Boladearo: 3+5+7/3 =15/3 = 5.
My oga at the top am i correct.
this question is from this year mathematics cowbell competition, am i right

yeah

echibuzor:
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.1) Standard deviation is still d (Option E)
2) Actual Value of x-y is -25, but id value is 25 (Option B)

No. 1 is correct but explain ur answer in question 2.

busuyem:

No. 1 is correct but explain ur answer in question 2.

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The difference between two consecutive odd numbers is 2, with this info, u can use the Sum of arithmetic Progression to get the sum of odd numbers from 1 - 50, which is 625, Then u do the same for the sum of even numbers between 1 - 50, which is 650... The difference is 25.. U got it now?

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busuyem: solve these questions:
1) If the standard deviation of a, b and c is d, what is the standard deviation of a+1, b+1 and c+1?
A. 3d B. 2d C. d/2 D. d+1 E. d
2)If the sum of the odd integers frm 1 to 50 is x and the sum of the even integers frm 1 to 50 is y, what id the value of x-y?
A. 50 B. 25 C. 0 D. 75 E. 15

...1.increase or decrease of each element in a data by a constant number or element has no effect on variance and standard deviation...so the answer is still d...2...for odd...1,3,5...and the nth term is 2n-1...so,we need to find the summation of that...therefore €(summation)2n-1.and from the rule,it becomes €2n-€1...therefore 2€n-€1...and €n=n(n+1)/2 while €1=n...therefore back to 2€n-€1...which is equal to n(n+1)-n...which is equal to n^2...therefore the formula for the sum of odd numbers is n^2...and there are 25 odd numbers in 50...so it will be 25 square to give 625(odd numbers)...therefore y=625 and using the same procedure...for the even numbers the sum is n(n+1) and there are 25 even numbers in 50...therefore 25*26=650...therefore x=650...therefore x-y=25

Calculusf(x):
...1.increase or decrease of each element in a data by a constant number or element has no effect on variance and standard deviation...so the answer is still d...2...for odd...1,3,5...and the nth term is 2n-1...so,we need to find the summation of that...therefore €(summation)2n-1.and from the rule,it becomes €2n-€1...therefore 2€n-€1...and €n=n(n+1)/2 while €1=n...therefore back to 2€n-€1...which is equal to n(n+1)-n...which is equal to n^2...therefore the formula for the sum of odd numbers is n^2...and there are 25 odd numbers in 50...so it will be 25 square to give 625(odd numbers)...therefore y=625 and using the same procedure...for the even numbers the sum is n(n+1) and there are 25 even numbers in 50...therefore 25*26=650...therefore x=650...therefore x-y=25

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.Great minds think alike...

rashywire: pls help luk @ dis
Ade cn do a certain job in 3hrs, Obi cn do d same job in 5hrs, and Audu cn do d same job in 7hrs. hw lng would d job take d three of dem working 2gether?

...ade can do it in 3 hours...therefore for 1 hour...ade will do 1/3 of the job...and for 1 hour obi will do 1/5 of the job while audu will do 1/7 of the job for just one hour...combining the work all of them will do in one hour...therefore it's 1/3 + 1/5 + 1/7=71/105...therefore for one hour they will do 71/105 of the job...and for the whole job to be completed they will do 105/71 as this;
1hr~~>71/105 therefore,
xhr~~>1...
Cross multiply from there to get 71x/105=1...therefore x=105/71 hours

Calculusf(x):
...ade can do it in 3 hours...therefore for 1 hour...ade will do 1/3 of the job...and for 1 hour obi will do 1/5 of the job while audu will do 1/7 of the job for just one hour...combining the work all of them will do in one hour...therefore it's 1/3 + 1/5 + 1/7=71/105...therefore for one hour they will do 71/105 of the job...and for the whole job to be completed they will do 105/71 as this;
1hr~~>71/105 therefore,
xhr~~>1...
Cross multiply from there to get 71x/105=1...therefore x=105/71 hours

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You are correct sir, I saw this question on another thread earlier today, I think this is a variation problem... All the same.. Good Job..

echibuzor:
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You are correct sir, I saw this question on another thread earlier today, I think this is a variation problem... All the same.. Good Job..

...thank you general...

Calculusf(x):
...ade can do it in 3 hours...therefore for 1 hour...ade will do 1/3 of the job...and for 1 hour obi will do 1/5 of the job while audu will do 1/7 of the job for just one hour...combining the work all of them will do in one hour...therefore it's 1/3 + 1/5 + 1/7=71/105...therefore for one hour they will do 71/105 of the job...and for the whole job to be completed they will do 105/71 as this;
1hr~~>71/105 therefore,
xhr~~>1...
Cross multiply from there to get 71x/105=1...therefore x=105/71 hours

You're right boss. I think I've done it for him @unilag's admission thread.
So to continue from where you stopped:
x = 1.478873. . . . . . .which is 1hrs 28mins 44secs

What's the centre of gravity of an equilateral triangle of side 24?...

