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Previous Year Question (PYQs)
Each family in a locality has at most two adults, and no family has fewer than 3 children. Considering all the families together, there are more adults than boys, more boys than girls, and more girls than families. Then the minimum possible number of families in the locality is
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Let nf be number of families, ng be number of girls, nb be number of boys and na be number of adults. number of adults more than number of boys, number of boys more than number of girls and number of girls more than number of families. hence we have, na > nb > ng > nf ..........................(1) we check the above relation (1) with different values for number of families, starting with nf = 1 Let nf = 1 , then nb + ng = 3 , if ng =2 then nb = 1 . But number of boys is greater than number of girls.due to this contradiction, nf ≠ 1
Let nf = 2 , then nb + ng = 6 , if ng =3 then nb = 3 . But number of boys is greater than number of girls.due to this contradiction, nf ≠ 2 Let nf = 3 , then nb + ng = 9 , if ng =4 then nb = 5 . number of adults na = 6. hence the relation na > nb > ng > nf is satisfied for the above numbers, 6>5>4>3
hence minimum number of families is 3
Let nf = 2 , then nb + ng = 6 , if ng =3 then nb = 3 . But number of boys is greater than number of girls.
due to this contradiction, nf ≠ 2
Let nf = 3 , then nb + ng = 9 , if ng =4 then nb = 5 . number of adults na = 6.
hence the relation na > nb > ng > nf is satisfied for the above numbers, 6>5>4>3
hence minimum number of families is 3
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