- When we perform any arithmetic operation on the whole numbers and we do not need an exact answer, then we can round off the whole numbers.
- The arithmetic operations are addition, subtraction, multiplication and division.
- Before performing any arithmetic operations on the whole numbers, we round off the whole numbers and then we can perform these operations.

Sum is the answer to an addition problem.

Difference is the answer of a subtraction problem.

- A product is an answer of a multiplication problem.
- A quotient is an answer of a division problem.
- To estimate sums, differences, products and quotients, we can round off the numbers that we are working with to the nearest power of ten.

**Example**: A hotel needs to arrange \(992\) light bulbs for the rooms. If in each room, there are \(8\) bulb holders, estimate the number of rooms in the hotel.

- First, we calculate the estimation of total light bulbs.

Total light bulbs \(=992\)

Tens place digit, \(9\) is greater than \(5\), so we will round the number up to \(1,000\).

There are \(8\) bulbs holders in each room.

\(\therefore\) Each room should have \(8\) bulbs.

So, estimated number of rooms in the hotel are:

\(1,000\div8\)

\(=125\)

Thus, \(125\) rooms are there in the hotel.

A \(30,000\)

B \(40,000\)

C \(20,000\)

D \(10,000\)

- Front-End Estimation method is a useful method of estimating when adding and subtracting the numbers greater than \(1000\).

- When number is greater than \(1000\), front end estimation method is applicable.

- Retain digits of the two highest place values in the number.
- Put zeros on other place values.

Example: \(5324+5683\)

- Each number is greater than \(1000\) so, Front-End Estimation method is applicable.
- Consider \(5324\).

Keep \(5\) and \(3\) which are the digits of two highest place values and insert zero for the other place values so,

\(5324\) becomes \(5300\).

Consider \(5683\).

Keep \(5\) and \(6\) which are the digits of two highest place values and insert zeros for the other place values so, \(5683\) becomes \(5600\).

Rewrite the problem and perform the operation.

\(5300+5600=10900\)

A \(100\)

B \(300\)

C \(200\)

D \(400\)

- While using the Front-End Estimation method with decimals, we separate the whole number and the decimal part from decimal number and then combine them together after performing operations.

\(\to\) Write the Front digits of the numbers being added or subtracted and perform operation.

\(\to\) Round off the decimal part of the numbers being added or subtracted and then perform operation.

\(\to\) Then combine the result.

Example: \(5.20+9.66\)

\(\to\) Front digits of the numbers being added are \(5\) and \(9\).

so, \(5+9=14\)

\(\to\) Round off the decimal parts.

\(0.20\) becomes \(0.2\) (no change).

\(0.66\) becomes \(0.7\).

So, \(0.2+0.7=0.90\)

Combine the result,

i.e. \(14+0.90=14.90\)

A \(7.50\)

B \(8.00\)

C \(8.50\)

D \(7.00\)

- Compatible numbers are numbers that are close in value to the actual numbers and which make it easy to perform mental arithmetic.
- The word compatible means "well-matched".
- Compatible numbers are numbers that are friendly with each other.

For example: \(15\) and \(5\) are compatible numbers when it come to division.

- Compatible numbers are useful in estimating the sum, difference, product or quotient.
- Compatible number are easy to add, subtract, multiply or divide.
- By rounding off the whole numbers, fractions or decimals, to the compatible numbers, we can easily perform the division operation.

Example: Estimate the value of \(29\div6.5\)

- To estimate the quotient, firstly we round off the numbers \(29\) and \(6.5\).
- \(6.5\) can be rounded to \(7\).
- \(29\) can be rounded to \(30.\) But \(30\) is not exactly divisible by \(7\) so, we round down \(29\) to \(28\) because \(28\) and \(7\) are compatible numbers.
- Now we can easily perform division.
- Now rewrite the problem using rounded values, so \(28\div7=4\)

- To perform any operation on a fraction, first we round off the fraction to the nearest half and then perform the operation.
- To estimate the sums, differences, products and quotients of the simple fraction or mixed fraction, we use the rounding rules of simple fraction or mixed fraction that we discussed earlier.

We can round off a simple fraction to the nearest half which can be \(0,\;\dfrac{1}{2}\) or \(1.\)

We can round off a mixed fraction to the nearest whole number.

**Example:** Julie needs \(2\dfrac{1}{5}\) cups of sugar and \(1\dfrac{3}{4}\) cups of flour for her cake. The total estimated quantity of sugar and flour Julie needs is:

We are given the exact quantity of sugar \(=2\dfrac{1}{5}\) cups

- First, we calculate the estimated quantity of sugar.

\(\dfrac{1}{5}\) is less than \(\dfrac{1}{2}\), so it becomes zero and whole number \(2\) remains same.

- So, the estimated quantity of sugar is \(2\) cups.
- The exact quantity of flour \(=1\dfrac{3}{4}\) cups
- Now, we calculate the estimated quantity of flour.
- \(\dfrac{3}{4}\) is greater than \(\dfrac{1}{2}\), so it becomes \(1\) and whole number \(1\) becomes \(1+1=2\)
- So, the estimated quantity of flour is \(2\) cups.
- Thus, total estimated quantity of sugar and flour is

\(2+2=4\) cups

A \(27\; liters\)

B \(47\; liters\)

C \(57\; liters\)

D \(37\; liters\)

- To perform any arithmetic operation on decimal numbers, first we round off the decimal numbers and then perform the operations.
- We can round off the decimals to the nearest whole number and then we can perform the arithmetic operations such as addition, subtraction, multiplication and division to find a reasonable solution of our problem.
- To find an estimated answer of sums, differences, products and quotients, we use the rounding rules.

**Example:**

A road was \(4.8\) miles long. After extension, it became \(23\) times longer than the previous road. Estimate the length of the extended road.

Given the original length of road is \(4.8\) miles.

- First, we calculate the estimated length of the road before extension.

The digit at the tenths place is \(8\) which is greater than \(5\).

So, the digit at ones place i.e. \(4\) is rounded up to \(5\).

- Now, we calculate the estimated value of \(23\).

The digit at the ones place is \(3\) which is less than \(5\). So, we will round the number down to \(20\).

- Thus, the estimated length of the road after extension is,

\(5×20=100\) miles

A \($2,070\)

B \($2,000\)

C \($1,980\)

D \($2,050\)