SOLUTION: Introduce variables and write two equations A collection of dimes and quarters is worth $15.25. There are 103 coins in total. How many of each kind of coin is there?

Question 1201833: Introduce variables and write two equations

A collection of dimes and quarters is worth $15.25. There are 103 coins in total. How many of each kind of coin is there?
Found 3 solutions by josgarithmetic, ikleyn, math_tutor2020:
Answer by josgarithmetic(39613)   (Show Source): You can put this solution on YOUR website!
d for how many dimes and q for quarters, if for example you want TWO variables.
Follow exactly the description as written and form two equations. One accounting for number of coins and other for how much in dollars.

Answer by ikleyn(52754)   (Show Source): You can put this solution on YOUR website!
.

Let x be the number of dimes and y be the number of quarters. Write equations as you read the problem x + y = 103 (the total number of coins) 5x + 25y = 1525 (total number of cents).

At this point, the assignment is completed.

Since you do not ask solve equations, I do not solve them.

====================

Regarding writing equations, there are two approaches among the teachers.

One part of teachers presents it to pupils as some miracle, which requires
extraordinary intellectual efforts.

The other part of teachers presents it as a routine exercise, saying "write equations as you read the problem".

Both approaches have equal rights to exist, but my personal opinion
is that after considering 1-2-3-5 problems as a miracle
a student should later move to the second view and consider it as a routine.

In 98% cases, school Math problems setup is really routine, as in this problem;
but in 2% of cases, it may really require some intellectual efforts.

Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Answer:
70 dimes
33 quarters

70 dimes + 33 quarters = 70*10 + 33*25 = 1525 cents = 15.25 dollars

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Work Shown:

d = number of dimes
q = number of quarters
These variables are placeholders for nonnegative whole numbers 0,1,2,3,...

d+q = 103 coins total
q = 103-d
It will be used in a substitution step later

10d = value of the dimes only (cents)
25q = value of the quarters only (cents)
10d+25q = total value (cents)
10d+25q = 1525
10d+25( q ) = 1525
10d+25( q ) = 1525
10d+25( 103-d ) = 1525 ... substitution
10d+25*103+25*(-d) = 1525
10d+2575-25d = 1525
-15d+2575 = 1525
-15d = 1525-2575
-15d = -1050
d = -1050/(-15)
d = 70 dimes

q = 103-d
q = 103-70
q = 33 quarters

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Another approach:

The 25 cents in $15.25 tells us that there's at least one quarter.
It is impossible to reach this subtotal amount with dimes alone.

If we had 1 quarter and 103-1 = 102 dimes, then we have:
1 quarter + 102 dimes = 1*25 + 102*10 = 1045 cents = $10.45
We come up short of the goal $15.25

Let's replace 1 dime with 1 quarter.
Difference in coin value = 25-10 = 15 cents
Each time we do this replacement of "1 dime for 1 quarter", the total value goes up by 15 cents aka $0.15
This is the net change.

Because we want the $10.45 to go to $15.25, we must increase by $4.80
(since 15.25-10.45 = 4.80 )

n = number of increases of 15 cents each
0.15n = total increase needed
0.15n = 4.80
n = 4.80/0.15
n = 32
We need 32 increases of 15 cents each to go from $10.45 to $15.25
Meaning we need to do the replacement of "1 dime for 1 quarter" a total of 32 times.

Add 32 to the quarter count, and subtract 32 from the dime count.
1+32 = 33 quarters
102-32 = 70 dimes
which are the final answers.

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Check:
70 dimes + 33 quarters
= 70*10 + 33*25
= 700 cents + 825 cents
= 1525 cents
= 1525/100
= 15.25 dollars
The answers are confirmed.

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