Simplification

Simplification mean find the most simplest form of given question with the help of different operations of maths in sequence .

1) Rule :

VBODMAS

V→ Bar (Virnaculum)

B→ Bracket [{( )}]

O→ Of ×

D→ Division ÷

M→ Multiplication ×

A→ Addition +

S→ Subtraction -

We must use the following order strictly

1) At first we will solve Bar.

2) We will remove Bracket in order ( ) , { } and [ ] .

3) We will solve Multiplication.

3) We will solve Division .

4) We will solve Multiplication .

5) We will solve Addition .

6) We will solve Subtraction .

Example

1) 8 ÷ 2 ( 3 + 1 )

= 8 ÷ 2 ( 4 )

= 8 ÷ 8

= 1

2) 100 + 25 ÷ 5 - 15 of 5 + 3 × 2

= 100 + 25 ÷ 5 - 15 × 5 + 3 × 2

= 100 + 25 ÷ 5 - 75 + 3 × 2

= 100 + 25 ÷ 5 - 75 + 6

= 111 - 75

= 36 _____

3)Simplify : b - [ b - ( a + b ) -{ b - ( b - a - b ) } + 2a ]

Solutuion : _____

= b - [ b - ( a + b ) - { b - ( b - a - b ) } + 2a ]

= b - [ b - ( a + b ) - { b - ( b - a + b ) } + 2a ]

= b - [ b - a - b - { b - 2b + a } + 2a ]

= b - [ - a - { a - b } + 2a ]

= b - [ - a - a + b + 2a ]

= b - [ - 2a + b + 2a ]

= b - b

= 0

4) What value will replace the question mark in the following equation ?

1 1 1 2

4---- + 3--- + ? + 2---- = 13----- .

2 6 3 5

Solution : Let

9 19 7 67

--- + ---- + x + --- = -----

2 6 3 5

67 9 19 7

x = ------ - ---- - ----- - -----

5 2 6 3

x = 17/ 5

Ans missing fraction 17/5

5) If 2x + 3y + z = 55, x + z - y = 4 and y- x + z = 12, then what are the values of x, y and z ?

Solution :

The given equations are :

2x + 3y + z = 55...............(i)

x + z - y = 4......................(ii)

y- x + z = 12.....................(iii)

(i) - (ii) we get

x + 4y = 51........................(iv)

(i) - (iii) we get

3x + 2y = 43.......................(v)

(v)×2 - (iv) we get

5x = 35

x = 7

put in (iv) we get

4y = 44

y = 11

put x = 7 and y = 11 in (i) we get

z = 8

6) Simplify :

99--------- × 245

Solution :

= ( 100 - 1/49) × 245 = 4899 × 245/49

= 4899 × 5

= 24495

7) A man divides ₹ 8600 among 5 sons, 4 daughters and 2 nephews. If each daughter receives four times as much as each nephew, and each son receives five times as much as each nephew, how much does each daughter receive ?

Solution :

Let the share of each nephew be ₹ x .

the share of each daughter be ₹ 4x.

the share of each son be ₹ 5x

ATQ

5 × 5x + 4 × 4x + 2 × x = 8600

25x + 16x + 2x = 8600

43x = 8600

x = 200

the share of each daughter be ₹ 4× 200 = ₹ 800

8) Arun and Sajal are friends. Each has some money. If Arun gives ₹ 30 to sajal, then Sajal will have twice the money left with Arun. But, if Sajal gives ₹ 10 to Arun, then Arun will have thrice as much as is left with Sajal. How much money does each have ?

Solution :

Let Arun has ₹ x ,

Sajal has ₹ y ,

then

ATQ

2 ( x - 30 ) = y + 30

2x - 60 = y + 30

2x - y = 30 + 60

2x - y = 90................(i)

x + 10 = 3 ( y - 10 )

x + 10 = 3y - 30

x - 3y = - 30 - 10

x - 3y = - 40..............(ii)

from (ii) x = - 40 + 3y......(iii)

Put in (i)

2 ( - 40 + 3y ) - y = 90

- 80 + 6y - y = 90

5y = 80 +90

y = 170/ 5

y = 34

put in (iii)

x = - 40 + 3 × 34

x = - 40 + 102

x = 62

Arun has ₹ 62

Sajal has ₹ 34.