Calculusf(x):
What's the centre of gravity of an equilateral triangle of side 24?...

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This should be a geometric construction problem now? Or not...

echibuzor:
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This should be a geometric construction problem now? Or not...

...it's not bro,it just requires something called the position of centre of gravity of a triangle which is two-third of its altitude...

Calculusf(x):
...it's not bro,it just requires something called the position of centre of gravity of a triangle which is two-third of its altitude...

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I have absolutely no idea of what you are talking about.. And by that I am Assuming u are an engineer of some sorts?

Most of us r always bn encouraged nd most tyms luv 2 post ..but our engagement often times prevents us...2 al d general nd guruz n d houx..."hat's off" 4 al....sti following

rashywire: pls i need d answer b4 3pm.
QUESTION
A MAN IS X YRS OLD WHILE HIS SON IS Y YRS OLD. D SUM OF THEIR AGES IS TWICE D PRODUCT OF THEIR AGE. IF D PRODUCT OF THEIR AGE IS 675, FIND D AGE OF D MAN

You guyz're right, there was a mistake in his question. Here it is jare:

A man is x years old, his son is y years old. The sum of their ages is twice the difference of their ages. If the product of their ages is 675, find the age of the man.

@rashywire, pls try to reveiw your questions before posting.

Ortarico:

You guyz're right, there was a mistake in his question. Here it is jare:

A man is x years old, his son is y years old. The sum of their ages is twice the difference of their ages. If the product of their ages is 675, find the age of the man.

@rashywire, pls try to reveiw your questions before posting.

x + y = 2 (x-y)---------------(1)
x + y = 2x - 2y
-x = -3y
x= 3y ------------------- (1a)

x.y = 675 ---------------------(2)
substituting

3y.y = 675
y^2 = 675/3 = 225

y = sq rt (225) = 15

x = 15 x 3 = 45

Calculusf(x):
...it's not bro,it just requires something called the position of centre of gravity of a triangle which is two-third of its altitude...

I thnk the altitude can be found using pythagoras theorem since all its sides are equal
x^2=24^2-12^2
x^2=576-144
x^2=432
x=20.7...i cnt find a calculator
2/3rd of that shuld be 13.8
Is that what you mean?

yemmytcm: learn from what ? I should learn that 4 is a prime number ?...[sighs]..it i s well

IT WAS 5 THAT EMEKA WANTED TO WRITE NOT 4....THE PRIME NUMBERS BETWEEN 1-10 ARE FOUR IN NUMBER WHICH INCLUDE 2,3,5,7....

Calculusf(x):
What's the centre of gravity of an equilateral triangle of side 24?...

If its an equilateral triangle, dat means all d sides ae equal. Taking one side of d triangle, u will have d adjascent side as 12, d hypotenuse as 24, while d oder side is unknown. Let it b y.
Using pathagoreas theorem
24^2 - 12^2=576-144
=sqrt(432)=12surd(3).
Therefore d centre of gravity will be
12surd(3)/2 =6surd(3).
D ans is 6surd(3)

2nioshine: Most of us r always bn encouraged nd most tyms luv 2 post ..but our engagement often times prevents us...2 al d general nd guruz n d houx..."hat's off" 4 al....sti following

Boss......i greet u sir

Pls solve this; 1. Thirty men take 20 days to complete a job working 9 hours a day. How many hour a day should 40 men work to complete the job?

2. A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work n 23 days. The number of days after which A left the work was

yetunde01: Pls solve this; 1. Thirty men take 20 days to complete a job working 9 hours a day. How many hour a day should 40 men work to complete the job?

2. A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work n 23 days. The number of days after which A left the work was

1. Your first q is not clear as no of days is missing
But i will assume the same 20 day period

30 men use 20 days working 9hrs
The quantum of work is thus 20 * 9 * 30 = 5400 man hours

Dividing this by 40 men and 20 days leads to 5400/800 = 54/8 = 6.75 hours daily


2.
A's work rate comprises 1/45 of the job daily
B's work rate comprises 1/40 of the job daily

Together they work 17/360 of thejob (add the two fractions)
Now they work for x no of days
This means that 17[i]x[/i] /360 of the job will be done

Then B works for 23 days which means he does 23/40 of the job
17x/360 + 23/40 = 1
Solve for x

Using 360 as a LCM

17x + 207 = 360
17x = 153
x = 9 days

They worked togeter for 9 days

biolabee:

1. Your first q is not clear as no of days is missing
But i will assume the same 20 day period

30 men use 20 days working 9hrs
The quantum of work is thus 20 * 9 * 30 = 5400 man hours

Dividing this by 40 men and 20 days leads to 5400/800 = 54/8 = 6.75 hours daily


2.
A's work rate comprises 1/45 of the job daily
B's work rate comprises 1/40 of the job daily

Together they work 17/360 of thejob (add the two fractions)
Now they work for x no of days
This means that 17[i]x[/i] /360 of the job will be done

Then B works for 23 days which means he does 23/40 of the job
17x/360 + 23/40 = 1
Solve for x

Using 360 as a LCM

17x + 207 = 360
17x = 153
x = 9 days

They worked togeter for 9 days

Thanks. help with this 1.After allowing a discount of 11.11% ,a trader still makes a gain of 14.28 % .at how many precent above the cost price does he mark his goods?
After allowing a discount of 11.11% ,a trader still makes a gain of 14.28 % .at how many precent above the cost price does he mark his goods?

