An agricultural research station evaluated 300 soil plots for the presence of three specif… — GRE General Test

An agricultural research station evaluated $300$ soil plots for the presence of three specific mineral deficiencies: Nitrogen ( $N$ ), Phosphorus ( $P$ ), and Potassium ( $K$ ). The survey revealed the following data:
- Exactly $60$ plots exhibited none of the three deficiencies.
- $140$ plots exhibited Nitrogen deficiency ( $N$ ).
- $130$ plots exhibited Phosphorus deficiency ( $P$ ).
- $120$ plots exhibited Potassium deficiency ( $K$ ).
- Exactly $70$ plots exhibited exactly two of the three deficiencies.

Which of the following statements MUST be true? Select all such statements.

Exactly $40$ plots exhibited all three mineral deficiencies.Cevap
Exactly $130$ plots exhibited exactly one mineral deficiency.Cevap
The number of plots exhibiting at least two mineral deficiencies is $110$ .Cevap
D
Exactly $170$ plots exhibited at most two mineral deficiencies.
E
The number of plots with Nitrogen deficiency only is $70$ .

Cevap

The statements confirming that exactly 40 plots exhibited all three deficiencies, exactly 130 plots exhibited exactly one deficiency, and 110 plots exhibited at least two deficiencies are all correct.

The system of set equations shows that

n_3 = 40

plots have all three deficiencies,

n_1 = 130

plots have exactly one deficiency, and

n_2 + n_3 = 70 + 40 = 110

plots have at least two deficiencies. Therefore, the statements asserting

40

plots for all three deficiencies,

130

plots for exactly one deficiency, and

110

plots for at least two deficiencies are all guaranteed to be true.

Adım Adım Çözüm

Determine the total number of plots exhibiting at least one deficiency.

The total number of plots with at least one deficiency is

300 - 60 = 240

Plots with no deficiencies are excluded from the set union

|N \cup P \cup K|

Set up the inclusion-exclusion equations for set membership.

Let

n_1

be the number of plots with exactly one deficiency,

n_2 = 70

be the number of plots with exactly two deficiencies, and

n_3

be the number of plots with all three deficiencies.
Equation 1 (Total elements in union):

n_1 + n_2 + n_3 = 240 \implies n_1 + 70 + n_3 = 240 \implies n_1 + n_3 = 170

.
Equation 2 (Sum of individual set cardinalities):

|N| + |P| + |K| = n_1 + 2n_2 + 3n_3 \implies 140 + 130 + 120 = n_1 + 2(70) + 3n_3 \implies 390 = n_1 + 140 + 3n_3 \implies n_1 + 3n_3 = 250

Each element in an individual set sum is counted once for single-set membership, twice for double-set membership, and three times for triple-set membership.

Solve the system of linear equations for

n_1

and

n_3

Subtract Equation 1 from Equation 2:

(n_1 + 3n_3) - (n_1 + n_3) = 250 - 170 \implies 2n_3 = 80 \implies n_3 = 40

.
Substitute

n_3 = 40

into Equation 1:

n_1 + 40 = 170 \implies n_1 = 130

This yields

40

plots with all three deficiencies and

130

plots with exactly one deficiency.

Evaluate each option statement against the computed set cardinalities.

1) All three deficiencies (

n_3

) =

40

(True).
2) Exactly one deficiency (

n_1

) =

130

(True).
3) At least two deficiencies (

n_2 + n_3

) =

70 + 40 = 110

(True).
4) At most two deficiencies = Total -

n_3 = 300 - 40 = 260 \neq 170

(False).
5) Nitrogen deficiency only depends on how the

70

dual-deficiency plots are split between

N \cap P

N \cap K

, and

P \cap K

, which is not uniquely determined by the given data (False).

Only statements meeting the exact numerical bounds and logical guarantees must be true.

Anahtar Kavram

Three-Set Principle of Inclusion-Exclusion and Partitioning into Disjoint Regions

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