Simplifications

In this section we are going to learn all about VBODMAS rule, what is it, where to use it along with some mind-blowing formulas which will transfer you in a human calculator.

E.g.  221 ÷ 13 × √576 + (10)2 = ?

We are completely clueless about where to start right now, so let’s start from left

⇒ 221 ÷ 13 × 24 + 100 = ?

⇒ 221 ÷ 13 × 124 = ?

⇒ 221 ÷ 1612 = ?

⇒ 7.290 (Approx.)

Which is wrong so how to approach these types of questions?

Is there a specific method to approach them?

Yes, VBODMAS rule.

VBODMAS RULE

Full form of VBODMAS is

For every question we have to do each step from up to bottom

Let’s take an example and try to solve it.

E.g.  221 ÷ 13 × √576 + (10)2 = ?

Step- 1: First try to Solve √576 and (10)2

⇒ 221÷13×24+100= ?

Step-2: Division

⇒ 221 ÷ 13 × 24 + 100 = ?

⇒ 17 × 24 + 100 = ?

Step-3: Multiplication

⇒ 17 × 24 + 100 = ?

⇒ 408 + 100 = ?

Step-4: Addition

⇒ 408 + 100 = ?

⇒ 508 (Ans).

Tip- While solving brackets first open small () then curly {} and last big brackets []

E.g.  [(2211 ÷ 67)2 – 21 × √(256 )] ÷ (549 – 213) = ? ÷ 1344

Let’s apply VBODMAS rule here

Step-1: Let’s Solve √256

⇒ [(2211÷67)2-21×16]÷(549-213)=?÷1344

Step-2: Start opening the Brackets

⇒ [(2211 ÷ 67)2 – 21 × 16] ÷ (549 – 213) = ? ÷ 1344

⇒ [(33)2 – 21 × 16] ÷ 336 = ? ÷ 1344

Step-3: Now we have to first simplify equation in bracket then open it by VBODMAS

⇒ [1089 – 21 × 14] ÷ 336 = ? ÷ 1344

⇒ [1089 – 336] ÷ 336 = ? ÷ 1344

Step-4: Now open the bracket

⇒ [1089 – 294] ÷ 336 = ? ÷ 1344

⇒ 753 ÷ 336 = ? ÷ 1344

Step-5: Division

⇒ 753 ÷ 336 = ? ÷ 1344

⇒ 753 × 1344 ÷ 336 = ?

⇒ 753 × 4

⇒ 3012 (Ans).

E.g.  1/6 of 355 of 1/5 of 2160 + √3969 – 448.98 = ?

Step -1: first solve Vinculum

1/5 of 2160

⇒ 1/5 × 2160

⇒ 432

And

1/6 of 355

⇒ 355/6

So, we got,

⇒ 355/6 of 432 + 37 – 448.98

Step-2: Solve Of

⇒ 355/6 of 432 + 37 – 448.98

⇒ 355/6 × 432 + 37 – 448.98

Step-3: Division

⇒ 355 × 72 + 37 – 448.98

Step-4: Multiplication

⇒ 355 × 72 + 37 – 448.98        

⇒ 25560 + 37 – 448.98

Step-5: Addition

⇒ 25560 + 37 – 448.98

⇒ 25597 – 448.98

Step-6: Subtraction

⇒ 25597 – 448.98

⇒ 25,148.02 (Ans).

E.g.  28/9 × 264/12 ÷ 17/5 + 13/17

Step-1: By BODMAS rule we would firstly (D) Divide

⇒ 28/9 × 12 ÷ 17/5 + 13/17

⇒ 28/9 × 60/17 + 13/17

Step-2: Multiplication

⇒ 28/9 × 60/17 + 13/17

⇒ 400/51 + 13/17

Step-3 Addition

⇒ (400 + 39)/51

⇒ 439/51 (Ans).

As we saw by using this concept by would lead to correct answer however question is how much tedious.

 APPROXIMATION

We use approximation in our daily life every day, we all know how to do it, as example we say 1000 for 999, what we do we simply ignore a comparably small portion.

In mathematical expressions which include division and multiplication of decimal values of large numbers we stuck. It becomes quite complex to solve these problems, so for solving these we use approximation. We just Round-off the numbers.

When we approximate the final result obtained is not equal to the exact result, but it is very close to the exact result.

