NewStats: 3,261,548 , 8,174,354 topics. Date: Thursday, 29 May 2025 at 03:51 PM 3r2f5z

Pls gurus in the house help me to solve this SET world problem
1)In a class of 35 students,19 take history and 12 take economics. If 5 takes both subjects,how many take neither?
2)In a class of 36 students,29 do mathematics and 20 do chemistry. If 5 students do neither,how many students do chemistry but not mathematics?
3)In a school of 750 students,320 are girls. 559 students do some kind of sport. If 101 girls do no sport,how many boys also do no sport?

I must comend u guys first for this grt work.
I'm new on this thread and I think God hs Brot me to d ryt plc.
my story goes dis way,

while in sec school,I thought I will study medical course as such wouldn't need much maths.i wasn't interested in math anymore.
I dropped futhermath in my ss2 even thou I was abv average.
as life changes,I now hv opportunity to go abroad for studies(which God hs done).and the course I want to pursue now requires excellent knowledge of maths(aerospace engr).
aldo I'm NT doom in math's,Bt I must confess DAT my math knowledge is not up to DAT.
I jus hav four month to d begin in of my course,I hv started self study.n I must it DAT I hv already seen d changes,Bt I want to b more than a GURU in math's.

my question is,how can I become overgud in math's?
what r d steps?
what must I do?
thankx

dprinz99:
I must comend u guys first for this grt work.
I'm new on this thread and I think God hs Brot me to d ryt plc.
my story goes dis way,

while in sec school,I thought I will study medical course as such wouldn't need much maths.i wasn't interested in math anymore.
I dropped futhermath in my ss2 even thou I was abv average.
as life changes,I now hv opportunity to go abroad for studies(which God hs done).and the course I want to pursue now requires excellent knowledge of maths(aerospace engr).
aldo I'm NT doom in math's,Bt I must confess DAT my math knowledge is not up to DAT.
I jus hav four month to d begin in of my course,I hv started self study.n I must it DAT I hv already seen d changes,Bt I want to b more than a GURU in math's.

my question is,how can I become overgud in math's?
what r d steps?
what must I do?
thankx

it requires constant practice. "practice makes perfect". first start with the known (simple ones) then advance to the unknown (difficult ones). don't feel discourage if u don't get the practice questions right at first. my favorite quote then I was "if u leave maths for one day, maths will leave u for 1 week" so keep on practicing till u become morn than a guru in maths.

cc: MathsChic, et al.

saipn:
Pls gurus in the house help me to solve this SET world problem
1)In a class of 35 students,19 take history and 12 take economics. If 5 takes both subjects,how many take neither?
2)In a class of 36 students,29 do mathematics and 20 do chemistry. If 5 students do neither,how many students do chemistry but not mathematics?
3)In a school of 750 students,320 are girls. 559 students do some kind of sport. If 101 girls do no sport,how many boys also do no sport?

. (1)
35:::19plus12plus x
35:::31plus x
x::::35-31
x::::4

therefore,4 student take neither

petengine:

it requires constant practice. "practice makes perfect". first start with the known (simple ones) then advance to the unknown (difficult ones). don't feel discourage if u don't get the practice questions right at first. my favorite quote then I was "if u leave maths for one day, maths will leave u for 1 week" so keep on practicing till u become morn than a guru in maths.

cc: MathsChic, et al.

I'm grateful bro

saipn:
Pls gurus in the house help me to solve this SET world problem
1)In a class of 35 students,19 take history and 12 take economics. If 5 takes both subjects,how many take neither?
2)In a class of 36 students,29 do mathematics and 20 do chemistry. If 5 students do neither,how many students do chemistry but not mathematics?
3)In a school of 750 students,320 are girls. 559 students do some kind of sport. If 101 girls do no sport,how many boys also do no sport?

(1) In a class of 35 students,19 take history and 12 take economics. If 5 takes both subjects,how many take neither?
‎
Solution:

Total students = 35
History student = 19
Economics students = 12
Students taking both courses = 5‎
‎
Economics = (12 - 5)= 7
History = (19 - 5) = 14‎
Neither = 35 - ( 14 + 7 + 5) = 9

(2) ‎In a class of 36 students,29 do mathematics and 20 do chemistry. If 5 students do neither,how many students do chemistry but not mathematics?

