9x+14y+20z = 150
That gives us the conditions
x = 2m
y = 3m+10n+5
z = 4-3m-7n
Clearly, m>0, which means that n must be negative
Since 3m+10n+5 must be positive,
trying n = -1, we have
3m-5 > 0
so m>2
So let's try m=3
x = 2m = 6
y = 3m + 10n + 5 = 10n+14 so n = -1 and y=4
z = 4-3m-7n = 4-9+7 = 2
That's not nearly enough coupons.
So now try increasing m until x+y+z = 150
Recently, packets of Amblers crisps have had ‘money-off’ coupons inside them. Some coupons
are worth 9p, some are worth 14p and some are worth 20p.
George has been collecting these coupons. He has more 14p coupons than 9p coupons, and
more 9p coupons than 20p coupons. The total value of all of his coupons is exactly £1.50.
How many coupons has George collected?
I have no idea how to solve this. I get that it's 9x+14y+20z = 150, but not the rest.
1 answer
1 answer