Question
Three numbers βXβ, βYβ and βZβ are such
that βXβ is one-fourth of βYβ and βZβ is equal to the average of βXβ and βYβ. If the sum of these three numbers is 750, then find the second smallest number.Solution
Let the value of βYβ be β4xβ So, value of βXβ = 4x/4 = βxβ So, value of βZβ = (4x + x)/2 = 5x/2 = β2.5xβ ATQ; x + 4x + 2.5x = 750 Or, 7.5x = 750 Or, x = 100 So, the second smallest number = βZβ = 2.5x = 2.5 Γ 100 = 250
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