complicate piece buh am followinq thouqh.

tnx

goofyone:
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.

you ain't a bad teacher .more power to ur elbow

This is making sense!

Love it

...

Question
convert(59/64)ten to base two.

sheddyboy:
Question
convert(59/64)ten to base two.

Don't get the question please

Repeat in a simplified way

LogoDWhiz:

Don't get the question please

Repeat in a simplified way

it's a question i got from comprehensive maths, it goes this way: convert (59/64)in base ten to base two.

sheddyboy:
it's a question i got from comprehensive maths, it goes this way: convert (59/64)in base ten to base two.

First and foremost, you need to know that this is not within the scope of utme, then onto the question,
To convert 59/64 to base 2, split 59 to span across the multiples of 2, you're gonna get (32 + 16 + 8 + 2 + 1)/64
That will give:
32/64 + 16/64 + 8/64 + 2/64 + 1/64
==> 1/2 + 1/4 + 1/8 + 1/32 + 1/64
==> 2^-1 + 2^-2 + 2^-3 + 2^-5 + 2^-6
==> therefore, 59/64 = 0.111011(the zero between them is for the missing 2^-4 in the expansion.

I hope this helps. To the mods, I'm sorry I had to jump in, I was just bored.

To summarize the important rules you want to follow while attending to mathematics questions in an exam like JAMB:

1. Work from the answers by eliminating unlikely results
2. When working from the answers, start with the simplest and work in increasing order of difficulty
3. Simplify question & answers by substituting integers starting from n=0, or n= 1 or -1, or n= 2 or -2, etc
4. If you have eliminated options A to C and are EXTREMELY confident about your calculations, don't bother with attempting to eliminate the last option D; just go with it. If you are left with more time at the end of your maths session, you can always come back to double-check.

So, let's try a few of these tricks on the next question, still on NUmber Bases

Simplify 213₄ x 23₄

A. 10311₄
B. 10321₄
C. 12231₄
D. 13211₄

Now, when JAMB sets a question they almost always leave room for you to solve that question in little time (at least from my own JAMB experience). You may think some JAMB questions take minutes to solve, but that's not true. Only very few questions actually take a long time to solve. Most other questions you can solve quickly and accumulate more time to yourself for the other questions that will take more minutes to solve. In effect, the time allotted for the maths test is more than often enough! Surprising? Well, not so much.

I'll show you how to solve this question above in less time than you actually think.

Take a look at the options, first (First rule!). Always look at the options first. Don't just jump at the question. If you look at the "units", you'll see that they all have "1". That's a giveaway. Since JAMB knows that's the first number you'll obtain from solving the question, they have made everything the same to make the question a little harder for you. But if you take a look at the "tens", you'll see that only two options carry the same number, "1". That's a good way for you to eliminate two options immediately while solving your question. Then you can finally go to the "hundreds" to completely eliminate three options and choose your answer.

So, how do we proceed?

Let's do the multiplication. The key here is to realize that you don't have to complete the calculation. It's sufficient to just get your units and tens, and then check your options to see which answers you can eliminate (or better still, which answer is certainly correct!).

213₄
23₄ x
---------------


So, let's multiply first 3₄ x 3₄, that'll give 9₁₀. COnvert that to base 4, to give 9₁₀ = 21₄. You can quickly do this by

4|9 ^
4|2 r 1 |
4|0 r 2 |

21₄

So right down 1 and "carry" 2. From your knowledge of multiplication, you know that the "1" will stand alone in the "units" end. Since all the options have "1" in their units, you can't eliminate anyone.

It's critical to note here that if you had had options like below:


A. 10313₄
B. 10322₄
C. 12231₄
D. 13214₄

Don't even waste your time going further. Just choose the option with its "units" as 1, in this case C, and move on to the next question

But that's not the case, so let's quickly solve for the number in the "tens" position


213₄
23₄ x
---------------
xxx1
sss
--------------
xxxx1
--------------

Can you see the pattern of the solution? Okay, let's move on to get the "tens".

