# Approximation – Aptitude Question and Answers

Last Updated : 21 Mar, 2023

Approximation is a mathematical process that involves finding an estimate or approximation of a given value or expression. It is a valuable skill, particularly in Quantitative Aptitude tests, and is frequently used in competitive exams such as the SBI Clerk Mains, UPSC, PSC, and Railways. In fact, up to 10% of the total marks in these examinations may be dedicated to approximation questions. These questions are typically straightforward and can be solved quickly with a bit of practice. This article will provide an overview of approximation questions, their importance in competitive exams, and some tips and tricks for tackling them successfully. By the end of this article, you’ll have a better understanding of how to approach approximation questions and improve your chances of success in these exams.

Practice Quiz

Practice Approximation Aptitude Quiz Questions

## Tips and Tricks to Solve Approximation Questions

The Quantitative Aptitude segment of the competitive exams comprises a vital part called Approximation. Candidates need to practice these questions while preparing for the examination.

### Significance of Approximation in Exams?

• You can expect 5-10 Approximation inquiries in each bank test, particularly in the prelims stage.
• These inquiries are very simple to settle assuming that you have polished well on your Calculation speed.
• You can settle every one of the inquiries within 3-4 minutes with 100 percent exactness.
• These inquiries can build your possibilities by clearing the test.

### What is Approximation?

An approximation is a cycle by which complex mathematical articulation are separated into less difficult structures by performing different numerical activities as indicated by the BODMAS rule.

In numerical expression, which incorporates division and multiplication of decimal values of huge numbers, it turns out to be very difficult to settle these expressions. thus, to decrease this difficulty we use Approximation techniques. in the Approximation strategy, we need not compute the exact value of an expression, yet we work out the closest Values (round-off values). at the point when we use the Approximation strategy, the eventual outcome acquired isnâ€™t equivalent to the specific outcome yet it is extremely close either minimal less or minimal more to the last result.

### BODMAS:

B â†’  Brackets

O â†’ Of

D â†’ Division

M â†’ Multiplication

S â†’ Subtraction

### Order of BODMAS Rule

Here are the ways to carry out the BODMAS rule in Approximation questions:

Step 1: Identify the brackets, start computation by solving brackets in this order (), {} and [] should be followed.

Step 2: Calculate the Powers and roots.

Step 3: Calculate Division and Multiplication (from left to right as both Multiplication and Division have the same priority)

Step 4: Calculate Addition and  Subtraction (from left to right as both Addition and  Subtraction have the same priority).

## Rules to Solve the Mathematical Expressions by Approximation:

### Rule 1:

To solve complex problems, take the closest value of the number given in the expression. for example, 77.8 is round off to 78;

33.02 is round off to 33 etc.

EX1:  19%of (399.88/20X400)+30=?

20/100x(400/20×400)+30=?

1/5x(8000)+30=1600+30=1630

### Rule 2:

To solve problems with large numbers involved in multiplication, we can consider the approximate value of the large numbers by increasing or decreasing the round off values making the computation easy. for example, 239×111 is approximated to 240×110.

EX2: 192×397+560×5/7+729.80=?

192×397+560×5/7+730=?

190×400+400+730=76000+1130=77130

### Rule 3:

To solve problems with large numbers involved in the division, we can consider the approximate value of the large numbers by increasing or decreasing the round off values making the computation easy. for example, 6198.36/38.69 is approximated as 6200/40.

EX3:862.5/18.64=?

860/20=43

## Sample Questions on Approximation

### Q1. Approximate the value of 1.2 x 2.8 using rounding to one decimal place.

Solution

To approximate, we can round each number to one decimal place: 1.2 â‰ˆ 1.2 and 2.8 â‰ˆ 2.8 Then, we can multiply: 1.2 x 2.8 â‰ˆ 3.36 Therefore, 1.2 x 2.8 approximates to 3.4 when rounded to one decimal place.

### Q2: Approximate the value of 0.75 Ã· 0.3 using rounding to two decimal places.

Solution

To approximate, we can divide 0.75 by 0.3 using a calculator and round the answer to two decimal places: 0.75 Ã· 0.3 â‰ˆ 2.50 Rounded to two decimal places, 2.50 â‰ˆ 2.50. Therefore, 0.75 Ã· 0.3 approximates to 2.50 when rounded to two decimal places.

### Q3. Estimate the value of âˆš(6.98 + 2.03).

Solution

We can use the linear approximation, which gives us âˆš(6.98 + 2.03) â‰ˆ âˆš7 + 1/14(2.03/âˆš7) = 2.85. Therefore, the approximate value is 2.85.

### Q4. Approximate the value of e^(-0.01).

Solution

Using the Taylor series expansion of e^x, we have e^(-0.01) â‰ˆ 1 – 0.01 + 0.005 = 0.995. Therefore, the approximate value is 0.995.

### Q5. Estimate the value of ln(1.1).

Solution

Using the linear approximation, we have ln(1.1) â‰ˆ 0.1/1.1 = 0.0909. Therefore, the approximate value is 0.0909.

### Q6. 235.8 Ã· 0.0032 + 12.6 Ã— 4.9 = ?

Solution

Using the order of operations, we perform the division first: (73500) + 12.6 Ã— 4.9 =? Next, we perform the multiplication: (73500) + 61.74 =? Adding, we get: 73561.74 =? Therefore, the solution is 73561.74.

### Q7. 75.002 Ã— 1.98 – 89.997 + âˆš(32.1 Ã— 17.9) = ?

Solution

Using the order of operations, we perform the multiplication first: (148.502) – 89.997 + âˆš(574.59) =? Next, we perform the square root: (148.502) – 89.997 + 23.977 =? Adding and subtracting, we get: 82.482 =? Therefore, the solution is 82.482.

### Q8. 15.996 Ã— 7.998 + 18.002 Ã· 2.998 = ?

Solution

Using the order of operations, we perform the multiplication first: (127.949008) + 18.002 Ã· 2.998 =? Next, we perform the division: (127.949008) + 6.006004 =? Adding, we get: 133.955012 =? Therefore, the solution is 133.955012.

### Q9. âˆš(16.003 Ã— 23.998) – 12.997 Ã— 1.998 + 31.999 Ã· 3.998 = ?

Solution

Using the order of operations, we perform the multiplication first: âˆš(384.023994) – 25.960006 + 8.000000 =? Next, we perform the square root: 19.599649 – 25.960006 + 8.000000 =? Adding and subtracting, we get: 1.639643 =? Therefore, the solution is 1.639643.

### Q10. 43.999 Ã· 1.997 Ã— 2.998 + 13.002 Ã— 1.998 – 19.003 = ?

Solution

Using the order of operations, we perform the division first: (44.0661) Ã— 2.998 + 13.002 Ã— 1.998 – 19.003 =? Next, we perform the multiplication: (132.023838) + 25.962996 – 19.003 =? Adding and subtracting, we get: 139.983834 =? Therefore, the solution is 139.983834.

Test your knowledge of Approximation in Quantitative Aptitude with the quiz linked below, containing numerous practice questions to help you master the topic:-

Previous
Next