Compute the following differences: ( 125 × 75 ) - ( 126 × 74 )

Given difference is ( 125 × 75 ) - ( 126 × 74 ) Therefore, ( 125 × 75 ) can be written as ( 100 + 25 ) ( 100 - 25 ) Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 100 + 25 ) ( 100 - 25 ) = ( 100 ) 2 - ( 25 ) 2 ⇒ ( 100 + 25 ) ( 100 - 25 ) = 10000 - 625 ⇒ ( 100 + 25 ) ( 100 - 25 ) = 9375 Now, ( 126 × 74 ) can be written as ( 100 + 26 ) ( 100 - 26 ) Using identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 100 + 26 ) ( 100 - 26 ) = ( 100 ) 2 - ( 26 ) 2 ⇒ ( 100 + 26 ) ( 100 - 26 ) = 10000 - 676 ⇒ ( 100 + 26 ) ( 100 - 26 ) = 9324 Now substituting the values in the given equation ( 125 × 75 ) - ( 126 × 74 ) = 9375 - 9324 = 51 Hence, the value of ( 125 × 75 ) - ( 126 × 74 ) is 51 .

Compute the following differences: ( 124 × 76 ) – ( 126 × 74 )

Given: ( 124 × 76 ) – ( 126 × 74 ) Therefore, ( 124 × 76 ) can be written as ( 100 + 24 ) ( 100 - 24 ) Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 100 + 24 ) ( 100 - 24 ) = ( 100 ) 2 - ( 24 ) 2 ⇒ ( 100 + 24 ) ( 100 - 24 ) = 10000 - 576 ⇒ ( 100 + 24 ) ( 100 - 24 ) = 9424 Now, ( 126 × 74 ) can be written as ( 100 + 26 ) ( 100 - 26 ) Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 100 + 26 ) ( 100 - 26 ) = ( 100 ) 2 - ( 26 ) 2 ⇒ ( 100 + 26 ) ( 100 - 26 ) = 10000 - 676 ⇒ ( 100 + 26 ) ( 100 - 26 ) = 9324 Now substitute the values in the given equation ⇒ ( 124 × 76 ) – ( 126 × 74 ) = 9424 - 9324 ⇒ ( 124 × 76 ) – ( 126 × 74 ) = 100 Hence, the difference is 100 .

Compute the following difference: ( 224 × 176 ) – ( 226 × 174 )

Given: ( 224 × 176 ) – ( 226 × 174 ) ( 224 × 176 ) can be written as ( 200 + 24 ) ( 200 - 24 ) Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 200 + 24 ) ( 200 - 24 ) = ( 200 ) 2 - ( 24 ) 2 ⇒ ( 200 + 24 ) ( 200 - 24 ) = 40000 - 576 ⇒ ( 200 + 24 ) ( 200 - 24 ) = 39424 Now ( 226 × 174 ) can be written as ( 200 + 26 ) ( 200 - 26 ) Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 200 + 26 ) ( 200 - 26 ) = ( 200 ) 2 - ( 26 ) 2 ⇒ ( 200 + 26 ) ( 200 - 26 ) = 40000 - 676 ⇒ ( 200 + 26 ) ( 200 - 26 ) = 39324 Now substituting the values in given equation: ( 224 × 176 ) – ( 226 × 174 ) = 39424 - 39324 = 100 Hence, the required answer is 100 .

Compute the following differences: ( 10 . 3 × 9 . 7 ) – ( 10 . 7 × 9 . 3 )

Given: ( 10 . 3 × 9 . 7 ) – ( 10 . 7 × 9 . 3 ) Therefore, ( 10 . 3 × 9 . 7 ) can be written as ( 10 + 0 . 3 ) ( 10 - 0 . 3 ) Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 10 + 0 . 3 ) ( 10 - 0 . 3 ) = ( 10 ) 2 - ( 0 . 3 ) 2 ⇒ ( 10 + 0 . 3 ) ( 10 - 0 . 3 ) = 100 - 0 . 09 ⇒ ( 10 + 0 . 3 ) ( 10 - 0 . 3 ) = 99 . 91 Now ( 10 . 7 × 9 . 3 ) can be written as ( 10 + 0 . 7 ) ( 10 - 0 . 7 ) Therefore, Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 10 + 0 . 7 ) ( 10 - 0 . 7 ) = ( 10 ) 2 - ( 0 . 7 ) 2 ⇒ ( 10 + 0 . 7 ) ( 10 - 0 . 7 ) = 100 - 0 . 49 ⇒ ( 10 + 0 . 7 ) ( 10 - 0 . 7 ) = 99 . 51 Now substituting the values in the given equation ( 10 . 3 × 9 . 7 ) – ( 10 . 7 × 9 . 3 ) = 99 . 91 - 99 . 51 = 0 . 4

Compute the following differences: ( 11 . 3 × 10 . 7 ) – ( 11 . 7 × 10 . 3 )

Given: ( 11 . 3 × 10 . 7 ) – ( 11 . 7 × 10 . 3 ) Therefore, ( 11 . 3 × 10 . 7 ) can be written as ( 11 + 0 . 3 ) ( 11 - 0 . 3 ) Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 11 + 0 . 3 ) ( 11 - 0 . 3 ) = ( 11 ) 2 - ( 0 . 3 ) 2 ⇒ ( 11 + 0 . 3 ) ( 11 - 0 . 3 ) = 121 - 0 . 09 ⇒ ( 11 + 0 . 3 ) ( 11 - 0 . 3 ) = 120 . 91 Now ( 11 . 7 × 10 . 3 ) can be written as ( 11 + 0 . 7 ) ( 11 - 0 . 7 ) Using the identity ( x + y ) × ( x - y ) = ( x ) 2 - ( y ) 2 ⇒ ( 11 + 0 . 7 ) ( 11 - 0 . 7 ) = ( 11 ) 2 - ( 0 . 7 ) 2 ⇒ ( 11 + 0 . 7 ) ( 11 - 0 . 7 ) = 121 - 0 . 49 ⇒ ( 11 + 0 . 7 ) ( 11 - 0 . 7 ) = 120 . 51 Now substituting values in question ( 11 . 3 × 10 . 7 ) – ( 11 . 7 × 10 . 3 ) = 120 . 91 - 120 . 51 = 0 . 40

E M B I B E

Standard 8 Mathematics Part - 1>Chapter 4 - Identities>Exercise 5

Kerala Board Solutions for Chapter: Identities, Exercise 5: Exercise 5

Author:Kerala Board

Kerala Board Mathematics Solutions for Exercise - Kerala Board Solutions for Chapter: Identities, Exercise 5: Exercise 5

Attempt the free practice questions on Chapter 4: Identities, Exercise 5: Exercise 5 with hints and solutions to strengthen your understanding. Standard 8 Mathematics Part - 1 solutions are prepared by Experienced Embibe Experts.