### Permutation involving sum of digits

There is a certain variety of questions that involve getting a bunch of numbers using permutation, and then doing some operations on the numbers we get. The questions can get

### How to work on rate problems

Rate questions, so far as I can remember, have been a staple of almost every standardized test  ever. And, rest assured, dear reader, you will see them on the CAT.

### Stuggling with solutions

One of the most misleading parts of the whole CAT experience is the process of reading the solution to a math problem in the Quant section. When you try the

### Cyclicity in Remainders

Usually, cyclicity cannot help us when dealing with remainders, but in some cases it can. Let’s look at the cases in which it can, and we will see why it

### Cyclicity of numbers

The first thing you need to understand is that when we multiply two integers together, the last digit of the result depends only on the last digits of the two integers.

### Using differences of Squares

Whenever we’re given unpleasant numbers on the CAT, it’s worthwhile to think about the characteristics of round numbers in the vicinity. Like, 3599 is the same as 3600 – 1.

### Problems with reversed digits

The CAT asks a fair number of questions about the properties of two-digit numbers whose tens and units digits have been reversed. Because these questions pop up so frequently, it’s worth

### Grammatical structure of conditional sentences

Let us start out with some basic ideas on conditional sentences (though I know that most of you will be comfortable with these): A conditional sentence (an if/then sentence) has two clauses – the “if

### Making rate questions easier to solve

The goal, when preparing for the CAT, isn’t to internalize hundreds of strategies; it’s to absorb a handful that will prove helpful on a variety of questions. Try the following

### Calculating Probability of intersecting events

We know our basic probability formulas (for two events), which are very similar to the formulas for sets: P(A or B) = P(A) + P(B) – P(A and B) P(A) is