Let’s try one,

E.g. 2? = 32.01 + 128.01 × 1023.99 + 7.99

Solving by BODMAS rule, as learned earlier

⇒ 2?  = 2.01 + 128.01 × 1023.99 + 7.99

⇒ 2?  = 32.01 + 131,080.9599 + 7.99

⇒ 2? = 32.01 + 131,080.9599 + 7.99

⇒ 2?  = 131,120.9599

 E.g. 10% of 1350 + ? = 365

⇒ 10/100 × 1350 + ? = 365

⇒ 135 + ? = 365

⇒ ? = 365 – 135

⇒ ? = 230

 E.g. 78 × 98 – 25% of 1376

⇒ 78 × 98 – 25/100 × 1376

⇒ 7644 – 34400/100

⇒ 7644 – 344

⇒ 7300 (it requires tedious calculation)

After this tedious calculation (128.01 × 1023.99) without calculator, we are not left with time, in exam hall we will prefer to leave this question.

In Approximation, to solve the complex mathematical expression, take the nearest value of numbers given in the expression. Try to make unit digit 0’ in mostly cases.

E.g. 2? = 32.01 + 128.01 × 1023.99 + 7.99

Let’s try to round off to nearest integer

⇒ 2? = 32 + 128 × 1024 + 8

Or we can say,

⇒ 2?  = 25 + 27 × 210 + 23

⇒ 2? = 25 + 217 + 23

⇒ 2? = 23 (22 + 214 + 1)

⇒ 2? = 8 (16389)

⇒ 2? = 13112

⇒ ? = 17

By calculator we would get 17.004

Here it is, we got the solution.

 E.g. 393 × 197 + 5600 × 5/4 + 8211.80 = ?

⇒ 393 × 197 + 5600 × 5/4 + 8211.80 = ?

⇒ 390 × 200 + 5600 × 5/4 + 8200 = ?

⇒ 390 × 200 + 5600 × 5/4 + 8200 = ?                              (By BODMAS rule)

⇒ 390 × 200 + 7000 + 8200 = ?

⇒ 78000 + 7000 + 8200 = ?

⇒ ? = 93200 (Ans).

We got 92,632.8 by calculator. It is quite near to original value.

Yeah it is effective.

 Tip- For finding 10% of a number simply move the decimal to one digit left.

For finding 25% simply divides the number by 4

Shout-cut method for Percentage

E.g. 10% of 1350 + ? = 365

⇒ ? = 365 – 10% of 1350

⇒ ? = 365 – 135.0                                                                                                             (By table)

⇒ ? = 230.0 (Ans).

E.g. 78 × 98 – 25% of 1376

⇒ 78 × 98 – 344                      (Using approximation and table)

⇒ 8000 – 344

⇒ 7656 (Ans).

This can be done in mind without the use of pen and paper.

E.g. 34.02% of 550.09 + ? = 297.07 + √728.95

⇒ 34.02% of 550.09 + ? = 297.07 + √728.95

⇒ 34% of 550 + ? = 300 + √729

⇒ (25 + 10 – 1)% of 550 + ? = 300 + 27

⇒ 25% of 550 + 10% of 550 – 1% of 550 + ? = 327

⇒ 137.5 + 55 – 5.5 + ? = 327

⇒ 132 + ? = 327

⇒ ? = 327 – 132

⇒ ? = 195 (Ans).

E.g. (? + 9.97) × 12.8 = 20.12% of 1319.97

⇒ (? + 10.00) × 13.0 = 20.00% of 1320.00                                                   (Using approximation)

⇒ (? + 10.00) × 13.0 = 1/5 × 1320.00

⇒ (? + 10.00) × 13.0 = 264

⇒ (? + 10.00) = 260/13          (Using approximation)

⇒ ? = 20 – 10

⇒ ? = 10 (Ans).

More on Simplifications

• Concept behind Simplification:

What do we do when we are stuck in a complex situation? We try to break the complexity in fragments and try to go through a definite rule so that the probability of any wrong becomes almost zero.

Simplification does the same for us in Mathematics.

Suppose we have an arithmetical expression: 12 – [26 – {2 + 5 × (6 – 3)}]

How we are going to solve it?

Let’s start solving the operations randomly.

⇒ 12 – [26 – {2 + 5 × (6 – 3)}]

⇒ 12 – [26 – {7 × (6 – 3)}]

⇒ 12 – [19 × (6 – 3)]

⇒ 12 – [19 × 3]

⇒ 12 – 57 = – 45

But it is wrong. Why? Because if we carry out the operations randomly it will give a different solution every time.