We then multiply 3 by 1, to give 3 and then add to the 2 we carried. So we have 5. 5 is greater than 4, so we need to convert 5₁₀ to base 4. That'll give 11.

And just a side note. When converting from one base to another, you should know some really fast facts. 4₁₀ to base 4 is 10. 5₁₀ in base 5 is 10. 5₁₀ to base 4 is 11....do you see the trend? That number which starts a base is 10, the way 4₁₀ is 10 and 5₁₀ to base 4 is 11. This is just quick fact, you should know to help you in this kind of calculations

So we put the "11" in there, by putting 1 and carrying 1

213₄
23₄ x
---------------
xx11
sss
--------------
xxxx1
--------------

we carry "1" for the next computation which should be 3 x 2.

Now, this is where you want to stop with computing the hundreds, cos you don't need to. Compute the lower "tens" instead by multiplying 2 (in 23₄) x 3 (in 213₄) in the numbers below

213₄
23₄ x
---------------
xx11
sss
--------------
xxxx1
--------------

When you do that, you get 6. Convert that to base 4 to get 12₄. This means you need to write down "2" and carry 1. Do that to have the result below

213₄
23₄ x
---------------
xx11
xx2
--------------
xxx31
--------------

You can see that we now have a number ending with "31". Look at the options again, you see anything?


A. 10311₄
B. 10321₄
C. 12231₄
D. 13211₄

option C looks more like it. This is the part where you want to choose your answer and just move ahead, provided you are confident of your calculations, which in any case you should be.

Indeed, C is the correct answer. The full calculation is shown below

213₄
23₄ x
---------------
..1311
1032
--------------
12231 C
--------------

But you needn't go this far in order to solve the problem. That's JAMB for you!

goofyone:
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.

dis is wat we call crude method.

As a final note on Number Bases, I'm going to treat one more example:

The sum of four numbers is 1214₅. What is the average expressed in base five?

A. 114
B. 141
C. 401
D. 411

To solve this question, you could come up first with the expression

w+x+y+z = 1214₅

Average will now be:

(w+x+y+z)/4 = 1214₅/4

Typically what you should do next is divide 1214₅ by 4 in base 5.

Again, you need to look at the options first. Look, look, look. Cannot be overemphasized.

Looking at the options, you should realize that you need to multiply one of those options by 4 in base 5 and the answer should be 1214₅.

Alternatively you could just convert 1214₅ to base 10, divide by 4 and then convert your answer back to base 5. Whichever you do, just make sure to keep your answer simple.

When I mean "simple", I mean do not simplify unnecessarily. Simplify only when required

I'll show you what I mean.

Let's try the second approach first:

1214₅/4
(1 x 5³ + 2 x 5² + 1 x 5¹ + 4 x 5⁰)/ 4

Now, don't simplify the numerator unnecessarily, except you have a calculator at hand. I don't know how they do JAMB these days, though I hear it's without calculators.

Simplify this to
(1 x 5³ + 2 x 5² + 1 x 5¹)/4 + 5⁰
You get that?

Then simplify to

5(1 x 5² + 2 x 5¹ + 1)/4 + 5⁰

The expression in the bracket is easily simplified to (25+10+1)=36
36/4 = 9

5(9) + 5⁰
5(5 + 4) + 5⁰
5(5) + 4(5) + 5⁰
5² + 4 x 5¹ + 5⁰

which simplifies to 141₅. And you obtain your answer without having to handle large numbers. JAMB probably had this in mind when they set the question.

Now using the first method, which I consider the better sha.

Look at the options again

A. 114
B. 141
C. 401
D. 411

Which one of these do you think you can multiply by 4 in base 5 to get 1214₅?

Let's go from the simplest option to the hardest (One of our rules!)

