Question

A total of ₹ 8,600 was invested in two accounts. One account earned ${\displaystyle 4\frac{3}{4}}$% annual interest and the other earned ${\displaystyle 6\frac{1}{2}}$% annual interest. If the total interest for one year was ₹ 431.25, how much was invested in each account? (Use determinant method)

Answer 1

Let ₹ x and ₹ y be the amounts invested in the two accounts.

Interest or first account = ${\displaystyle 4\frac{3}{4}}$% x

= ${\displaystyle \frac{19}{4}\times \frac{1}{100}\times x=\frac{19}{100}x}$

Interest for second ccount = ${\displaystyle 6\frac{1}{2}}$%

= ${\displaystyle \frac{13}{2}\frac{\times }{100}}$ y

= ${\displaystyle \frac{13}{200}}$ y

According to the problem,

x + y = 8600

${\displaystyle \frac{19}{400}x+\frac{13}{200}y}$ = 431.25

Multiplying the second equation by 400,

19x + 26y = 172500

Here ${\displaystyle \Delta =\left|\begin{array}{cc}1& 1\\ 19& 26\end{array}\right|}$

= 26 – 19

= 7 ≠ 0

So the system has a unique solution

${\displaystyle {\Delta }_x=\left|\begin{array}{cc}8600& 1\\ 172500& 26\end{array}\right|}$

= 223600 – 172500

= 51100

${\displaystyle {\Delta }_y=\left|\begin{array}{cc}1& 8600\\ 19& 172500\end{array}\right|}$

= 172500 – 163400

= 9100

∴ By Cramer's rule,

x = ${\displaystyle \frac{{\Delta }_x}{\Delta }=\frac{51100}{7}}$ = 7300

y = ${\displaystyle \frac{{\Delta }_y}{\Delta }=\frac{9100}{7}}$ = 1300

Hence the amount invested at ${\displaystyle 4\frac{3}{4}}$% is ₹ 7300 and amount invested at ${\displaystyle 6\frac{1}{2}}$% is ₹ 1300.