Well here it goes!!

Question 1

Let x,y and z be the three boxes

Average of the three boxes = x+y+z/3=7 (or) x+y+z=21

we know that the median is 9 so lets assume y to be the median

so, x+9+y=21 which means x+y=12

Now lets plug in some answers and see which one satisfies all the conditions.

If x=4 then y=8, but then the median would be 8.

If x=3 then y=9 and so x,y and z would be 3,9 and 9 which has a median of 9 and a mean of 7. Thus the answer is 3.


Question 2

This sure is a tricky question. You should read the question properly to understand what has been asked. The question says that "for each additional day the "total" fine would either be doubled or increased by $0.30, whichever is less

According to the problem, the total fine would be calculated as follows..

1st day- $0.10

2nd day-Either $0.10+$0.30 or 2*$0.10 whichever is less so the fine would obviously be 2*$0.10=$0.20

3rd day-Either 0.20+0.30 or 2*0.20 so the fine would be 2*0.20=0.40

4th day-Either 0.40+0.30 or 2*0.40 and so the total fine amount would be 0.40+0.30=0.70

The trick is not to add the fine for all the days.

Hope my explaination is understandable.


OK..Movin on to Question 3

You can actually find this problem in OG 11

Notice that 14n/60 can be written as 7n/30

For 7n/30 to be an integer, n has to be a multiple of 30 and hence n is less than 200 the possible values for n is 30,60,90,120,150 and 180.

All these values for n have only three prime factors 2,3 and 5.

Hey nice explns moneyman, keep up the good work

one more approach for the last one

7n/30 is int & n < 200

since 30 =5*3*2 none of which divide 7 so n must have all these plus it can more prime factors

so now the next prime factor in the list is 7 but if n has 7 then 3*2*5*7 =210 not possible

so 3,2,5

Thanks Samir!! Your approach makes a lot of sense