of logarithm
(d) logarithm of any positive number to a given base.
(e) change of bases in logarithm and application.
(f) relationship between indices and logarithm
(g) surds

4. Sets:
(a) types of sets
(b) algebra of sets
(c) venn diagrams and their applications.

Lets roll

SECTION II: ALGEBRA
1. Polynomials:
(a) change of subject of formula
(b) factor and remainder theorems
(c) factorization of polynomials of degree not exceeding 3.
(d) multiplication and division of polynomials
(e) roots of polynomials not exceeding degree 3
(f) simultaneous equations including one linear, one quadratic
(g) graphs of polynomials of degree not greater than 3

2. Variation:
(a) direct
(b) inverse
(c) joint
(d) partial
(e) percentage increase and decrease.

3. Inequalities:
(a) analytical and graphical solutions of linear inequalities.
(b) quadratic inequalities with integral roots only.

4. Progression:
(a) nth term of a progression
(b) sum of A. P. and G. P.

5. Binary Operations:
(a) properties of closure, commutativity, associativity and distributivity.
(b) identity and inverse elements.

6. Matrices and Determinants:
(a) algebra of matrices not exceeding 3 x 3.
(b) determinants of matrices not exceeding 3 x 3.
(c) inverses of 2 x 2 matrices [excluding quadratic and higher degree equations].

SECTION III: GEOMETRIC AND
TRIGONOMETRY
1. Euclidean Geometry:
(a) angles and lines
(b) polygon; triangles, quadrilaterals and
general polygon.
(c) circles, angle properties, cyclic,
quadrilaterals and intersecting chords.
(d) construction.

2. Mensuration:
(a) lengths and areas of plane geometrical
figures.
(b) length s of arcs and chords of a circle.
(c) areas of sectors and segments of circles.
(d) surface areas and volumes of simple
solids and composite figures.
(e) the earth as a sphere, longitudes and
latitudes

3. Loci:
locus in 2 dimensions based on geometric
principles relating to lines and curves.

4. Coordinate Geometry:
(a) midpoint and gradient of a line segment.
(b) distance between two points.
(c) parallel and perpendicular lines
(d) equations of straight lines.

5. Trigonometry:
(a) trigonometric ratios of angels.
(b) angles of elevation and depression and
bearing.
(c) areas and solutions of triangle
(d) graphs of sine and cosine
(e) sine and cosine formulae.

SECTION IV: CALCULUS
1. Differentiation:
(a) limit of a function;
(b) differentiation of explicit algebraic and
simple trigonometric functions – sine, cosine
and tangent.

2. Application of differentiation:
(a) rate of change
(b) maxima and minima

3. Integration:
(a) integration of explicit algebraic and simple trigonometric functions.
(a) area under the curve.

SECTION V: STATISTICS
1. Representation of data:
(a) frequency distribution
(b) histogram, bar chart and pie chart.

2. Measures of Location:
(a) mean, mode and median of ungrouped and grouped data – (simple cases only)
(b) cumulative frequency

3. Measures of Dispersion:
range, mean deviation, variance and standard deviation.

4. Permutation and Combination

5. Probability

TUTORS
Dejt4u
raayah
logoDwhiz
goofyone.

The tutors above shall take the topic one by one.. Starting from section one..

Methodology:
the topic will be introduced here and some necessary concepts and tips shall be explained..thereafter, some example will be solved wit 'unique', 'very short' and 'very simple' methods by all the available tutors.. Tutors shall provide his/her example and he/she shall provide solutions accordingly..

There will be room for questions and observations from the viewing students and other observers..

The class proper shall begin shortly..
Stay tuned

Before we move one..any observation or correction on d above guidelines??

dejt4u:
The class proper shall begin shortly..
Stay tuned

okay sir! Cc: Mizmycoli, fynestboi, seun , obinoscopy, oma4u. Please can you just help us to place this thread on the front page to bring in students to this class? Urgently please.......

...following...

MizMyColi:
Sweerie

I'm not a mod, how much more super mod.

imo
Just continue with what you doing, okay?
This is a laudable one and in due time, it would be moved to FP.

okay! Thanks alot.

From the time table released, maths is only 3times a week which is equivalent to 9hours and we have only 13weeks.. So we need to be moving at the speed of light.. We will just be brief on some simple topics..