9) Solve the following pair of equations

7x - 15y = 2.................(1)

x + 2y = 3.....................(2)

Solution :

We can write (2) x = 3 - 2y ............(3)

Substitute the value of x in equation (1) we get

7(3 - 2y) - 15y= 2

21 - 14y - 15y = 2

- 29y = -19

y = 19/29

Substituting this value in (3) we get

x = 3 - 2( 19/29 ) = 49/29

Answer x = 49/29 , y = 19/29

10) The difference between two numbers is 26 and one number is three times the other. Find them.

Solution :

Let first no be x ,

Second no be y ,

ATQ

x - y = 26 ..................(1)

x = 3y ...................(2)

By Substituting x = 3y in (1)

3y - y = 26

y = 26/2

y = 13

By putting this value of y in equation (2), we get

x = 3×13

x = 39

Thus , the required numbers are 39 and 13.

11) Solve 2x + 3y = 11 & 2x - 4y = -24 and hence find the value of ' m' for which y = mx + 3.

Solution :

2x + 3y = 11.................(1)

2x - 4y = -24.................(2)

We can write (2)

x = 2y - 12....................(3)

By putting this value of x in equation (1) we get

2 ( 2y - 12 ) + 3y =11

4y - 24 + 3y = 11

7y = 11 + 24

y = 35/5

y = 5

By putting y = 5 in equation (1)

2x + 3(5) = 11

2x + 15 = 11

x = 11- 15/ 2

x = - 4/2

x = -2

Thus the required solution of the given system of equations is x = - 2 and y = 5 .

Now, since y = mx + 3

5 = m(-2) + 3

5 = - 2m + 3

5 - 3 = - 2m

2 = - 2m

m = -1

Thus, the required value of value of m = -1.

12) Five years hence, the age of Jacob will be three times that of his son, Five years ago, Jacob's age was seven times that of his son. What are their present age ?

Solution :

Let the present age of Jacob be x years

the present age of his son are y years

After 5 years

age of Jacob be ( x + 5 ) years

age of his son ( y + 5 ) years

ATQ

( x + 5 ) = 3 ( y + 5 )

x + 5 = 3y + 15

x - 3y = 15 - 5

x - 3y = 10...............(1)

5 years ago

age of Jacob be ( x - 5 ) years

age of his son ( y - 5 ) years

ATQ

(x - 5 ) = 7 ( y - 5 )

x - 5 = 7y - 35

x - 7y = - 35 + 5

x - 7y = -30.............(2)

By equation (2)

x = 7y - 30................(3)

Substituting this value of x in equation (1), we get

( 7y - 30 ) - 3y = 10

7y - 3y = 10 + 30

4y = 40

y = 10

put in (3)

x = 7 × 10 - 30

x = 40

Thus, the required present ages of Jacob and his son are 40 years and 10 years respectively .

13) 2 - [ 2 - { 2 - 2 ( 2 + 2 ) } ] = ?

Solution :

= 2 - [ 2 - { 2 - 2 ( 4 ) } ]

= 2 - [ 2 - { 2 - 2× 4 } ]

= 2 - [ 2 - { 2 - 8 } ]

= 2 - [ 2 + 6 ]

= 2 - 8

= - 6

14) 100 × 10 -100 + 2000 ÷ 100 = ?

Solution :

= 100 × 10 -100 + 2000 ÷ 100

= 1000 - 100 + 20

= 1020 - 100

= 920

15) What mathematical operation should come at the place of '' ? '' in the equation :

2 ? 6 - 12 ÷ 4 + 2 = 11

Solution :

Let ? = x

11= 2 x 6 - 12 ÷ 4 + 2

11= 2 x 6 - 3 + 2

11= 2 x 6 - 1

11 + 1 = 2 x 6

12 = 2 x 6

So, x must be replaced by ' × ' .

16) If 45 - [ 28 - { 37 - ( 15 - * ) } ] = 58, then * is equal to :

Solution :

Let * = x

45 - [ 28 - { 37 - ( 15 - x ) } ] = 58

45 - [ 28 - { 37 - 15 + x } ] = 58

45 - [ 28 - { 22 + x } ] = 58

45 - [ 28 - 22 - x ] = 58

45 - [ 6 - x ] = 58

45 - 6 + x = 58

39 + x = 58

x = 58 - 39

x = 19

So, x must be replaced by 19. ____

17) Simplify : 18 - [ 5 - { 6 + 2 ( 7 - 8 - 5 ) } ]

Solution :

=18 - [ 5 - { 6 + 2 ( 7 - 8 + 5 ) } ]

=18 - [ 5 - { 6 + 2 ( 4 ) } ]

=18 - [ 5 - { 6 + 8 } ]

=18 - [ 5 - 14 ]

