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.
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
To solve this problem, let's denote the number of 9p coupons as x, the number of 14p coupons as y, and the number of 20p coupons as z.
We have three equations based on the given information:
1) y > x
This means that the number of 14p coupons is greater than the number of 9p coupons.
2) x > z
This means that the number of 9p coupons is greater than the number of 20p coupons.
3) 9x + 14y + 20z = 150
This equation represents the total value of all the coupons, which is £1.50.
Now, let's solve these equations step by step:
Step 1: Simplify equation 3:
Divide the equation by 2 to make the coefficients smaller:
4.5x + 7y + 10z = 75
Step 2: Rearrange equation 3:
Rearrange the equation to isolate one variable in terms of the others:
7y = 75 - 4.5x - 10z
7y = -4.5x - 10z + 75
y = (-4.5/7)x - (10/7)z + (75/7)
Step 3: Analyze possible values for x, y, and z:
By examining the equation obtained in step 2 (y = (-4.5/7)x - (10/7)z + (75/7)), we can see that for x and y to be integers, (-4.5/7)x - (10/7)z must be a fraction with a denominator of 7.
Step 4: Determine possible values for x and y:
By checking different values of x, we can find that x = 10 and y = 8 would satisfy the condition that y is greater than x.
Step 5: Calculate the value of z using equation 1:
Since x = 10 and y = 8, we can substitute these values into equation 1 to determine the value of z:
10 > z
We can conclude that z can be 9 or smaller.
Step 6: Test possible values of z:
Substitute the values of z into equation 3 and check if the total value of all coupons equals £1.50:
For z = 0:
4.5x + 7y + 10z = 4.5(10) + 7(8) + 10(0) = 45 + 56 + 0 = 101 (not equal to 150)
For z = 1:
4.5x + 7y + 10z = 4.5(10) + 7(8) + 10(1) = 45 + 56 + 10 = 111 (not equal to 150)
For z = 2:
4.5x + 7y + 10z = 4.5(10) + 7(8) + 10(2) = 45 + 56 + 20 = 121 (not equal to 150)
For z = 3:
4.5x + 7y + 10z = 4.5(10) + 7(8) + 10(3) = 45 + 56 + 30 = 131 (not equal to 150)
For z = 4:
4.5x + 7y + 10z = 4.5(10) + 7(8) + 10(4) = 45 + 56 + 40 = 141 (not equal to 150)
For z = 5:
4.5x + 7y + 10z = 4.5(10) + 7(8) + 10(5) = 45 + 56 + 50 = 151 (not equal to 150)
Step 7: Determine the correct values for x, y, and z:
After testing all possible values of z, we can conclude that z = 4 does not satisfy the equation. Therefore, the valid solution is for z = 3.
Hence, the number of 9p coupons (x) is 10, the number of 14p coupons (y) is 8, and the number of 20p coupons (z) is 3. Therefore, George has collected a total of 10 + 8 + 3 = 21 coupons.
To solve this problem, we can use a mathematical approach called "Diophantine Equations," which deals with finding integer solutions to equations. In this case, we'll solve the system of equations to find the number of coupons George has collected.
Let's define three variables: x represents the number of 9p coupons, y represents the number of 14p coupons, and z represents the number of 20p coupons.
We have three conditions given in the problem:
1) George has more 14p coupons than 9p coupons: y > x
2) George has more 9p coupons than 20p coupons: x > z
3) The total value of all his coupons is exactly £1.50: 9x + 14y + 20z = 150
Now, let's start solving the equations step by step:
1) George has more 14p coupons than 9p coupons: y > x
This condition gives us the following possibilities:
y = x + 1, y = x + 2, y = x + 3, ...
2) George has more 9p coupons than 20p coupons: x > z
Similarly, this condition gives us the following possibilities:
x = z + 1, x = z + 2, x = z + 3, ...
3) The total value of all his coupons is exactly £1.50: 9x + 14y + 20z = 150
From this equation, we can see that both x, y, and z must be multiples of 3, as their sum needs to be divisible by 3.
We can start by assuming a value for z (let's say z = 1) and calculate the corresponding values for x and y. If these values satisfy all the conditions, we have found a valid solution. If not, we can try different values for z and check if we get valid solutions.
1) z = 1
From x > z, we have: x > 1
From 9x + 14y + 20z = 150, we get:
9x + 14y + 20 = 150
9x + 14y = 130
The possible values that satisfy both conditions are:
x = 2, y = 7
x = 3, y = 9
x = 4, y = 11
...
We need to check if any of these values also satisfy y > x. In this case, x = 2 and y = 7 satisfy all conditions. Therefore, George has collected 2 coupons worth 9p, 7 coupons worth 14p, and 1 coupon worth 20p.
Hence, George has collected a total of 10 coupons.