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Responding to a pm (Whenever you would like to discuss a particular question, kindly put it in the forum and send me a pm of the link)
Question: At a certain party, ratio of women to men is 20:11. If 7 men arrive the party, the ratio would be 3:2. how many men are currently at the party?
Since the initial ratio is 20:11, we can say that the number of women is 20x and the number of men is 11x. If 7 men arrive, the ratio changes.
\(\frac{20x}{(11x + 7)} = \frac{3}{2}\)
Now you solve for x by cross multiplying and you get x = 3 So 11x = 33 = number of men currently at the party
As for your query:
"But I have another way to solve this problem that old ratio of men to total minus new ratio of men to total = 7 which is (11/31 - 2/5) x= 7 thenwe should find x from here and multiply with 11/31 and find current men to total situation. Right? But when I solved this it comes up as x=155. Something I am doing wront. But what it is?Could you please help me."
No. of men currently = 11/31 of total = (11/31)*x No of men if 7 men arrive will be 2/5 of NEW total = (2/5)*(x+7) (I hope you see your mistake)
The difference between these two numbers is of 7 men.
So (2/5)*(x+7) - (11/31)*x = 7 62*(x+7) - 55x = 7*5*31 x = 93
No of men currently = (11/31)(93) = 33
Isn't this a really long way of arriving at the answer? Try to understand the method given above.
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I get to x:3 ,but then how can we calculate the number of men by using the old ratio? Aren't we supposed to use the new ratio given?
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We are supposed to use the old ratio because the question asks us to find the number of men currently at the party. " If 7 men arrive the party, the ratio would be 3:2" is a hypothetical statement. It doesn't actually happen. It is a "If A happens, B will happen" kind of statement. The question doesn't say that 7 men actually arrived at the party.
I found this question a little bit confusing - I think it would have been clear if it said 7 "additional" men arrived. Thats when it makes sense to use the ratio 3/2 to equate to 20x/(11x+7)
A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
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