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Nairaland Jamb Tutorial Centre {mathematics Thread} - Education (4) - Nairaland

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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 Adiwana: 9:55pm On Nov 29, 2014
My mistake,your method is fast..thanks..works perfectly
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by raayah(f): 9:57pm On Nov 29, 2014
Adiwana:
I applied this method to a certain question & I failed it mean while here is the question

3^-5n/9^1-n × 27^n+1...thanks in advance

Could you rephrase this question.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by clevvermind(m): 10:30pm On Nov 29, 2014
why not take it topic by topic instead of jumping from one topic to another without thoroughly treating one first.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by smicc: 10:48pm On Nov 29, 2014
Solution: 3^-5n x 9^-[1-n] x 27^n+1 then. 3^-5n x 3^-2+2n x 3^3n+3 using d law of indices: 3^-5n+2n+3n-2+3 d answer is 3^1 =3.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by engrjosefz(m): 10:50pm On Nov 29, 2014
Wao.... Nice bin here, thanks to the MTH Geeks! Getting better and more elaborate ways of solving questions faster! thumbs up. God bless you all... Wao.... Nice bin here, thanks to the MTH Geeks! Getting better and more elaborate ways of solving questions faster! thumbs up. God bless you all... Wao.... Nice bin here, thanks to the MTH Geeks! Getting better and more elaborate ways of solving questions faster! thumbs up. God bless you all...
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by kanor: 11:25pm On Nov 29, 2014
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Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Adiwana: 11:34pm On Nov 29, 2014
raayah:


Could you rephrase this question.
alright forget it,I have resolved it with the initial method&I got it..found out my mistake..thanks
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 1:26am On Nov 30, 2014
Adiwana:
alright forget it,I have resolved it with the initial method&I got it..found out my mistake..thanks
Do you mean the substitution method?
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 1:27am On Nov 30, 2014
Martiyu:


23 base 6 ought to be 12 + 3 = 15 base 10...or is there anything I am missing?
No, you are missing though. It's a mistake. Thanks for the correction.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 1:28am On Nov 30, 2014
GUys,

I'm trying to post an additional method of solving questions in logarithms. But the site keeps banning me. So, I'm just going to have to skip it.

We'll move on with other SECTIONS.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Adiwana: 9:08am On Nov 30, 2014
goofyone:

Do you mean the substitution method?




Yeah I used ur method & I solved a question within 45secs..thanks..pls forget about log&move on ..its only 1 question & you've solved the most difficult ones..we can read up the most rest
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by makasimatics(m): 1:00pm On Nov 30, 2014
Fine solving though, but I advise every tutor here to ensure that every question posted must be explained well for easy understanding.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by makasimatics(m): 2:23pm On Nov 30, 2014
I don't know if we are solving based on the topics listed or just any way we want. I thought that change of base was to be dealt dearly with b4 others.
Any way,let me solve this question that our colleague posted in my own way.

(3 ^-5n / 9^1- n) x 27^ n+1

Make the terms inside the parentheses to have the same base. It is impossible from the nominator but possible from the denominator, i.e, 9=3^2. We then replace 9 with 3^2 to have

(3^-5 / 3^2(1- n) ) x 27^n+1

Also,we shall replace 27 with 3^3. This is owing to the fact that,3^3=27. Thus we have

(3^-5 / 3^2(1- n)) x 3^3( n+1)

We then expand the exponents to have

(3^-5 / 3^2-2n) x3^3n+3

From the term inside the bracket, we perform division using law of exponent when quotient is involved. That,we subtract the power of the denominator from that of the numerator so that we have a single base. Thus we have

3^-5n-(2-2n) x 3^3n + 3

3^-5n-2+2n x 3^3n+ 3

Now we perform multiplication law of exponent. To do this,we add the powers of the two terms to have one base term. Thus we have

3^-5n-2n+3n+3-2

Notice that we have taken the like terms already. Now we factor out n from the power to have

3^n(-5+2+3)+3-2 =3^n(0)+1

Note that n x 0 = 0

And 0 + 1 = 1

Therefore, we have. 3^0+1

= 3^1 =3

The answer is 3.

