Peter had 48 quarters and 11 dimes
<h3>Further explanation</h3>Simultaneous Linear Equations could be solved by using several methods such as :
If we have two linear equations with 2 variables x and y , then we need to find the value of x and y that satisfying the two equations simultaneously.
Let us tackle the problem!
Let :
<em>Number of quarters ( 25 cent coins ) = x</em>
<em>Number of dimes ( 10 cent coins ) = y</em>
<em>When he counted his quarters and dimes, he found they had a total value of $13.10.</em>
<h2>0.25x + 0.10y = 13.10</h2><em>The number of quarters was 15 more than 3 times the number of dimes.</em>
<h2>x = 15 + 3y</h2>If we would like to use the Substitution Method , then second equations above could be substituted into first equations.
0.25x + 0.10y = 13.10
0.25 (15 + 3y) + 0.10y = 13.10
3.75 + 0.75y + 0.10y = 13.10
0.85y = 13.10 - 3.75
0.85y = 9.35
y = 9.35 / 0.85
<h3>y = 11</h3>At last , we could find the value of x by substituting this y value into one of the two equations above :
x = 15 + 3y
x = 15 + 3(11)
x = 15 + 33
<h3>x = 48</h3><h3>Learn more</h3>Grade: High School
Subject: Mathematics
Chapter: Simultaneous Linear Equations
Keywords: Simultaneous , Elimination , Substitution , Method , Linear , Equations
Answer: Peter had 48 quarters and 11 dimes.
Step-by-step explanation:
Let be "q" the number of quarters and "d" the number of dimes.
We know that $13.10 in cents is 1,310 cents. Then, we can set up the following system of equations:
Applying the Substitution method, we can substitute the second equation into the first one and solve for "d":
Finally, we must substitute the value of "d" into the second equation to find the value of "q". Then:
Alright, lets gets started.
I suppose , we are asked to find simple interest for above question.
Principal P = $1,250
Rate r = 5.7 %
Time t = 4 years
The formula for simple interest =
Hence simple interest
Simple interest
Simple interest = 285 $
Hence interest earned is $ 285. : Answer
Hope it will help :)