IF ronaldo , bale , isco collect 150%, 100% , 75% of X respectively and messi's salary in 14 weeks 5days is what ronaldo pays as tax per annual . provided that tax is 45% of income , express the amount each player will take home in terms of X

If you don't understand what this OP is rambling about,just type "Halleluyah!"

am too busy calculating the amount of all this celebrities.... when will people start calculating mine? I don't even care about how much any of my club players are sold or bought.....f************k

Solving problems accurately in Mathematics is about specificity. So @OP next time be specific about what kind of year because leap year considered makes a difference in result.

Now,
Let 1 year = 365 days
Let annual salary = AS = X
Let Ronaldo's annual salary = RAS = 150X
Let Bale's annual salary = BAS = 100X
Let Isco's annual salary = IAS = 75X
Let Messi's annual salary = soccergod
Let Annual tax = AT = 45% annual salary


Ronaldo's annual tax = RAT = 45% of RAS = 0.45 * 150X = 67.5X

14weeks 5days = 103 days
Messi annual salary = soccergod (in 365 days)
If Messi earns 67.5X = 103days,
Then 1 day = 67.5X /103
=> 365 days = (67.5X / 103) * 365 = 239.19903X
soccergod = 239.19903X = 239.20X (approx. 2 dp)

Isco Annual tax = IAT = 45% of IAS = 0.45 * 75X = 33.75X
Bale's Annual tax = BAT = 45% of BAS = 0.45 * 100X = 45X
Messi Annual tax = Fuk-da-govt = 45% of soccergod = 0.45 * 239.20X = 107.64X

Therefore,
Take home
1) Ronaldo = 150X - 67.5X = 82.5X
2) Bale = 100X - 45X = 55X
3) Isco = 75X - 33.75X = 41.25X
4) Messi = 239.20X - 107.64X = 131.56X.

This ur question get as e be. Very funny question. Not clear, implicit; ambiguous!
Lets make these assumptions anyway:

1.Players are paid wages weekly
2. There'r 52 weeks in year.
3. 7 days a week.
4. Income is taxed weekly.

Income before tax:
ronaldo, 1.5x
bale; x
isco, 0.75x

income after tax:
ronaldo; 0.825x
bale; 0.55x
isco; 0.4125x

ronaldo's annual tax = 52*0.675x = 35.1x = messi's wages in 14 weeks and 5 days.

14weeks+5 days= 103 days.
Messi's income per day= 35,1x/103
messi's income per week= (35.1x/103)*7= 2.39x

therefore, messi's income after tax=0.55*2.39x= 1.3145x

so in terms of x, income after tax for each repective player per week is;
ronaldo's = 0.825x
bale= 0.55x
isco=0.4125x
messi= 1.3145x

i had to make some assumptions. I even assume ur question.

Let
Ronaldo =R = 1.5X
Bale =B = 1.0X
Isco =I = 0.75X
Messi =M = 2.4X

N.B
For Messi
0.45*R = M(in 14weeks 5 days= 103 days)
0.45*1.5X =M(in 103 days)
0.45*1.5X*365/103 = M( in 365 days)
M = 2.4X(approx)

There take home pay after tax(55% of earnings),
R= 0.55*1.5X =0.825X.
B = 0.55*1.0X =0.55X.
I = 0.55*0.75 = 0.4125X.
M = 0.55*2.4 = 1.32X.

paranorman: This ur question get as e be. Very funny question. Not clear, implicit; ambiguous!
Lets make these assumptions anyway:

1.Players are paid wages weekly
2. There'r 52 weeks in year.
3. 7 days a week.
4. Income is taxed weekly.

Income before tax:
ronaldo, 1.5x
bale; x
isco, 0.75x

income after tax:
ronaldo; 0.825x
bale; 0.55x
isco; 0.4125x

ronaldo's annual tax = 52*0.675x = 35.1x = messi's wages in 14 weeks and 5 days.

