People often ask – how do we go from 48 to 51 in Quant? This question is very hard to answer since we don’t have a step by step plan – do theory from here – do questions from there – take a test from here – read posts from there etc. Today and in the next few weeks, we will discuss how to go from 48 to 51 in Quant.

Above Q48, the waters are pretty choppy! Questions are hard less because of the content and more because they look so unique – even though they’re testing the same concepts. Training yourself to see familiarity in the obscure is difficult, and that happens from seeing a lot of problems. There is barely any scope for making silly mistakes – you must run through all simple questions quickly and neatly, leaving you plenty of time to think through the tougher ones. It’s important to have enough time and keep your cool, which is easier to do if you have more time.

The question for today is: how do you handle simple questions quickly? We have mentioned many times that most GMAT Quant questions do not need Algebra. We can easily solve them by just analyzing while reading the question stem!

Here is how we can do that:

Question: 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

Solution: This is a pretty simple non-tricky PS question. To solve it, most people use an algebraic method which looks something like this.

Girls in school A : Girls in school B = 4 : 3

So number of girls in school A = 4n and number of girls in school B = 3n

Since in school A, 40% students are girls and 60% are boys, number of boys is 6n.

Since in school B, 60% students are girls and 40% students are boys, number of boys is 2n.

If we transfer 20 boys from school A, we are left with 6n – 20 and when 20 boys are added to school B, we get 2n + 20 boys in school B.

(6n – 20)/(2n + 20) = 5/3

You get n = 20

Boys at school A after transfer = 6*20 – 20 = 100

Boys at school B after transfer = 2*20 + 20 = 60

Difference = 40

Answer (B)

This method gives you the correct answer, obviously, but it does take quite a bit of time. On the other hand, this is what should go through your mind while reading the question if you are focused on using logic:

“School A is 40% girls and school B is 60% girls.”

School A – 40% girls

School B – 60% girls

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

When we read this line, we should take a step back to the previous line with the % figures. We see that school A has more girls than school B (4:3) but its % of girls is lower (only 40% compared to 60% in B). This means that school A has more students than school B. Say something like school A has 200 students while school B has 100 (use easy numbers). So school A has 80 girls while school B has 60 girls. This gives us a ratio of 4:3. (If you do not get 4:3 on your first try, you should tweak the assumed numbers a bit but you should stick to simple numbers.) Then verify the rest of the data against these numbers and get your answer.

School A has 120 boys and school B has 40 boys. Transfer 20 boys from school A to school B to get 100 boys in school A and 60 boys in school B giving us a difference of 40 boys.

This takes lesser time but requires some ingenuity. That could be the difference between Q48 and Q51.

Hope this gave you some ideas. Try the reasoning approach on other simple questions. With practice, you can save a ton of time!

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Why is it that since in school B, 60% students are girls and 40% students are boys, number of boys is 2n ???

We are given that Girls in school A : Girls in school B = 4 : 3

In number terms, number of girls in school A = 4n and number of girls in school B = 3n

We are also given that in school B, 60% students are girls and 40% students are boys.

We have assumed that number of girls in school B is 3n. So 3n is 60% of the total students in school B. Then 40% of total students in school B i.e. the number of boys in school B must be 2n.

If you want to see the calculation, 60% * Total = 3n

Total = 5n

40% * Total = 40/100 * 5n = 2n

Notice that since Girls(A):Girls(B) = 40%:60% and also 4:3, the Total(A):Total(B) must be 2:1. We can begin with Total(A)=100 and Total(B)=50, which fails the Boys(A):Boys(B) 5:3 test – this should lead us to trying doubling the Totals, and indeed 200 and 100 does offer the solution. However,I’m not convinced under exam duress, this should be the approach, since there are too many constraints – satisfy 5:3 and the difference 20 – to deal with and I guess its open to debate if this is indeed pragmatic, since we’re not having to deal with quadratic equations to solve.

I don’t understand whi it can’t be simple as, School A has 60% boys, which is 60/100 and School B has 40% boys which is 40/100. Why can’t you just take 20 boys from School A, leaving 40 boys and add them to School B which is 60. The difference being 20.Obviously that’s not the right answer but why is it that, if the questions is telling me I have 60% boys in School A, why can’t I just take that as simply 60/100? Isn’t that what 60% percent means? Same thing for School B.

You are assuming that both the schools have equal number of total students (100 each). Think – is this a valid assumption?