₦airaland Forum

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

Where are d math gurus in d house?
This's just basic math

Wealthgang:
Where are d math gurus in d house? This's just basic math

i will give it a look

We r coming

disgraceful!

7a.)n =0 or 1

b) x = 1 or 4

3) The father is 60 years old while the son is 12years

(Q3) 60 and 12.
(Q7a) n=0 or 1
(Q7b) x=4 or 1
(Q9a) 94

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
20x² - 100x + 80 = 0
20         20      20


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
[s][/s]

3²ⁿ⁺¹ - 4(3ⁿ⁺¹) + 9 = 0

Note that
3²ⁿ⁺¹ = 3¹.3²ⁿ = 3¹.(3ⁿ)²

while
4(3ⁿ⁺¹) = 4(3¹.3ⁿ)

Now
3(3ⁿ)² - 4 x 3(3ⁿ) + 9 = 0
3(3ⁿ)² - 12(3ⁿ) + 9 = 0
Let us call 3ⁿ = y

therefore 3y² - 12y + 9 = 0

Divide by 3
3yⁿ - 12y + 9 = 0
3       3       3

we will have
y² - 4y + 3 = 0
y² - 3y - y + 3 = 0
(y² - 3y)(-y +3)
y(y - 3) - 1(y - 3)
(y - 1)(y - 3)
y=1 or 3

therefore when y=1
3ⁿ = y
n = 0

when y = 3
3ⁿ = y
n = 1

n= 0 or 1    Answer

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

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
. You missed sir. The correct equation for 4 years ago is (5x-4)(x-4)=448
Op

5x^2-20x-4x+16=448

Wealthgang:
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

will try harder next time

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) into 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

essay123:
5x^2-20x-4x+16=448

Yeah...that's VERY true Essay123! Thanks for the correction.

TITOBIGZ:
We r coming

We r waiting

Elxandre:
disgraceful!

Na so we see am o...
Remember dat teacher in Edo who couldn't read well?

Mypeople2:
7a.)n =0 or 1

Mypeople2:
b) x = 1 or 4

Mypeople2:
3) The father is 60 years old while the son is 12years

Nailed it. Respek Boss!

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

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

Xflint:
3²ⁿ⁺¹ - 4(3ⁿ⁺¹) + 9 = 0
Note that
3²ⁿ⁺¹ = 3¹.3²ⁿ = 3¹.(3ⁿ)²
while
4(3ⁿ⁺¹) = 4(3¹.3ⁿ)
Now
3(3ⁿ)² - 4 x 3(3ⁿ) + 9 = 0
3(3ⁿ)² - 12(3ⁿ) + 9 = 0
Let us call 3ⁿ = y
therefore 3y² - 12y + 9 = 0
Divide by 3 we will have
y² - 4y + 3 = 0
y² - 3y - y + 3 = 0
(y² - 3y)(-y +3)
y(y - 3) - 1(y - 3)
(y - 1)(y - 3)
y=1 or 3
therefore when y=1
3ⁿ = y
n = 0
when y = 3
3ⁿ = y
n = 1
n= 0 or 1 Answer

A casket for these two questns. They've been murdered .
Respek boss! Tnx 4 showing d working too.

Laive:
Let the son's age be:x

Therefore, father's age:5x

Splendid!

Laive:
Four years ago, if we multiply their ages, we would get:

5x = 448

Four years ago, d father was (5x-4)yrs while d son was (x-4)yrs old
The product of their ages 4yrs ago equaled 448.
=> (5x-4)(x-4) = 448
5x² - 24x + 16 = 448
5x² - 24x - 432 = 0
This quadratic equation has d solution to d problem

essay123:
5x^2-20x-4x+16=448

I stopped solving this thing at this point.

Just couldn't bother to factorize Or find any roots

Wealthgang:
Nailed it. Respek Boss!

Where are his workings pls

integritylady:
will try harder next time

I know u can solve it. Seen some of ur handwork

Mypeople2:
3) The father is 60 years old while the son is 12years

pls Mypeople2, show us some workings...

Wealthgang:
Splendid!



Four years ago, d father was (5x-4)yrs while d son was (x-4)yrs old
The product of their ages 4yrs ago equaled 448.
=> (5x-4)(x-4) = 448
5x² - 24x + 16 = 448
5x² - 24x - 432 = 0
This quadratic equation has d solution to d problem

My friend, complete your solvings.

Wealthgang:
Splendid!



Four years ago, d father was (5x-4)yrs while d son was (x-4)yrs old
The product of their ages 4yrs ago equaled 448.
=> (5x-4)(x-4) = 448
5x² - 24x + 16 = 448
5x² - 24x - 432 = 0
This quadratic equation has d solution to d problem

Ok! Learnt something new! But the quadratic equation does not say much about their present ages. Why?

Elxandre:
I stopped solving this thing at this point.

Just couldn't bother to factorize Or find any roots

Looks scary but its actually easy

@Mypeople. No workings Sir

Elxandre:
My friend, complete your solvings.

I'm d examiner here

Wealthgang:
Splendid!



Four years ago, d father was (5x-4)yrs while d son was (x-4)yrs old
The product of their ages 4yrs ago equaled 448.
=> (5x-4)(x-4) = 448
5x² - 24x + 16 = 448
5x² - 24x - 432 = 0
This quadratic equation has d solution to d problem

After this, use quadratic equation general formular [ -b +_ √(b^2-4ac)]/2a.You are going to get .X = 12 or -7.2.Disregard the negative value of x since someone's age cannot be negative. So the son's age(x) is 12 .Remember the father's age is 5x....so 5 times 12 = 60.Therefore the father is 60 years old while the son is 12 years old

Laive:
Ok! Learnt something new! But the quadratic equation does not say much about their present ages. Why?

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