The sum of the squares of three positive numbers that are consecutive multiples of  5 is 725. Find the three numbers.

This question is similar to Ex 4.2, 4 Chapter 4 Class 10 - Quadratic Equations

[Class 10] Sum of squares of three positive numbers that consecutive - CBSE Class 10 Sample Paper for 2022 Boards - Maths Basic [Term 2]

part 2 - Question 10 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Basic [Term 2] - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 10 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Basic [Term 2] - Solutions of Sample Papers for Class 10 Boards - Class 10

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Question 10 The sum of the squares of three positive numbers that are consecutive multiples of 5 is 725. Find the three numbers. Given three positive numbers that are consecutive multiples of 5 Since the number is a multiple of 5 Let First Number = 5x Now, Since other 2 numbers are consecutive multiples of 5, they will have a difference of 5 ∴ Second Number = 5x + 5 Third Number = (5x + 5) + 5 = 5x + 10 Given that Sum of the squares of three positive numbers that are consecutive multiples of 5 is 725 (5x)2 + (5x + 5)2 + (5x + 10)2 = 725 52 × x2 + 52 (x + 1)2 + 52 (x + 2)2 = 725 52 × [x2 + (x + 1)2 + (x + 2)2] = 725 25 × [x2 + (x + 1)2 + (x + 2)2] = 725 x2 + (x + 1)2 + (x + 2)2 = 725/25 x2 + (x + 1)2 + (x + 2)2 = 29 x2 + x2 + 12 + 2x + x2 + 22 + 4x = 29 x2 + x2 + x2 + 2x + 4x + 1 + 4 = 29 3x2 + 6x + 5 = 29 3x2 + 6x + 5 − 29 = 0 3x2 + 6x − 24 = 0 3(x2 + 2x − 8) = 0 x2 + 2x − 8 = 0 Solving by splitting the middle term x2 + 4x − 2x − 8 = 0 x(x + 4) − 2(x + 4)= 0 (x − 2) (x + 4) = 0 ∴ x = 2, x = −4 Since we need a positive number ∴ x = 2 Thus, First Number = 5x = 5 × 2 = 10 Second Number = 5x + 5 = 5 × 2 + 5 = 15 Third Number = 5x + 10 = 5 × 2 + 10 = 20

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo