by **HallsofIvy** » Tue Feb 11, 2020 10:15 am

Looks like a pretty standard algebra problem where the first thing you need to do is change from "sentences" to "equations"! The problem is to determine "How many football stickers were there altogether?" so let "N" be the total number of football stickers.

"Alec has seven more than one quarter of the stickers"

One quarter of the stickers is N/4. Seven more than that is N/4+ 7.

So Alec has N/4+ 7 stickers.

"Katie has three less than one sixth of the stickers"

"One sixth of the stickers" is N/6. 3 less than that is N/6- 3.

So Katie has N/6- 3 stickers.

"John has four fifths of the amount Alec has"

Alec has N/4+ 7 stickers so Alec has (4/5)(N/4+ 7)= N/5+ 28/5 stickers.

"Nick has nine more than John"

John has N/5+ 28/5 stickers so Nick has N/5+ 28/5+ 9= N/5+ 28/5+ 45/5= N/5+ 73/5 stickers.

Since these four people have all N stickers, the sum of each of these numbers must be N:

(N/4+ 7)+ (N/6- 3)+ (N/5+ 28/5)+ (N/5+ 73/5)= N

[N/4+ N/6+ N/5+ N/5]+ [7- 3+ 28/5+ 73/5]= N

To add the fractions, we need to get a common denominator:

4= 2*2 and 6= 2*3 so to get all the prime factors we need 2*2*3*5= 60.

N/4= 15N/60, N/6= 10N/60, and N/5= 12N/60

N/4+ N/6+ N/5+ N/5= (15N+ 10N+ 12N+ 12N)/60= 49N/60.

28/5+ 73/5= 101/5. 7- 3+ 28/5+ 73/5= 4+ 101/5= (20+ 101)/5= 121/5

49N/60+ 121/5= N

Subtract 49N/60 from both sides:

121/5= 11N/60 so 11N= (121)(12) and then N= 11(12)= 132.

There are a total of 132 football stickers.

Check:

"Alec has seven more than one quarter of the stickers"

1/4 of 132 is 33. Alec has 33+ 7= 40 stickers.

"Katie has three less than one sixth of the stickers"

1/6 of 132 is 22. Katie has 22- 3= 19 stickers.

"John has four fifths of the amount Alec has"

Alec has 40 stickers. 4/5 of 40 is 4(8)= 32. John has 32 stickers.

"Nick has nine more than John"

Nick has 32+ 9= 41 stickers.

Altogether there are 40+ 19+ 32+ 41= 132 stickers as needed!