- 99% of 4444 + 101% of 6666 =
- (108.999)² - (102.001)²=?
- 5983.987 + 59832.999 – 598.873 = ?
- ? % of 759.96 + 932.99 = 1237.01
- ?² x 8 - 320 = 513 ...
- 66.05 17.95 – 38.99 18.12 = (60+?) 6 ...
- (278% of 695) ÷ 543 =?
- 5275 of 105% + 99.07 × 17.889 =? ...
- 123% of 3825 +2745 =11800 - ? ...
- 783 ÷ 42.59 × 25.86 =? ...
- (14.56)² × √840 =? ...
- 956.41 of 45.06% = ?
- 90.004% of 9500 + 362 = ?
- 31% of 3300 +659 = ?
- 456 x 99.999 + 654 = ?
- 999.99 + 99.99 + 99= ?
- (124.99)² = ?
- – (8.002)³ + (30.001)² - (4.01)⁴ =?
- (91.004)2 - (40.003)2 - (52.9)2 = ?
- `sqrt(1297)` + 189.99 =?
- 21% of 6600 + 453.987 =?
- 49.99% of 5400 + 923=?
- √1567.763`xx` √4400 ÷ 326.9827 + 1223.293 = ?
- 40.93√? + √6625 + √6920 + √? = 205.7542`xx` 7.654
- ³√? `xx` 32.87 + 59.83 `xx` 28.7665 – 48.8745 `xx` 21.642 = 1085.344
- 499.99 + 1999 ÷ 39.99 × 50.01 = ?
- `[(7.99)^2 - (13.001)^2 + (4.01)^3]^2=` ?
- `(601/49)xx(399/81)` ÷ `(29/201)` = ?
- (√8648 ×7) ÷ (3.9 ×5.99) = ?
- 9.8 × 77.9 ÷ (2.3)2 = ?
- (√845 ×19.932+ √4230 ×14.385)/(√1765 ×4.877 ) = ?
- 5.55% of 8120 – 66.66% of 540 = ? – 28% of 5500
- 159.98% of 4820 + 90.33% of 2840 = ? + 114.99% of 1980
- 25, 28, 26, 29, 27, ?
- 362, 452, 550, 656, 770, 892, ?
- 3, 4, 10, 33, 136, ?
- 56237.05 + 10616.99 - 137.25 + 1795.33 = ?
- (√4623.9 + √484.2) – √2303.97 ÷ √1296.4 × √35.98 ÷ √15.99 = ?
- 39.9% of 1720 + 80.2% of 630 = 89.9% of 1280 + ?

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Approximations form one of the most common topics in the section on quantitative aptitude. Exams like SSC CGL, SSC CPO, RRB NTPC, SSC CHSL, SBI PO, SBI Clerk and more. If you are preparing for any exams aforementioned, it is advisable to have a firm grip on this topic. Let’s understand the concept of approximation in detail. Questions from Approximation are frequently asked in banking examinations. If practiced well, one can score full marks in it in very less time. 5 to 10 questions can be expected from this topic in upcoming examinations of SBI Clerk Mains, RBI Assistant Mains, SBI PO prelims and IBPS PO Prelims.

Approximation Definition

An approximation is anything that is similar, but not exactly equal, to something. A number can be approximated by rounding. Before conducting the operations, one can approximate a calculation by rounding the values within it. Approximation includes estimation, rounding to powers of 10, decimal places and significant figures.

The type of questions that can be framed from this topic are based on finding the significant value, limits of accuracy, estimation, lower and upper bounds, along various others. To understand how to solve these questions it is recommended to firstly understand the BODMAS rule along with the methods of approximation. Let’s take a look at it.

BODMAS Rule

This rule specifies the correct order in which operations should be performed in order to get the value of a given expression. ‘B' represents ‘Brackets,' ‘O' stands for ‘Order,' ‘D' stands for ‘Division,' ‘M' stands for ‘Multiplication,' ‘A' stands for ‘Addition,' and ‘S' is for ‘Subtraction'. To acquire a correct answer to any arithmetic question, these procedures must be done in the order listed.

According to BODMAS rule, if an expression contains brackets ((), {}, []) we have first to solve or simplify the bracket followed by ‘order’ (that means powers and roots, etc.), then division, multiplication, addition and subtraction from left to right. Solving the problem in the wrong order will result in a wrong answer.

The “O” in the BODMAS full form is also called “Order”, which refers to the numbers which involve powers, square roots, etc.

Example 2: Simplify: 16 + (9 – 2 × 4)

Solution: 16 + (9 – 2 × 4)

= 16 + (9 – 8)

= 16 + 1

= 17

Therefore, 16 + (9 – 2 × 4) = 17

Methods of Approximation

- Rounding numbers to the nearest 10, 100, 1,000

To approximate to the nearest ten value look at the digit in the tens column. To approximate to the nearest hundred value look at the digit in the hundreds column. For the nearest thousand, look at the digit in the thousands column.

Then follow the steps:

- Draw a vertical line to the right of the required place value digit
- Take a look at the digit after that
- Increase the previous digit by one if the number is 5 or greater
- Keep the previous digit the same if it's 4 or less
- Any spaces to the right of the line should be filled with zeros

Example: Round 3,752 to the nearest 10, 100 and 1,000.

375|2 to the nearest 10 is 3,750

37|52 to the nearest 100 is 3,700

3|752 to the nearest 1,000 is 4,000

- Rounding to decimal places

When rounding with decimal places (dp), the degree of accuracy required is usually specified. However, there are some calculations where the degree of accuracy is more evident. Calculations involving money should be specified to two decimal places to represent pence.

