A man invested a total of N50000 in two companies. If these companies pay dividends of 6% and 8% respectively, how much did he invest at 8% if the total yield is N3700?

A.

N15 000

B.

N29 600

C.

N27 800

D.

N21 400

E.

N35 000

Correct answer is E

Total yield = N3,700

Total amount invested = N 50,000

Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest

then yield on x = \(\frac{6}{100}\) x and yield on y = \(\frac{8}{100}\)y

Hence, \(\frac{6}{100}\)x + \(\frac{8}{100}\)y

= 3,700.........(i)

x + y = 50,000........(ii)

6x + 8y = 370,000 x 1

x + y = 50,000 x 6

6x + 8y = 370,000.........(iii)

6x + 6y = 300,000........(iv)

eqn (iii) - eqn(2)

2y = 70,000

y = \(\frac{70,000}{2]\)

= 35,000

Money invested at 8% is N35,000