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26 Jun 2016, 13:33
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aditi2013
Thank you ccooley & VeritasPrepKarishma. Your explanation helps a lot in understanding the concept better.
I also tried one more thing with the question to check my understanding.
What if the question mentioned that the combined class has 17 boys and we need to determine number of students of Class A.
So, the question stem gives following :
Class A, Boys : Girls = 3: 4
Class B, Boys : Girls = 4:5
Combined Class, Boys : Girls = 17 : 22
Total number of boys in Combined Class = 17
Find the number of students in Class A?
Using the weighted average formula, Ratio of total number of students in Class A : Total in Class B = 7 : 6 (Though the calculation took time, is there an alternative method to find this step? )...(#)
No. of Boys in combined class => \(\frac{3}{7}\)*7n + \(\frac{4}{9}\)*6n = 17
n= 3
From (#), 7n = 21 students in Class A.
Hope this is correct.
Is there any other method to solve this?
Thanks in advance.
I also tried one more thing with the question to check my understanding.
What if the question mentioned that the combined class has 17 boys and we need to determine number of students of Class A.
So, the question stem gives following :
Class A, Boys : Girls = 3: 4
Class B, Boys : Girls = 4:5
Combined Class, Boys : Girls = 17 : 22
Total number of boys in Combined Class = 17
Find the number of students in Class A?
Using the weighted average formula, Ratio of total number of students in Class A : Total in Class B = 7 : 6 (Though the calculation took time, is there an alternative method to find this step? )...(#)
No. of Boys in combined class => \(\frac{3}{7}\)*7n + \(\frac{4}{9}\)*6n = 17
n= 3
From (#), 7n = 21 students in Class A.
Hope this is correct.
Is there any other method to solve this?
Thanks in advance.
Use the unknown multiplier again, in my opinion. I tend to go straight for that approach whenever a problem asks me to do algebra with 'pieces' of ratios. A ratio itself isn't very friendly to algebra - how do you 'add' 3:4 and 4:5? - but the unknown multiplier lets you bring algebra into the equation.
Class A has 3a boys, and 4a girls.
Class B has 4b boys, and 5b girls.
You don't know a or b at the moment - they're just unknown variables.
But, you do know that there are 17 boys and 22 girls when you combine the two classes. So,
3a + 4b = 17
4a + 5b = 22
That's just a system of linear equations, and you can solve with elimination or substitution, as you please. Elimination:
12a + 16b = 17*4
12a + 15b = 22*3
b = 17*4 - 22*3 = 68 - 66 = 2
So, class B has 4(2)=8 boys, and 5(2)=10 girls.
26 Jun 2016, 13:35
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By the way, you and anyone else who finds this line of inquiry interesting ought to write up a ratios 'tip sheet' for yourself. Run through some ratios problems, and see if you can sort them into categories. One place to start is with the variants you've listed here, which are variations on which information the problem gives you, and which information the problem asks for. Depending on what info you're given and asked for, the solution path will look a little different. If you try all of the possible varieties ahead of time, and write them out in your notes, you'll be better prepared for the ratio problems you might see on the GMAT.
26 Jun 2016, 18:38
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EvaJager
The information the ratio of boys to girls in the combined class would be 17 to 22 is superfluous.
You could also say that the information that class A has one more boy is superfluous, and you could also say the same about the information that class A has 2 more girls.
Either way, there is definitely redundant information in this question. Of the 3 pieces of information, only 2 are needed to solve this problem.