Try 114 x 4 and you see that 4 x 4 in the "units" will give 16. 16 to base 5 is 31₅. You can easily compute this. This will leave 1 in the "units" level. Since we are looking for 4 in the "units" level, that obviously is a wrong choice.

Try 141 x 4 and you see that you can indeed have 4 in the units level once you multiply 4 x 1. But options C & D, 401 & 411 also give you 4 in the "units" level. Confusion! How about the "tens" level? Since you are not carrying anything, that's a good thing (trust me, JAMB also factored that in to allow for this method). 401 at the "tens" level will give you 0. Remember at that "tens" level you are looking for 1 (1214₅). 411, on the other hand, will give you 4 at the "tens" level. Again, bad choice.

So, our answer is indeed looking like 141₅ (B). But can we quickly confirm that? Multiply 4 x 4 at the "tens" level to give 16. 16 to base 5 is 31. If you write down 1 and "carry" 3, you immediately see that you have 1 at the "tens" level.

This is the stage where you want to just choose "B" and move on!. All of this you should have done in less than a minute.

One rule you should have learnt from our last example is to not simplify what you don't HAVE TO simplify. Keep your calculations free-flowing, so that something can always cancel another thing at the end. Example?

Say you have (39 +36 + 38 +10) x 5/(3) at a stage in your calculation

And you are doing an exam like JAMB.

Never first add the numerator, multiply by 5 and then start dividing by 3. NEVER DO THAT!!!

Instead, simplify your equation like this

5 x (39/3 + 36/3 + 38/3 + 10/3)

5 x (13 + 12 + ~13 + 10/3) = 5 x(25 + ~13 + 10/3)

You did 38/3 = ~13 because it's in fact close enough

And then move on....

If you need to get an answer from that expression immediately, maybe because you need to compare it with the options, then you can add up. And while adding up, work with small numbers first.

= 5 x ( ~38 + (9 +1)/3)
= 5 x (~38 + 3 + 1/3)
= 5 x (~41 + 0.3333)
= 5 x (~41.33) = ~ 5 x 41 = ~205

You should find an option close enough.

Using a calculator actually gives you 205 for that expression!
Of course, you do this only because you don't have a calculator. And you should be able to immediately run through this in very little time.

That's that for number bases. I hope you guys enjoyed it. Any questions? Feel free to ask.

I'll treat an example from algebra, which is supposed to be the next module. The intent is to make one of the methods clear.
But just to note: you may not be able to apply this method to every question. The deal is to make you aware of these methods so that you could use a few of them in your exams. Just be cautious as you work. After this example in factorization, I'll see if I can treat a few questions under the remainder topics in SECTION I as outlined by dejt4u. I bet you, there are lot more tricks you'd learn in this section.

2. Fractions, Decimals, Approximations
and Percentages:
3. Indices, Logarithms and Surds:
4. Sets:

So, let me treat this question:

Factorize completely 4abx - 2axy -12b2x + 6bxy

A. 2x(a - 3b)(2b - y)
B. 2x(3b - a)(2b - y)
C. 2x(a - 3b)(y - 2b)
D. 2x(2b - a)(3b - y)

In order to solve this, you might want to start factorizing and trying to look for what's common and all that. Another wise method to use would be to use our method of substitution. Substitute a = 1, b = 1, and x =1

That'll give you

Factorize completely 4 -2 - 12 + 6 = -4

A. 2(1-3)(2-1) = -4
B. 2(3-1)(2-1) = 4
C. 2(1-3)(1-2) = 4
D. 2(2-1)(3-1)=4

You can immediately see that A is the answer.

Except you are fast enough to know how to quickly factorize, you may just want to adopt this method. In fact, something tells me it's not a coincidence only option A has a -4

As an example, see how well you can apply the above method to this question

Factorize completely X²+2XY+Y²+3X+3Y-18

A. (x+y+6)(x+y-3)
B. (x-y-6)(x-y+3)
C. (x-y+6)(x-y-3)
D. (x+y-6)(x+y+3)

Or to this:

Divide 4x³-3x+1 by 2x-1

A. 2x2-x+1
B. 2x2-x-1
C. 2x2+x+1
D. 2x2+x-1

Xtarxhyne:
walks in wit my writing materials....

the last time i checked, you aren't a "Jambite" so what are Чou doing here?

