Question

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.

Answer 1

Number of luxury gift tins of biscuits L sold is 18.

Number of mini- packs of 2 biscuits M sold is 80.

Number of normal packets N sold is 2.

Explanation:

Let L, M, N represent the number of Luxury, Mini, and Normal packets sold.

Total earned by selling luxury gift tins of biscuits = 5L

Total earned by selling normal packets = £1N

Total earned by selling 2 biscuits at 10p = 0.1M

Marianne tells Amy that she has received exactly 100 donations in total:

Equation becomes:

`L+N+M=100` .... (1)

total earning is £100, so another equation becomes:

`5L+N+0.1M=100` .......(2)

Given is that stock of £1 packets is very low compared with the other items:

so, N<L and N<M

Putting y = 0 in equation 1 and 2

`L+M=100` or
`L=100-M` .....(3)

`5L+0.1M=100` ......(4)

Replacing L by 100-M in the equation 4

`5(100-M)+0.1M=100`

`500-5M+0.1M=100`

4.9M=400

M=81.63

As, L=100-M

`L=100-81.63=18.37`

L=18.37

We can round L and M to 18 and 81 respectively and solve for N from 1 and 2 equations.

`L+M+N=100`

=>
`18+N+81=100`

=>
`N=1`

`5(18)+N+0.1(81)=100`

`90+N+8.1=100`

=> N=1.9

We will repeat the same process with N = 2 and get the following result.

Number of luxury gift tins of biscuits L sold is 18

Number of mini- packs of 2 biscuits M sold is 80

Number of normal packets N sold is 2

Answer 2

Answer:

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

Answer:
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