# 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

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

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

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#### shastri.chandoo

##### New member
4/9= Females. 5/9= Males.
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:

71 female children
85 male children

161 female children
85 male children

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.
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

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.

#### Jomo

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

71 female children
85 male children

155 female children
85 male children

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

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.

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).
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

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
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

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

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.

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:

71 female children
85 male children

161 female children
85 male children

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 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 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.