My co-tutors, where art thou!
Pls move in here nd lets bring this our first topic down in a moment..lol

walks in wit my writing materials....

dejt4u:
ladies and gentlemen, we are now starting with Section1 Subsection 1

SECTION I: NUMBER AND
NUMERATION.
1. Number bases:
(a) operations in different number bases from 2 to 10;
(b) conversion from one base to another including fractional parts.

I believe everybody is familiar with this elementary maths topic..so I dnt think we shld waste any time here at all..
In case you got any question here.. Simply ask so that our generals can solve it for you.. Thanks

Some questions on the topic shall be posted shortly

I am so sorry that I the class couldnt continue in d morning.. My phone battery was down nd it suddenly went off..

There will be make up class later

Following....
Ayamlaykorn

dejt4u:
I am so sorry that I the class couldnt continue in d morning.. My phone battery was down nd it suddenly went off..

There will be make up class later

ok thanks

dejt4u:

I believe everybody is familiar with this elementary maths topic..so I dnt think we shld waste any time here at all..
In case you got any question here.. Simply ask so that our generals can solve it for you.. Thanks

Express 473.5625 base 10 in binary

Superstar007:


Express 473.5625 base 10 in binary

.
2 | 473
2 | 236 r 1
2 | 118 r 0
2 | 59 r 0
2 | 29. r 1
2 | 14. r 1
2 | 7. r 0
2 | 3. r 1
2 | 1. r 1
2 | 0. r 1
For the whole part (ie the part before the decimal point), we have 111011001 base two.
Then taking the part after the decimal point,
0.5625
1). U multiply it by 2
then we have 0.5625 × 2= 1.125
U take the number before the decimal point and then we have 0.1

2). Taking 1.125, u ignore the figure before the decimal point and multiply it by two,then we have;
0.125 × 2 = 0.25
U take the number before d decimal point (that's 0) and join it with 0.1 we got in step 1,we have 0.10

3). Taking 0.25,as usual,ignore the number before the decimal point,in dis case,it is zero,so its d same thing,then we have;
0.25 × 2 = 0.50
U take the number before the decimal (that's 0) and join it with 0.10 we had in step 2,we now have 0.100

4). Taking 0.5
0.50 × 2 = 1.0
U take the number before the decimal (that's 1) and join it with 0.100 we had in step 3,we now have 0.1001.

As this point,u cant go further cos the number after d decimal point is 0.
The answer will be 111011001.1001 base2. Goodluck.If its wrong,don't hesitate to inform me plz.

This is beautiful. I am learning.

you guys are good.

2muchopoTBdope:
.
2 | 473
2 | 236 r 1
2 | 118 r 0
2 | 59 r 0
2 | 29. r 1
2 | 14. r 1
2 | 7. r 0
2 | 3. r 1
2 | 1. r 1
2 | 0. r 1
For the whole part (ie the part before the decimal point), we have 111011001 base two.
Then taking the part after the decimal point,
0.5625
1). U multiply it by 2
then we have 0.5625 × 2= 1.125
U take the number before the decimal point and then we have 0.1

2). Taking 1.125, u ignore the figure before the decimal point and multiply it by two,then we have;
0.125 × 2 = 0.25
U take the number before d decimal point (that's 0) and join it with 0.1 we got in step 1,we have 0.10

3). Taking 0.25,as usual,ignore the number before the decimal point,in dis case,it is zero,so its d same thing,then we have;
0.25 × 2 = 0.50
U take the number before the decimal (that's 0) and join it with 0.10 we had in step 2,we now have 0.100

4). Taking 0.5
0.50 × 2 = 1.0
U take the number before the decimal (that's 1) and join it with 0.100 we had in step 3,we now have 0.1001.

As this point,u cant go further cos the number after d decimal point is 0.
The answer will be 111011001.1001 base2. Goodluck.If its wrong,don't hesitate to inform me plz.

