The club has 53 members. 5/3 of girls, 3/5 of boys are from

[pugnog007]

Hey! I just cant figure this out. I've worked on it for hours. :roll:
Any help will be awsome, brilliant, helpful, mind easing, and wonderful thanks.

"The club has 53 members. 5/7 of the girl members and 3/5 of the boy members are from a lower grade. The total number of lower grade members is 35. How many members of the club are boys?"


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[Denis]

boys = b, girls = g

b + g = 53

3b/5 + 5g/7 = 35

If you can't finish that off, you need classroom help.

 

[The Highlander]

Hey! I just cant figure this out. I've worked on it for hours. :roll:
Any help will be awesome, brilliant, helpful, mind easing, and wonderful thanks.

"The club has 53 members. 5/7 of the girl members and 3/5 of the boy members are from a lower grade. The total number of lower grade members is 35. How many members of the club are boys?"

While searching out my own previous post (about the 1% Club), I happened across this thread and thought there might well be some viewers who, despite @Denis' prompt, wouldn't see exactly how to go about progressing from there, so I thought I would just finish it off for the sake of completeness and the benefit of anyone else who comes across it and wants to know how the solution is obtained (using "Simultaneous Equations").

Starting from where @Denis left off, we have two equations where: \(\displaystyle b\) is the total number of boys in the club and \(\displaystyle g\) is the total number of girls.

Thus...
[math]b + g = 53[/math]and
[math]\footnotesize\frac{3}{5}\normalsize b + \footnotesize\frac{5}{7}\normalsize g = 35[/math]
We can now give the fractions the same denominators by multiplying top and bottom by the same number(s)...

[math]\footnotesize\frac{3\times 7}{5\times 7}\normalsize b + \footnotesize\frac{5\times 5}{7\times 5}\normalsize g = 35[/math][math]\implies \footnotesize\frac{21}{35}\normalsize b + \footnotesize\frac{25}{35}\normalsize g = 35[/math][math]\implies \footnotesize\frac{21b}{35}\normalsize + \footnotesize\frac{25g}{35}\normalsize = 35[/math][math]\implies \footnotesize\frac{21b + 25g}{35}\normalsize = 35[/math][math]\implies 21b + 25g = 35^2 = 1225[/math]

(Both sides of the previous equation multiplied by 35)​

and if we multiply (both sides of) our first equation by 25, we now have...

[math]25b + 25g = 1325[/math]and
[math]21b + 25g = 1225[/math]
Subtracting the lower equation from the upper, we get...

[math]25b + 25g = 1325[/math][math]21b + 25g = 1225[/math][math]           4b     = 100 \implies b = \underline{\underline{25}}[/math]
Which is the answer we were asked for (The total number of boys in the club).

To check our result....

If \(\displaystyle b = 25\), then \(\displaystyle g = 28\) (because \(\displaystyle b + g = 53\), the total club membership)

and
[math]\footnotesize \frac{3}{5}\normalsize 25 + \footnotesize\frac{5}{7}\normalsize 28 = \footnotesize\frac{3\times25}{5}\normalsize + \footnotesize\frac{5\times28}{7}\normalsize = \footnotesize\frac{75}{5}\normalsize + \footnotesize\frac{140}{7}\normalsize = 15 + 20 = 35[/math].
Thus we have the correct numbers of "lower grade" pupils too.

So that all works out nicely and the correct answer is that there are 25 boys (in total) who are members of the club.

 

[Steven G]

[math]\footnotesize \frac{3}{5}\normalsize 25 + \footnotesize\frac{5}{7}\normalsize 28 = \footnotesize\frac{3\times25}{5}\normalsize + \footnotesize\frac{5\times28}{7}\normalsize = \footnotesize\frac{75}{5}\normalsize + \footnotesize\frac{140}{7}\normalsize = 15 + 20 = 35[/math].
Thus we have the correct numbers of "lower grade" pupils too.