So, how do we go about it?

Yes, VBODMAS

To simplify an expression which contains various mathematical operations, we have a ‘VBODMAS’ rule which goes in a proper sequence and chronology. The words in VBODMAS denote different mathematical operations.

So, the operations are to be applied from top to bottom (insides the brackets too).

Let’s see some examples from RRB NTPC Quant questions.

• Expression: 125 – 73 + 48 – 137 + 99 = ?

So, we have two operations in the expression. Since, addition comes before subtraction in VBODMAS.

Step 1: Addition

Add all the positive integers in one and all the negative integers in one

⇒ 125 – 73 + 48 – 137 + 99 = ?

⇒ (125 + 48 + 99) – (73 + 137) = ?

⇒ 272 – 210 = ?

Step 2: Subtraction

⇒ 272 – 210=?

⇒62 (Ans).

Note:- In case of brackets the order of removing the brackets while applying BODMAS rule is:

1.         (  )

2.         {  }

3.         [  ]

Expression: 12 – [26 – {2 + 5 × (6 – 3)}] = ?

We solved this question at the very beginning and the result was -45. Let’s see what happens when we apply VBODMAS.

Step 1: Bracket

⇒ 12 – [26 – {2 + 5 × (6 – 3)}]        

(Firstly, we have to solve the equations in bracket and then open it.)

⇒ 12 – [26 – {2 + 5 × 3}]

Step 2: Multiplication inside Bracket                                                                                                                                                                    

⇒ 12 – [26 – {2 + 5 × 3}]

⇒ 12 – [26 – {2 + 15}]

Step 3: Addition inside Bracket                                                                                                                       

⇒ 12 – [26 – {2 + 15}]

⇒ 12 – [26 – 17]

Step 4: Subtraction inside Bracket

⇒ 12 – [26 – 17]

⇒ 12 – 9

Step 5: Subtraction

⇒ 12 – 9

⇒ 3 (Ans).

In the steps 2, 3 and 4 we have applied the VBODMAS inside the brackets too.

• Fractions:

A number which can be expressed in p/q form, where q≠0, and p, q are positive integers, is called a fraction. Here, p is called the numerator and q is called the denominator. We will go through the types of fractions very quickly.

1.         Proper Fraction- When numerator is less than its denominator. E.g.: 2/3, 4/7, 22/23, etc.

2.         Improper Fraction- When numerator is greater than its denominator. E.g.: 3/2, 7/4, 23/22, etc.

3.         Mixed Fraction- When a fraction is represented in integers and fraction both. E.g.: 1½, 1¾, etc.

• How to convert mixed fraction to improper fractions?

Note: Only improper fractions can be converted into mixed fraction and vice versa.

1.         Improper to Mixed.

Step1: Divide the numerator by denominator to get the maximum possible quotient and the remainder.

E.g.: 7/4 is an improper fraction. Divide 7 by 4, we get the maximum possible quotient 1 and the remainder 3.

Step2: While writing mixed fraction, we keep the quotient at the integer place, remainder at numerator place and denominator remains the same. Here in previous example, Quotient = 1, Remainder = 3 and Denominator = 4. So, the mixed fraction is 1¾.

2.         Mixed to Improper.

While converting mixed fraction into improper fraction, denominator remains the same as that of mixed fraction and numerator of improper fraction = Denominator × Integer + Remainder

E.g.: Covert 1¾ to improper fraction

Here, d = 4

Integer or quotient, q = 1

Numerator, n = 3

Numerator, n’= 4 × 1 + 3 = 4 + 3 = 7

Denominator, d’= 4

So, the required fraction = n’/d’ = 7/4.

• Addition and Subtraction in Fraction:

Step 1: If we have mixed fraction, convert it into improper fraction.

E.g.:

 (21/3 – 11/2)

= 7/3 – 3/2     

 [Use the conversion technique to convert the fractions]

Step 2:

Case 1: If the fractions are like (denominators are same), then we can directly add or subtract the numerators.

E.g., Suppose the fractions in above problem is 7/2 – 3/2

Here the denominators are same, so we can directly add or subtract the numerators.

⇒ 7/2 – 3/2                     [New n = 7 – 3 = 4}

⇒ 4/2 = 2              [Always convert the fraction into its simplest form]

Case 2: If the fractions are unlike (denominators are different), then we need to take the LCM of the denominators.

