At a regional agricultural exhibition, 150 farmers registered their crop cultivation. 85 f… — JAMB UTME

At a regional agricultural exhibition, $150$ farmers registered their crop cultivation. $85$ farmers grow cassava ( $C$ ), $70$ grow maize ( $M$ ), and $60$ grow yam ( $Y$ ). $35$ farmers grow both cassava and maize, $25$ grow both maize and yam, and $30$ grow both cassava and yam. If $15$ farmers grow all three crops, how many farmers grow none of these three crops?

Answer: 10 farmers

Answer

The number of farmers who grow none of the three crops is 10.

Using the Principle of Inclusion-Exclusion for three sets, the number of farmers growing at least one of cassava, maize, or yam is given by

85 + 70 + 60 - 35 - 25 - 30 + 15 = 140

. Since there are

150

farmers in total, the number of farmers who grow none of these three crops is

150 - 140 = 10

Step-by-Step Solution

Calculate the cardinality of the union of the three sets using the Principle of Inclusion-Exclusion.

|C \cup M \cup Y| = 85 + 70 + 60 - 35 - 25 - 30 + 15 = 140

Summing the three individual set counts overcounts elements in pairwise intersections, which must be subtracted. The central triple intersection is then added back because it was subtracted once too often.

Subtract the size of the union from the size of the universal set.

|(C \cup M \cup Y)'| = 150 - 140 = 10

The complement of the union represents the set of farmers outside all three crop categories.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

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