You are correct although your method is kinda cumbersome. I think I prefer this method:
firstly, convert the decimal part to fraction I.e 5625/1000 then reduce it to its lowest term to have 9/6. So, the figure is now 473.9/16(ignore the dot, that's a mixed fraction)

2) solve for 473

2 | 473
2 | 236 r 1
2 | 118 r 0
2 | 59 r 0
2 | 29. r 1
2 | 14. r 1
2 | 7. r 0
2 | 3. r 1
2 | 1. r 1
2 | 0. r 1
I.e 111011001 base two

3) solving 9/16, we find the multiples of 16 to have 8, 4, 2 and 1. We then make 16 the denominator of these figures I.e 8/16, 4/16, 2/16, 1/16. To get 9/16 we eliminate 4/16 and 2/16. So, we are left with 8/16 and 1/16 which we can add to get 9/16. After reducing it to the lowest term, we have 1/2 and 1/16. Using indices to keep them in a common base, 1/2^1 and 1/2^4.

4. Converting to base two, we have
1*2^-1 + 0*2^-2 + 0*2^-3 + 1*2^-4
Note: since 4/16 and 2/16 were eliminated that's why they are multiplied by 0.

5) Therefore, 9/6 = .1001two
473.5625 = (473)ten + (9/6)ten
111011001two +. 1001two
= 111011001.1001 base two.

Ask any question if you have one.

guys, I'm sorry for not being around. My timezone at the moment is not very favourable.
Nevertheless, I'll go about espousing my little simple methods of solving JAMB questions here. Hopefully, someone will find it useful.
Since we are treating number bases, let me start with a few number bases examples.

Please take caution as you go through these solutions. You are supposed to learn from them, not to cram them.

First question:

What is the answer when 2434 base 6 is divided by 42 base 6?

A. 23 base 6
B. 35 base 6
C. 52 base 6
D. 55 base 6

To solve a question like this normally and fast, you'd have to first convert 2434 base 6 to base 10, then also convert 42 base 6 to base 10, divide your answers, and then convert back to base 6.

But there is a faster way to go around it, and I'll show you.

You see numbers in a base, say 6 for example, will not only have digits less than that number. Because of this, a number in a certain base is less than the same number in a higher base. In this case, 2434 base 6 is actually much less than 2434 to base 7, 8, 9 or 10. While 42 base 6 is also less than 42 in base 7, for example.

Since it is easy to work with base 10, we'll take this approach.

So 2434 base 6 is actually much less than 2434 to base 10 (and base 6 is very far from base 10). 42 to base 6 is also much less than 42 to base 10. Now, let's assume for a moment that we are actually dividing 2434 base 10 / 42 base 10. The answer will be about 57 or so. If you cannot quickly do this computation, a faster way is to immediately do 2436/42, instead of 2434/42, which divides cleanly to 58. You can also do 2400/40 should be very close to 2434/42.

Now, since you have that answer as 58 to base 10. You can then figure that if 2434 base 6 << 2434 base 10 and 42 base 6 << 42 base 10, then 2434 base 6 / 42 base 6 << 2434 base 10/42 base 10. This means that the options 52 base 6 & 55 base 6 can't be correct since they are so close to 58 base 10. This leaves you with 35 base 6 and 23 base 6. All of this you should have done in about 1 minutes or less.

Quickly convert both to base 10, 35 base 6 = 18 + 5 = 23 base 10 & 23 base 6 = 18 + 3 = 21 base 10. Then convert 2434 base 6 to 10 to give 598 base 10; also convert 42 base 6 to base 10 = 26 base 10. 598/26 = 23. Which means you answer is 35 base 6.

Alternatively, you could have converted everything in both question & answer to base 10 and then work directly in base 10.

What is the answer when 2434 base 6 is divided by 42 base 6?

A. 23 base 6
B. 35 base 6
C. 52 base 6
D. 55 base 6

This would mean converting all numbers to base 10 and treating the question like

What is the answer when 598 base 10 is divided by 26 base 10?

A. 23 base 6 = 18 + 3 = 21 base 10
B. 35 base 6 = 18 + 5 = 23 base 10
C. 52 base 6 = 30 + 2 = 32 base 10
D. 55 base 6 = 30 + 5 = 35 base 10

Using your usual knowledge of base 10 mathematics then, you can easily divide 598/26 which is 23 (B). If you can't immediately divide 598 by 26, then divide 600/25 = 24 ~ 23(B)

To solve this question, there is no point following the usual step first described: converting first to base 10, dividing your results and then converting your result to base 6. It'll waste your time.

