Thread: Marc and His Coins

1. Marc and His Coins

Marc Bryan is putting P1 and P5 coins in his piggy bank. The total amount inside is P145. The number of P1 is 1 more than 3 times the number of P5 coins. How many P1 and P5 coins are there?

2. Hello, magentarita!

Marc Bryan is putting 1p and 5p coins in his piggy bank.
The total amount inside is 145p.
The number of 1p coins is 1 more than 3 times the number of 5p coins.
How many 1p and 5p coins are there?

Let $\displaystyle x$ = number of 1p coins.
Let $\displaystyle y$ = number of 5p coins.

We are told that: .$\displaystyle x \:=\:3y+1 \quad\Rightarrow\quad x - 3y \:=\:1$ .[1]

The total value is: .$\displaystyle 1\!\cdot\!x + 5\!\cdot\!y \:=\:145 \quad\Rightarrow\quad x + 5y \:=\:145$ .[2]

Subtract [1] from [2]: .$\displaystyle 8y \:=\:144 \quad\Rightarrow\quad y \:=\:18$

Substitute into [1]: .$\displaystyle x - 3(18) \:=\:1\quad\Rightarrow\quad x \:=\:55$

Answer: .$\displaystyle \begin{Bmatrix}55\,\text{ 1p coins} \\ 18\,\text{ 5p coins} \end{Bmatrix}$

3. thanks

Originally Posted by Soroban
Hello, magentarita!

Let $\displaystyle x$ = number of 1p coins.
Let $\displaystyle y$ = number of 5p coins.

We are told that: .$\displaystyle x \:=\:3y+1 \quad\Rightarrow\quad x - 3y \:=\:1$ .[1]

The total value is: .$\displaystyle 1\!\cdot\!x + 5\!\cdot\!y \:=\:145 \quad\Rightarrow\quad x + 5y \:=\:145$ .[2]

Subtract [1] from [2]: .$\displaystyle 8y \:=\:144 \quad\Rightarrow\quad y \:=\:18$

Substitute into [1]: .$\displaystyle x - 3(18) \:=\:1\quad\Rightarrow\quad x \:=\:55$

Answer: .$\displaystyle \begin{Bmatrix}55\,\text{ 1p coins} \\ 18\,\text{ 5p coins} \end{Bmatrix}$
I like the way you set up equations from the given data in word problems.