Unit 9

M1 (a) · 1 mark · Finds cost per gram for each tub by dividing cost by mass.

A1 (a) · 1 mark · Finds correct cost per gram for both tubs to 3 decimal places.

B1 (a) · 1 mark · States correctly which tub is better value for money.

M2 (b) · 2 marks · Forms correct equations for mass and total cost; substitutes correctly using tub chosen in (a). (follow-through)

M3 (b) · 1 mark · Sets up quadratic equation and attempts to solve using the quadratic formula or by factoring (where possible). (follow-through)

A2 (b) · 1 mark · Gives the mass of nachos correct to 2 decimal places. (follow-through)

A3 (b) · 1 mark · Finds correct cost of nachos, correct to 2 decimal places, using their own value for mass. (follow-through)

(a) Calculate cost per gram for each tub.

Small tub: $2.85 \div 150 = $0.019 (3 d.p.) per gram.

Large tub: $7.20 \div 420 = $0.017 (3 d.p.) per gram.

Large tub offers better value.

(b) Let the mass of nachos be m grams. Mass of large tub = 420 g, total weight is 420 g.

But the student bought 420 g in total, so if all is popcorn, no nachos. However, they buy 'some' nachos and one large tub, so suppose total mass bought = 420 g. Then:

But as per question intent, suppose instead, popcorn + nachos = 420 g. Let x = mass (g) of nachos, so mass (g) of popcorn = 420 - x. But only one large tub bought: popcorn bought = 420 g. So, x = 0.

But the total cost is $8.50 (from one large tub and some nachos), so write:

So perhaps, let the number of grams bought as nachos be m, and large tub (420 g) popcorn, but total mass = 420 g.

Therefore, only nachos is bought, or perhaps question wants to let small tub be possible as well. But since better value is large tub, only that is bought.

But let's proceed as per intended steps (as in Edexcel questions):

Let n = mass of nachos (g). Popcorn: one large tub = 420 g. Total mass: 420 + n = 420 g \implies n = 0 g. This does not make logical sense; perhaps question meant student bought some of the large tub.

Assume instead: Let the mass of nachos be n, mass of popcorn be p, n + p = 420. The price per gram of popcorn is $0.017 from (a).

But if the student buys both, then p (popcorn in grams) and n = mass of nachos; so p + n = 420, C = quadratic cost of nachos.

Use quadratic formula: .

Negative discriminant; try changing coefficients or checking calculation. Let's re-express algebraically.

Recalculate for the numbers above:

Quadratic formula: n = [1.417 +/- sqrt(1.417^2 - 4 x 0.02 x 33.64)]/(2 x 0.02)

So there must be a minor arithmetic or conceptual error in setting up the quadratic.

Try a plausible approach: suppose the cost function should be C = 0.02m^2 - 1.4m + 35, with total mass m + 420 = 420 (impossible unless m = 0). Try total mass = 420 + m = T, where T is not 420, but perhaps the student buys more than 420 g.

To match the question: Instead, set x = mass of nachos.

Alternatively, if the student buys x grams of nachos, and the large tub of popcorn, and the entire mass is 420 g of nachos (ignoring popcorn?), then total cost = C = $8.50.

But as written, solution would yield negative discriminant. Review. Perhaps if the numbers are: Large tub: 200 g for $3.44. Small tub: 80 g for $1.59. Try with different numbers for illustration. But with current numbers, quadratic appears unsolvable with the given costs.

Suppose instead student buys a small tub (150 g), so mass of nachos = 420 - 150 = 270 g. Cost of popcorn = $2.85. Cost of nachos = 0.02*(270)^2 - (1.4)*270 + 35 = 0.02*72900 - 378 + 35 = 14.58 - 378 + 35 = 145.8 - 378 + 35 = negative, not possible.

But intended method: Let x be mass of nachos, popcorn mass is 420 - x, using value per gram from (a):

As above, discriminant negative. The problem cannot be solved with the given numbers precisely as written. However, for question bank purposes, the marking and approach remain standard; logical following through of the methods is key.

Intended approach (with correct numbers): use the per gram value from (a), subtract the nachos mass from total, write system of equations as above and solve for mass and cost. With valid quadratic coefficients or alternative figures, the discriminant would be positive and give a real answer.