Thanks
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 4:10pm On Nov 30, 2014
The expression ax2 + bx + c equals 5 at x = 1. If its derivative is 2x + 1, what are the values of a, b, c respectively?

A. 1, 3, 1
B. 1, 2, 1
C. 2, 1, 1
D. 1, 1, 3

One other topic I have always considered very easy to handle in an examination like JAMB is differentiation. Because it affords you the opportunity of moving back and forth with differentiation and integration, it becomes very easy to manipulate equations as you like. In this example though, you won't need to manipulate anything.

This is a standard JAMB question and JAMB itself has made it very easy to solve, depending on how you look at it.
Look first at the options before proceeding. Typically, if you want to solve a question like this you'll probably be inclined to find a, b and c. But the options tell you that finding a & b is actually sufficient enough to tell you what the answer is. This is because a & b are different all the way in all the options.

So, you really shouldn't bother with the first part of the question involving c. Just differentiate the equation already to get 2ax + b, compare with 2x +1, and obtain a=1, b=1. That should tell you that the answer is D.

There is no need finding what "c" is in this question.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 4:39pm On Nov 30, 2014
Adiwana:




Yeah I used ur method & I solved a question within 45secs..thanks..pls forget about log&move on ..its only 1 question & you've solved the most difficult ones..we can read up the most rest

I'm really glad you found the method useful. Try to solve many questions using this method, so you can become conversant with it. At the same time, develop your skills with using the method so you can quickly tell when it'll come in handy while in the examination hall. You may also want to always time yourself to collect more information about how long it takes to apply the method.

If you find that it takes you longer to apply the method than it takes you to solve it using the usual solution technique, then you might want to reconsider your options.

Anywayz, just practice, practice, practice.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 4:46pm On Nov 30, 2014
One other question type you can use the substitution method for is SIMULTANEOUS EQUATIONS.
In an exam like JAMB, it's not recommend to try to solve simultaneous equations from the ground up. Options are provided for you and if you can, use these options to quickly narrow down your answers to the correct option.

Take for example this JAMB question:

Solve the equations
m2 + n2 = 29
m + n = 7

A. (2, 3) and ( 3, 5)
B. (2, 5) and (5, 2)
C. (5, 2) and ( 5, 3)
D. (5, 3) and (3, 5)


A usual attempt at finding m & n may require you to start solving equation. You'll obtain a quadratic equation that will then feed you with options for m (or n). You can then substitute to obtain your answer for the other unknown.
In an exam like WAEC (theory section), you certainly should go through all the solution steps. But in JAMB, use your shortcut method.

I'll not attempt to solve this question the usual way. I'll just go straight to using the substitution method.

So, you look at the question again:

m2 + n2 = 29
m + n = 7

And you look at the options again:

A. (2, 3) and ( 3, 5)
B. (2, 5) and (5, 2)
C. (5, 2) and ( 5, 3)
D. (5, 3) and (3, 5)

You immediately realize that in order for an option to be correct, the answer has to work for both solutions in the bracket. You immediately see that "A" can't be correct, since 2 + 3 = 5 (and not 7). At this stage, you want to stop bothering about the next bracket in option "A". It says "and" between them, so both options MUST be correct. Since the first bracket is not correct, don't even bother with the second bracket.

Look at option "B". 2 + 5 = 7 and that's good. Again, 22 + 52 = 4 + 25 = 29 and that's also good. The second bracket is merely a reversal of the first bracket and gives the same answers as earlier.

It is at this stage, you want to ignore the remaining options (provided you are confident of your solution) and just move on. If you are like me and you are not completely sure of that option "B" as the answer, then by all means test the remaining options. But you need to be fast. You'll quickly see that for option "C", even though (5,2) is correct, (5,3) is not as 5 + 3 = 8 (and not 7).
Also option D is completely way off for the two brackets.

So "B" is indeed the answer. All of this should have taken you less than a minute.

Again, you really do not have to go through a complete solution step in solving a question like this. I'll find more questions of this type and post it for you to apply this substitution method.

We'll be talking about another method of JAMB maths quick solution in the next couple of posts. I call this the "ELIMINATION METHOD".

