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HARDEST MATHS Question Ever by XXX5(m): 11:56am On Dec 20, 2014
A finite set "S" is defined by positive integers "x"...in which 5≤x≤50.
Two numbers exist in the set such that their sum is greater than the upper limit of the set by 5. The digits "057" is the mirrored image of the product of the two numbers (r&n). Calculate the solution of "nPr".



HINT: n>r,
S={5,6,...,r,...,27,...,n,...50}
nPr=n!/(n-r)!

3 Likes 4 Shares

Re: HARDEST MATHS Question Ever by Nobody: 12:21pm On Dec 20, 2014
Grabs my pen , be right back .


Question has been solved , came back late but this question is the hardest question ever for you alone.

5 Likes 2 Shares

Re: HARDEST MATHS Question Ever by Nastydroid(m): 12:39pm On Dec 20, 2014
Re: HARDEST MATHS Question Ever by Nobody: 12:41pm On Dec 20, 2014
30 and 25, 30p25 = 2.210441e30

10 Likes

Re: HARDEST MATHS Question Ever by bhaliz44(m): 1:26pm On Dec 20, 2014
Ask omadioha angry

23 Likes 1 Share

Re: HARDEST MATHS Question Ever by Lordseyad(m): 1:39pm On Dec 20, 2014
Simple question
Re: HARDEST MATHS Question Ever by XXX5(m): 2:09pm On Dec 20, 2014
juanmata:
30 and 25

incomplete answer...calculate nPr
Re: HARDEST MATHS Question Ever by dejt4u(m): 2:42pm On Dec 20, 2014
XXX5:
A finite set "S" is defined by positive integers "x"...in which 5≤x≤50.
Two numbers exist in the set such that their sum is greater than the upper limit of the set by 5. The digits "057" is the mirrored image of the product of the two numbers (r&n). Calculate the solution of "nPr".



HINT: n>r,
S={5,6,...,r,...,27,...,n,...50}
nPr=n!/(n-r)!

(n, r) = (30, 25) since n+r = 55 (that is, their sum is greater than the upper limit of the set by 5)..
30P25 = 30!/5! = 2.21 E 30

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Re: HARDEST MATHS Question Ever by osamaBUSH(m): 2:44pm On Dec 20, 2014
d answer dey back of d textbook. go check am.

43 Likes 6 Shares

Re: HARDEST MATHS Question Ever by Nobody: 2:50pm On Dec 20, 2014
XXX5:


incomplete answer...calculate nPr
30p25 = 2.210441e30

9 Likes

Re: HARDEST MATHS Question Ever by rattlesnake(m): 3:20pm On Dec 20, 2014
XXX5:
A finite set "S" is defined by positive integers "x"...in which 5≤x≤50.
Two numbers exist in the set such that their sum is greater than the upper limit of the set by 5. The digits "057" is the mirrored image of the product of the two numbers (r&n). Calculate the solution of "nPr".



HINT: n>r,
S={5,6,...,r,...,27,...,n,...50}
nPr=n!/(n-r)!
mad

1 Like

Re: HARDEST MATHS Question Ever by fr3do(m): 3:42pm On Dec 20, 2014
Elantracey:



lol , I love mathematics

Engineer!
Re: HARDEST MATHS Question Ever by fr3do(m): 3:50pm On Dec 20, 2014
30!/(30-25)!
Re: HARDEST MATHS Question Ever by fr3do(m): 3:50pm On Dec 20, 2014
We Nigerians know book o
Re: HARDEST MATHS Question Ever by Chuukwudi(m): 3:59pm On Dec 20, 2014
XXX5:
A finite set "S" is defined by positive integers "x"...in which 5≤x≤50.
Two numbers exist in the set such that their sum is greater than the upper limit of the set by 5. The digits "057" is the mirrored image of the product of the two numbers (r&n). Calculate the solution of "nPr".
HINT: n>r,
S={5,6,...,r,...,27,...,n,...50}
nPr=n!/(n-r)!


If you need assistance with your assignments why not tell us! If you call this difficult, then you've not seen anything.

16 Likes

Re: HARDEST MATHS Question Ever by Nobody: 4:34pm On Dec 20, 2014
Elantracey:
Grabs my pen , be right back .


Question has been solved , came back late but this question is the hardest question ever for you alone.

