leahcimzil:
lol but this ain't Maths nah...I used to think i knew maths until i got into the Uni and started doing, Abstract Algebra and Advanced Calculus

Laive:

Let me try one Question at a time.
Solution
Let the son's age be:
.
x
.
Therefore, father's age:
.
5x
.
Four years ago, if we multiply their ages, we would get:
.
5x = 448
.
From here, the son's age becomes:
.
x = 448/5
x = 89.6

.
Therefore, 4 year ago, son's age was:
.
89.6
.
4 years have passed, we'll add extra 4 years to the son's age:
.
89.6 + 4 = 93.6
.
Therefore, his son is currently 93.6 years old (more like 93years and 6 months).
.
From the question, we were told that the man is 5 times older than his son.
.
Therefore, we will multiply the son's age by 5 to get the man's age 4 years ago:
.
5 x 93.6 = 448
.
The man was 448 years in the last 4 years.
.
4 years have passed, his present age will be:
448 + 4 = 452
.
Op

Xflint:
7b) solution

Log(x² + 4) = 2 + log^x - log²⁰
log(x² + 4) = log¹⁰⁰ + log^x - log²⁰
log(x² + 4) = log (100 X x/20)

Now, you cross multiply
given you 20(x² + 4) = 100 X x
20x² + 80 = 100x

now divid by 20
[strike]20[/strike]x²/[strike]20[/strike] - [strike]100[/strike]x/[strike]20[/strike] + 80/[strike]20[/strike] = 0

x² - 5x + 4 = 0
x² - 4x - x + 4 = 0
(x² - 4x)(x + 4)
x(x - 4) -1(x - 4)
(x - 1)(x - 4)
x = 1 or 4

the strike is not working here just check
http://forum.mathsboard.com/t594-solve-this-neco-past-questions-2016-2017#1517

boss096:
your log sucks!

Laive:

Let me try one Question at a time.
Solution
Let the son's age be:
.
x
.
Therefore, father's age:
.
5x
.
Four years ago, if we multiply their ages, we would get:
.
5x = 448
.
From here, the son's age becomes:
.
x = 448/5
x = 89.6

.
Therefore, 4 year ago, son's age was:
.
89.6
.
4 years have passed, we'll add extra 4 years to the son's age:
.
89.6 + 4 = 93.6
.
Therefore, his son is currently 93.6 years old (more like 93years and 6 months).
.
From the question, we were told that the man is 5 times older than his son.
.
Therefore, we will multiply the son's age by 5 to get the man's age 4 years ago:
.
5 x 93.6 = 448
.
The man was 448 years in the last 4 years.
.
4 years have passed, his present age will be:
448 + 4 = 452
.
Op

BLoomfrancs:


Guy you no dey shame to post this rubbish here. People are talking about building robots that can fight in battles like ground soldiers you are here posting nonsense.

Laive:

Let me try one Question at a time.
Solution
Let the son's age be:
.
x
.
Therefore, father's age:
.
5x
.
Four years ago, if we multiply their ages, we would get:
.
5x = 448
.
From here, the son's age becomes:
.
x = 448/5
x = 89.6

.
Therefore, 4 year ago, son's age was:
.
89.6
.
4 years have passed, we'll add extra 4 years to the son's age:
.
89.6 + 4 = 93.6
.
Therefore, his son is currently 93.6 years old (more like 93years and 6 months).
.
From the question, we were told that the man is 5 times older than his son.
.
Therefore, we will multiply the son's age by 5 to get the man's age 4 years ago:
.
5 x 93.6 = 448
.
The man was 448 years in the last 4 years.
.
4 years have passed, his present age will be:
448 + 4 = 452
.
Op

Laive:

Let me try one Question at a time.
Solution
Let the son's age be:
.
x
.
Therefore, father's age:
.
5x
.
Four years ago, if we multiply their ages, we would get:
.
5x = 448
.
From here, the son's age becomes:
.
x = 448/5
x = 89.6

.
Therefore, 4 year ago, son's age was:
.
89.6
.
4 years have passed, we'll add extra 4 years to the son's age:
.
89.6 + 4 = 93.6
.
Therefore, his son is currently 93.6 years old (more like 93years and 6 months).
.
From the question, we were told that the man is 5 times older than his son.
.
Therefore, we will multiply the son's age by 5 to get the man's age 4 years ago:
.
5 x 93.6 = 448
.
The man was 448 years in the last 4 years.
.
4 years have passed, his present age will be:
448 + 4 = 452
.
Op

Horlawale1:
Am a big lover of mathematics, but lately am losing interest.

Sincere question, how can this be use to solve economy situation, or how is this applicable in real life.

While some School Cert holder are discovering new inventions in China, we are here looking for 'X'
Who X help

Nigeria education sector needs review.
My opinion

Kaybee7000:
N=3 or 1 Mr man

cosmic22:
calculating for 4 years ago

man-5x

son-x

product of ages

5x * x = 448

5x squared = 448

x squared=448/5

x = √89.6

x = 9.465 ( boys age )

(fathers age) 5x = 47.325

calculating present ages

approximate ages

9 + 4 = 13 years old (boys age)
47 + 4 = 51 years old ( fathers age)

Elxandre:
disgraceful!

