Comparing options

Comparing options

A lot of  test takers complain about insufficient time. This is understandable as far as the Verbal section is concerned. We all have different reading speeds and that itself accounts for a lot of time issues in the Verbal section. Obviously then there are other factors – your comfort with the language, your comprehension skills, your conceptual understanding of the Verbal question types, etc.

However, timing issues should not arise in the Quant section. Your reading speed has very little effect on the overall timing scheme because most of the time during the Quant section is spent in solving the question. So if you are falling short on time, it means the methods you are using are not appropriate. We have said it before and will say it again – most Quant questions can be done in under one minute if you just look for the right thing.

For example, of the four listed numbers below, which number is the greatest and which is the least?

2/3

2^2/3^2

2^3/3^3

Sqrt(2)/Sqrt(3)

Now, how much time you take to solve this depends on how you approach this problem. If you get into ugly calculations, you will end up wasting a ton of time.

2/3 = .667

2^2/3^2 = 4/9 = .444

2^3/3^3 = 8/27 = .296

Sqrt(2)/Sqrt(3) = 1.414/1.732 = .816

So we know that the greatest is Sqrt(2)/Sqrt(3) and the least is 2^3/3^3. We got the answer but we wasted at least 2-3 mins in getting it.

We can do the same thing very quickly. We know that the squares/cubes/roots etc of numbers vary according to where the numbers lie on the number line.

2/3 lies in between 0 and 1, as does 1/4.

The Sqrt(1/4) = 1/2, which is greater than 1/4, so we know that the Sqrt(2/3) will be greater than 2/3 as well.

Also, the square and cube of 1/4 is less than 1/4, so the square and cube of 2/3 will also be less than 2/3. So the comparison will look like this:

(2/3)^3 < (2/3)^2 < 2/3 < Sqrt(2/3)

That is all you need to do! We arrived at the same answer using less than 30 secs.

Using this technique, let’s solve a question:

Which of the following represents the greatest value?

(A) Sqrt(3)/Sqrt(5) + Sqrt(5)/Sqrt(7) + Sqrt(7)/Sqrt(9)

(B) 3/5 + 5/7 + 7/9

(C) 3^2/5^2 + 5^2/7^2 + 7^2/9^2

(D) 3^3/5^3 + 5^3/7^3 + 7^3/9^3

(E) 3/5 + 1 – 5/7 + 7/9

Such a question can baffle someone who believes in calculating everything. We know better than that!

Note that the base values in all the options are 3/5, 5/7 and 7/9. This should hint that we need to compare term to term and not the entire expressions. Also, all values lie between 0 and 1 so they will behave the same way.

Sqrt(3)/Sqrt(5) is the same as Sqrt(3/5). The square root of a number between 0 and 1 is greater than the number itself.

3^2/5^2 is the same as (3/5)^2. The square (and cube) of a number between 0 and 1 is less than the number itself.

So, the comparison will look like this:

(3/5)^3 < (3/5)^2 < 3/5 < Sqrt(3/5)

(5/7)^3 < (5/7)^2 < 5/7 < Sqrt(5/7)

(7/9)^3 < (7/9)^2 < 7/9 < Sqrt(7/9)

This means that out of (A), (B), (C) and (D), the greatest one is (A).

Now we just need to analyse (E) and compare it with (B).

The first term is the same, 3/5.

The last term is the same, 7/9.

The only difference is that (B) has 5/7 in the middle and (E) has 1 – 5/7 = 2/7 in the middle. So (E) is certainly less than (B).

We already know that (A) is greater than (B), so we can say that (A) must be the greatest value.

A quick recap of important number properties:

Case 1: N > 1

N^2, N^3, etc. will be greater than N.

The Sqrt(N) and the CubeRoot(N) will be less than N.

The relation will look like this:

… CubeRoot(N) < Sqrt(N) < N < N^2 < N^3 …

Case II: 0 < N < 1

N^2, N^3 etc will be less than N.

The Sqrt(N) and the CubeRoot(N) will be greater than N.

The relation will look like this:

… N^3 < N^2 < N < Sqrt(N) < CubeRoot(N)  …

Case III: -1 < N < 0

Even powers will be greater than N and positive; Odd powers will be greater than N but negative.

The square root will not be defined, and the cube root of N will be less than N.

CubeRoot(N) < N < N^3 < 0 < N^2

Case IV: N < -1

Even powers will be greater than N and positive; Odd powers will be less than N.

The square root will not be defined, and the cube root of N will be greater than N.

N^3 < N < CubeRoot(N) < 0 < N^2

Note that you don’t need to actually remember these relations, just take a value in each range and you will know how all the numbers in that range behave.

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