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?
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Correct Answer: Option E
Explanation:
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
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