Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.1 - Solving Quadratic Equations by Completing the Square - Exercise Set - Page 765: 69


Rate of interest r = 11.11%

Work Step by Step

Amount (A) = 1000 Principal (P) = 810 Time (t) = 2 We have to find the rate of interest i.e. r. Step-1: Substitute the given values in the formula. $A=P(1+r)^t$ $1000=810 (1+r)^2$ Step-2: We now solve the equation for r. $1000=810 (1+r)^2$ Step-3: Divide both sides of the equation by 800 $\frac{1000}{810}=(1+r)^2$ Step-4: Use the square root property $±\sqrt\frac{1000}{810}=1+r$ Step-5: Simplify the left hand side by dividing the numerator and denominator by the Highest Common Factor(H.C.F) of 1000 and 810 H.C.F of 1000 and 810 is 10 $±\sqrt\frac{1000\div10}{810\div10}=1+r$ $±\sqrt\frac{100}{81}=1+r$ $±\frac{10}{9}=1+r$ Step-6: Subtract both the sides with 1 and further simplify the left hand side $±\frac{10}{9}−1=r$ $\frac{±10-9}{9}=r$ Therefore, $ r=\frac{1}{9},\frac{−19}{9}$ Rate of Interest cannot be negative, hence, $r=\frac{1}{9}=0.1111=$11.11%.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.