Elxandre:
disgraceful!

GeneralShepherd:


Wrong....

Man's age - 5X

Son's age - X

Four years ago the product of their ages (not the son's age) was 448

(x-4) (5x-4) = 448
5x^2 - 24x - 432 =0
Using the quadratic formula

X= 12 or X= -7.199

As we know age cannot be negative so the son's age is 12 years.

The father's age is 60 years

Wealthgang:

It does. One of d solutions will give u X which is d present age of d boy. The father's present age will simply be 5X.
Lemme solve it:
By factorization,
5x² - 24x - 432 = 0
5x² - 60x + 36x - 432 = 0
5x (x - 12) + 36 (x - 12) = 0
(5x + 36)(x -12) = 0
When 5x + 36 = 0, x = -36/5 = -7.2
When x - 12 = 0, x = 12.
Boy's age = 12yrs and not -7.2 since age cannot be negative.
Father's age = 5x = 5 x 12 = 60yrs.

By quadratic eqn general formula, Mypeople2 has handled that.
x = [-b ± √(b²-4ac)]/2a
5x² - 24x - 432 = ax² + bx + c
Comparing, a = 5, b = -24, c = -432
x = [-(-24) ± √(-24² - 4 x 5 x -432)]/2 x 5
= [24 ± √9216]/10
= (24 ± 96)/10
= (24 + 96)/10 or (24 - 96)/10
= 12 or -7.2. (Again, discard -7.2)

Now let's check if answers are correct.
Father's 5 times boy's age: 60 = 5 x 12
4years ago,
(60 - 4)(12 - 4) = 56 x 8 = 448

Damfash:
Q9a. I'll try to make the solution very simple. Let the two-digit number be XY. X is the tens digit while Y is the unit digit. X=Y+5........(i). Note: A two-digit number is ten times the tens digit plus the unit digit. E.g 54=(5*10)+4, 63=(6*10)+3. Let's proceed! (10X+Y)+14=3*X*Y. So, we have 10X+Y+14=3XY........(ii). Sub. equation (i) in equation (ii). 10(Y+5)+Y+14=3Y(Y+5)... 10Y+50+Y+14=3Y^2+15Y.... 3Y^2+4Y-64=0. Y=4 or Y=-5.33. Discard the second value. If Y=4, X=Y+5=4+5=9. The two digit number is 94.@ Wealthgang

Wealthgang:
Check Out The Recent NECO Math Questions Some School Teachers Couldn't Solve


Same order they appeared in d question paper.

Q3) A man is 5 times older than his son. Four yrs ago, d product of their ages was 448. Find their present ages.

Q7a) Find d value of n if
3²ⁿ⁺¹ - 4(3ⁿ⁺¹) + 9 = 0

7b) Solve d eqn:
log(x²+4) = 2 + logx - log20

Q9a) A two-digit number is such that its tens digit is greater than its unit digit by 5. If d number is 14 less than 3 times d product of its digits, find d number.

Let's see how many Nairalanders can give it a shot. Anyone who answers correctly will get a reward.

Cc: Xflint, Boyoski09, truth12, themall, Danosaur007, Jerrypolo, 8BitGee, integritylady, Laive, lilmonarch

valarinz:

Ma oga, u no know book oo, lol

Laive:
Too many critics, so I had to remove the thread. All the same, I HAVE LEARNT THE RIGHT WAY OF DOING IT. Thanks!

come and see metuselah!!

Wealthgang:
Check Out The Recent NECO Math Questions Some School Teachers Couldn't Solve


Same order they appeared in d question paper.

Q3) A man is 5 times older than his son. Four yrs ago, d product of their ages was 448. Find their present ages.

Q7a) Find d value of n if
3²ⁿ⁺¹ - 4(3ⁿ⁺¹) + 9 = 0

7b) Solve d eqn:
log(x²+4) = 2 + logx - log20

Q9a) A two-digit number is such that its tens digit is greater than its unit digit by 5. If d number is 14 less than 3 times d product of its digits, find d number.

Let's see how many Nairalanders can give it a shot. Anyone who answers correctly will get a reward.

Cc: Xflint, Boyoski09, truth12, themall, Danosaur007, Jerrypolo, 8BitGee, integritylady, Laive, lilmonarch

Wealthgang:
Check Out The Recent NECO Math Questions Some School Teachers Couldn't Solve


Same order they appeared in d question paper.

Q3) A man is 5 times older than his son. Four yrs ago, d product of their ages was 448. Find their present ages.

Q7a) Find d value of n if
3²ⁿ⁺¹ - 4(3ⁿ⁺¹) + 9 = 0

7b) Solve d eqn:
log(x²+4) = 2 + logx - log20

Q9a) A two-digit number is such that its tens digit is greater than its unit digit by 5. If d number is 14 less than 3 times d product of its digits, find d number.

Let's see how many Nairalanders can give it a shot. Anyone who answers correctly will get a reward.

Cc: Xflint, Boyoski09, truth12, themall, Danosaur007, Jerrypolo, 8BitGee, integritylady, Laive, lilmonarch