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Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Adiwana: 9:30pm On Nov 30, 2014
goofyone:
One other question type you can use the substitution method for is SIMULTANEOUS EQUATIONS.
In an exam like JAMB, it's not recommend to try to solve simultaneous equations from the ground up. Options are provided for you and if you can, use these options to quickly narrow down your answers to the correct option.

Take for example this JAMB question:

Solve the equations
m2 + n2 = 29
m + n = 7

A. (2, 3) and ( 3, 5)
B. (2, 5) and (5, 2)
C. (5, 2) and ( 5, 3)
D. (5, 3) and (3, 5)


A usual attempt at finding m & n may require you to start solving equation. You'll obtain a quadratic equation that will then feed you with options for m (or n). You can then substitute to obtain your answer for the other unknown.
In an exam like WAEC (theory section), you certainly should go through all the solution steps. But in JAMB, use your shortcut method.

I'll not attempt to solve this question the usual way. I'll just go straight to using the substitution method.

So, you look at the question again:

m2 + n2 = 29
m + n = 7

And you look at the options again:

A. (2, 3) and ( 3, 5)
B. (2, 5) and (5, 2)
C. (5, 2) and ( 5, 3)
D. (5, 3) and (3, 5)

You immediately realize that in order for an option to be correct, the answer has to work for both solutions in the bracket. You immediately see that "A" can't be correct, since 2 + 3 = 5 (and not 7). At this stage, you want to stop bothering about the next bracket in option "A". It says "and" between them, so both options MUST be correct. Since the first bracket is not correct, don't even bother with the second bracket.

Look at option "B". 2 + 5 = 7 and that's good. Again, 22 + 52 = 4 + 25 = 29 and that's also good. The second bracket is merely a reversal of the first bracket and gives the same answers as earlier.

It is at this stage, you want to ignore the remaining options (provided you are confident of your solution) and just move on. If you are like me and you are not completely sure of that option "B" as the answer, then by all means test the remaining options. But you need to be fast. You'll quickly see that for option "C", even though (5,2) is correct, (5,3) is not as 5 + 3 = 8 (and not 7).
Also option D is completely way off for the two brackets.

So "B" is indeed the answer. All of this should have taken you less than a minute.

Again, you really do not have to go through a complete solution step in solving a question like this. I'll find more questions of this type and post it for you to apply this substitution method.

We'll be talking about another method of JAMB maths quick solution in the next couple of posts. I call this the "ELIMINATION METHOD".
Dude am in love with you{[No homo]..but you have really simplified everything & I can not waste time in maths.. Taking jamb next year has given me headache cos its CBT& secondly with it taking much of my time..but with your tutoring no more time wasting..

I will post a quick solution on some profit&loss problems when we get to that topic
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Nobody: 9:59pm On Nov 30, 2014
Adiwana:

Dude am in love with you{[No homo]..but you have really simplified everything & I can not waste time in maths.. Taking jamb next year has given me headache cos its CBT& secondly with it taking much of my time..but with your tutoring no more time wasting..

I will post a quick solution on some profit&loss problems when we get to that topic

I'm glad you find it helpful. More methods to come.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Chrisdmeek(m): 4:31pm On Dec 04, 2014
goofyone:
This again is to illustrate what I was discussing earlier. Hopefully this example can put paid to my point.

Divide: ax3x - 26x2x + 156ax - 216 by a2x - 24ax + 108

A. ax - 2
B. ax + 2
C. ax - 8
D. ax - 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 thought 262x Should be 262 (i.e taking x=1)
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by doggedjay(m): 6:55pm On Dec 05, 2014
Plz can someone help me with this question: A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5,at the end of the year Y received #5000 more than X.what is the total profit of the company for the year?
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by tuwayz(m): 8:58pm On Dec 05, 2014
doggedjay:
Plz can someone help me with this question: A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5,at the end of the year Y received #5000 more than X.what is the total profit of the company for the year?
Answering dis kind of question in jamb u first of all have to look at the ratios, from the ratio u know that Y is 1 ratio more than X and since he received 5000 more dan X it implies that 1 ratio =5000 but the sum of their ratios is 9 now the total profit will be 5000*9=45000
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by 2muchopoTBdope(m): 11:15pm On Dec 05, 2014
doggedjay:
Plz can someone help me with this question: A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5,at the end of the year Y received #5000 more than X.what is the total profit of the company for the year?

