A newspaper carrier has $6.20 in change. He has four more quarters than dimes but five times as many nickels as quarters. How many of each type does he have?

Question

This one was another that was entirely too complicated for me. I get confused on this question.

Answer 1

The key to solving word (story) problems is to learn how to translate the words and phrases into concise, precise math expressions (variables and operations).  Make a table and add to it as you do problems:
       is              means           =
     more than                        +
       twice                              2*x
       sum                               +

       of                               usually means multiply

     times as many                multiply
      ...
 
Assign variables.  Try to assign variables to the values the problem asks you to find.
In this problem, there are three variables.

   Let Q = number of quarters

         D = number of dimes

         N = number of nickels

 

Note: not mentioned is that a quarter is worth 25 cents, a dime is worth 10 cents and a nickel is worth 5 cents.

Also, the number of cents = (the value of the coin)*(the number of coins)

 

Let's do everything in cents  (I have trouble with decimals)

 

Translate:

  "has $6.20 in change"    means         (assume he has only nickels, dimes and quarters)

     25Q + 10D + 5N  = 620                   (eq1)

  "He has four more quarters than dimes"  means

               Q    =    4   +   D                    (eq2)

  "five times as many nickels as quarters"   means

          N                           = 5Q              (eq3)

 

Now, the easy math:

 

     N = 5Q        (eq3)

     N = 5(4+D)              (substitute value of Q from eq2)

     N = 20 + 5D

 

Now, eq1 can be written (with substitution):

      25(4+D)   +   10(D)    +   5(20 + 5D)  =  620

       100 + 25D  + 10D  + 100 + 25D = 620             (just expand expressions)

              60D + 200 = 620                           (collect terms)

                60D = 420                (subtract 200 from both sides)

                    D = 7               (divide both sides by 60)

 

Put that value of D into eq2:

                 Q = 4 + 7

                 Q = 11

 

Put that value of Q into eq3:

                    N = 5(11)

                    N = 55

 

Checking (very important):

  Is      25(11) + 10(7) + 5(55)  = 620       ?

             275  + 70   +  275  +  620   ?

               620 = 620   ?  yes

  Is   11 = 4 + 7   ?

        11  =  11   ? yes

  Is   55 = 5*11    ?

        55 = 55  ?   yes

Answer 2

The key to solving this type of problem is to understand that we're being given two different types of numbers here: the count of coins versus the value of the coins.

 

Let's work with the count of coins first: there are four more quarters than dimes. (For example, if we have 4 dimes there must be 8 quarters.) To write an equation we have to add onto the smaller amount to find the larger amount. Let D represent the number of dimes and Q represent the number of quarters. D is smaller than Q so Q = D + 4.

 

There are 5 TIMES more nickels as quarters. This time we MULTIPLY the smaller amount to find the larger amount. (For example, if we have 3 quarters there must be 5 times 3 nickels, so that would be 15 nickels.) If we let N represent the number of nickels then we can write N = 5Q.

 

Now we need to work with the VALUE of the coins. A quarter is worth 25 cents, so we multiple the COUNT of quarters, Q, by 25 to get the VALUE of the quarters, 25 times Q is written 25Q. Similarly the value of nickels is 5N and the value of dimes is 10D.

 

Adding each of these gives us the total value of the coins in CENTS: 25Q + 5N + 10D. What is the value of $6.20 in CENTS? 620 cents! So, 25Q + 5N + 10D = 620.

 

We can't solve this equation yet because there are 3 different variables. We can use substitution to change Q's and N's into D's.

 

Q = D + 4 so we change 25Q to 25(D+4).

N = 5Q so we change 5N to 5(5Q). Multiply 5 times 5 times Q is 25Q. 

 

Now we can substitute into the value equation 25Q   +        5N + 10D = 620. 

We get............................................................ 25(D+4) + 25Q + 10D = 620.

We have to substitute Q = D+4 once again:     25(D+4) + 25(D+4) + 10D = 620.

 

Now we can solve for D. First, expand 25(D+4) to 25(D) +25(4) = 25D + 100.

 

So 25D + 100 + 25D + 100 + 10D = 620

 

Combine like terms:

 

25D + 25D + 10D is 60D and 100 + 100 is 200 so 60D + 200 = 620.

 

Subtract 200 from both sides: 60D + 200 - 200 = 620 - 200. This gives   60D = 420.

 

Divide both sides by 60, so D = 420/60 = 42/6 = 7. There are 7 dimes.

 

Since there are 4 more quarters than dimes... Q = D + 4 = 7 + 4 = 11.  There are 11 quarters.

 

Since there are 5 times more nickels than dimes... N = 5Q = 5(11) = 55. There are 55 nickels.

 

Lets make sure this works:

 

55 nickels is 5(55) = 275 cents

11 quarters is 25(11) = 275 cents

7 dimes is 10(7) = 70 cents

 

275 + 275 + 70 = 620 cents which is $6.20.  It works!

Answer 3

Take your time.

What coin is most talked about? The quarters. Let’s see how we can describe the nickels and dimes in terms of quarters.

Quarters are 4 more that dimes. Let’s write that two ways. I’m not sure which way we will be going on this so I want both relationships.

Quarters are 4 more that dimes.
Q-4= D
D+4= Q

Nickels are most prevalent. They are 5 times the amount of quarters.
N=5Q
Q=N/5

Again I want both relationships.

So I have dimes in terms of quarters and
Nickels in terms of quarters.

Let’s write the total cash amount in terms of dollars
$.25* Q +$.10*D +$ .05* N= $6.20

We can describe the nickels and dimes in terms of quarters
$.25*Q+ $ .10*(Q-4) +$.05*(5Q) =6.20
Simplify;

$.25Q + $.10Q+ -$.40 +$.25Q = $ 6.20
Add $.40 to both sides
$.25Q + $.10Q +$.25Q = $6.60

Add all the Q’s
$.60Q= $ 6.60
You have 11 quarters

Now how many dimes Q-4=7
Now many nickels? Q*5=55
Now sum up the value of all the coins


$.25*11 + $ .10*7 + $.05*55 =????
$2.75 + $.70 +$2.75 = $6.20

The key here is that all coins revolved around the quarter.

 

 

 

 

Answer 4

Let:

 

n be nickel, 5 cents

d be dime, 10cents

q be quarter, 25cents

 

$6.20 = 6.20*100=620cents

 

 

q=d+4  ( more means addition)

n=5q  ( 5 times as many nickels as quarters means multiplication)

n=5(d+4) (substituting for q)

 

To make up a change of 620 cents, you need to add all the coins :

 

5n+10d+25q =620

 

Now we can express the same equation in terms of d:

 

5*5(d+4) +10d+25(d+4) = 620

25(d+4)+10d+25(d+4) = 620

 

50(d+4) + 10d = 620

50d+200+10d=620

60d=420

d=7 dimes

 

q=d+4=7+4=11

n=5q=5*11=55

 

So, the answer is:

 

55 nickels

7 dimes

11 quarters

 

 

You can prove it yourself.