by Ron Angelo A. Gelacio

(Paranaque, Metro Manila , Philippines)

Ricky, Carl and Jerome have a total of n candies. Ricky ate 1/4 of the candies. Carl ate 1/3 of the remaining candies. Jerome ate 1/5 of the remaining candies.

Then they grouped the remaining candies into 4 groups. If there were 10 candies in each group, how many candies did the three have in first?**Solution**

Ricky ate 1/4 of the candies

This means that Ricky ate 1/4 of n

1/4 of n = 1/4 × n = n/4

Carl ate 1/3 of the remaining candies

The remaining candies is n - n/4

n - n/4 = 4n/4 - n/4 = 3n/4

This means that Carl ate 1/3 of 3n/4

1/3 of 3n/4 = 1/3 × 3n/4 = 3n/12 = n/4

Jerome ate 1/5 of the remaining candies

The remaining candies is 3n/4 - n/4

3n/4 - n/4 = 2n/4 = n/2

This means that Jerome ate 1/5 of n/2

1/5 of n/2 = 1/5 × n/2 = n/10

Then they group n/10 into 4 groups.

This means that the remaining candies is divided by 4

n/10 divided by 4 = n/40

There are 10 candies in each group

n/40 = 10

Multiply both sides by 40

n = 10 × 40 = 400

The number of candies in each group is 400.

Since there are 4 groups, the total number of candies is 400 × 4 = 1600

Ricky ate 1/4 of 1600 or 1/4 × 1600 = 1600/4 = 400

1600 - 400 = 1200

Carl ate 1/3 of 1200 or 1/3 × 1200 = 1200/3 = 400

1200 - 400 = 800

Jerome ate 1/5 of 800 or 1/5 × 800 = 800/3 = 160

800 - 160 = 640

640 divided by 4 = 160

At the end each group had 160 candies