A school's stationery order has \(\xi\) representing 90 items. The Venn diagram shows the number of folders, set \(F\), and labels, set \(L\), in terms of \...
Question 1 Report
A school's stationery order has \(\xi\) representing 90 items. The Venn diagram shows the number of folders, set \(F\), and labels, set \(L\), in terms of \(x\).
- Write an equation in \(x\) from the diagram and solve it. (2)
- Find \(n(F)\). (1)
- Find \(n(F \cup L)\). (1)
- Supplier 1 charges \(\pounds2\) for each item in \(F\). Supplier 2 charges \(\pounds1.50\) for each item in \(F \cup L\). Decide which supplier is cheaper for these items, and by how much. (2)
Answer Details
In a two-set Venn diagram, every region (both sets, each set only, and outside both) must add up to the total in the universal set, giving an equation whose solution unlocks every other value on the diagram.
(a) The four regions of the diagram, \(4x\) (folders only), \(x\) (both), \(2x\) (labels only), and \(3x\) (outside both sets), must sum to the total of 90 items:
\[4x+x+2x+3x=90\] \[10x=90\]giving \(x=9\). [2 marks]
(b) \(n(F)\) is everything inside the \(F\) circle, the "folders only" region plus the overlap:
\[n(F)=4x+x=5x=5(9)=45\] [1 mark](c) \(n(F \cup L)\) is everything inside either circle, which is the total minus the region outside both circles, \(3x\):
\[n(F \cup L)=90-3x=90-3(9)=90-27=63\] [1 mark](d) Supplier 1 charges for every item in \(F\): \(45\times\pounds2=\pounds90\). Supplier 2 charges for every item in \(F \cup L\): \(63\times\pounds1.50=\pounds94.50\). Since \(\pounds90 \lt \pounds94.50\), Supplier 1 is cheaper, by \(94.50-90=\pounds4.50\). [2 marks]
Solving for \(x\) from the "all regions sum to the total" equation is the standard first move for any algebraic Venn diagram; every later part then follows by substituting this single value of \(x\) into the relevant region or combination of regions.
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