# Question

There were some 50-cent and 20-cent coins.There were 80 coins altogether.When 50% of the 50-cent coins and 25% of the 20-cent coins were taken away,there were 46 coins left.How many 20-cent coins were there at first?

Source: Rivervale Primary

This is actually an algebra question, which wants you to formulate two equations and solve them.

Let 2 x be the number of 50-cent coins initially . Let 4 y be the number of 20-cent coins initially.

(I use 2 x and 4 y because in the later part, it is easy to remove 50% and 25% respectively)

Initially, there were 80 coins altogether. Hence,

2 x + 4 y = 80 — (1)

When 50% of the 50-cent coins and 25% of the 20-cent coins were taken away, we are left we 46 coins. Hence,

x + 3 y = 46 — (2)

Now that we have 2 algebraic equations, we can solve for x and y.

Elimination Method

(Looking at both equations, I need to manipulate one or both equations to eliminate one term. This is done by taking the LCM of one of the terms. In this case, it is better to eliminate the x term, since I can just multiply equation 2 by 2, instead of having to manipulate both equations.)

(2) x 2:

2 x + 6 y = 92 — (3) (Remember to multiply both sides)

(3) – (1):

(2 x + 6 y) – (2 x + 4 y) = 92-80

2 y = 12

(Initial number of 20-cent coins is 4 y)

4 y = 12 x 2

= 24//

Substitution Method

(Since equation 2 has a convenient x term by itself, I can choose to rewrite the equation and substitute the x term inside the first equation)

Rewrite equation 2 as:

x = 46 – 3 y

Substitute x = 46 – 3 y into (1):

2 (46 – 3 y) + 4 y = 80

92 – 6 y + 4 y = 80 (Remember to expand the brackets properly)

2 y = 12

4 y = 12 x 2

= 24//

To double check:

Sub 4 y = 24 into (1), 2 x = 56

56/2 = 28

24/4 x 3 = 18

28 + 18 = 46

(If you sub in the x and y terms for any equation, it should be correct)