So, next example will be on Surds (on of SECTION I).

We are going to try this question:

If the operation * on the set of integers is defined by p * Q = √(pq), find the value of 4 * ( 8 * 32).

A. 8
B. 3
C. 16
D. 4

Looks like an innocent question.

you should know the concept of binary operations so you can easily solve this question. The concept I want to explore here is the essence of NOT SIMPLIFYING IF YOU DON'T HAVE TO

8 * 32 should give √(8 x 32)
Some people will then multiply 8 by 32 to get 256. A smart student may already know that the square root of 256 is 16. But working with large numbers is not that easy. What would I recommend? Keep your equations simple.

Instead of multiplying and then finding the square root of √(8 x 32). Choose to simplify the expression even further. Write it as:

√(8 x 8 x 4) = √(8² x 2²) = 8 x 2 = 16

Then compute 4 * 16 = √(4 x 16).

Again, don't multiply 4 by 16 and then start looking for its square root. Instead proceed like this:

√(4 x 16) = √(2² x 4²) = 2 x 4 = 8

Then choose your answer calmly as 8 (A). I hope that's clear.

dejt4u:
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..

i hope you people will finish what you've started.

goofyone:
So, next example will be on Surds (on of SECTION I).

We are going to try this question:

If the operation * on the set of integers is defined by p * Q = √(pq), find the value of 4 * ( 8 * 32).

A. 8
B. 3
C. 16
D. 4

Looks like an innocent question.

you should know the concept of binary operations so you can easily solve this question. The concept I want to explore here is the essence of NOT SIMPLIFYING IF YOU DON'T HAVE TO

8 * 32 should give √(8 x 32)
Some people will then multiply 8 by 32 to get 256. A smart student may already know that the square root of 256 is 16. But working with large numbers is not that easy. What would I recommend? Keep your equations simple.

Instead of multiplying and then finding the square root of √(8 x 32). Choose to simplify the expression even further. Write it as:

√(8 x 8 x 4) = √(8² x 2²) = 8 x 2 = 16

Then compute 4 * 16 = √(4 x 16).

Again, don't multiply 4 by 16 and then start looking for its square root. Instead proceed like this:

√(4 x 16) = √(2² x 4²) = 2 x 4 = 8

Then choose your answer calmly as 8 (A). I hope that's clear.

Excuse me Sir, whats your recommended textßook?

goofyone:
That's that for number bases. I hope you guys enjoyed it. Any questions? Feel free to ask.

I'll treat an example from algebra, which is supposed to be the next module. The intent is to make one of the methods clear.
But just to note: you may not be able to apply this method to every question. The deal is to make you aware of these methods so that you could use a few of them in your exams. Just be cautious as you work. After this example in factorization, I'll see if I can treat a few questions under the remainder topics in SECTION I as outlined by dejt4u. I bet you, there are lot more tricks you'd learn in this section.



So, let me treat this question:

Factorize completely 4abx - 2axy -12b2x + 6bxy

A. 2x(a - 3b)(2b - y)
B. 2x(3b - a)(2b - y)
C. 2x(a - 3b)(y - 2b)
D. 2x(2b - a)(3b - y)

In order to solve this, you might want to start factorizing and trying to look for what's common and all that. Another wise method to use would be to use our method of substitution. Substitute a = 1, b = 1, and x =1

That'll give you

Factorize completely 4 -2 - 12 + 6 = -4

A. 2(1-3)(2-1) = -4
B. 2(3-1)(2-1) = 4
C. 2(1-3)(1-2) = 4
D. 2(2-1)(3-1)=4

You can immediately see that A is the answer.

