Question

In a class 70% passed in mathematics and 80 % passed in English. Together 40% failed in maths and English in a class of 133 members. How many students passed in both the subjects

Answer 1

= 0.7(133) + 0.8(133) - 2(0.4[133])
= 93 + 106 - 2(53)
= 199 - 106
= 93

Answer 2

Is that one of your options?

Answer 3

@acacia
could u explain me with venn diagrams becoz the answer which u gave could be found in yahoo answers

Answer 4

Yeah hold on

Answer 5

If 70% passed in math, that means 30% failed.
If 80% passed in english, that means 20% failed.
That totals 50% failed. So how can 40% fail ?

Answer 6

@skullpatrol ...can you take a peek at this.....please

Answer 7

@Jhannybean ...can you explain this to me please

Answer 8

@SithsAndGiggles ...help

Answer 9

@texaschic101, it's possible that some students are in math, and some in English, but not both.

Answer 10

In total, there is 133 students ....right ? In both classes total ?

Answer 11

40% failed in a class of 133.
(.40)(133) = 53.2 failed ?

Answer 12

anybody ?

Answer 13

@Suchandrarao, do all 133 students take both classes?

Answer 14

That is what I was wondering.....is there a total of 133 students in total ?

Answer 15

yes @SithsAndGiggles

Answer 16

.7(133) = 94
.8(133) = 107

107
94
----
201
-133
-----
68/2 = 34 in AnB

Answer 17

so 34 people passed in both subjects ?

Answer 18

well, i dint account for the failures

Answer 19

.4(133) = 54 students not acowingspaned for on the outsides

Answer 20

We divide the total by 2 since there are only 2 subjects, if there were 3, we'd divide our total by 3?

Answer 21

107 + 94 + 34 = 235 people in total ?

Answer 22

133
-54
----
79 people to account for passing i believe

Answer 23

Why did you delete,@SithsAndGiggles ?

Answer 24

deciphering this thing is half the battle .... im not sure if ive read it correctly yet or not tho

Answer 25

@Jhannybean I'm having my doubts.

Answer 26

@SithsAndGiggles
actually there are no options given here

Answer 27

im thinking 61 is passed AnB

Answer 28

I think this is what's going on here:
\[\begin{align*}n(Class)&=n(Pass)+n(Fail)\\
133&=\bigg[n(M_p)+n(E_p)-\color{red}{n(M\cap E_p)}\bigg]+\bigg[\color{blue}{n(M_f)}+\color{blue}{n(E_f)}-n(M\cap E_f)\bigg]
\end{align*}\]
The red is what you're looking for, but I don't think you can determine it without first figuring out the blues.

(By the way, \(n(\cdots)\) is the "number of elements in the set \(\cdots\).)

Answer 29

you can't do 70% of 133 or 30% of 133 because there is more then 133 students...right ?

Answer 30

We don't know the total number of students...there is more then 133....right ?

Answer 31

.3k + .2k - .4k = k (.1) failed math or english

Answer 32

10% of what failed math or english ?

Answer 33

"in a class of 133 members" so k = 133

Answer 34

\[\begin{align*}133&=\bigg[93+106-\color{red}{n(M\cap E_p)}\bigg]+\bigg[\color{blue}{n(M_f)}+\color{blue}{n(E_f)}-53\bigg]\\
13&=\color{red}{n(M\cap E_p)}-\color{blue}{n(M_f)}-\color{blue}{n(E_f)}
\end{align*}\]

Answer 35

is the question asking: how students passed math and english together?

or is it asking how many students passed math? How many students passed english?

Answer 36

if 10% (14 students): (failed math OR failed english)

-(failed math OR failed english) = (passed math AND passed english) = 133 - 14

Answer 37

if we let M' be the event that we fail math and E' we fail english...is the problem sayin

\(P(M'\cap E')=.4\) ?

Answer 38

where are you getting the 14 students ?

Answer 39

I think we need the number of students passed in maths and english i mean if am a student then i should be pass maths as well as english

Answer 40

from the setup that 30% failed math, and 20% failed english

that gives us a sum total of 50% failed both MandE, but it is stated that 40% failed MandE

P(fM or fE) = .3k + .2k - .4k = .1k

10% of 133 is 14 kids

Answer 41

oh....you rounded up ?

Answer 42

well, if you rnd down you get less than 10% so you always want to rnd up to cover it

Answer 43

\[.4=P(M'\cap E')\le P(M')=.3\]

Answer 44

In the chart you made, @amistre64 , what are the 1's signifying?

Answer 45

total of 100%...I think

Answer 46

that 100% of the students took math; and 100% of the students took english

Answer 47

or maybe that 100% of the students the subject are accounted for

Answer 48

thats what a class consists of 133 students

Answer 49

Oh i see.

Answer 50

That can't be the total number of students in both classes total....it doesn't equal

Answer 51

how so

Answer 52

oh god this question is making hell out of here

Answer 53

passing math : (.70)(133) = 93.1....rnd up 92
passing english : (.80)(133) = 106.4 ...rnd up 107
passing math and english : (.60)(133) = 79.8 ...rnd up 80
107 + 80 + 92 = 279.....that is more then 133

Answer 54

I'm going to type this again...

\[.4=P(\text{Fail math}\cap \text{Fail english})\le P(\text{Fail math})=.3\]

Answer 55

you see my avatar face.....this is what my real face looks like when I look at that...lol

Answer 56

i still claim uncertainty :)

Answer 57

I am so lost on this.....I am leaving it up to somebody that knows what they are doing...and that is not me...sorry

Answer 58

does anyone disagree with what i typed above?

Answer 59

i dont disagree with it :)

Answer 60

then we have \(.4\le .3\) which is ridiculous. Therefore the problem is ill posed

Answer 61

page 11.46, problem 120 has a similar kind of structure

http://books.google.com/books?id=mpxdqsZVMdEC&pg=RA3-PA47&dq=In+a+class+70%25+passed+in+mathematics+and+80+%25+passed+in+English.&hl=en&sa=X&ei=i__WUYHYGY2K9gTBg4C4Dg&ved=0CDMQ6AEwAQ#v=onepage&q=In%20a%20class%2070%25%20passed%20in%20mathematics%20and%2080%20%25%20passed%20in%20English.&f=false

the answer key gives "a" as the solution