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Nairaland Forum / Nairaland / General / Education / Nairaland Jamb Tutorial Centre {mathematics Thread} (20107 Views)
2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } / Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] / Nairaland Jamb Tutorial Centre. {NJTC} (2) (3) (4)
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Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by stuff46(m): 6:54am On Nov 27, 2014 |
complicate piece buh am followinq thouqh. |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by olowookerekemi(f): 8:03am On Nov 27, 2014 |
tnx |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by shakrullah(m): 4:31pm On Nov 27, 2014 |
goofyone:you ain't a bad teacher .more power to ur elbow |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by LogoDWhiz(m): 6:25pm On Nov 27, 2014 |
This is making sense! Love it |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by fr3do(m): 6:58pm On Nov 27, 2014 |
... |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by sheddyboy: 7:37pm On Nov 27, 2014 |
Question convert(59/64)ten to base two. |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by LogoDWhiz(m): 7:41pm On Nov 27, 2014 |
sheddyboy:Don't get the question please Repeat in a simplified way |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by sheddyboy: 8:17pm On Nov 27, 2014 |
LogoDWhiz:it's a question i got from comprehensive maths, it goes this way: convert (59/64)in base ten to base two. |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by mathefaro(m): 9:04pm On Nov 27, 2014 |
sheddyboy: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. 1 Like 2 Shares |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 9:07pm On Nov 27, 2014 |
To summarize the important rules you want to follow while attending to mathematics questions in an exam like JAMB: 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: 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! 3 Likes 5 Shares |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by ameer2: 9:32pm On Nov 27, 2014 |
goofyone:dis is wat we call crude method. |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 9:57pm On Nov 27, 2014 |
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. |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 10:09pm On Nov 27, 2014 |
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. |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 10:35pm On Nov 27, 2014 |
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 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 1 Like |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Neldrizzy(m): 11:06pm On Nov 27, 2014 |
Xtarxhyne:the last time i checked, you aren't a "Jambite" so what are Чou doing here? |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 11:09pm On Nov 27, 2014 |
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. 2 Likes 2 Shares |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by clevvermind(m): 12:16am On Nov 28, 2014 |
dejt4u:i hope you people will finish what you've started. |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by NACHIEVER: 12:18am On Nov 28, 2014 |
goofyone:Excuse me Sir, whats your recommended textßook? |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by raayah(f): 12:44am On Nov 28, 2014 |
goofyone: 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) |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 6:05am On Nov 28, 2014 |
Neldrizzy:LEARNING 1 Like |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nastydroid(m): 7:09am On Nov 28, 2014 |
*walks in with jotter and pencil* |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by 2muchopoTBdope(m): 7:34am On Nov 28, 2014 |
sheddyboy: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?? 1 Like 2 Shares |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by stuff46(m): 12:07pm On Nov 28, 2014 |
i thought if you miss a lesson you have missed it. Who is bringing us back to number bass. |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by omoadeleye(m): 12:10am On Nov 29, 2014 |
mathefaro: 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... |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by omoadeleye(m): 12:24am On Nov 29, 2014 |
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 |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 12:48am On Nov 29, 2014 |
all diz tin cum b lyk magic. oboi i don turn olodo for auz chai. i sabi am diz year may |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by mikeayus(m): 12:52am On Nov 29, 2014 |
[b][/b]nice thread...thumbs up |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by barackodam: 2:05am On Nov 29, 2014 |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Slatemsk(m): 2:43am On Nov 29, 2014 |
Nice thread |
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by mathefaro(m): 3:33am On Nov 29, 2014 |
omoadeleye: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 1 Like |
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