I'm sorry but I need to complain about how you are showing to solve this. Students need to learn must learn how to reduce before multiply fractions (or fractions by whole numbers). For the benefit of students who see this post:

[math]\footnotesize \frac{3}{5}\normalsize 25 + \footnotesize\frac{5}{7}\normalsize 28 = 3*5 + 5*4 = 15 + 20 = 35\ or\ even\ 3*5 + 5*4 = 5(3+4)=5*7=35[/math]

 

[The Highlander]

I'm sorry but I need to complain about how you are showing to solve this. Students need to learn must learn how to reduce before multiply fractions (or fractions by whole numbers). For the benefit of students who see this post:

[math]\footnotesize \frac{3}{5}\normalsize 25 + \footnotesize\frac{5}{7}\normalsize 28 = 3*5 + 5*4 = 15 + 20 = 35\ or\ even\ 3*5 + 5*4 = 5(3+4)=5*7=35[/math]

I'm afraid we're going to have to agree to disagree on that (or, failing that, just disagree on it).
Your "alternative" gives no indication/explanation of how you got from \(\displaystyle \footnotesize \frac{3}{5}\normalsize 25 \text{ to } 3\times 5\)
If you had suggested: \(\displaystyle \footnotesize \frac{3}{5}\normalsize 25=\footnotesize \frac{3}{\cancel{5}}\normalsize \cancel{25}~5 = 3\times 5\) then I might have thought your alternative suggestion acceptable but I still think my way is better. ?

 

[lookagain]

. Students need to learn must learn how to reduce ...

@ Steven G -- How about this? "Students are missing out on a helpful opportunity to reduce earlier on. Otherwise, the reducing
later option with larger numbers will become relatively more difficult, time-consuming, and be more apt for human error."

 

[The Highlander]

And what if you can't cancel?

What if your 'students' were faced with \(\displaystyle \footnotesize \frac{3}{5}\normalsize 26\) instead of \(\displaystyle \footnotesize \frac{3}{5}\normalsize 25\)?

Surely they "need to learn must learn" that they should do this: \(\displaystyle \footnotesize \frac{3}{5}\normalsize 26=\footnotesize \frac{3\times 26}{5}\normalsize =\footnotesize\frac{78}{5}= \large15\footnotesize\frac{3}{5}~\left(\text{ or }\large 15.6\right)\)

"...the reducing later option with larger numbers will become relatively more difficult, time-consuming, and be more apt for human error."

The only human error(s) that tend to crop up nowadays is pressing the wrong button on the calculator! ?

 

[Steven G]

Students need to learn how to cancel! This is not debatable! Consider this problem: 2365(5/2365). You really want students to do the multiplication first?! Students can't factor very well as it is and then they get bigger numbers by multiplying. Do you think that they can reduce more easily with the bigger numbers.
Student need to be fluent in basic arithmetic. I will never think that is wrong!

 

[Steven G]

The only human error(s) that tend to crop up nowadays is pressing the wrong button on the calculator! ?

@The Highlander : It is not funny that students use a calculator to do basic arithmetic. I think that companies like texas instruments are parasites for pushing calculators as much as they have.

 

[pka]

It needs to be pointed out that it has been 15 years, 7 months and 26 days since this post was first made and answered.

 

[Steven G]

It needs to be pointed out that it has been 15 years, 7 months and 26 days since this post was first made and answered.

Yes, I forgot that!

 

[stapel]

@The Highlander : It is not funny that students use a calculator to do basic arithmetic. I think that companies like texas instruments are parasites for pushing calculators as much as they have.

I think Texas Instruments (TI) has filled a "need" that was created elsewhere.

I can remember listening to a instructor (of math instructors) rave about how she'd gotten her students to estimate. "Suppose I ask them to estimate the cost of three items that cost $3.99 each. By estimating, they can tell me that the cost is about $12." But she skipped over the part where the kids weren't required to learn their basic multiplication tables. Her students didn't know that [imath]3\times 4=12[/imath]. Instead, they'd plugged [3][x][3][.][9][9][ENTER] into their graphing calculator, gotten an answer of 11.96, and rounding to the nearest whole number, being 12.

Instead of learning to estimate, based on assumed (but missing) skills, they instead were reinforced in the middle- and high-school "rule" that every answer is a whole number, or at least one of the basic fractions (so an answer of 0.49 would "really" be [imath]\frac{1}{2}[/imath]).

I can't fault Texas Instruments for seeing a need and filling it. TI didn't create the need; it merely exploited it.

 

[stapel]

It needs to be pointed out that it has been 15 years, 7 months and 26 days since this post was first made and answered.

