Lesson More complicated coin problems

More complicated coin problems

Problem 1

Let "d" be the number of dimes and "q" be the number  of quarters.
Notice that the number of pennies = d.

Then you have an equation for the number of coins

d + d + q = 31, 

and the equation for the "value"

d + 10d + 25q = 229.

Thus you have a system

 2d +   q =  31,   (1)
11d + 25q = 229.   (2)

From (1), express q = 31-2d and substitute it into (2). You will get a single equation for q

11d + 25*(31-2d) = 229.

Simplify and solve it for d.
Then find q.

Problem 2

Let "q" be the number of quarters, "d" be the number of dimes and "n" be the number of nickels.

Then what you have from the condition is 

d = q - 10  and  n = 2q.     (1)

Next, you have the "value" equation

5n + 10d + 25q = 665.        (2)

Substitute (1) into the "value" equation. You will get a single equation for q:

5*(2q) + 10*(q-10) + 25q = 665.

Simplify and solve it for q:

10q + 10q - 100 + 25q = 665,

45q = 665 + 100,

45q = 765,

q =  = 17.

Now from (1)  d = q - 10  = 17 - 10 = 7  and  n = 2q = 2*17 = 34.

Answer.  34 nickels,  7 dimes  and  17 quarters.

Problem 3

Let n = # of nickels, d = # of dimes, q = # of quarters.

Then immediately from the condition you have these system of three equations for three unknowns:

 n +   d +   q =  133,       (1)      ("he had 133 coins in all")
5n + 10d + 25q = 1470,       (2)      ("When he gets to the bank he receives a total of $14.70")
d = 3q.                      (3)

Next, we reduce this 3x3 system to the more simple 2x2 system of two equations for two unknowns:

For it, I substitute (replace) "d" in equations (1) and (2) by 3q based on equation (3). I will get

 n +     3q  +   q =  133,   (4)
5n + 10*(3q) + 25q = 1470.   (5)

or

 n +  4q =  133,             (6)
5n + 55q = 1470.             (7)

OK. So, you have now much simpler system (6), (7). To solve it, express n = 133-4q  from (6) and substitute it into (7). You will get

5*(133-4q) + 55q = 1470,  or

665 - 20q + 55q = 1470  --->  35q = 1470 - 665  --->  35q = 805  --->  q =  = 23.

So we just found the number of quarters. It is 23. There were 23 quarters.

Then the number of dimes is trice of it: there were 3*23 = 69 dimes.

Now the number of nickels is simply 133 - 69 - 23 = 41.

Check.  5*41 + 10*69 + 23*25 = 1470.  Correct!

Answer.  41 nickels, 69 dimes and 23 quarters.

Problem 4

Let "d" be the number of dimes.

Then the number of quarters is 5d,
and the number of nickels is d-28, according to the condition.

Hence, the total in cents is

10d + 25*(5d) + 5*(d-28) = 4760,  or

10d + 125d + 5d - 140 = 4760,  or

140d = 4760 + 140,

d =  = 35.

So, there are 35 dimes in the collection.

Then the number of quarters is 5*35 = 175 and the number of nickels is 35 - 28 = 7.

Check.  5*7 + 10*35 + 25*175 = 4760.   Correct !

Answer. 7 nickels, 35 dimes and 175 quarters.