Jeff invested $6000 for 1 year.?
He invested portion of it (a) 9% and the rest (a) 11%. At the winding up of the year he earn $624 surrounded by interest. How much did he invest at respectively rate. work problem out.
Answers:
If he'd invested it adjectives at 9% he'd own gotten $540. If he'd invested it adjectives at 11% he'd hold gotten $660.
Range is $120 (660-540=120).
He got $624, and 624 - 540 = 84.
So he be 84/120 of the course between 9 and 11.
So 84/120 of the money be at 11, and the rest - 36/120 be at 9.
So, 84/120 * 6000 = $4200 - amount at 11%
36/120 * 6000 = $1800 - amount at 9%.
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Answers:
If he'd invested it adjectives at 9% he'd own gotten $540. If he'd invested it adjectives at 11% he'd hold gotten $660.
Range is $120 (660-540=120).
He got $624, and 624 - 540 = 84.
So he be 84/120 of the course between 9 and 11.
So 84/120 of the money be at 11, and the rest - 36/120 be at 9.
So, 84/120 * 6000 = $4200 - amount at 11%
36/120 * 6000 = $1800 - amount at 9%.