=18 + 9

=27

18) The value of 1÷ [ 1 + 1 ÷ { 1 + 1 ÷ ( 1 + 1 ÷ 2 ) } ] is

Solution :

= 1÷ [ 1 + 1 ÷ { 1 + 1 ÷ ( 1 + 1 ÷ 2 ) } ]

= 1÷ [ 1 + 1 ÷ { 1 + 1 ÷ ( 1 + 1/2 ) } ]

= 1÷ [ 1 + 1 ÷ { 1 + 1 ÷ ( 3/2 ) } ]

= 1÷ [ 1 + 1 ÷ { 1 + 1 × 2/3 } ]

= 1÷ [ 1 + 1 ÷ { 1 + 2/3 } ]

= 1÷ [ 1 + 1 ÷ { 5/3 } ]

= 1÷ [ 1 + 1 × 3/5 ]

= 1÷ [ 1 + 3/5 ]

= 1÷ [ 8/5 ]

= 1× 5/8

= 5/8

19) Evaluate :

8 - [ 5 - ( - 3 + 2 ) ] ÷ 2

----------------------------------

| 5 - 3 | - | 5 - 8 | ÷ 3

Solution:

8 - [ 5 - ( - 1 ) ] ÷ 2

= ----------------------------------

| 2 | - | - 3 | ÷ 3

8 - [ 5 + 1 ] ÷ 2

= ----------------------------------

2 - 3 ÷ 3

8 - 6 ÷ 2

= ----------------------------------

2 - 1

8 - 3

= ----------------------------------

2 - 1

= 5/1

= 5

20) 1 + 1/2 + 1/4 + 1/7 + 1/14 + 1/ 28 is equal to :

Solution :

= 1 + 1/2 + 1/4 + 1/7 + 1/14 + 1/ 28

28 + 14 + 7 + 4 + 2 + 1

= -------------------------------------

= 56/28

= 2

21) 3 1 2

1--------- + 5 -------- + 3 --------- = ?

4 3 5

Solution :

3 1 2

= 1--------- + 5 -------- + 3 ---------

4 3 5

= 1 × 4 + 3 5 × 3 + 1 3 × 5 + 2

----------------- + ------------------ + ----------------------

4 3 5

7 16 17

= -------- + -------- + ----------

4 3 5

105 + 350 + 204

= ----------------------------

629

= ------------

= 10---------

22) 1 1 1

20----------+ 30---------- - 15---------- = ?

2 3 6

Solution :

1 1 1

= 20----------+ 30---------- - 15----------

2 3 6

= 41 + 91 - 91

2 3 6

= 214

= 107

= 2

35----------

23) 3 of 4 of 5 of 21 of 504 = ?

5 7 9 24

= 3 × 4 × 5 × 21 × 504

5 7 9 24

= 84

24) 11.6 + 9.28 ÷ 0.464 - 0.2828 ÷ 0.07 = ?

Solution :

= 11.6 + 9.28 ÷ 0.464 - 0.2828 ÷ 0.07

= 11.6 + 9280 - 2828

464 7

= 11.6 + 20 - 4.04

= 27. 56

25) If 2 = x + 1 then the value of x is

1 + 1

3 + 1

Solution :

x = 2 - 1

1 + 1

3 + 1

x = 2 - 1

1 + 1

x = 2 - 1

1 + 4

x = 2 - 1

x = 2 - 13

x = 21

26) What is the value of 9 - 5 ÷ ( 8 - 3 ) × 2 + 6 ?

Solution :

We use BODMAS

= 9 - 5 ÷ ( 8 - 3 ) × 2 + 6 [ solve bracket ]

= 9 - 5 ÷ 5 × 2 + 6 [ multiply ]

= 9 - 5 ÷ 10 + 6 [ division ]

= 9 - 2 + 6 [ addition ]

= 15 - 2 [ subtraction ]

= 13

Thus, value is 13.

27) The 6th term of an AP is 6 and the 16th term is 14. What is the 27th term ?

Solution :

Exercise

1) Find x if x - x = 4.

5 6

a) -120 b) -100 c) 100 d) 120

2) Which of the following values of x and y satisfy the following equations I and II ?

I. 3x + y = 19 II. x - y = 9

a) - 7, - 2 b) - 7, 2 c) 7, - 2 d) 7, 2

3) If a + b = 5 and 3a + 2b = 20 , then ( 3a + b ) will be :

a) 10 b) 15 c) 20 d) 25

Answer

1) d 2) c 3) d

---------X-------

Comments

Anonymous said…

Very good work...keep it up!

7 December 2020 at 22:27

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