Solution:

Total students = 36
Mathematics students = 29
Chemistry students = 20
Neither = 5
‎
Both = ((29 - x) + (20 - x) + x + 5) - 36 = 18
Chemistry = 20 - 18 = 2

(3) In a school of 750 students,320 are girls. 559 students do some kind of sport. If 101 girls do no sport,how many boys also do no sport?

Solution:

‎Total students = 750‎
Total Sport student = 559‎
Girls population = 320
Boys population = (750 - 320) = 430‎
Non-sport activity girls = 101‎

Sports active girls = (320 - 101) = 219
Sports active boys = (559 - 219) = 340
Non-sport activity boys = (430 - 340) = 90‎

saipn:
Pls gurus in the house help me to solve this SET world problem
1)In a class of 35 students,19 take history and 12 take economics. If 5 takes both subjects,how many take neither?
2)In a class of 36 students,29 do mathematics and 20 do chemistry. If 5 students do neither,how many students do chemistry but not mathematics?
3)In a school of 750 students,320 are girls. 559 students do some kind of sport. If 101 girls do no sport,how many boys also do no sport?

1)
35=(19-5)+(12-5)+5+x
35=14+7+5+x
35-26=x
X=9
2)
36=(29-x)+(20-x)+x+5
36=54-x
36-54=-x
-18=-x
X=18
thus chemistry only gives 20-18=2
3)solved already by mathemagician1

Mathemagician1:

(1) In a class of 35 students,19 take history and 12 take economics. If 5 takes both subjects,how many take neither?
‎
Solution:
Total students = 35
History student = 19
Economics students = 12
Students taking both courses = 5‎
‎
Economics = (12 - 5)= 7
History = (19 - 5) = 14‎
Neither = 35 - ( 14 + 7 + 5) = 9
(2) ‎In a class of 36 students,29 do mathematics and 20 do chemistry. If 5 students do neither,how many students do chemistry but not mathematics?
Solution:
Total students = 36
Mathematics students = 29
Chemistry students = 20
Neither = 5
‎
Both = (29 + 20) - (36 - 5) = 8
Chemistry = 20 - 8 = 12
(3) In a school of 750 students,320 are girls. 559 students do some kind of sport. If 101 girls do no sport,how many boys also do no sport?
Solution:
‎Total students = 750‎
Total Sport student = 559‎
Girls population = 320
Boys population = (750 - 320) = 430‎
Non-sport activity girls = 101‎
Sports active girls = (320 - 101) = 219
Sports active boys = (559 - 219) = 340
Non-sport activity boys = (430 - 340) = 90‎

saipn:
Pls gurus in the house help me to solve this SET world problem
1)In a class of 35 students,19 take history and 12 take economics. If 5 takes both subjects,how many take neither?
2)In a class of 36 students,29 do mathematics and 20 do chemistry. If 5 students do neither,how many students do chemistry but not mathematics?
3)In a school of 750 students,320 are girls. 559 students do some kind of sport. If 101 girls do no sport,how many boys also do no sport?

The solutions of question 2 above is all wrong. The question is chemistry only

stlaibrowne:
The solutions of question 2 above is all wrong. The question is chemistry only

i got mine naa

ladokuntlad:

i got mine naa

Oh yes boss the other two foul foul foul

stlaibrowne:


Oh yes boss the other two foul foul foul

my set is becoming rusty. Nice work ladokuntla and stlaibrowne

1/3plus(5/7(9/10-1plus3/4))
its fraction.
I hv tried by stil no ryt anxa
plx help

dprinz99:
1/3plus(5/7(9/10-1plus3/4))
its fraction.
I hv tried by stil no ryt anxa
plx help

1/3 + ( 5/7 ( 9/10 - 1 + 3/4 ))‎
Using BODMAS‎
1/3 + ( 5/7 ( 9/10 - 1/4 ))‎
1/3 + ( 5/7 ( 13/20 ))
1/3 + ( 13/28)
1/3 + 13/28
= 67/84‎

Help solve this (3/4 --2/5) - 1/3 of 3/10 all divided by 2 -- (1/1 3/7) moreso i want sir to teach me sets

.