Except you are fast enough to know how to quickly factorize, you may just want to adopt this method. In fact, something tells me it's not a coincidence only option A has a -4

For the factorization method, another method is to multiply and open the brackets individually. It may take some time but its also efficient. Option A is the right answer.

Solving this my way would be 4abx - 2axy -12b2x + 6bxy

separating equations (4abx-2axy)-(12b^2x-6bxy)
The Minus sign in front of the second bracket changes the signs of 12b^2x and 6bxy.

factorizing 2ax(2b-y)-6bx(2b-y)
Factorizing is identifying the common factors. 2ax is common to the first bracket and 6bx is common to the second

together (2ax-6bx) (2b-y)
Since (2b-y) is common, we take one of it ad multiply it 2ax-6bx

Further factorization 2x is common to both 2ax and -6bx.
so factorizing by 2x, we have:

2x(a-3b)(2b-y)
Compact form
4abx - 2axy -12b2x + 6bxy
(4abx-2axy)-(12b^2x-6bxy)
2ax(2b-y)-6bx(2b-y)
(2ax-6bx)(2b-y)
2x(a-3b)(2b-y)

Neldrizzy:
the last time i checked, you aren't a "Jambite" so what are Чou doing here?

LEARNING

*walks in with jotter and pencil*

sheddyboy:
Question
convert(59/64)ten to base two.

59/64 = 0.921875
0.921875 × 2 = 1.84375
forget the number before the decimal point in dis next step
0.84375 × 2 = 1.6875
forget the number before the decimal point as usual
0.6875 × 2 = 1.375
As usual,
0.375 × 2 = 0.75
As usual,
0.75 × 2 = 1.5
As usual, forget the number before the decimal point and continue
0.5 × 2 = 1.00
At this point,u dont go further cos the number after d decimal is zero. Now ur answer will come from those numbers u ignored(ie all the numbers before the decimal point. Therefore,from the first multiplication, the answer is .111011 base two. I hope u get it??

i thought if you miss a lesson you have missed it.

Who is bringing us back to number bass.

mathefaro:
First and foremost, you need to know that this is not within the scope of utme, then onto the question,
To convert 59/64 to base 2, split 59 to span across the multiples of 2, you're gonna get (32 + 16 + 8 + 2 + 1)/64
That will give:
32/64 + 16/64 + 8/64 + 2/64 + 1/64
==> 1/2 + 1/4 + 1/8 + 1/32 + 1/64
==> 2^-1 + 2^-2 + 2^-3 + 2^-5 + 2^-6
==> therefore, 59/64 = 0.111011(the zero between them is for the missing 2^-4 in the expansion.

I hope this helps. To the mods, I'm sorry I had to jump in, I was just bored.

pls oo, i accpted am dull, just help me try and do the workings on that 59, on how you split it to 2 to gt 32 16...

am very happy for this thread, for all the tutors that take their time to make this thread awesome, i prayed God will take time to make your life awesome too, pls i just wanna make a suggestion on behalf of we average learners, to all tutors that is here and other interested enthusiast that is following this thread, if you gat any simple procedure for any related questions bring solved already, you can try and contribute, at least if one methods is too hard for we to decipher we re gonna get on with another contributed method , thanks in anticipation

all diz tin cum b lyk magic. oboi i don turn olodo for auz chai. i sabi am diz year may

[b][/b]nice thread...thumbs up

Xtarxhyne:
walks in wit my writing materials....

You like maths ni, I think say you hate am??

Nice thread

omoadeleye:



pls oo, i accpted am dull, just help me try and do the workings on that 59, on how you split it to 2 to gt 32 16...

we already know that the numbers in the powers of 2 are 1, 2, 4, 8, 16, 32, 64, 128 and so on, but for us to split 59, we should know that we can only use numbers less than 59, so we started from 32, adding the smaller consecutive powers of 2 until we get 59.
but like I said earlier, it's beyond the scope of UTME, so don't expect this kind of question