It really doesn’t matter whether you were a Arts, Science or Commerce student in your Grad’s School. Questions of GMAT Maths only analyses your quant skills based upon the fundamental topics which you have leant in your school. Here is a list of GMAT Arithmetic Formulas to help you to get a hold of this section. Arithmetic in GMAT Quant contains typical questions from the following topics –

- Combinatorics – Factorials, Combinations, Permutations
- Fractions
- Percents
- Number Properties
- Operations
- Probability
- Statistics – Mean, Median, Mode, Standard Deviation

## Combinatorics

The branch of mathematics which deals with collections of objects of similar characteristics is known as Combinatorics. Let’s solve a sum based on Identical Objects.

**1) How many different unique combinations of letters can be created by rearranging the letters in “MATHEMATICS”?**

Solution – In the word, “MATHEMATICS” – S, E, H, I, C – These five alphabets are occurring only once, whereas, M, A, T – are occurring twice.

Hence, Total number of alphabets, \(n = 5 \times 1 + 3 \times 2 = 11\)

As we need to find out only “Unique Combinations”, for repeated Items –

\(\frac{11!}{1! \; 1! \; 1! \; 1! \; 1! \; 2! \; 2! \; 2!} = \frac{11!}{8}\)

Answer = \(\frac{11!}{8}\)

## Fraction – Multiplication and Division

**2) \(\frac{\frac{2}{6} \times \frac{3}{2}}{\frac{7}{8} – \frac{1}{3}} = ?\)**

Solution: Numerator = \(\frac{2}{6} \times \frac{3}{2}\)

= \(\frac{2(3)}{6(2)}\)

= \(\frac{6}{12}\)

= \(\frac{1}{2}\)

Denominator = \(\frac{7}{8} – \frac{1}{3}\)

= \(\frac{7(3)}{8(3)} – \frac{1(8)}{3(8)}\)

= \(\frac{21}{24} – \frac{8}{24}\)

= \(\frac{13}{24}\)

Hence,

\(\frac{\frac{2}{6} \times \frac{3}{2}}{\frac{7}{8} – \frac{1}{3}}\)

= \(\frac{\frac{1}{2}}{\frac{13}{24}}\)

= \(\frac{1}{2} \times \frac{24}{13}\)

= \(\frac{12}{13}\)

## Percent Change

**3) The rate of Apple has increased from 75 cents per pound to 81 cents per pound. What is the percent increase in the cost of the item?**

Solution: As \(Percent \; Change = \frac{New \; Value – Old \; Value}{Old \; Value} \times 100\)

\(Percent \; Change = \frac{81 – 75}{75} \times 100\)

\(= \frac{6}{75} \times 100\)

\(= 0.08 \times 100\)

\(= 8 \%\)

## Number Properties

**4) In the prime factorization of 4320, how many 2s are there?**

Solution: \(4320 = 2 \times 2160\)

\(4320 = 2 \times 2 \times 1080\)

\(4320 = 2 \times 2 \times 2 \times 540\)

\(4320 = 2 \times 2 \times 2 \times 2 \times 270\)

\(4320 = 2 \times 2 \times 2 \times 2 \times 2 \times 135\)

\(4320 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 45\)

\(4320 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 15\)

\(4320 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5\)

Therefore, there are five 2s in the factorization of 4320.

## Combination of Multiple Operations

**5) Simplify \(\frac{(8-4)(10-2)}{2^{2}}\)**

Solution:

By Parenthesis, 8 – 4 = 4 and 10 – 2 = 8

As per Exponent’s rule, \(2^{2}\) = 4

Now, \(= \frac{4 \times 8}{4}\)

\(= 8\)

## Probability

Important Probability Formula = \(Probability \; of \; Occurrence \; of \; an \; Event = P(E) = \frac{n(E)}{n(S)}\)

Where, S = Sample and E = Event

**6) Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?**

Solution: Here, S = {1, 2, 3, 4, …., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

Therefore, \(P(E) = \frac{n(E)}{n(S)} = \frac{9}{20}\)

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