14weeks+5 days= 103 days.
Messi's income per day= 35,1x/103
messi's income per week= (35.1x/103)*7= 2.39x

therefore, messi's income after tax=0.55*2.39x= 1.3145x

so in terms of x, income after tax for each repective player per week is;
ronaldo's = 0.825x
bale= 0.55x
isco=0.4125x
messi= 1.3145x

i had to make some assumptions. I even assume ur question.

making assumptions without telling you shouldn't be problem . Thanks for solving it all are correct except that of messi I got 1.3112X

McWhillion: Let
Ronaldo =R = 1.5X
Bale =B = 1.0X
Isco =I = 0.75X
Messi =M = 2.4X

N.B
For Messi
0.45*R = M(in 14weeks 5 days= 103 days)
0.45*1.5X =M(in 103 days)
0.45*1.5X*365/103 = M( in 365 days)
M = 2.4X(approx)

There take home pay after tax(55% of earnings),
R= 0.55*1.5X =0.825X.
B = 0.55*1.0X =0.55X.
I = 0.55*0.75 = 0.4125X.
M = 0.55*2.4 = 1.32X.

nice bro

Vengeance:

Solving problems accurately in Mathematics is about specificity. So @OP next time be specific about what kind of year because leap year considered makes a difference in result.

Now,
Let 1 year = 365 days
Let annual salary = AS = X
Let Ronaldo's annual salary = RAS = 150X
Let Bale's annual salary = BAS = 100X
Let Isco's annual salary = IAS = 75X
Let Messi's annual salary = soccergod
Let Annual tax = AT = 45% annual salary


Ronaldo's annual tax = RAT = 45% of RAS = 0.45 * 150X = 67.5X

14weeks 5days = 103 days
Messi annual salary = soccergod (in 365 days)
If Messi earns 67.5X = 103days,
Then 1 day = 67.5X /103
=> 365 days = (67.5X / 103) * 365 = 239.19903X
soccergod = 239.19903X = 239.20X (approx. 2 dp)

Isco Annual tax = IAT = 45% of IAS = 0.45 * 75X = 33.75X
Bale's Annual tax = BAT = 45% of BAS = 0.45 * 100X = 45X
Messi Annual tax = Fuk-da-govt = 45% of soccergod = 0.45 * 239.20X = 107.64X

Therefore,
Take home
1) Ronaldo = 150X - 67.5X = 82.5X
2) Bale = 100X - 45X = 55X
3) Isco = 75X - 33.75X = 41.25X
4) Messi = 239.20X - 107.64X = 131.56X.

taking assumptions shouldn't be problem . you are correct anyways

I've this for you guys....

A contractor agreeing to finish a work in 150 days employed 75 men each working 8 hours daily. After 90 days, only 2/7 of the work was completed. Increasing the number of men by …………. each working now for 10 hours daily, the work can be completed in time.


CarlOzeki, Snowland, Tinkybabe, hopelink1, paranorman, vengeance, kobonaire.... YOUR ATTENTIONS ARE NEEDED

McWhillion: I've this for you guys....
A contractor agreeing to finish a work in 150 days employed 75 men each working 8 hours daily. After 90 days, only 2/7 of the work was completed. Increasing the number of men by …………. each working now for 10 hours daily, the work can be completed in time.
CarlOzeki, Snowland, Tinkybabe, hopelink1, paranorman, vengeance, kobonaire.... YOUR ATTENTIONS ARE NEEDED

Let work to be done = W
Let total duration of work = TD = 150 days = 150 * 24hrs =
Let initial employees = IE = 75
Let work phase one = 90 days
Let work time phase one per employee per day = WT1/E/D = 8hrs
Work duration per employee per phase one = WD/E/1 = WT1/E/D * 90 = 8 * 90 = 720hrs = 30 days
Total work duration per IE per phase one = TWD/IE/1 = WD/E/1 * 75 = 2250 days

