Question 978672


Tiana invests ${{{1900}}} in one account  at rate {{{x}}} % and ${{{1400}}} in an account paying {{{3}}} % higher interest. 

At the end of one year she had earned ${{{372}}} in interest. 
{{{(x/100)1900+(x/100+3/100)1400=372}}}
{{{(x/1)19+((x+3)/1)14=372}}}
{{{19x+(x+3)*14=372}}}
{{{19x+14x+42=372}}}
{{{33x=372-42}}}
{{{33x=330}}}
{{{x=330/33}}}
{{{x=10}}}% 

and an account paying {{{3}}} % higher interest is {{{x+3=13}}}% 

 she invest at {{{10}}}%  and {{{13}}}% 

check: 
{{{10}}}%  of  ${{{1900}}} is   ${{{(10/100)1900=10*19=190}}}
{{{13}}}%  of  ${{{1400}}} is   ${{{(13/100)1400=13*14=182}}}

and {{{190+182=372}}} which confirms our solution