Marianne has been collecting donations for her biscuit stall at the school summer fayre. there are some luxury gift tins of biscuits to be sold at £5 each, normal packets at £1 each, and mini- packs of 2 biscuits at 10p each. she tells amy that she has received exactly 100 donations in total, with a collective value of £100, and that her stock of £1 packets is very low compared with the other items. amy wants to work out how many of each item marianne has. show how she can do it.

Number of luxury gift tins of biscuits sold at £5 each: 18
Number of normal packets sold at £1 each: 2
Number of mini- packs of 2 biscuits sold at 10p each: 80

Solution:

Number of luxury gift tins of biscuits sold at £5 each: x
Amount received by the number of luxury gift tins of biscuits sold: £5 x Number of normal packets sold at £1 each: y
Amount received by the number of normal packets sold: £1 y
Number of mini- packs of 2 biscuits sold at 10p each: z
Amount received by the number of 2 biscuists sold: 10p z =£0.1z

She tells Amy that she has received exactly 100 donations in total:
(1) x+y+z=100

A collective value of £100:
Total amount receibed: £5 x + £1 y+ £0.1z= £100 → £ (5x+1y+0.1z)= £100 → 5x+y+0.1z=100 (2)

Her stock of £1 packets is very low compared with the other items:
y<<x
y<<z
Then we can despice the value of "y" and solve for "x" and "z":
y=0:
(1) x+y+z=100→x+0+z=100→x+z=100 (3)
(2) 5x+y+0.1z=100→5x+0+0.1z=100→5x+0.1z=100 (4)

Solving the system of equations (3) and (4) using the substitution method:
Isolating z in equation (3): Subtrating x from both sides of the equation:
(3) x+z-z=100-x→z=100-x

Replacing z by 100-x in the equation (4):
(4) 5x+0.1z=100→5x+0.1(100-x)=100
Eliminating the parentheses:
5x+0.1(100)-0.1x=100→5x+10-0.1x=100
Adding similar terms on the left side of the equation:
4.9x+10=100
Solving for x: Subtrating 10 from both sides of the equation:
4.9x+10-10=100-10→4.9x=90
Dividing both sides of the equation by 4.9:
4.9x/4.9=90/4.9→x=18.36734693

Replacing x=18.36734693 in the equation:
z=100-x→z=100-18.36734693→z=81.63265306

We can round x and z: x=18 and z=81 and solve for y in equations (1) and (2):
(1) x+y+z=100→18+y+81=100
Adding similar terms on the left side of the equation:
99+y=100
Subtracting 99 from both sides of the equation:
99+y-99=100-99→y=1

(2) 5x+y+0.1z=100→5(18)+y+0.1(81)=100→90+y+8.1=100→98.1+y=100→98.1+y-98.1=100-98.1→y=1.9→y=2 different to 1

Suppose y=1. Replacing y=2 in equations (1) and (2):
(1) x+y+z=100→x+1+z=100→x+1+z-1=100-1→x+z=99
(2) 5x+y+0.1z=100→5x+1+0.1z=100→5x+1+0.1z-1=100-1→5x+0.1z=99

Repeating the process:
z=99-x
5x+0.1z=99→5x+0.1(99-x)=99→5x+9.9-0.1x=99→4.9x+9.9=99→4.9x+9.9-9.9=99-9.9→4.9x=89.1→4.9x/4.9=89.1/4.9→x=18.18367346
z=99-18.18367346→z=80.81632653
x=18 and z=80
(1) x+y+z=100→18+y+80=100→y+98=100→y+98-98=100-98→y=2
(2) 5x+y+0.1z=100→5(18)+y+0.1(80)=100→90+y+8=100→y+98=100→y+98-98=100-98→y=2


Number of luxury gift tins of biscuits sold at £5 each: 18
Number of normal packets sold at £1 each: 2
Number of mini- packs of 2 biscuits sold at 10p each: 80