After 90days, workdone by 75 employees = 2W/7
=> 2W/7 = 2250 days; W = 7875 days


[Remainder work to be done / Phase two]
Let balance days of work phase 2= BDW2 = 150 - 90 = 60 days
Let remainder work to be done = RW
Now,
Remainder work to be done = 5W/7 = 5625 days

Let WT2/E/D = 10hrs
Let number of additional workers = AW = z
Let total employees = TE = 75 + z
Work duration per employee per phase 2 = WD/E/2 = WT2/E/D * BDW2 = 10 * 60 = 600hrs = 25 days

=> Total work duration per employee per TE per phase two = TWD/TE/2 = WD/E/2 * TE = 5625 days

=> 25 days * (75 + z) = 5625 days;
75 + z = 225;
z = 225 - 75 = 150

Therefore additional workers required to finish the job done on schedule = 150 men

Vengeance:




Let work to be done = W
Let total duration of work = TD = 150 days = 150 * 24hrs =
Let initial employees = IE = 75
Let work phase one = 90 days
Let work time phase one per employee per day = WT1/E/D = 8hrs
Work duration per employee per phase one = WD/E/1 = WT1/E/D * 90 = 8 * 90 = 720hrs = 30 days
Total work duration per IE per phase one = TWD/IE/1 = WD/E/1 * 75 = 2250 days

After 90days, workdone by 75 employees = 2W/7
=> 2W/7 = 2250 days; W = 7875 days


[Remainder work to be done / Phase two]
Let balance days of work phase 2= BDW2 = 150 - 90 = 60 days
Let remainder work to be done = RW
Now,
Remainder work to be done = 5W/7 = 5625 days

Let WT2/E/D = 10hrs
Let number of additional workers = AW = z
Let total employees = TE = 75 + z
Work duration per employee per phase 2 = WD/E/2 = WT2/E/D * BDW2 = 10 * 60 = 600hrs = 25 days

=> Total work duration per employee per TE per phase two = TWD/TE/2 = WD/E/2 * TE = 5625 days

=> 25 days * (75 + z) = 5625 days;
75 + z = 225;
z = 225 - 75 = 150

Therefore additional workers required to finish the job done on schedule = 150 men

Nice one bro

Vengeance:

Solving problems accurately in Mathematics is about specificity. So @OP next time be specific about what kind of year because leap year considered makes a difference in result.

Now,
Let 1 year = 365 days
Let annual salary = AS = X
Let Ronaldo's annual salary = RAS = 150X
Let Bale's annual salary = BAS = 100X
Let Isco's annual salary = IAS = 75X
Let Messi's annual salary = soccergod
Let Annual tax = AT = 45% annual salary


Ronaldo's annual tax = RAT = 45% of RAS = 0.45 * 150X = 67.5X

14weeks 5days = 103 days
Messi annual salary = soccergod (in 365 days)
If Messi earns 67.5X = 103days,
Then 1 day = 67.5X /103
=> 365 days = (67.5X / 103) * 365 = 239.19903X
soccergod = 239.19903X = 239.20X (approx. 2 dp)

Isco Annual tax = IAT = 45% of IAS = 0.45 * 75X = 33.75X
Bale's Annual tax = BAT = 45% of BAS = 0.45 * 100X = 45X
Messi Annual tax = Fuk-da-govt = 45% of soccergod = 0.45 * 239.20X = 107.64X

Therefore,
Take home
1) Ronaldo = 150X - 67.5X = 82.5X
2) Bale = 100X - 45X = 55X
3) Isco = 75X - 33.75X = 41.25X
4) Messi = 239.20X - 107.64X = 131.56X.

Nice 1 exactly ma ans

Vengeance:

Solving problems accurately in Mathematics is about specificity. So @OP next time be specific about what kind of year because leap year considered makes a difference in result.