2.If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours a day; how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type.

yetunde01:

Thanks. help with this

1.After allowing a discount of 11.11% ,a trader still makes a gain of 14.28 % .at how many precent above the cost price does he mark his goods?

2.If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours a day; how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type.

No problem though i prefer to know where you gor stuck
this way you learn rather than just cramming what you are told

1.
Let SP be the final selling price and CP the cost price, SP' will be the original selling price
SP = (100+ 14.28)% = 114.28% of CP
However SP = (100-11.11%) of original selling price = 88.89% of SP'

thus 114.28 * CP = 88.89 * SP'

SP'/CP = 114.28/88.89 = 1.26

Thus he marks his gods by 26%

2. 24 tonnes services 9 engines for 8 hours/day
3 engines in the old batch here use the same fuel as 4 engines in the new batch
Thus using the new engines as the pilot case we have 4 + (9-3) = 4+6 = 10 engines comparable to the new batch

10 new engines working 8 hrs/day (10 * 8 new engine-hours) consume 24 tonnes
Thus 80 new engine-hours consume 24 t
1 new-engine-hour consume 24/80t
Since there are 8 ne engines running 13 hours daily
These requires 8 * 13 = 104 new-engine-hours
104 new-engine-hours require (24/80) * 104 t = 31.2 tonnes

Nice work @General biolabee!...

doubleDx: Nice work @General biolabee!...

Oga sir... I am humble o///

biolabee:

Oga sir... I am humble o///

Well done Sir

need ur assistance guys:: Integrate sqareroot(x^2+4x)dx

zzz zzz zzz

Mbahchiboy: need ur assistance guys:: Integrate sqareroot(x^2+4x)dx

...NOTE:£...REPRESENTS THE INTEGRAL FUNCTION...the question is £sqrt x^2+4x.dx...make it a perfect square...so it becomes £sqrt x^2+4x+4-4.dx...which equals to £sqrt(x+2)^2 -2^2.dx...{NOTE...when in form of sqrtx^2+a^2,use substitution x=asinh@...in form of sqrta^2-x^2,use substitution x=asin@ and in form of sqrtx^2-a^2,use substitution x=acosh@}so,which one are we going to use now...x=acosh@...so back to where we stopped...£sqrt(x+2)^2-2^2.dx...then we substitute x+2=2cosh@.equation 1...substitute that,then it becomes £sqrt(2cosh)^2-4dx=£sqrt4cosh^2@-4dx=£sqrt4(cosh^2@-1)dx=£2sqrt(cosh^2@-1)dx...then let's go to hyperbolic functions...cosh^2@-sinh^2=1...therefore cosh^2@-1=sinh@...substitute it to where we stopped...£2sqrt(cosh^2@-1)dx...=£2sqrtsinh^2@.dx...=£2sinh@.dx...so,from equation 1...x+2=2cosh@...dx=2sinh@d@...then substitute that to where we stopped...£2sinh@.2sinh@d@...£4sinh^2@d@.equation x...from hyperbolic identities...cosh^2@+sinh^2@=cosh2@...(i) and from the first identity of hyperbolic functions...cosh^2@-sinh^2@=1...cosh^2@=sinh^2@+1...substitute that to (i)...cosh^2@+sinh^2@=cosh2@...then 1+2sinh^2@=cosh2@...therefore sinh^2@=(cosh2@-1)/2...then take that to equation x=£4sinh^2@.d@...£4.(cosh2@-1)2.d@...£2cosh2@-2d@...integrate normally to get...sinh2@-2@...equation 2...from hyperbolic identity...sinh2@=2sinh@cosh@...therefore it becomes 2sinh@cosh@-2@...equation 3...but sinh^2@=cosh^2@-1(as we did earlier)...and from equation 1...x+2=2cosh@,cosh@=(x+2)/2...substitute that...sinh^2@=(x+2)^2/4 -1...find the l.c.m and do it to get (x^2+4x)/4.{therefore sinh@=sqrt(x^2+4x)/2...and from equation 1...(x+2)/2=cosh@...therefore @=arccosh(x+2)/2}...back to equation 3...2sinh@cosh@-@...substitute those in bracket...to get 2*sqrt(x^2+4x)/2*(x+2)/2 - 2.arccosh(x+2)/2 + c...which becomes (x+2)sqrt(x^2+4x)/2-2arccosh(x+2)/2+c...IT'S SIMPLE...JUST SIT AND STUDY IT...

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