Q:

A bank loaned out $21,500 , part of it at the rate of 11% annual interest, and the rest at 10% annual interest. The total interest earned for both loans was $2,340.00 . How much was loaned at each rate? ____ was loaned at 11%____ was loaned at 10%

Accepted Solution

A:
Answer:The loaned amount at 11 % is $ 19,000The loaned amount at 10 % is $ 2,500Step-by-step explanation:Given as :The total loan amount = $21,500The total interest earn = $2,340.00The rate of interest are 11 %  and 10 %Let The loan amount at 11 % rate = $Pand The loan amount at 10 % rate = $21,500 - $PLet The loan took for 1 yearNow, From Simple Interest method Simple Interest = [tex]\dfrac{\textrm Principal\times \textrm Rate\times \textrm Time}{100}[/tex][tex]SI_1[/tex] = [tex]\dfrac{P_1\times R_1\times \textrm Time}{100}[/tex]Or, [tex]SI_1[/tex] = [tex]\dfrac{P\times 11\times \textrm 1}{100}[/tex]Similarly[tex]SI_2[/tex] = [tex]\dfrac{21,500 - P\times 10\times \textrm 1}{100}[/tex]∵  [tex]SI_1[/tex] +  [tex]SI_2[/tex] =  $2,340Or, [tex]\dfrac{P\times 11\times \textrm 1}{100}[/tex] + [tex]\dfrac{21,500 - P\times 10\times \textrm 1}{100}[/tex] = $2,340Or, 11 P - 10 P + 215000 = 234000Or, P = 234000 - 215000∴ P = $ 19,000And $21,500 - $ 19,000 = $ 2,500Hence The loaned amount at 11 % is $ 19,000And     The loaned amount at 10 % is $ 2,500    Answer