Now,
Let 1 year = 365 days
Let annual salary = AS = X
Let Ronaldo's annual salary = RAS = 150X
Let Bale's annual salary = BAS = 100X
Let Isco's annual salary = IAS = 75X
Let Messi's annual salary = soccergod
Let Annual tax = AT = 45% annual salary


Ronaldo's annual tax = RAT = 45% of RAS = 0.45 * 150X = 67.5X

14weeks 5days = 103 days
Messi annual salary = soccergod (in 365 days)
If Messi earns 67.5X = 103days,
Then 1 day = 67.5X /103
=> 365 days = (67.5X / 103) * 365 = 239.19903X
soccergod = 239.19903X = 239.20X (approx. 2 dp)

Isco Annual tax = IAT = 45% of IAS = 0.45 * 75X = 33.75X
Bale's Annual tax = BAT = 45% of BAS = 0.45 * 100X = 45X
Messi Annual tax = Fuk-da-govt = 45% of soccergod = 0.45 * 239.20X = 107.64X

Therefore,
Take home
1) Ronaldo = 150X - 67.5X = 82.5X
2) Bale = 100X - 45X = 55X
3) Isco = 75X - 33.75X = 41.25X
4) Messi = 239.20X - 107.64X = 131.56X.

Nice 1 exactly ma ans

McWhillion: I've this for you guys....

A contractor agreeing to finish a work in 150 days employed 75 men each working 8 hours daily. After 90 days, only 2/7 of the work was completed. Increasing the number of men by …………. each working now for 10 hours daily, the work can be completed in time.

Job X was to be completed by 75 men working 8hours/day => 75*8=600ManHours/day.

However @600ManHrs/day only 2/7 of X was completed.
Implying that to complete X, the number of man hour/days required = 600/(2/7) = 2100 ManHour/days

Since number of men have been increased by say Y and the daily work rate increased. The new ManHours/day = (75*10 + 10Y) = (750+10Y)
And it has been stated that this new ManHour/day rate will be able to complete the remaining part of the project (5/7X).
Since we know that the project X requires 2100 man hours
=> 5/7th of project requires 5/7*2100= 1500 ManHours/day

Thus 750 +10Y = 1500
Or Y = 75
It will thus require 75 additional men (all working at the new 10hr/day rate) to complete the project on time.

kobonaire:

Job X was to be completed by 75 men working 8hours/day => 75*8=600ManHours/day.

However @600ManHrs/day only 2/7 of X was completed.
Implying that to complete X, the number of man hour/days required = 600/(2/7) = 2100 ManHour/days

Since number of men have been increased by say Y and the daily work rate increased. The new ManHours/day = (75*10 + 10Y) = (750+10Y)
And it has been stated that this new ManHour/day rate will be able to complete the remaining part of the project (5/7X).
Since we know that the project X requires 2100 man hours
=> 5/7th of project requires 5/7*2100= 1500 ManHours/day

Thus 750 +10Y = 1500
Or Y = 75
It will thus require 75 additional men (all working at the new 10hr/day rate) to complete the project on time.

Its 150 men extra bro.

McWhillion:

Its 150 men extra bro.

how my guy?? ... pls show me the mistake in my calculation

kobonaire: how my guy?? ... pls show me the mistake in my calculation

Bro, you didn’t put the first 90 days and the leftover 60 days into consideration
If you were to include it in your equations, it will look like this:
The first phase.
@600 man hrs/day reps 2/7= 2100 man hrs/day
Which translates to working (2100*90) man hrs for the 90days = 189,000 man hrs (for the first phase of work)

The second phase.
@(75 +Y)*10 man hrs/day reps 5/7 = (750+10Y) man hrs/ day
Which translates to working ((750+10Y)*60) man hrs for the 60 days = ( 45,000 +600Y) man hrs (for the second phase of work)
From your eqs,(5/7 of the whole project = 5/7*189,000 = 135,000 man hrs for 60 days)

Thus your eq. becomes 45,000 + 600Y = 135,000(instead of 750+10Y= 1500)
Y= 150….
(Bro, your workings nearly burst my head o... you too dey think, naso i be too, i never find a simple way of solving maths problems)

McWhillion:
Bro, you didn’t put the first 90 days and the leftover 60 days into consideration.....

