A bag contains 29 coins. Some of them are 5p coins and the rest are 2p coins.?

Question

The value of the coins is 1=(100p). Find the number of each type of coin.

Answer 1

Let x be the number of 5p coins, and let y be the number of 2p coins. Since there are only 5p and 2p,
x + y = 29
y = 29 - x

The amount each 5p is worth is 5, so the amount of all the 5p coins is 5x. The amount of all the 2p coins is 2y.

5x + 2y = 100

Substitute (29 - x) in for y.
5x + 2(29 - x) = 100
5x + 58 - 2x = 100
3x = 42
x = 14
So there are fourteen 5p coins. Since 29 - 14 is 15, there are fifteen 2p coins.

Answer 2

Let x be the number of 5p coins. So (29-x) is the number of 2p coins.

Balance by the values,
5x+2(29-x)=100

Solve for x,
x = 14 5p coins

29-x = 15 2p coins

Answer 3

F = # 5 p coins
T = # 2 p coins

There are two unknowns, so we need two equations.

First the coins add up to a total of 29 coins.

F + T = 29

Next the value of the coins adds up to 100 p.

5F + 2T = 100

Multiply the first equation by -2 and add it to the second one

5F + 2T = 100
-2F - 2T = -58
3F = 42

so F = 14
since the total number of coins is 29, there must be 15 of the other coin.

14 5 p coins
15 2 p coins

Check yourself, 14 x 5 = 70, 15 x 2 = 30, they do add up to 100.

Answer 4

This problem is first converted into mathematical equation.
x+y = 29 (1)
5x +2y = 100 (2)
multiply eq 1 by -5
-5x -5y = -145 (3)

add eq 2 and 3
-3y = -45
so y = 15,x= 29-15 = 14

So there were 14 5p coins and 15 2p coins.

Answer 5

let x be the number of 5p coins and let y be the number of 2p coins

x + y = 29
5x + 2y = 100

multiply the first equation by 2 and subtract the second:

(2-5)x + (2-2)y = 2(29) - 100
-3x = -42
x = 14

from x + y = 29
y = 29 - x
y = 29 - 14
y = 15

Answer 6

let t= number of 2p coins
let f = number of 5p coins

t+f =29
2t +5f=1
2(29-f)+5f =1
58+3f =1
3f =42
f=14 5p coins
t=15 2p coins

Answer 7

2x + 5y = 100
x+y = 29

x = 29-y

2(29-y) + y = 100

58 - 2y + 5y = 100

3y = 42

y = 14
x = 15

14 5p pieces
15 2p pieces