Anklebiters Childcare bought t toys. The toys came in packs of 10. The next day,

Question

Anklebiters Childcare bought t toys. The toys came in packs of 10. The next day, at the start of the day there were n children at the centre. The children were evenly divided into groups of 4, and each group was given 6 toys. There were 2 toys left unused. Two toys were destroyed by the children’s brutal style of play and had to be thrown out. Later on, m more children were dropped at the centre. The children were then divided into groups of 6, with 7 toys per group. There were 6 (intact) toys not being used at this time. Anklebiters had purchased the minimum number of toys that is consistent with the above facts. How many children were at Anklebiters at the end of the day?

Explanation / Answer

Initial number of children = n

number of groups = n/4

6*(n/4) + 2 = t

3n + 4 = 2t

Since t is a multiple of 10, t = 10k

where k is an integer

3n + 4 = 20k

3n = 20k - 4

To satisfy above equation,

k = 2, 5

t = 20, 50

n = 12, 32

Total children = n + m

number of groups = (n + m)/6

Toys available = t - 2 (Since 2 toys were destroyed)

7*(n + m)/6 + 6 = t - 2

7n + 7m + 48 = 6t

Firstly, put t = 20, n=12

m becomes negative, which is not possible

Now, put t = 50, n=32

7m = 28

m = 4

Hence, n = 32, m = 4

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