Question

Four children have small toys. The first child has 1/10 of the toys, the second child has 12 more toys than the first, the third child has one more toy of what the first child has and the fourth child has double the third child. How many toys are there?

Answer 1

Let the total toys be x.

The first has (1/10)x =  x/10


The third child has one more toys than the first =  x/10  + 1

The fourth has double of the third:  =  2* (x/10 + 1)  =  2x/10 + 2


Since we know that the second child has 12 more toys than the first

Therefore the Second  minus First = 12.


First let us find the second.

1st + 2nd + 3rd + 4th = x

x/10   +  2nd +  (x/10 + 1) + (2x/10 + 2) = x

2nd + x/10  + x/10 + 1 + 2x/10 + 2 = x

2nd + x/10 + x/10 + 2x/10 + 1 + 2  = x

2nd + 4x/10 + 3 = x

2nd = x - 4x/10 - 3

2nd =  6x/10  - 3

But recall

2nd - 1st = 12

6x/10 - 3 - x/10 = 12

6x/10 - x/10 = 12 + 3

5x/10 = 15

5x =  15 * 10

x  = 15*10 /5

x = 30

So there were 30 toys in all.