# Thread: Just can’t work it out after 3 days of trying

1. ## Just can’t work it out after 3 days of trying

Hi all I have been trying to work this out for 3 days and seem to be missing something and get a different answer no matter what do.

Sent from my iPad using Tapatalk

2. ## Re: Just can’t work it out after 3 days of trying

$m + 3m + (3m-10) + \dfrac{3m-10}{4} = 65$

$\dfrac{31}{4}m - \dfrac{50}{4} = 65$

$31m - 50 = 260$

$31m = 310$

$m = 10$

10 men
30 women
20 girls
5 boys

3. ## Re: Just can’t work it out after 3 days of trying

Originally Posted by romsek

$m + 3m + (3m-10) + \dfrac{3m-10}{4} = 65$

$\dfrac{31}{4}m - \dfrac{50}{4} = 65$

$31m - 50 = 260$

$31m = 310$

$m = 10$

10 men
30 women
20 girls
5 boys
Thanks Romsek

I am only new to algebra so please excuse my ignorance but what does $stand for and the \defrac as this is not terminology I have been taught so far . I really appreciate your help Sent from my iPad using Tapatalk 4. ## Re: Just can’t work it out after 3 days of trying That's Latex ($\LaTeX$) the typesetting language used to render mathematical notation on the site. It's not mathematical notation. Your error was in trying to remove the denominator from the fraction. You did it wrong, but also it was the wrong thing to do. You should first look to group like terms. In this case you have terms in$m$and constant terms, so you need to combine all the terms in$m$into a single term and all the constant terms into a single term. Remember that$\frac{3m+10}{4}=\frac{3m}{4} + \frac{10}{4} = \frac34 m + \frac52$. 5. ## Re: Just can’t work it out after 3 days of trying Originally Posted by Archie That's Latex ($\LaTeX$) the typesetting language used to render mathematical notation on the site. It's not mathematical notation. Your error was in trying to remove the denominator from the fraction. You did it wrong, but also it was the wrong thing to do. You should first look to group like terms. In this case you have terms in$m$and constant terms, so you need to combine all the terms in$m$into a single term and all the constant terms into a single term. Remember that$\frac{3m+10}{4}=\frac{3m}{4} + \frac{10}{4} = \frac34 m + \frac52$. Thanks Archie I seem to be getting more confused by the minute. I am stuck on romsek’s second line as I can see how he got to there. Sent from my iPad using Tapatalk 6. ## Re: Just can’t work it out after 3 days of trying Originally Posted by bartman72 Thanks Romsek I am only new to algebra so please excuse my ignorance but what does$ stand for and the \defrac as this is not terminology I have been taught so far .

Sent from my iPad using Tapatalk
I guess your browser isn't rendering the LaTex

7. ## Re: Just can’t work it out after 3 days of trying

Originally Posted by romsek
I guess your browser isn't rendering the LaTex

Hi Romsek

I am have trouble understanding how you got to the 31m and 50

Sent from my iPad using Tapatalk

8. ## Re: Just can’t work it out after 3 days of trying

Originally Posted by bartman72
Hi Romsek

I am have trouble understanding how you got to the 31m and 50

Sent from my iPad using Tapatalk
I multiplied both sides of the equation by 4.

9. ## Re: Just can’t work it out after 3 days of trying

There are 4 unknown values, the number of men, m, the number of women, w, the number of boys, b, and the number of girls, g. We are given 4 pieces of information so can form 4 equations to solve for those 4 values.

"Three times as many women as men are on the tour": w= 3m.
"There are 10 less girls than the number of women": g= w- 10.
"The number of boys is a quarter of the number of girls": b= g/4.
"There are 65 people booked in for the tour": m+ w+ b+ g= 65.

I would put everything in terms of m: w= 3m, g= w- 10= 3m- 10, b= g/4= (3m- 10)/4= 3m/4- 10/4= 3m/4- 5/2.

Then m+ w+ b+ g= m+ 3m+ 3m- 10+ 3m/4- 5/2= 65

(1+ 3+ 3+ 3/4)m= 65+ 10+ 5/2
(4+ 12+ 12+ 3)m/4= 155/2
31m= 310 so m= 10.

Then w= 3m= 30, g= 3m- 10= 20, and b= 3m/4- 5/2= 30/4- 5/2= 15/2- 5/2= 10/2= 5.

There were 10 men, 30 women, 5 boys, and 20 girls.

Check:
"Three times as many women as men are on the tour": yes, 30= 3(10)
"There are 10 less girls than the number of women": yes, 20= 30- 10.
"The number of boys is a quarter of the number of girls": yes, 5= 20/4.
"There are 65 people booked in for the tour": yes, 10+ 30+ 20+ 5= 65.

(Checking is important! The first time I did this problem, I accidently swapped two numbers and only caught the error after checking.)
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5567417740821838, "perplexity": 4019.693605411732}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525402.30/warc/CC-MAIN-20190717201828-20190717223828-00306.warc.gz"}