School A is 40% girls and school B is 60% girls. The ratio

Assume the common variable for the ratio to be x.
As ratio of girls of school A (Ga) to girls of school B (Gb) is 4:3. Therefore, Ga/Gb = 4/3
i.e. there are Ga=4x, Gb=3x
Now, as school A has 40% girls,
Ga=40/100 * (Total students at school A)
Substuting Ga=4x.
Total students at school A = 10x
Similarly, Total students at school B = 5x

From this, we know that boys at school A (Ba) = 6x
boys at school B (Bb) = 2x

As 20 boys leave school A and join school B
new Ba = 6x - 20
new Bb = 2x + 20

It is also given that,
(new Ba)/(new Bb) = 5/3
i.e (6x - 20)/(2x + 20)=5/3
x=20

Hence difference between number of boys after transfer,
[6(20)-20] - 2(20)+20] = 40
Ans : B

Hope this helps

School A...................................................................... School B

Boys.............. Girls.......................................Boys................ Girls

\(\frac{60a}{100}............... \frac{40a}{100} ........................................... \frac{40b}{100} .................... \frac{60b}{100}\)

The ratio of the number of girls at school A to the number of girls at school B is 4:3

\(\frac{40a}{100} / \frac{60b}{100} = \frac{4}{3}\)

a = 2b

20 boys transferred from school A to school B

\(\frac{120b}{100} - 20 = \frac{6b-100}{5}\) ............ (1)

\(\frac{40b}{100} + 20 = \frac{2b+100}{5}\) ............. (2)

Ratio (1) : (2) = 5/3

\(\frac{6b-100}{2b+100} = \frac{5}{3}\)

Solving b = 100

We require to find \(\frac{6b-100}{5} - \frac{2b+100}{5}\)

= 100 - 60 = 40

Answer = B

I am getting number of students in A as 200 so my answer reduces to 60 instead of 40.
My approach is as following :

.4A / .6B = 4/3 hence A = 2B where A and B are total number of students in A and B school respectively.
.6A - 20 / .4B + 20 = 5/3 the new ratio ... when I solve this equation I left with A= 200.

200 - 100 - 40 = 60

Can somebody please identify issue with my solution.
Thanks

Till here, everything is fine. A = 200 and B = 100 (since A = 2B)
Number of boys in A = .6A = 120
Number of boys in B = .4B = 40

When 20 are transferred, number of boys in A = 100 and number of boys in B is 60.
Difference = 100 - 60 = 40

Hi karishma,

Thanks for your explanation.

Even I started of with the reasoning and using smart numbers.

I see the number of students that you initially took were 200 and 100.

I started of with 100 and 50 which did not work.

So in such cases should one go for the next set of smart numbers or should switch to the equation method??

During practice, always start with the more intuitive method. Stick to it for a while. Even if you spend a few mins and are unable to get the answer, you can later switch to algebra. With practice, these instances of "not getting the answer" will reduce. You will automatically zero in on the right numbers with some tweaking (for example, is your answer greater than or smaller than what is required? Adjust the assumed numbers accordingly etc).

In the actual exam, know where you stand. If, during recent practice, you were able to get most answers without algebra, then go ahead and use the intuitive methods. Even if you have to rely on algebra in some questions after wasting 2-3 mins on it, you will have plenty and more extra time. If you were almost always forced to resort to algebra during practice, then might as well stick with algebra in the actual exam.

Looking at the answers they are in multiples of 20.

So I will use substitution on this one:

Girls
School A to School B
4:3
Multiply y 20
School A= 80 girls
School B = 60 girls

From the difference of percentages we can calculate the number of boys
School A: 60% is 120
School B: 40% is 40

So if 20 are transferred from A to B then we will have 100 in School A and 60 in School B.

Difference is 100-60=40.
Choose Answer C

Responding to a pm:



Girls in A:Girls in B = 4:3
Assume actual numbers with a variable:
Girls in A = 4n, Girls in B = 3n

Since in school A, girls are 40% and boys are 60%,
40% of Total students in A = 4n
Total students in A = 4n *100/40 = 10n
So number of boys in school A = 10n - 4n = 6n

Using the same logic for school B, since in school B, girls are 60%,
60% of Total students in B = 3n
Total students in = 3n * 100/60 = 5n
So the number of boys in school B = 5n - 3n = 2n

I hope you understand from where I got 6n and 2n. I did this so that a single variable represents all the relevant quantities.

Hi All,

Why can't we go backwards in such type of questions where two ratios are given and final answer really depends on the final ratio.