Let’s continue with the previous problem,

⇒ 7/3 – 3/2

1.         Taking LCM of 2 and 3,

LCM= 6

2.         Now divide the LCM by denominator and multiply the numerator by the quotient

Here, required fraction = {(7 × 2) – (3 × 3)}/6

3.         Now apply the VBODMAS rule in the numerator and convert the fraction into its simplest from (if required).

Required fraction = {(7 × 2) – (3 × 3)}/6 = (14 – 9)/6 = 5/6

• Simplest form of a fraction:

We say a fraction is in its simplest form when it cannot be reduced any smaller. To convert the fraction in its simplest from, we need to divide the Numerator and Denominator by its HCF. Let’s see with an example-

36/60

Here, the HCF of the Numerator and Denominator = 12

Dividing the Numerator and Denominator by HCF we get, 3/5. Now, HCF becomes 1 so this is the simplest form of the fraction.

This is time consuming. We have an easy way also. We don’t need to find HCF. If we know the table and basic division then we can do everything in our mind by cancellation method.

Let’s do this example by cancellation method (do everything in your mind)

36/60 – Look for any Common Factor (CF) in it. Suppose we see that both are divisible by 2. Cancel out the numerator and denominator by 2 in your mind. We get 18/30. Look for CF again. If we know table of 6 we can clearly say that 6 is common. Now cancel out again by 6. We get 3/5. Can we see any more CF? No. So, this is the simplest form.

Understand how knowledge of table is beneficial for us. If we knew table of 12 in our mind while solving this problem then we could have directly said the simplest from is 3/5.

• Multiplication of fraction:

Multiplication of fraction is just an extended form of cancellation method that we just learnt. What we need to do now is cancel all the numbers to their simplest form and then multiply numerator with numerator and denominator with denominator. We can also multiply first and then do the simplification but that will be time consuming. Let’s solve a problem from RRB NTPC quant questions:  12/13×285/6.

Here, if we multiply first then we will have to multiply 285 by 12 and that is going to take time. That’s why we will try to cancel out first.

What we see is 12 is divisible by 6. So we will cancel out both 12 and 6 by 6. We get, 2/13×285/1. Now we can multiply 285 by 2 and that will save some time for us.

So we have, 570/13. Can it be simplified further? No, because 13 is a prime number and we are not getting any CF. We will practice some more multiplication problem in division because after a point of time division of fraction is changed into multiplication.

• Division of fractions:

Division of fraction is a combination of reciprocal and multiplication of fractions.

Reciprocal of a fraction is called reverse fraction. Suppose, we have a fraction a/b then its reciprocal is b/a.

Reciprocal Rule: While dividing a fraction by another, we can replace the division sign with multiplication sign if we take the reciprocal of the divisor.

Let’s see some examples from RRB NTPC quant questions,

• E.g. 1: 18/26÷90/52

So, here we will use the reciprocal rule to convert into multiplication.

Step 1:- Use Reciprocal Rule

⇒ 18/26÷90/52

⇒ 18/26×52/90

Step 2:- Multiplication

⇒ 18/26×52/90                [Look for CF]

Here, we see 52 is divisible by 26 and 90 comes in table of 18.

⇒ 1/1×2/5                  [Cancelling out 90 and 18 by 18 and 26 and 52 by 26]

⇒ 2/5 Ans.

• E.g. 2: Compute. 15872/32/4 = ?

⇒ 15872/32/4 = ?

We can write as 15872/32/4 as 15872/32 ÷ 4,

Using Reciprocal rule,

⇒ 15872/128 = ?

Since, the dividend is large. So, dividing directly by 128,

⇒ 15872/128

⇒ 124/1= 124 (Ans).

• E.g. 3: 5/28÷28/35÷20/112

⇒ 5/28 × 35/28 × 112/20        [Using Reciprocal Rules]

⇒ 1/28 × 5/4 × 112/4           [Cancelling 20 by 5 and 35 and 28 by common factor 7]

⇒ 1/2 × 5/4 × 8/4                [Cancelling 112 and 28 by common factor 14]

⇒ 1/1 × 5/4 × 1/1                 [Cancelling 8 by (4×2) in denominator]

⇒ 5/4 Ans.

