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Given:
The rate of interest on three accounts are 7%, 8%, 9%.
She has twice as much money invested at 8% as she does in 7%.
She has three times as much at 9% as she has at 7%.
Total interest for the year is $150.
To find:
Amount invested on each rate.
Solution:
Suppose x be the amount invested at 7%. Then,
The amount invested at 8% = 2x
The amount invested at 9% = 3x
Total interest for the year is $150.
Multiplying both sides by 100, we get
Dividing both sides by 50, we get
So, the amount invested at 7% is .
The amount invested at 8% is
The amount invested at 9% is
Hence, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.
Answer:
<h2>There are 13 quarters, 5 nickels and 20 dimes.</h2>Step-by-step explanation:
Givens
- Max has $5.50.
- He has eight more quarters than he does nickels.
- He had four times as many dimes as he has nickels.
From these statements, we can deduce algebraical expressions to solve the problem. Let's call quarters,
nickels and
dimes.
(8 more quarters than nickels)
(he has four times as many dimes as he has nickels).
(Max has a total of $5.50, one quartes is $0.25, one nickel is $0.05 and one dime is $0.10).
Let's replace the first two expressions into the third equation
There are 5 nickels.
Now, we use this value to find the number of quarters
There are 13 quarters.
Also,
There are 20 dimes.
Therefore, there are 13 quarters, 5 nickels and 20 dimes.