To round to the nearest decimal place, use the following method:

- If rounding to one decimal place, look at the first digit after the decimal point; look at the second digit for two decimal places
- Draw a vertical line to the right of the required place value digit
- Take a look at the digit after that
- If the next digit is 5 or more, increase the previous digit by one; if it's 4 or less, keep the previous digit the same; remove additional numbers to the right of the line

Example: Round 742.368 to 1 decimal place, then round it to 2 decimal places:

742.3|68 to 1 decimal place is 742.4

742.36|8 to 2 decimal places is 742.37

Note that the answer should have the same number of decimal places as asked in the approximation.

- Rounding to significant figures

The method of rounding to a significant figure is frequently used because it may be applied to any number, no matter how large or small. When a newspaper reports that a lottery winner has won 1.5 billion, it is rounded to the nearest significant sum. The number is rounded up to the most significant figure.

To round up to a significant number:

- Take a look at the first non-zero digit if rounding to the nearest significant figure
- If rounding to two significant figures look at the digit after the first non-zero digit
- After the required place value digit, draw a vertical line to the next digit
- If the subsequent digit is 5 or more, increase the previous digit by one; if it is 4 or less, keep the previous digit as it is
- Fill all spaces to the right of the line with zeros, stopping if there is a decimal point

Example: Round 62,789 to 1 significant figure, then 2 significant figures.

6|2789 to 1 significant figure is 60,000

62|789 to 2 significant figures is 63,000

Note that the number of significant figures in the question is the maximum number of non-zero digits in the answer

- Truncation

Truncation is a method of approximating numbers. It is easier than rounding, but does not always give the best approximation to the original number. To truncate a number, find an estimate for the number without doing any rounding.

To truncate a number, the digits are usually skipped past a given point in the number, with zeros filled in as needed to make the truncated number roughly the same size as the original.

- Skip all the digits after the first decimal place to truncate a number to one decimal point
- Skip all the digits after the second decimal place to truncate a number to two decimal places
- Skip all the digits after the first three significant figures to reduce a number to three significant figures (the first non-zero digit and the next two digits)

Example: Truncate 4. 692 to 1 decimal place and then to 2 decimal places.

4.6|92 truncated to 1 decimal place is 4.6

4.69|2 truncated to 2 decimal places is 4.69

- Estimating calculations

In this case, before performing the calculation, round the numbers in the question. Numbers are usually rounded to one significant figure. The 'approximately equal to' sign, ≈, is used to illustrate that values have been rounded.

Example: Estimate the value of 21*76.

Rounding to 1 significant figure gives: 20*80 = 1600

Therefore: 21*76 ≈ 1600

- Limits of accuracy

When a number or measurement has been rounded, it's necessary to think about what the exact value could have been. If a lobby unit is 700 mm wide to the nearest 10 mm, it might be any width between 695 and 705 mm, therefore it may not fit into a gap that is exactly 600 mm wide. Limits of accuracy are used to describe all of the possible values that a rounded number could have.

The lowest value that would round up to the estimated value is called the lower limit.

The upper limit is the least amount that would result in the next estimated value.

Example: The lower limit for a mass of 80 kg, rounded to the closest 10 kg, is 75 kg. (Because 75 kg is the smallest mass, this is rounded up to 80 kg to the nearest 10 kg.) Because 85 kg is the smallest mass that would round up to 80 kg, the upper limit is 85 kg.

This can be represented as an error interval, which is the difference between the highest and lowest values (represented by inequality symbols):

75Kg ≤ weight ≤ 85Kg

All measurements are approximate. Everything that can be measured has a little margin of error, which means the measured value is extremely close to – but not exactly the same as – the true value. This could be due to human error or a minor measuring instrument sizing error. For example, someone measured his height to be 161 cm, their actual height may be 159.2 cm or 162.3 cm.

When a number is rounded, the original value may fall between a set of measures known as the limits of accuracy.

When more than one number in a calculation has been rounded, the calculation should use the upper and lower bounds for each value to identify the greatest and least possible values that the calculation can accept.

Few Approximation Questions/ Approximation Solutions are provided for your help below. Similar questions can be asked in the exams

Directions (Ques.1 - 15): What approximate value should come in place of question mark (?) in the following questions? (You are not needed to calculate the exact value)

1.324.99 of 48.07% =?

- 162 2)156 3) 174 4) 138 5) 144

Ans. 2

Explanation: 325 × (48/100) =?

? = 156

1845 of 893527 =?

- 281 2)291 3)271 4)251 5)261

Ans. 1

Explanation: 1845 × (48/315) =?

? = 281.14

? = 281 approx

(18.79)² × √730 =?

- 9570 2)9750 3)9700 4)9500 5)10500

Ans.2

Explanation: (19)² × √730 =?

361 × 27 =?

? = 9747

(16.072)² + (22.202)² + (35.986)² + (38.8761)² =?

3560 2)3650 3)3750 4)3570 5)3590

Ans. 1

Explanation: (16)² + (22)² + (36)² + (39)² =?

256 + 484 + 1296 + 1521 =?

? = 3557

(42.42)² + (39.549)² - (11.764)² - (45.47)² =?

1010 2)1195 3)2155 4)2210 5)2020

Ans. 2

Explanation: (42)² + (40)² - (12)² - (45)² =?

1764 + 1600 – 144 – 2025 =?

3364 – 2169 =?

?= 1195

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