Using identities, evaluate. 78 × 82

Given 78 × 82 = ( 80 - 2 ) ( 80 + 2 ) We know that, p 2 - q 2 = ( p + q ) p - q Let p = 80 , q = 2 ⇒ 80 - 2 80 + 2 = ( 80 ) 2 - ( 2 ) 2 ⇒ 80 - 2 80 + 2 = 6400 - 4 = 6396 Hence, the required answer is 6396 .

Using identities, evaluate. 1 . 05 × 0 . 95

Given 1 . 05 × 0 . 95 = ( 1 + 0 . 05 ) ( 1 - 0 . 05 ) We use the identity p 2 - q 2 = ( p + q ) p - q Let p = 1 , q = 0 . 05 ⇒ ( 1 + 0 . 05 ) ( 1 - 0 . 05 ) = ( 1 ) 2 - ( 0 . 05 ) 2 ⇒ ( 1 + 0 . 05 ) ( 1 - 0 . 05 ) = 1 - 0 . 0025 = 0 . 9975 Hence, the required answer is 0 . 9975 .

Using ( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b , find 103 × 104

Given, 103 × 104 The above numbers can be written as = ( 100 + 3 ) ( 100 + 4 ) By using the identity: ( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b . Here x = 100 , a = 3 and b = 4 ⇒ ( 100 + 3 ) ( 100 + 4 ) = ( 100 ) 2 + ( 3 + 4 ) × 100 + 3 × 4 = 10000 + 700 + 12 = 10712 ∴ 103 × 104 = 10712 .

Using ( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b , find 5 . 1 × 5 . 2

Given: 5 . 1 × 5 . 2 The above numbers can be written as 5 + 0 . 1 5 + 0 . 2 By using the identity: ( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b Here, x = 5 , a = 0 . 1 and b = 0 . 2 ⇒ 5 + 0 . 1 5 + 0 . 2 = 5 2 + 0 . 1 + 0 . 2 × 5 + 0 . 1 × 0 . 2 = 25 + 1 . 5 + 0 . 02 = 26 . 52 ∴ 5 . 1 × 5 . 2 = 26 . 52 .

Using ( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b , find 103 × 98

Given, 103 × 98 The above numbers can be written as = ( 100 + 3 ) ( 100 - 2 ) By using the identity: ( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b . Here, x = 100 , a = 3 and b = - 2 ⇒ ( 100 + 3 ) ( 100 - 2 ) = ( 100 ) 2 + ( 3 - 2 ) × 100 + 3 × ( - 2 ) = 10000 + 100 - 6 = 10094 ∴ 103 × 98 = 10094 .

Using ( a + b ) ( x + b ) = x 2 + ( a + b ) x + a b , find 9 . 7 × 9 . 8

Given, 9 . 7 × 9 . 8 The above numbers can be written as = ( 9 + 0 . 7 ) ( 9 + 0 . 8 ) By using the identity: ( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b . Here, x = 9 , a = 0 . 7 and b = 0 . 8 ⇒ ( 9 + 0 . 7 ) ( 9 + 0 . 8 ) = ( 9 ) 2 + ( 0 . 7 + 0 . 8 ) × 9 + 0 . 7 × ( 0 . 8 ) = 81 + 13 . 5 + 0 . 56 = 95 . 06 ∴ 9 . 7 × 9 . 8 = 95 . 06 .

E M B I B E

START UP MATHEMATICS>Chapter 5 - Algebraic Expressions and Identities: Part I>SOME NCERT TEXTBOOK QUESTIONS

Monicaa Abhijit Solutions for Chapter: Algebraic Expressions and Identities: Part I, Exercise 10: SOME NCERT TEXTBOOK QUESTIONS

Author:Monicaa Abhijit

Monicaa Abhijit Mathematics Solutions for Exercise - Monicaa Abhijit Solutions for Chapter: Algebraic Expressions and Identities: Part I, Exercise 10: SOME NCERT TEXTBOOK QUESTIONS

Attempt the practice questions on Chapter 5: Algebraic Expressions and Identities: Part I, Exercise 10: SOME NCERT TEXTBOOK QUESTIONS with hints and solutions to strengthen your understanding. START UP MATHEMATICS solutions are prepared by Experienced Embibe Experts.