Yes. If I could figure out how to split threads, so the new posts could be moved to a new thread, I'd do it. But I can't seem to locate that functionality...? ?‍

 

[Steven G]

I can't fault Texas Instruments for seeing a need and filling it. TI didn't create the need; it merely exploited it.

This is why I do not support the capitalistic system we live under. In general, we never do anything for the good of the people, but rather we do what will make the most money. Unfortunately public education for K-12 students isn't profitable. There are enough rich people who send their kids to very good private school and these kids fill the need for educated workers after graduating college.

 

[The Highlander]

It needs to be pointed out that it has been 15 years, 7 months and 26 days since this post was first made and answered.

Except that it wasn't really "answered"! All that was in the thread was (helpful) hint on how to get started which is why I went to the bother of writing up a full answer; as I said in my original post, for the sake of completeness and for the benefit of anyone else who came across the post and didn't know how to proceed from that hint.
(I don't believe that breaks any "rules" and it adds to the usefulness of the site. ?)

We have, indeed, gone way off topic due to the introduction of teaching preferences on how to deal with fractional multiplication(s) (and social philosophies, by gum!).

It would appear that there is no room for agreement (to disagree) after all so can I just say that I wholeheartedly support the need for students to be thoroughly conversant with their times tables and it that it is good practice to encourage them at every opportunity to look for ways to simplify arithmetic (and algebraic) problems whenever possible.

Again, harping back to those halcyon days (well over half a century ago) when I was in primary school, we spent a substantial part of every school day chanting out (as a class) those very tables. Unfortunately, these days the pupils are too busy finger painting or practising for school shows to devote much time to that! ?

Consider this problem: 2365(5/2365). You really want students to do the multiplication first?

No! Of course not. I would expect all but the very poorest student to "see" (immediately) that the 2365's cancel each other out. And I'm not "against" canceling; of course I would encourage students to seek any opportunities to "reduce" wherever they arise.

But, say (again) instead of your \(\displaystyle 2365\times\footnotesize\frac{5}{2365}\) a student is faced with \(\displaystyle 2365\times\footnotesize\frac{5}{2362}\).

Would you really expect them to waste any time hunting around for common factors that would allow for any cancelling to take place?

I would expect them to reach immediately for their "TI" device; I know I would if faced with that sum!
Except I can't afford Texas Instruments' outrageous prices; I get my calculators from Poundland!
(And, IMNSHO, they're just as good! ?)

 

[Steven G]

But, say (again) instead of your \(\displaystyle 2365\times\footnotesize\frac{5}{2365}\) a student is faced with \(\displaystyle 2365\times\footnotesize\frac{5}{2362}\).

Would you really expect them to waste any time hunting around for common factors that would allow for any cancelling to take place?

If in the end the student was required to reduce, then yes, I would advice them to reduce first!

 

[Steven G]

This is my opinion and although some times my opinion can be changed this is not going to be one. I also believe that you know that I am correct so there is not need to continue with this (old) thread.
I also have no problem at all with you responding to an old thread to help students.

Here is a question to all the tutors: Do you think that students use the forum to find similar problems from old threads or just post their problems right away?

 

[stapel]

Here is a question to all the tutors: Do you think that students use the forum to find similar problems from old threads or just post their problems right away?

Considering the existence of the Needed: Similar Threads Mod thread, I would guess not (especially not those who post their entire homework set with no work show, tell us that the bus is coming in fifteen minutes, and that they want it nicely hand-written so they can print it out and hand it in.)

(Yes, I've seen exactly that sort of post.)

 

[lev888]

This is why I do not support the capitalistic system we live under. In general, we never do anything for the good of the people, but rather we do what will make the most money. Unfortunately public education for K-12 students isn't profitable. There are enough rich people who send their kids to very good private school and these kids fill the need for educated workers after graduating college.

This is an interesting topic for a separate discussion. IMO, capitalism is the best system possible. It allows people to do for themselves what _they_ think is good for them and not what some bureaucrat in a state or federal department considers good for "the people".
And "I will never think that is wrong!"

 

[lev888]

This is my opinion and although some times my opinion can be changed this is not going to be one. I also believe that you know that I am correct so there is not need to continue with this (old) thread.
I also have no problem at all with you responding to an old thread to help students.

Here is a question to all the tutors: Do you think that students use the forum to find similar problems from old threads or just post their problems right away?

I think, as of a couple of years ago, it's more likely that AIs use old threads for training.