# 4/9 of audience is female; 1/3 of adults are female; if 85 male children, then how many persons in audience?

#### shastri.chandoo

##### New member
At the school's Divali's Program, 4/9 of the audience was female and the rest were male. 1/3 of the adults are female. 1f there were 85 male children how many persons altogether attended the Program?

#### Jomo

##### Elite Member
What have you tried? If you show us your work then we can see where you are making a mistake. We really hate showing work and it turns out that the student knew that part already.

I would make a table like this and fill it in!

------------------Males------Female---------totals

Adult-------------------- -----1/3

Child-------------85

totals--------------------------4/9

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#### shastri.chandoo

##### New member
4/9= Females. 5/9= Males.
1/3= adult F(y)
2/3= adult M(x).
85= young M.
5/9= × + 85
45= 9x + 765
9x= 765 - 45= 720
x = 80 = 2/3
y = 40 = 1/3
5/9 =80 + 85= 165
4/9 = 132
audience = 297

#### Otis

##### Senior Member
It seems like this exercise has many possible answers. Here are just three examples:

5 adult females
10 adult males
71 female children
85 male children

155 adult females
310 adult males
161 female children
85 male children

390 adult females
780 adult males
302 female children
85 male children

Did I goof?

#### shastri.chandoo

##### New member
How did you arrive at that?

#### Jomo

##### Elite Member
4/9= Females. 5/9= Males.
1/3= adult F(y)
2/3= adult M(x).
85= young M.
5/9= × + 85
45= 9x + 765
9x= 765 - 45= 720
x = 80 = 2/3
y = 40 = 1/3
5/9 =80 + 85= 165
4/9 = 132
audience = 297
5/9= × + 85, x = 80 = 2/3 and y = 40 = 1/3 are not correct. Do you really think that 80 = 2/3 and that 40 = 1/3. I don't believe that you do, so why write it??????

Here is an updated table:

------------------Males------Female---------totals

Adult-------------2/3---------1/3=3/9

Child-------------85(1/3)------1/9

totals--------------------------4/9

So (1/3)m = 85. Then m= 255
Since male child is 85, we have adult male = 170

So total males is 255 which is (5/9)p. Hence p=459

Now f = 459-255= 204.

Please fill in the rest from here. PLEASE make sure that your equal signs are valid!

#### Jomo

##### Elite Member
It seems like this exercise has many possible answers. Here are just three examples:

5 adult females
10 adult males
71 female children
85 male children

155 adult females
310 adult males
155 female children
85 male children

390 adult females
780 adult males
302 female children
85 male children

Did I goof?

At least one of us goofed (I hope it is not both of us) as I got a unique answer

#### shastri.chandoo

##### New member
5/9= × + 85, x = 80 = 2/3 and y = 40 = 1/3 are not correct. Do you really think that 80 = 2/3 and that 40 = 1/3. I don't believe that you do, so why write it??????

Here is an updated table:

------------------Males------Female---------totals

Adult-------------2/3---------1/3=3/9

Child-------------85(1/3)------1/9

totals--------------------------4/9

So (1/3)m = 85. Then m= 255
Since male child is 85, we have adult male = 170

So total males is 255 which is (5/9)p. Hence p=459

Now f = 459-255= 204.

Please fill in the rest from here. PLEASE make sure that your equal signs are valid!
Ok. Thanks.

#### Otis

##### Senior Member
… I got a unique answer
I looked at your posts, but I'm not sure what some symbols represent.

Maybe I'm misreading the op. Here's what I'm thinking, for two examples (from post #4).
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

5 female adults
10 male adults
71 female children
85 male children

There are 76 females, out of 171 people.

(4/9)(171) = 76

There are 15 adults; one-third are female.

(1/3)(15) = 5
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

155 female adults
310 male adults
161 female children
85 male children

There are 316 females, out of 711 people.

(4/9)(711) = 316

There are 465 adults; one-third are female.

(1/3)(465) = 155
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

#### shastri.chandoo

##### New member
5/9= × + 85, x = 80 = 2/3 and y = 40 = 1/3 are not correct. Do you really think that 80 = 2/3 and that 40 = 1/3. I don't believe that you do, so why write it??????

Here is an updated table:

------------------Males------Female---------totals

Adult-------------2/3---------1/3=3/9

Child-------------85(1/3)------1/9

totals--------------------------4/9

So (1/3)m = 85. Then m= 255
Since male child is 85, we have adult male = 170

So total males is 255 which is (5/9)p. Hence p=459

Now f = 459-255= 204.

Please fill in the rest from here. PLEASE make sure that your equal signs are valid!
How did you arrive at (1/3)m = 85?
The problem talks about audience and then a subset which is the adults.

#### shastri.chandoo

##### New member
It seems like this exercise has many possible answers. Here are just three examples:

5 adult females
10 adult males
71 female children
85 male children

155 adult females
310 adult males
161 female children
85 male children

390 adult females
780 adult males
302 female children
85 male children

Did I goof?

Can you show me your working?

#### ksdhart2

##### Senior Member
I, too, was unable to arrive at a unique answer. To double check, I used a purely algebraic method and I made a list of all the things I knew. Breakdown by age:
• # of people: x
• # of adults: a
• # of children: x - a
Breakdown by sex:
• # of females: 4/9x
• # of males: 5/9x
Breakdown by age and sex:
• # of female adults: 1/3a
• # of female children: 4/9x - 1/3a
• # of male adults: 2/3a = 5/9x - 85
• # of male children: 85
The only new piece of information I gained from scouring the problem was $$\frac{2}{3}a = \frac{5}{9}x - 85 \implies x = \frac{6a}{5} + 153$$. This, then, means that we can arbitrarily decide there were any number of adults (cleanly divisible by 5 of course, since there can't be fractional audience members) and find the corresponding number of total audience members.

#### shastri.chandoo

##### New member
I, too, was unable to arrive at a unique answer. To double check, I used a purely algebraic method and I made a list of all the things I knew. Breakdown by age:
• # of people: x
• # of adults: a
• # of children: x - a
Breakdown by sex:
• # of females: 4/9x
• # of males: 5/9x
Breakdown by age and sex:
• # of female adults: 1/3a
• # of female children: 4/9x - 1/3a
• # of male adults: 2/3a = 5/9x - 85
• # of male children: 85
The only new piece of information I gained from scouring the problem was $$\frac{2}{3}a = \frac{5}{9}x - 85 \implies x = \frac{6a}{5} + 153$$. This, then, means that we can arbitrarily decide there were any number of adults (cleanly divisible by 5 of course, since there can't be fractional audience members) and find the corresponding number of total audience members.
I understand. the audience number is a function of the number of adults - Straight line equation x = (6/5)a + 153
This means that the examiner did left out the number of adults in the question.
Thank you and best regards.

#### Denis

##### Senior Member
Wrote looper!
Otis is correct....

A solution exists with total attendance = 171 (minimum);
then by jumps of 18: 189, 207, 225, ........ infinity!

An extra condition required; like number of female children.