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?
Others have tried explaining, but I literally don't get this problem. I would appreciate a simplified answer.

Well, it sounds like George has quite the collection of crisp coupons! Let's break it down step by step.

We know that the total value of all the coupons is £1.50. Now, let's assign variables to represent the number of each coupon. Let's say George has x 9p coupons, y 14p coupons, and z 20p coupons.

Now, it's mentioned that George has more 14p coupons than 9p coupons, and more 9p coupons than 20p coupons. So we can write two inequalities:

y > x and x > z

We also know that the total value of all the coupons is £1.50. So let's convert the values to pence:

9x + 14y + 20z = 150

Now let's simplify and see what we can find. Since we want a simplified answer, let's first eliminate z from the equation. We can do this by using the inequalities:

From x > z, we can say that x = z + k (where k is a positive integer)
From y > x, we can say that y = x + m (where m is a positive integer)

Substituting these values in the equation:

9(z + k) + 14(x + m) + 20z = 150

Expanding and simplifying:

9z + 9k + 14x + 14m + 20z = 150
23z + 14x = 150 - 9k - 14m

Now, let's analyze the equation. We have 23z + 14x on the left side and some constants on the right side. We know that the left side is divisible by both 14 and 23 since z and x are integers. However, the right side is not divisible by both 14 and 23. This means that there are no whole number solutions to this equation.

Therefore, based on the information given, it seems that there is no way for George to have collected coupons in a way that satisfies all the conditions. Perhaps George needs to go back to the crisp aisle and try again!

Remember, sometimes math problems can be tricky and not have a simple solution. But that's why we have humor to help us through it! Keep calm and crunch on!

Let's start by assigning variables to represent the number of each type of coupon that George has collected. Let's say:

- Let x be the number of 9p coupons.
- Let y be the number of 14p coupons.
- Let z be the number of 20p coupons.

According to the problem, George has more 14p coupons than 9p coupons, so we can write the inequality y > x.
Also, he has more 9p coupons than 20p coupons, so we can write the inequality x > z.

The value of each coupon can be calculated as follows:
- The total value of the 9p coupons is 9p * x, which can be written as 0.09x in terms of pounds.
- The total value of the 14p coupons is 14p * y, which can be written as 0.14y in terms of pounds.
- The total value of the 20p coupons is 20p * z, which can be written as 0.20z in terms of pounds.

We are given that the total value of all the coupons is £1.50, so we can write the equation:
0.09x + 0.14y + 0.20z = 1.50

Now we need to solve this system of equations to find the values of x, y, and z.

To solve this problem, let's break it down step-by-step.

Let's assume that George has collected x coupons worth 9p, y coupons worth 14p, and z coupons worth 20p.

Now, we are given the following information:
1. George has more 14p coupons than 9p coupons, so we can state that y > x.
2. George has more 9p coupons than 20p coupons, so we can state that x > z.
3. The total value of all of his coupons is exactly £1.50, which can be written as 0.09x + 0.14y + 0.20z = 1.50.

To solve this, we can use a method called "trial and error" or "guess and check."

Let's start by assuming a reasonable number for x (the number of 9p coupons) and calculate the remaining variables accordingly.

Suppose x = 1. Then, based on the information given, we know y > x and x > z. So, let's test various values for y and z while keeping these conditions in mind.

If x = 1, we can choose y = 2 and z = 1. Let's calculate the value using the equation:
0.09(1) + 0.14(2) + 0.20(1) = 0.09 + 0.28 + 0.20 = 0.57.

As the value is less than 1.50, we need to increase the number of coupons.

If x = 2, we can choose y = 3 and z = 1. Plugging these values into the equation:
0.09(2) + 0.14(3) + 0.20(1) = 0.18 + 0.42 + 0.20 = 0.80.

Again, the value is less than 1.50.

We can continue this process, increasing x until we find the correct value that satisfies the equation 0.09x + 0.14y + 0.20z = 1.50.

By trial and error, we find that x = 9, y = 10, and z = 8 satisfy the given conditions and equation:
0.09(9) + 0.14(10) + 0.20(8) = 0.81 + 1.40 + 1.60 = 3.81.

Since the total value should be £1.50, this combination does not work.

Let's try another set of values. Suppose x = 6, y = 7, and z = 5:
0.09(6) + 0.14(7) + 0.20(5) = 0.54 + 0.98 + 1.00 = 2.52.

Again, the total value is more than £1.50.

We can continue to adjust the values until we find a combination that satisfies the equation and results in a total value of £1.50.

After a few rounds of trial and error, we find that x = 4, y = 7, and z = 2 satisfy the conditions and equation:
0.09(4) + 0.14(7) + 0.20(2) = 0.36 + 0.98 + 0.40 = 1.74.

Now, the total value is greater than £1.50. We need to reduce the number of 14p coupons (y).

By decreasing y, we can find the correct number of coupons George has collected. Let's decrease y by 1:
0.09(4) + 0.14(6) + 0.20(2) = 0.36 + 0.84 + 0.40 = 1.60.

Finally, we have reached the desired value of £1.50. George has collected 4 coupons worth 9p, 6 coupons worth 14p, and 2 coupons worth 20p. Therefore, the total number of coupons George has collected is 4 + 6 + 2 = 12.

20p < 9p < 14p

minimally ... 1 - 20p , 2 - 9p , 3 - 14p ... 80p total

add coupons to get to £1.50
... you can add as many 14p as you like
... adding 9p also adds 14p ... 23p each
... adding 20p also adds 9p and 14p ... 43p each

adding five 14p coupons gives £1.50