Decimals deals with place values, operations with decimals and rounding. One should recognize whenever a number is written, then one should essentially creating a recipe for expressing some value.

**Example**: In the number 743

- Here

7 is in hundred’s place

4 is in ten’s place

3 is in one’s place

743 can be written as (7 x 100) + (4 x 10) + (3 x 1)

A positional system of notation in which the position of a number with respect to a point determines its value is known as place value.

The value of the digits is based on the number ‘10’ in the decimal system.

Each position in a decimal number has a value that is a power of 10.

- for decimal form, 82.743

Here

Before the decimal point

8 is in ten’s place

2 is in one’s place

After the decimal point

7 is in ten’s place

4 is in hundred’s place

3 is in thousand’s place

So, 82.743 is written as

(8 x 10) + (2 x 1) + (7 x \(\frac{1}{10}\)) + (4 x \(\frac{1}{100}\)) + (3 x \(\frac{1}{1000}\))

- 82.743196

After the decimal,

The first digit represents tenths

The second digit represents hundredths

The third digit represents thousands

The fourth digit represents ten thousands

The fifth digit represents hundred thousands

The sixth digit represents millionths

and so on…

If a number is considered and zeros are added to the last digit after the decimal point, the value of the number is not changed.

**Example**:

- 5.32 = (5 x 1) + (3 x \(\frac{1}{10}\)) + (2 x \(\frac{1}{100}\))

Now, add two zeros to the last digit after the decimal point

5.3200 = (5 x 1) + (3 x \(\frac{1}{10}\)) + (2 x \(\frac{1}{100}\)) + (0 x \(\frac{1}{1000}\)) + (0 x \(\frac{1}{10,000}\))

So, Adding zeros have no effect on the value of the number

**Operations with decimals**

**Adding and subtracting decimals**

To add and subtract the decimals, the rule is line up the decimals

After the decimal point, add additional zeros (or assume there are zeros) if required before adding or subtracting.

**Example 1**:

- 2.0756 + 17.5 + 0.083 = 2.0756 + 17.5000 + 0.0830 = 19.6586

**Example 2**:

- 14.248 – 3.18749 = 14.24800 – 3.18749 = 11.06051

**Multiplying decimals**

First, count the total number of digits that are to the right of each decimal.

Then, ignore the decimals and find the product.

Take the product and move the decimal place to the left based on the total number of digits that were originally to the right of each decimal.

**Example**:

- 23.456 x 7.8

Ignore the decimals and find the product

23456 x 78 = 1829568

There are 4 digits to the right of the decimal

So, move decimal 4 space to the left of 1829568

i.e. 182.9568

**Dividing decimals**

Move both decimals until the divisor becomes an integer.

Divide and keep the decimal in the same location.

**Example**:

- 43.548 ÷ 2.85

Take the divisor i.e. 2.85 and turn it into an integer by moving the decimal two spaces to the right i.e. 285

Since decimal in the divisor is moved to two spaces to the right, also take the dividend i.e. 43.548 and move the decimal two spaces to the right i.e. 4354.8

Now, divide 4354.8 by 285,

4354.8 ÷ 285 = 15.28

**Rounding Decimals**

Next digit is 0, 1, 2, 3, or 4 → round down

Next digit is 5, 6, 7, 8 or 9 → roundup

**Example 1**:

- 7.

First, locate the digit in tenth position i.e. 3

Since the tenth place after the decimal is ‘3’ and digit ‘8’ is to the right of ‘3’. It is rounded up to ‘4’

Now the given digit becomes equal to 7.4

**Example 2**:

- 15.02

First, locate the digit in thousandth position i.e. 3

Since the thousandth place after the decimal is ‘3’ and digit ‘1’ is to the right of ‘3’. It is rounded down to ‘3’

Now the given digit becomes equal to 15.023

Quantity A | Quantity B |
---|---|

The 25th digit to the right of the decimal point in D | 4 |

- A. Quantity A is greater.

B. Quantity B is greater.

C. The two quantities are equal.

D. The relationship cannot be determined from the information given.

**Solution**:

- By dividing 4 by 11, you get the decimal form D = 0.363636…, where the sequence of two digits “36” repeats without end. Continuing the repeating pattern, you see that the 1st digit, the 3rd digit, the 5th digit, and every subsequent odd-numbered digit to the right of the decimal point is 3.

Therefore, Quantity A, the 25th digit to the right of the decimal point, is 3. Since Quantity A is 3 and Quantity B is 4, the correct answer is Choice B.

**2. What is the 25th digit to the right of the decimal point in the decimal form of \(\frac{6}{11}\)?**

- A. 3

B. 4

C. 5

D. 6

E. 7

**Solution**:

- \(\frac{6}{11}\) = 0.5454…….

Every odd-numbered digit, to the right of the decimal point, is 5, thus 25th (odd) digit will also be 5.

Hence, option C is the answer.

**3. Quantitative Comparison**

Column A | Column B |
---|---|

0 .25% of .25 | 6.25 * \((10)^{-5}\) |

- A. Column A’s quantity is greater.

B. Column B’s quantity is greater.

C. The quantities are the same.

D. The relationship cannot be determined from the information given.

**Solution**:

- 0.25% of 0.25

0.0025 (0.25) — Move the decimal on 0.25% two places to the left.

0.000625

6.25 * \((10)^{-5}\) — Move decimal 5 places to the left.

0.0000625

Since 0.000625 > 0.0000625, the answer is column A is greater.

Option A is the right choice.

**4. Evaluate: \(\frac{((2.39)^2 + (1.61)^2)}{(2.39 – 1.61)}\)**

- A. 2

B. 4

C. 6

D. 8

**Solution**:

- Apply \(\frac{(a^2 + b^2)}{a – b}\) = \(\frac{(a + b)(a – b)}{(a – b)}\) = (a + b)

So, here (a + b) = (2.39 + 1.61) = 4.

**5. What decimal of an hour is a second?**

- A. 0.0025

B. 0.0256

C. 0.00027

D. 0.000126

**Solution**:

- Required decimal = \(\frac{1}{60 * 60}\) = \(\frac{1}{3600}\) = 0.00027