I do hope you understand this procedure. If you have any questions, please don't fail to ask.

Here is another related JAMB question:

Simplify (3.(2ⁿ⁺¹) - 4(2^n-1)) ÷(2ⁿ⁺¹ - 2ⁿ)

A. 2ⁿ⁺¹
B. 2^n-1
C. 4
D. 1/4

Solving this would involve simplifying denominator and numerator to obtain
(3.2(2ⁿ) - 2.(2ⁿ))/(2.(2ⁿ)-2ⁿ).
And then simplify to 6 -2 =4

A faster method would be to simply substitute n = 0. Then subsitute n=0 also into all the options to get
A. 2¹
B. 2^-1
C. 4
D. 1/4

As you can see, all the answers are unique, so you can go ahead with n=0. If answers are not unique, i.e. if there are one or or equal answers, then you could vary to n =1 or -1, depending on what you will get. Always start from 0, then 1, then -1, then 2, .....

Substitue 0 into the question, simply gives you

(3(2¹)-4(2^-1))/(2¹ - 1)
This gives (6-2)/(2-1) = 4

You get your answer quickly.

Here is another JAMB question I got from the internet.

Just a note: you guys would probably need to submit your own JAMB questions you are having difficulty with if you want to get the best from this tutorial program. For lack of that, I'll just keep treating the questions I glean off the internet

If P344₆ - 23P2₆ = 2PP2₆, find the value of the digit P.

A. 2
B. 3
C. 4
D. 5

Whatever you do, don't ever try to solve this question the usual way in an exam. You have to guess at it. The good thing with JAMB maths is that you have options A -D. Since your answer MUST be one of them, you can start from the options instead of from the question. So, here we go:

P344₆
23P2₆ -
------
2PP2₆

First, you should know that if you had had options with 6 or above, then those would have been automatically incorrect since they can't be answers under a base 6. Let's start from option A, let's try 2.

If P were 2, then in the "tens" 4 - P would be P, with P=2, which is fine. But then in the "hundreds" 3 -3 would not be P, but instead 0, which is not fine. A P =3 would not even allow the "tens" 4 - P to be correct. Since 4 -3 = 1.
A P = 4 would also not even allow the "tens" 4 - P to be correct, since 4 -4 = 0. If you are sure of these three trials, don't even bother going far. Just choose D, i.e. P = 5 and move to the next question.

All of this you should have done in less than 1 minute.

Indeed P = 5 is the correct answer. Since 4 -5 will allow you take 1 from the hundreds add to 4 to give 6+4=10, and then 10 -5 = 5, Then take 1 again from the "thousands" to give add 6 to 3 to give 8. Subtract 3 from 8 to give, again, 5 = P. Then 4 - 2 = 2.

The only reason you could have done this quick is because in the "units" with 4 -2 = 2 valid, you didn't even have to do anything anymore there.

The usual method of solving this question would have been to say

P344₆ - 23P2₆ = 2PP2₆ = P x 6³ + 3 x 6² + 4 x 6¹ + 4 x 6⁰ - 2 x 6³ - 3 x 6² - ..........**sighs**

Please don't even try this in an exam....

This again is to illustrate what I was discussing earlier. Hopefully this example can put paid to my point.

Divide: ax^3x - 26x^2x + 156a^x - 216 by a^2x - 24a^x + 108

A. a^x - 2
B. a^x + 2
C. a^x - 8
D. a^x - 6

In order to solve this question, the typical thing to do would be to simplify the numerator and also the denominator. Then something should cancel out and you should get one of the options listed above.

Never go through such stress in the examination hall. Perhaps when studying, no problem. For an exam like JAMB, you need all the shortcuts you can apply.

For a question like this, just make a = 1 & x = 1. Substitute that into both the question and the answer, and you should get

Divide: 1 - 26 + 156 - 216 by 1 - 24 + 108

A. 1 -2 = -1
B. 1 + 2 = 3
C. 1 - 8 = -7
D. 1 - 6 = -5

1 - 26 + 156 - 216 = -85 & 1 - 24 + 108 = +85

Dividing gives you -85/85 = -1 which means A is the answer.

All of this should take you less than a minute and you can move on to the next question.

I'll try to find more questions on number bases and treat these

carry on prof..............