BTGmoderatorDC wrote:School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?

A) 20
B) 40
C) 60
D) 80
E) 100

OA B

Source: Veritas Prep

Let A = TOTAL population of School A
Let B = TOTAL population of School B

School A is 40% girls and School B is 60% girls
So, number of GIRLS at School A = 0.4A
And number of GIRLS at School B = 0.6B

The ratio of the number of girls at School A to the number of girls at School B is 4:3
We can write: 0.4A/0.6B = 4/3
Cross multiply to get: (3)(0.4A) = (4)(0.6B)
Simplify to get: 1.2A = 2.4B
Divide both sides by 1.2 to get A = 2B

Let's re-examine the following: School A is 40% girls and School B is 60% girls
So, current number of BOYS at School A = 0.6A
And current number of BOYS at School B = 0.4B

If 20 boys transferred from School A to School B....
So, NEW number of BOYS at School A = 0.6A - 20
And NEW number of BOYS at School B = 0.4B + 20

...., the new ratio of boys at School A to boys at School B would be 5:3
So, we can write: (0.6A - 20)/(0.4B + 20) = 5:3
Cross multiply to get: (3)(0.6A - 20) = (5)(0.4B + 20)
Simplify to get: 1.8A - 60 = 2B + 100
Add 60 to both sides to get: 1.8A = 2B + 160
Subtract 2B from both sides to get: 1.8A - 2B = 160

We now have a system of equations to solve:

Divide both sides by 1.2 to get A = 2B
1.8A - 2B = 160

Take 1.8A - 2B = 100 and replace A with 2B to get: 1.8(2B) - 2B = 160
Expand: 3.6B - 2B = 160
Simplify: 1.6B = 160
Solve: B = 100
So, the (current) TOTAL population at School B is 160
Since we know that A = 2B, we can conclude that the (current) TOTAL population at School A = (2)(100) = 200

Now that we know the values of A and B, we can complete our calculations.
Current number of BOYS at School A = 0.6A = 0.6(200) = 120
And current number of BOYS at School B = 0.4B = 0.4(100) = 40

So, NEW number of BOYS (after the transfer) at School A = 100
And NEW number of BOYS (after the transfer) at School B = 60

What would the difference between the number of boys at School A and at School B be after the transfer?
100 - 60 = 40


Answer: B

Cheers,
Brent

BTGmoderatorDC wrote:School A is 40% girls and school B is 60% girls. The ratio of the number of girls at school A to the number of girls at school B is 4:3. if 20 boys transferred from school A to school B and no other changes took place at the two schools, the new ratio of the number of boys at school A to the number of boys at school B would be 5:3. What would the difference between the number of boys at school A and at school B be after the transfer?

A) 20
B) 40
C) 60
D) 80
E) 100

OA B

Source: Veritas Prep

If School A has 40% girls, it also has 60% boys, giving us a ratio of 4 : 6. We can let the number of girls and boys in School A be 4x and 6x, respectively. Similarly, we can let the number of girls and boys in School B be 6y and 4y, respectively. Therefore, we can express the ratio of girls at School A to girls at School B as 4x/6y, and we are given that this ratio is also equal to 4 : 3. Thus:

4x/6y = 4/3

For the boys, School A lost 20 boys, giving us (6x - 20) boys at School A, and School B gained those 20 boys, so School B now has (4y + 20) boys. The new ratio is given as 5 : 3. Thus:

(6x - 20)/(4y + 20) = 5/3

Solving the first equation, we have:

12x = 24y

x = 2y

Solving the second equation, we have:

18x - 60 = 20y + 100

18x = 20y + 160

9x = 10y + 80

Since x = 2y, we have:

9(2y) = 10y + 80

18y = 10y + 80

8y = 80

y = 10

Thus, x = 2(10) = 20. After the transfer of 20 boys, School A has 6x - 20 = 6(20) - 20 = 100 boys and School B has 4(10) + 20 = 60 boys. So, the difference is 100 - 60 = 40.

Answer: B

Hi All,

This is a layered algebra question that would take a number of "math steps" to solve. The math involved isn't too hard, but you have to do a lot of work to get the job done....

"School A is 40% girls (thus 60% boys) and School B is 60% girls (thus 40% boys)"

We can write these ratios as:

School A
B:G
3:2

School B
B:G
2:3

Since ratios are all about multiples, I'm going to add a "variable" into each ratio (since the schools are different, I need to use a different variable for each).

School A
B:G
3x:2x

School B
B:G
2y:3y

"The ratio of girls at School A to girls at School B is 4:3"

We can write this ratio as...

2x/3y = 4/3

And cross-multiply....

6x = 12y
x = 2y

"If 20 boys transferred from School A to School B...the new ratio of boys at School A to boys at School B would be 5:3"

We can write this ratio as....

(3x - 20)/(2y + 20) = 5/3

And cross-multiply...

9x - 60 = 10y + 100
9x = 10y + 160

We now have 2 variables and 2 equations, so we can solve using "System" math to solve for the two variables...

x = 2y
9x = 10y + 160

By substituting in, we get...

9(2y) = 10y + 160
18y = 10y + 160
8y = 160
y = 20

Plugging back in, we get...

x = 40

The question asks for the DIFFERENCE in the number of boys at School A and at School B AFTER the transfer:

Number at School A after the transfer = 3x - 20 = 120-20 = 100
Number at School B after the transfer = 2y + 20 = 40 + 20 = 60

The difference = 100 - 60 = 40

Final Answer: B

GMAT assassins aren't born, they're made,
Rich