Bros, you are on 100% point ...that was a flaw in my earlier solution and thanks for the explanation. You are a math whiz. Yeah, i tend to take a complex approach to solving these problems (when a simpler approach may suffice) .. I blame it on my engineering background. During my engineering school days, teacher go ask us to be solving everything with differential equations, laPlace transformations etc etc ... maybe my head don permanently scatter small

kobonaire: Bros, you are on 100% point ...that was a flaw in my earlier solution and thanks for the explanation. You are a math whiz. Yeah, i tend to take a complex approach to solving these problems (when a simpler approach may suffice) .. I blame it on my engineering background. During my engineering school days, teacher go ask us to be solving everything with differential equations, laPlace transformations etc etc ... maybe my head don permanently scatter small

Naso o, wetin person don know no dey leave am.
Are u a practicing engineer?

I ve this for you guys,

There are 3 persons Wale, Femi, and Dele. Wale lent cars to Femi and Dele as many as they had already. After some time Femi gave as many cars to Wale and Dele as many as they have. After sometime Dele did the same thing. At the end of this transaction each one of them had 24 cars. Find the cars each originally had.

CarlOzeki, Snowland, Tinkybabe, hopelink1,
paranorman, vengeance, kobonaire, meryl4merit.... YOUR ATTENTIONS ARE NEEDED!

McWhillion: I ve this for you guys,

There are 3 persons Wale, Femi, and Dele. Wale lent cars to Femi and Dele as many as they had already. After some time Femi gave as many cars to Wale and Dele as many as they have. After sometime Dele did the same thing. At the end of this transaction each one of them had 24 cars. Find the cars each originally had.

CarlOzeki, Snowland, Tinkybabe, hopelink1,
paranorman, vengeance, kobonaire, meryl4merit.... YOUR ATTENTIONS ARE NEEDED!

from the look of the question, Wale's got to ave the highest number of cars. Wale's got 39 cars, Femi 21 and poor guy Dele has got just 12.

paranorman:

from the look of the question, Wale's got to ave the highest number of cars. Wale's got 39 cars, Femi 21 and poor guy Dele has got just 12.

Correct Sir,
But if Dele who has 12 cars is poor then i don't mind being poor bro.

McWhillion: I ve this for you guys,

There are 3 persons Wale, Femi, and Dele. Wale lent cars to Femi and Dele as many as they had already. After some time Femi gave as many cars to Wale and Dele as many as they have. After sometime Dele did the same thing. At the end of this transaction each one of them had 24 cars. Find the cars each ... YOUR ATTENTIONS ARE NEEDED!

If each of them had A, B and C cars respectively

After the first exchange then each respectively has:- (A-B-C), 2B and 2C

After 2nd exchange: (2A-2B-2C), (3A-A-C), 4C

And final exchange: 4A -4B-4C, 6B-2A-2C, 7C-A-B

And since we are told that after the final exchange, each has 24 cars, we can thus form 3 equations for our 3 variables :-
4A -4B-4C = 24,
6B-2A-2C = 24 ,
7C-A-B = 24.

So solving these 3 equations yields
A=39 : Wale
B=21 : Femi
C=12 : Dele

@paranorman how were you able to compute yours, hope not mentally, then you must be a Math PhD

kobonaire:

If each of them had A, B and C cars respectively

After the first exchange then each respectively has:- (A-B-C), 2B and 2C

After 2nd exchange: (2A-2B-2C), (3A-A-C), 4C

And final exchange: 4A -4B-4C, 6B-2A-2C, 7C-A-B

And since we are told that after the final exchange, each has 24 cars, we can thus form 3 equations for our 3 variables :-
4A -4B-4C = 24,
6B-2A-2C = 24 ,
7C-A-B = 24.

