Is there a quick way to find out how many 10c coins?

Question

Kim has 20 coins which add to make $5. She only has 10c, 20c and 50c coins. She has more 50c coins than 10c coins.

How many 10c coins does Kim have?

Answer 1

I worked it out my working out the average value of each coin

500c/20 = 25c per coin

Then I worked out different groups that had this average coin value

1 x 50c + 5 x 20c = 150c
1 x 50c + 2 x 20c + 1 x 10c = 100c

You can now combine these groups in a way to get the required total


1 x 50c + 5 x 20c = 150c
1 x 50c + 5 x 20c = 150c
1 x 50c + 2 x 20c + 1 x 10c = 100c
1 x 50c + 2 x 20c + 1 x 10c = 100c

4 x 50c + 14 x 20c + 2 x 10c = $5
4 + 14 + 2 = 20

So Kim has 2 10c coins

Answer 2

Let's say Kim has (x) 10c coins,
(y) 20c coins and (z) 50c coins.

Then we can make these 2 equations:
(1) x + y + z = 20
(2) 10x + 20y + 50z = 500

Dividing (2) through by 10 gives:
(3) x + 2y + 5z = 50

Multiplying (1) through by 2 gives:
(4) 2x + 2y + 2z = 40

I did that so I could eliminate y.

Subtracting (4) from (3) gives:
-x + 3z = 10

Rearranging gives:
(5) x = 3z - 10

Here, it can be seen that z >= 4,
otherwise x will be negative.

We know from the question that x < z.
But (5) says that x = 3z - 10.

Therefore, 3z - 10 < z
or, 2z - 10 < 0
or, 2z < 10
or, z < 5

With z >=4 and z < 5, then z = 4.
Thus from (5), x = 3*4 - 10 = 2
and from (1), y = 20 - x - z = 14.

Answer: Kim has 2 10c coins.

Answer 3

There's no quick way, because the algebra takes so long to set up and solve, then eliminate the answers with negative numbers of some coins, or with less 50c coins than 10c coins.

By trial and error, the possible numbers of 50c, 20c, and 10c coins are:

4, 14, 2 = $2.00 + $2.80 + $0.20 = $5.00
5, 10, 5 = $2.50 + $3.00 + $0.50 = $5.00
6, 6, 8 = $3.00 + $1.20 + $0.80 = $5.00
7, 2, 11 = $3.50 + $0.40 + $1.10 = $5.00

and only the first case has more 50c coins than 10c coins.

Answer 4

Let the number of 10c coins be a, 20c coins b and 50c coins be c
From the info, you get three equations -
1) a+b+c= 20;
2) 10a+20b+50c=500 => a+2b+5c=50
3) c>a

from the 1st equation, you get that b=20-a-c
putting this in equation 2 you get
a+2(20-a-c)+5c=50
a+40-2a-2c+5c=50
-a+40+3c=50
3c-a=10

The only way the above can be true with c>a is when c=4 and a is 2

SO, a=2, b would be = 14 and c=4
10c coins are 2
20c coins are 14
50c coins are 4

Answer 5

20 coins of 20c will make $4

One dollar is missing:
Let's start removing coins of 20, and add 50c instead

If we remove 4 coins of 20c and add 4 * 50c We'll have 5.20

Now, let's remove 2 coins of 20c and add 2 coins of 10c and we'll have $5.20 - 2*20c + 2*10c = 5

4 coins of 50c = $2
2 coins of 10c = $0.20
14 coins of 20c = $2.80

Thus, we have 20 coins and $5.

If we want to keep this balance and number of coins, we'll have to replace each time 4 coins of 20c with one coin of 50c and 3 of 10c, which will make the number of 10c coins larger than or equal to the number of 50c coins,

because 4, the number of 50c coins + 1 = 5
and 2, the number of 10c coins + 3 = 5

Answer 6

88