Let us start with taking final Boys ratio of 5:3, for school A & B respectively. We know that no. of students in school A is significantly greater than B, so, we can take final numbers of boys in A & B to be 100 & 60 (5:3), also 100 sounds better and is easy for calculation purposes.

now Boys(A original)= 100 +20= 120 & Boys (B original)= 60-20=40). These new numbers of 120 & 40 are original percentage of boys in schools A & B respectively (60% & 40%). So, the number of girls in school A & B are 80 & 60 respectively (40% & 60%), which matches with the original ratio 4:3.

I prefer "picking numbers method" in such question more useful as it will eliminate any silly mistakes we may commit while solving algebraic equations and eventually get stuck in getting/marking the right answer.

Experts let me know, if you also suggest to pick numbers in such question types.

Regards
Yash

A : Girls = 0.4 A, Boys = 0.6 A
B: Girls = 0.6 B, Boys = 0.4 B

Now, 0.4A/0.6B = 4/3 -> A = 2B

So,
A= 2B : Girls = 0.8B, Boys = 1.2B
B: Girls = 0.6 B, Boys = 0.4 B

After transfer of 20 boys from A to B
A= 2B : Girls = 0.8B, Boys = 1.2B-20
B: Girls = 0.6 B, Boys = 0.4 B+20

(1.2B-20)/(0.4B+20) = 5/3 -> B = 100

Difference in no. of boys between Class A and B after transfer = (1.2B-20)-(0.4B+20) = 0.8B -40 = 0.8*100 - 40 = 40

Answer B

Yes back solving also works most of the times, unless data is very complex..
For GMAT problems is a very good way to solve the problems.

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:

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

let a and b=total students at A and B respectively
.4a/.6b=4/3
b=a/2
substituting,
(.6a-20)/(.4a/2+20)=5/3
a=200; b=100
.6*200=120 boys originally at A
120-20=100 boys remaining at A
.4*100=40 boys originally at B
40+20=60 boys currently at B
100-60=40 boys difference between A and B
B

Can I approach the problem in the following way?

Since 4 in 4:3 corresponds to 40% of school A and 3 in 4:3 corresponds to 60%, then 60% in school A (males) would be 6 (if 4 is 40%, 6 is 60%), and 40% males in school B would be 2 (if 3 is 60%, then 2 is 40%). So ratio is 6:2. The final answer is strikingly correct in this case, when I approached the first step in this way.

Did I miss something? Or am I correct? Thanks in advance!

Total students A = x, Total students B = y,
girls A = 0.4x, B = 0.6y
boys A = 0.6x, B = 0.4y

\(\frac{0.4x}{0.6y}=4/3…1.2x=2.4y…x=2y\)

\(\frac{0.6x-20}{0.4y+20}=5/3…\frac{0.6(2y)-20}{0.4y+20}=5/3…3(1.2y-20)=5(0.4y+20)…y=100\)

\((0.6x-20) - (0.4y+20) = 0.6(2y)-20-0.4y-20=120-20-40-20=40\)

Answer (B)

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

My Approach― 01(Details):
Let, the number of girls in School A= 4x
and the number of boys in School B= 3x

Total Students in School A;
Total Students*40% = 4x
Total Students= (100/40)*4x= 10x

Total Students in School B;
Total Students*60%= 3x
Total Students= (100/60)*3x= 5x

So, the number of boys in School A= 60% of 10x=6x
The number of boys in School B= 40% of 5x=2x

Initial ratio of Boys in School A to School B= 6:2= 12:4
Final ratio of Boys in School A to School B= 5:3= 10:6

Here, (6-4) or 2 ≡ 20
(10+6) or 16≡ 160

After 2o students transfer, boys in School A= (10/16)*160 = 100
and boys in School B= (6/16)*160= 60
Now the difference between the number of boys at school A and at school B= 100-60=40

Answer: B. 40

My Approach― 02(Shortcut):
40% of T in A= 4x (Here T means total Students)
T in A= 10x
Again, 60% of T in B= 3x
T in B= 5x
Now, Boy in A= 60% of 10x= 6x
Boy in B= 40% of 5x= 2x

Initial Ratio of Boys in A to B= 6:2= 12:4
Final Ration of Boys in A to B= 5:3= 10:6
Here, 6-4 ≡ 20
or, 2≡ 20
or, 16≡ 160 (sum of the final ratio of 10 and 6 is 16)

Now, Boys in A= (10/16)*160= 100
and Boys in B= (6/16)*160= 60
Here difference= 100-60= 40

Answer: B. 40