So solving these 3 equations yields
A=39 : Wale
B=21 : Femi
C=12 : Dele

@paranorman how were you able to compute yours, hope not mentally, then you must be a Math PhD

nope. Follwed a similar steps to urs. Na phone i dey use. Stressful to type solution when 'm not on my PC.

paranorman:

nope. Follwed a similar steps to urs. Na phone i dey use. Stressful to type solution when 'm not on my PC.

oya, please type out the outline .... mine took me a while to compute and I am looking for a more efficient solution.

[size=2pt]

Tinkybabe: let the no of cars each had initially be

Wale= z ,Femi =y ,Dele = x

STEP 1:Wale gave Femi and Dele y and x cars respectively
ie
Wale= z-(x+y)
Femi=y+y= 2y
Dele= x+x=2x

Let the no of cars left with Wale be t
Hence
T= z-(x+y) equ i

STEP 2:Femi gave Wale[b] t[/b] cars and Dele 2x cars , hence has left
2y -(t + 2x)

STEP 3: Dele gave Wale 2t cars and Femi 2y-(t+2x) cars ,therefore he is left with

4x-[2y-(t+2x) +2t]

So in total each has
Wale ; 4t = 24 equ 1
Femi ; 2[2y-(t+2x)]= 24 equ 2
Dele ; 4x-[2y-(t+2x) +2t] = 24 equ 3

from equ 1 , t= 6

from equ 2 , 2y-(t+2x) = 12
2(y-x) = 12+6 = 18
y-x = 9
y = 9+x equ ii

from equ 3, use y = 9+x
4x-[2(9+x)-6-2x+2(6) =24
4x -[(18-6)+12]= 24
4x = 48
x= 12
y= 9+12 = 21
{from equ i; t = z-(y+x)}
6=z-(21+12) ; z= 6+33
z = 39

Therefore initially,
Wale had 39 cars
Femi had 21 cars
Dele had 12 cars

[/size]nice work - nothing hotter than a chic who digs mathematics

kobonaire:

If each of them had A, B and C cars respectively

After the first exchange then each respectively has:- (A-B-C), 2B and 2C

After 2nd exchange: (2A-2B-2C), (3A-A-C), 4C

And final exchange: 4A -4B-4C, 6B-2A-2C, 7C-A-B

And since we are told that after the final exchange, each has 24 cars, we can thus form 3 equations for our 3 variables :-
4A -4B-4C = 24,
6B-2A-2C = 24 ,
7C-A-B = 24.

So solving these 3 equations yields
A=39 : Wale
B=21 : Femi
C=12 : Dele

@paranorman how were you able to compute yours, hope not mentally, then you must be a Math PhD

Nice one...

Tinkybabe: let the no of cars each had initially be

Wale= z ,Femi =y ,Dele = x

STEP 1:Wale gave Femi and Dele y and x cars respectively
ie
Wale= z-(x+y)
Femi=y+y= 2y
Dele= x+x=2x

Let the no of cars left with Wale be t
Hence
T= z-(x+y) equ i

STEP 2:Femi gave Wale[b] t[/b] cars and Dele 2x cars , hence has left
2y -(t + 2x)

STEP 3: Dele gave Wale 2t cars and Femi 2y-(t+2x) cars ,therefore he is left with

4x-[2y-(t+2x) +2t]

So in total each has
Wale ; 4t = 24 equ 1
Femi ; 2[2y-(t+2x)]= 24 equ 2
Dele ; 4x-[2y-(t+2x) +2t] = 24 equ 3

from equ 1 , t= 6

from equ 2 , 2y-(t+2x) = 12
2(y-x) = 12+6 = 18
y-x = 9
y = 9+x equ ii

from equ 3, use y = 9+x
4x-[2(9+x)-6-2x+2(6) =24
4x -[(18-6)+12]= 24
4x = 48
x= 12
y= 9+12 = 21
{from equ i; t = z-(y+x)}
6=z-(21+12) ; z= 6+33
z = 39

Therefore initially,
Wale had 39 cars
Femi had 21 cars
Dele had 12 cars

Very nice one there

sorry ,