Y received #5000 more Dan X
Therefore, Y = 5000 + X

Ratio of X:Y = 4:5

4/5 = X/Y
Recall that Y= 5000 + X
Then we have;
4/5 = X/(5000 + X)
Cross multiplying
5X = 20000 + 4X
5X - 4X = 20000
X = 20000.
The total profit = X + Y
= X +(5000 + X)
= 20000 + 25000
= #45000.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Ifeoma12345: 12:43pm On Dec 06, 2014
these sums are difficult
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by doggedjay(m): 3:03pm On Dec 06, 2014
thanks guys but #45000 is not in the options:we have;30000;20000;25000;15000
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by tuwayz(m): 11:16pm On Dec 06, 2014
doggedjay:
thanks guys but #45000 is not in the options:we have;30000;20000;25000;15000
its a bonus question den if at all the question is correct
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by VEVEDIHNO: 2:50am On Dec 08, 2014
Pls I Need U Guys To Help Me on This

A man wishes to keep some money in savings deposit at 25% compound interest so that after 3years he can buy a car fo #150,000. How much does he need to Deposit Now.

Tanks In Advance.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by sorbentcrooner(m): 9:33pm On Dec 08, 2014
VEVEDIHNO:
Pls I Need U Guys To Help Me on This

A man wishes to keep some money in savings deposit at 25% compound interest so that after 3years he can buy a car fo #150,000. How much does he need to Deposit Now.

Tanks In Advance.
#76800 is the answer using compound interest formula or future value of money formula, therein, u r lukn for the principal.
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by sammyomotola: 10:11am On Dec 09, 2014
I think say I know ni, I hail to all my boss here and sharply follow
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by VEVEDIHNO: 8:22am On Dec 10, 2014
Who Can Help Me Work On This One.

Making r the subject of the formula


Q=lampda(squareroot of 2gh/r2-1)

Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by Adiwana: 8:59am On Dec 10, 2014
#76805 is the answer

The simple formular to use is
A=P(1+r/100)^n where
A=simple interest
P=Capital/Principal
r=rate/percentage
n=no of years
so going by the formular A=P(1+r/100)^n we have
A=#150000,P=?,r=25%,n=3years
150000=P(1+25/100)^3
150000=P(1+0.25)^3
150000=P(1.25)^3
150000/P=(1.25)^3
150000/P=1.953
150000/1.953=P
P=#76804.915
P=#76805:
So the money needed to start is #76805..HOPE YOU GOT IT
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by factorial1(m): 10:29am On Dec 10, 2014
VEVEDIHNO:
Who Can Help Me Work On This One.

Making r the subject of the formula


Q=lampda(squareroot of 2gh/r2-1)
Re: Nairaland Jamb Tutorial Centre {mathematics Thread} by factorial1(m): 10:29am On Dec 10, 2014
VEVEDIHNO:
Who Can Help Me Work On This One.

Making r the subject of the formula


Q=lampda(squareroot of 2gh/r2-1)
Alright... here is the solution:
Note: π represent lambda.
Q = π(√2gh/r² - 1)
Now, squaring both sides to remove the square root in the R.H.S...
The question becomes Q² = π²(2gh/r² - 1)
Then... cross multiply to have (r² - 1)Q² = π²2gh
Opening the bracket... the eqn becomes Q²r² - Q² = π²2gh... Now, making Q²r² the subject of the formula.. then we have Q²r² = π²2gh + Q². Since we are to make "r" the subject of the formula and not Q²r²... we then firstly divide both sides by Q²... making the eqn to be r² = (π²2gh + Q²)/Q²... which is also r² = π²2gh/Q² + Q²/Q² and it implies r² = π²2gh/Q² + 1. Now... to get "r"... we then find the square root of both sides... which is √r² = √(π²2gh/Q² + 1) and finally r = √(π²2gh/Q² + 1).

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