The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to

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.
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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.
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Your solution is correct. That is how I would solve it too.

Note that in the original question, you don't need to use the ratio 17:22. Your modified question needed it. In an actual GMAT question, in such a case, the given ratio would be easier. I have rarely found a "tough to work with ratio" in even practice GMAT questions, let alone actual GMAT questions (old or current). Hence, don't be concerned about the difficult calculations here.
Also, I advise my students to know the multiplication tables till 20. They add value by helping one speed up in some questions.
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The ratio of boys to girls in Class A is 3 to 4.
Let B = number of boys in class A
Let G = number of girls in class A
We get: B/G = 3/4
Cross multiply to get: 4B = 3G

Class A has one more boy and two more girls than class B
So B - 1 = number of boys in class B
So G - 2 = number of girls in class B

The ratio of boys to girls in Class B is 4 to 5
We get: (B - 1)/(G - 2) = 4/5
Cross multiply to get: 5(B - 1) = 4(G - 2)
Expand: 5B - 5 = 4G - 8

So, we now have the following system to solve for G:
4B = 3G
5B - 5 = 4G - 8

Take 4B = 3G and solve for B to get: B = 3G/4

Take 5B - 5 = 4G - 8 and replace B with 3G/4
We get: 5(3G/4) - 5 = 4G - 8
Expand: 15G/4 - 5 = 4G - 8
Multiply both sides by 4 to get: 15G - 20 = 16G - 32
Solve to get: G = 12
Answer:

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Some great Solutions above.
Here's what i think of it now =>

The combined ratio given to us is just painfully useless.
WE can get the values of x and y from these two equations alone=>
3x=4y+1
4x=5y+2

Girls in class A => 12

Hence E
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We can create the ratios:

Class A:

Boys : girls = 3x : 4x

Class B:

Boys : girls = 4y : 5y

Combining, we have:

(3x + 4y)/(4x + 5y) = 17/22

22(3x + 4y) = 17(4x + 5y)

66x + 88y = 68x + 85y

3y = 2x

Also, we are given that class A has one more boy and two more girls than class B. So we have:

3x = 4y + 1 and 4x = 5y + 2

Since 3y = 2x, we know that 6y = 4x, and therefore we have:

6y = 5y + 2

y = 2

Since class A has 4x girls and 4x = 5y + 2, we have 5(2) + 2 = 12 girls in class A.

Answer: E
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Given:
1. The ratio of boys to girls in Class A is 3 to 4.
2. The ratio of boys to girls in Class B is 4 to 5.
3. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22.

Asked: If Class A has one more boy and two more girls than class B, how many girls are in Class A?

1. The ratio of boys to girls in Class A is 3 to 4.
bA : gA = 3:4
bA = 3k ; gA = 4k; cA = 7k

2. The ratio of boys to girls in Class B is 4 to 5.
bB : gB = 4:5
bB = 4m ; gB = 5m : cB = 9m

3. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22.
bA + bB : gA + gB = 3k + 4m : 4k + 5m = 17:22
66k + 88m = 68k + 85m
2k = 3m
k = 1.5m

If Class A has one more boy and two more girls than class B, how many girls are in Class A?
bA = bB + 1
3k = 4m + 1
4.5m = 4m + 1
m = 2
k = 3

gA = 4k = 12

IMO E
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Boys A / Girls A = 3 / 4 = (x + 1)/ (y + 2)
Boys B / Girls B = 4 / 5 = x / y

(2x + 1)/ (2y + 2) = 17 / 22

split the numerator and denominator: (2x + 1) = [2(8) + 1] = 17 , (2y + 2) = [2(10) + 2] = 22

Hence y = 10, Girls in class A = (y + 2) = 12

The trick is really just taking x for class A and y for class B.

Given that the relationship between boys and girls from Class A and B is given, we can quickly come to this:

6x-1/8x-2 = 17/22

x = 3

Class A girls --> 4 * 3 = 12

(I also quickly checked this by plugging 12 into all equations to come up with a satisfying resolution - only doing this if i have time on the real test though)