 Let’s take some more examples  

Problem 1: (21/3 – 11/2) of 3/5 + 12/5 ÷ 21/3

Step 1: Convert mixed fractions

⇒ (7/3 – 3/2) of 3/5 + 7/5 ÷ 7/3 [n= denominator× integer+ remainder, denominator remains same]

Step 2: Bracket

We have unlike fractions, so taking LCM and then dividing the LCM with denominators and multiplying with quotients in numerator

⇒ (7×2/6 – 3×3/6) of 3/5 + 7/5 ÷ 7/3

⇒ (14 – 9)/6 of 3/5 + 7/5 ÷ 7/3

⇒ 5/6 of 3/5 + 7/5 ÷ 7/3

Step 3: Of

⇒ 5/6 of 3/5 + 7/5 ÷ 7/3                [Look for CF]

We see that 5 is divisible by 5 and 6 is divisible by 3,

⇒ 1/2 of 1/1 + 7/5 ÷ 7/3                 [Using cancellation method]

⇒ ½ + 7/5 ÷ 7/3

Step 4: Division

⇒ ½ + 7/5 × 3/7                                   [Reciprocal Rule]

Step 5: Multiplication

Cancelling, 7 by 7 we get,

⇒ ½ + 1/5 × 3/1                                   [Cancellation method]

⇒ ½ + 3/5

Step 6: Addition

⇒ (1×5)/10 + (3×2)/10                                 [Same as Step 2]

⇒ 5/10 + 6/10

⇒ 11/10 or 11/10 (Ans).

 Problem 2:  Simplify: (2/7 + 3/5) ÷ (2/5 + 2/7)

Solving the brackets,

LCM of 5 and 7= 35

Diving the LCM by denominators and multiplying in numerators by quotient,

⇒ {(2 × 5) + (3 × 7)}/35 ÷ {(2 × 7) + (2 × 5)}/35

⇒ (10 + 21)/35 ÷ (14 + 10)/35

⇒ 31/35 ÷ 24/35

Using reciprocal rule,

⇒ 31/35 × 35/24

Cancelling out 35 by 35,

⇒ 31/1 × 1/24

⇒ 31/24 (Ans).

• Comparison of Fractions:

Fractions can be compared by the following techniques:

1. By converting them to like fractions: We can use the same technique as used in addition and subtraction to compare the fractions

Important: Avoid applying this method when the denominators are small or prime numbers.

Let’s use this method in RRB NTPC examples,

E.g. 1: Which is the largest among 5/8, 2/3, 7/9, 3/5?

LCM of 8, 3, 9 and 5= 360

Dividing the denominators and multiplying by quotient in numerators,

⇒ 5/8 = (5 × 45)/360 = 225/360

⇒ 2/3 = (2 × 120)/360 = 240/360

⇒ 7/9 = (7 × 40)/360 = 280/360

⇒ 3/5 = (3 × 72)/360 = 216/360

Now, we can get the largest fraction by comparing the numerators.

Hence, 280/360 i.e. 7/9 is the largest fraction.

E.g. 2: Arrange in ascending order: 1/2,2/3,7/12

LCM of 2, 3 and 12 = 12

Dividing the denominators and multiplying by quotient in numerators,

⇒ ½ = (1 × 6)/12 = 6/12,

⇒ 2/3 = (2 × 4)/12 = 8/12,

⇒ 7/12 = (7 × 1)/12 = 7/12

Now, we can get the required order by comparing the numerators.

Required order = 6/12, 7/12, 8/12 = 1/2, 7/12, 2/3

This method is long and time consuming. Let’s see a quick method.

 2. By changing fractions to decimals: We can use the values of fractions from the given table to compare fractions:

Important: Apply this method when the denominators and numerators both are small.

 Applying in same examples as above,

E.g. 1: Which is the largest among 5/8, 2/3, 7/9, 3/5?

⇒ 5/8 = 5 × 1/8 = 5 × 0.125 = 0.625

⇒ 2/3 = 2 × 1/3 = 2 × 0.33 = 0.66

⇒ 7/9 = 7 × 1/9 = 7 × 0.11 = 0.77

⇒ 3/5 = 3 × 1/5 = 3 × 0.20 = 0.60

Hence, 0.77 i.e. 7/9 is the largest.

E.g. 2: Arrange in ascending order: 1/2,2/3,7/12

⇒ ½ = 1 × ½ = 1 × 0.50 = 0.50

⇒ 2/3 = 2 × 1/3 = 2 × 0.33 = 0.66

⇒ 7/12 = 7 × 1/12 = 7 × 0.083 = 0.581

Required order = 0.50, 0.581, 0.66 =1/2, 7/12, 2/3.