Having trouble with this ratio problem

Yammy

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Joined
Nov 5, 2013
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2
I have never had to solve a ratio problem where I have been given these circumstances, if you can help me get started with perhaps an equation or two to find the differences that would be great.

Problem:

The ratio of men to women on a bus is 5/7. Two women and one man board the bus and the ratio is then 7/10. How many men and women were on the bus before the other three passengers boarded?

I have not done much with ratios so I am having trouble figuring out an equation to figure this out. Any help would be appreciated. Thank you ^_^
 

Subhotosh Khan

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Staff member
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Jun 18, 2007
Messages
18,360
I have never had to solve a ratio problem where I have been given these circumstances, if you can help me get started with perhaps an equation or two to find the differences that would be great.

Problem:

The ratio of men to women on a bus is 5/7. Two women and one man board the bus and the ratio is then 7/10. How many men and women were on the bus before the other three passengers boarded?

I have not done much with ratios so I am having trouble figuring out an equation to figure this out. Any help would be appreciated. Thank you ^_^
Assume:

# of men initially = 5*x

# of women initially = 7*x

# of men later = 5*x + 1

# of women later = 7*x + ??

Now continue.....
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,998
I have never had to solve a ratio problem where I have been given these circumstances, if you can help me get started with perhaps an equation or two to find the differences that would be great.

Problem:

The ratio of men to women on a bus is 5/7. Two women and one man board the bus and the ratio is then 7/10. How many men and women were on the bus before the other three passengers boarded?

I have not done much with ratios so I am having trouble figuring out an equation to figure this out. Any help would be appreciated. Thank you ^_^
This is what I would have done (I'm not as good as Subhotosh Kahn at avoiding fractions):
Let x be the number of men on the bus initially and y the number of women. "The ratio of men to women on a bus is 5/7": \(\displaystyle \dfrac{x}{y}= \dfrac{5}{7}\) "Two women and one man board the bus and the ratio is then 7/10": So the number of men goes from x to x+ 1 and the number of women from y to y+ 2. \(\displaystyle \dfrac{x+ 1}{y+ 2}= \dfrac{7}{10}. Multiply both sides of the first equation by 7y: 7x= 5y. Multiply both sides of the second equation by 10(y+ 2): 10(x+ 1)= 7(y+ 2). That is the same as 10x+ 10= 7y+ 14 so 10x= 7y+ 4. Can you solve 7x= 5y and 10x= 7y+ 4 for x and y?\)
 

Yammy

New member
Joined
Nov 5, 2013
Messages
2
This is what I would have done (I'm not as good as Subhotosh Kahn at avoiding fractions):
Let x be the number of men on the bus initially and y the number of women. "The ratio of men to women on a bus is 5/7": \(\displaystyle \dfrac{x}{y}= \dfrac{5}{7}\) "Two women and one man board the bus and the ratio is then 7/10": So the number of men goes from x to x+ 1 and the number of women from y to y+ 2. \(\displaystyle \dfrac{x+ 1}{y+ 2}= \dfrac{7}{10}. Multiply both sides of the first equation by 7y: 7x= 5y. Multiply both sides of the second equation by 10(y+ 2): 10(x+ 1)= 7(y+ 2). That is the same as 10x+ 10= 7y+ 14 so 10x= 7y+ 4. Can you solve 7x= 5y and 10x= 7y+ 4 for x and y?\)
\(\displaystyle

Thank you so much! I should have been able to figure that out but I was skipping the cross multiplying and didn't know where to go. For anyone who wants to know I changed those final equations to 7x - 5y = 0 and 10x - 7y = 4 and then solved using determinants. The answer ended up being x = 20 and y = 28

HallsofIvy you're a life saver ^_^\)
 

Subhotosh Khan

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Jun 18, 2007
Messages
18,360
Assume:

# of men initially = 5*x

# of women initially = 7*x

# of men later = 5*x + 1

# of women later = 7*x + 2

Now continue.....
my way:

(5x+1)/(7x+2) = 7/10

10(5x+1) = 7(7x+2)

50x + 10 = 49x + 14

x = 4

Number males originally = 20

Number females originally = 28
 
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