[quote author=ladokuntlad post=40275175] is it like dis? NOPE
(3/4-(-2/5))-1/3 of 3/10 all dived by (2-(-1/(1 3/7)) ) i made a mistake when typing is like this (3/4 - 2/5) - 1/3 of 3/10 all divided by 2 - (1/1 3/7)

Mathemagician1:


1/3 + ( 5/7 ( 9/10 - 1 + 3/4 ))‎
Using BODMAS‎
1/3 + ( 5/7 ( 9/10 - 1/4 ))‎
1/3 + ( 5/7 ( 13/20 ))
1/3 + ( 13/28)
1/3 + 13/28
= 67/84‎

I got that Bt its NT the ryt ansa ooo

Simplify as much as possible log_ca * log_cb

dprinz99:
I got that Bt its NT the ryt ansa ooo

67/27?

jackpot:
Simplify as much as possible

log_ca * log_cb

check ur question again.
because dis has already been simplified.
maybe dia ought to b a little modification in yaa log

except u looking for dis

log_ca * log_cb= log_c(a+b)
wich is very wrong

seuntosyn:
i made a mistake when typing is like this (3/4 - 2/5) - 1/3 of 3/10 all divided by 2 - (1/1 3/7)

((3/4-(2/5))-(1/3x3/10))/(2-(1/(1 3/7)) )

((3/4-2/5)-(1/10))/(2-(1/(10/7)) )

(7/20-1/10)/(2-7/10)
(5/20)/(13/10)
=5/20 x 10/13
=5/26

According to ur new question

dprinz99:

I got that Bt its NT the ryt ansa ooo

then d mistake is from the way u typed d question.
Try tak a snapshot of the question and it here

ladokuntlad:
check ur question again. because dis has already been simplified. maybe dia ought to b a little modification in yaa log
except u looking for dis log_ca * log_cb= log_c(a+b) wich is very wrong

okay.
What of the following:

Simplify as much as possible: (log_c a) / (log_cb)

jackpot:
okay.

What of the following:


Simplify as much as possible: (log_c a) / (log_cb)

Log_c(a-b)

jackpot:
Simplify as much as possible

log_ca * log_cb

[logalogb]/[logclogc]

jackpot:
okay.
What of the following:
Simplify as much as possible: (log_c a) / (log_cb)

loga base b

jackpot:
okay.

What of the following:


Simplify as much as possible: (log_c a) / (log_cb)

(log_c a) / (log_cb)
=(loga/logc)/(logb/logc)
=(loga/logc)*(logc/logb)
=(loga/logc)*(logc/logb)
=loga/logb
=log_b a
QED

stlaibrowne:


Log_c(a-b)

Wrong.
try insert a value and c for ursef.
no such rule exist

thankyouJesus:
[logalogb]/[logclogc]loga base b

kudos to you and ladokuntlad

stlaibrowne:
The solutions of question 2 above is all wrong. The question is chemistry only

thanks bro

ladokuntlad:
1) 35=(19-5)+(12-5)+5+x 35=14+7+5+x 35-26=x X=9 2) 36=(29-x)+(20-x)+x+5 36=54-x 36-54=-x -18=-x X=18 thus chemistry only gives 20-18=2 3)solved already by mathemagician1

thanks ladokuntlad

ladokuntlad:
1) 35=(19-5)+(12-5)+5+x 35=14+7+5+x 35-26=x X=9 2) 36=(29-x)+(20-x)+x+5 36=54-x 36-54=-x -18=-x X=18 thus chemistry only gives 20-18=2 3)solved already by mathemagician1

thanks ladokuntlad

Mathemagician1:


(1) In a class of 35 students,19 take history and 12 take economics. If 5 takes both subjects,how many take neither?
‎
Solution:

Total students = 35
History student = 19
Economics students = 12
Students taking both courses = 5‎
‎
Economics = (12 - 5)= 7
History = (19 - 5) = 14‎
Neither = 35 - ( 14 + 7 + 5) = 9

(2) ‎In a class of 36 students,29 do mathematics and 20 do chemistry. If 5 students do neither,how many students do chemistry but not mathematics?

Solution:

Total students = 36
Mathematics students = 29
Chemistry students = 20
Neither = 5
‎
Both = ((29 - x) + (20 - x) + x + 5) - 36 = 18
Chemistry = 20 - 18 = 2

(3) In a school of 750 students,320 are girls. 559 students do some kind of sport. If 101 girls do no sport,how many boys also do no sport?

Solution:

‎Total students = 750‎
Total Sport student = 559‎
Girls population = 320
Boys population = (750 - 320) = 430‎
Non-sport activity girls = 101‎

Sports active girls = (320 - 101) = 219
Sports active boys = (559 - 219) = 340
Non-sport activity boys = (430 - 340) = 90‎

thanks mathematicians