So you like mathematics too? cool shocked
Re: HARDEST MATHS Question Ever by Nobody: 4:38pm On Dec 20, 2014
Dapo777:


So you like mathematics too? cool shocked


lol , I love mathematics
Re: HARDEST MATHS Question Ever by SELENAqueensy(f): 6:40pm On Dec 20, 2014
I just hate math with my life

17 Likes 1 Share

Re: HARDEST MATHS Question Ever by PHAYOL81: 7:25pm On Dec 20, 2014
Dis is exactly y i hate MADmatics....it looks really crazy i cant underrstand a thng.Tell d answr and i'd still found it difficult 2 get it yet.#effect of MADmatics#

9 Likes

Re: HARDEST MATHS Question Ever by goalburner(m): 7:29pm On Dec 20, 2014
ok
Re: HARDEST MATHS Question Ever by Emodeee: 7:37pm On Dec 20, 2014
@elantracey na mouth u get. Oya solve am
Re: HARDEST MATHS Question Ever by Obinoscopy(m): 7:43pm On Dec 20, 2014
30 × 29 × 28 × 27 × 26

1 Like

Re: HARDEST MATHS Question Ever by Nobody: 7:51pm On Dec 20, 2014
SELENAqueensy:
I just hate math with my life
If I'm right u hate maths and u hate ur life? Or is it a grammatical sulphuri-volcanous blunder?

10 Likes

Re: HARDEST MATHS Question Ever by naturalwaves: 8:23pm On Dec 20, 2014
There is nothing hard about this sir. I have solved it. Find the solution attached below......

119 Likes 9 Shares

Re: HARDEST MATHS Question Ever by Nobody: 9:42pm On Dec 20, 2014
Emodeee:
@elantracey na mouth u get. Oya solve am
she's an electrical student, mathematics is life
Re: HARDEST MATHS Question Ever by Emodeee: 10:01pm On Dec 20, 2014
bataderemi:

she's an electrical student, mathematics is life

Sheeee u fit solve am?
Re: HARDEST MATHS Question Ever by Nobody: 10:08pm On Dec 20, 2014
Yep..... A claim I obviously can't prove but which u can't refute either

1 Like

Re: HARDEST MATHS Question Ever by Nobody: 10:58pm On Dec 20, 2014
Emodeee:
@elantracey na mouth u get. Oya solve am
it has already being solved , of what use is solving it again?
Re: HARDEST MATHS Question Ever by XXX5(m): 11:21pm On Dec 20, 2014
Solution:
Upper limit of the set = 50
n + r = 50 + 5
n + r = 55
n = 55 - r
from St 2.
n × r = 750
but n = 55 - r
hence,
r(55 - r) = 750
r² - 55r + 750 = 0
r² - 30r - 25r + 750 = 0
r(r - 30) -25(r - 30) = 0
r = 30 Or 25.
but n + r = 55
for r = 25
n = 55 - 25 = 30.
for r = 30
n = 55 - 30 = 25.
but n + r > 50 by 5.
if r = 25 and n = 30
25 + 30 = 55 > 50 by 5.
and from the question, n > r.
so the right evaluation to be used is
r = 25 and n = 30.
Now, nPr
nPr = n!/(n - r)!
nPr = 30!/(30 - 25)!
nPr = 30!/5!

27 Likes 1 Share

Re: HARDEST MATHS Question Ever by Nobody: 11:30pm On Dec 20, 2014
Elantracey:



lol , I love mathematics

Wow nice. Didn't know you were a nerd
Re: HARDEST MATHS Question Ever by Bsmartt(m): 11:33pm On Dec 20, 2014
Please why is it that the alphabet mostly used in mathematics are 'a' 'b' 'x' 'y' 'n' and sometimes 'r'. But at the long run the answers are not more than 2 digits. Mathematics is very interesting and technical.

3 Likes

Re: HARDEST MATHS Question Ever by UnimkeAk(m): 11:46pm On Dec 20, 2014
XXX5:
Solution:
Upper limit of the set = 50
n + r = 50 + 5
n + r = 55
n = 55 - r
from St 2.
n × r = 750
but n = 55 - r
hence,
r(55 - r) = 750
r² - 55r + 750 = 0
r² - 30r - 25r + 750 = 0
r(r - 30) -25(r - 30) = 0
r = 30 Or 25.
but n + r = 55
for r = 25
n = 55 - 25 = 30.
for r = 30
n = 55 - 30 = 25.
but n + r > 50 by 5.
if r = 25 and n = 30
25 + 30 = 55 > 50 by 5.
and from the question, n > r.
so the right evaluation to be used is
r = 25 and n = 30.
Now, nPr
nPr = n!/(n - r)!
nPr = 30!/(30 - 25)!
nPr = 30!/5!

Niggaz b lyk what is 30! ?? grin

! = Factorial

9 Likes

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