There are 4 consecutive odd numbers (x₁, x₂, x₃ and x₄) and three consecutive even numbers (y₁, y₂ and y₃). The average of the odd numbers is 6 less than the average of the even numbers. If the sum of the three even numbers is 16 less than the sum of the four odd numbers, what is the average of x₁, x₂, x₃ and x₄?

Solution(By Examveda Team)

According to given information
Average of odd numbers = Average of even numbers - 6
$$ \Rightarrow \frac{{{x_1} + {x_2} + {x_3} + {x_4}}}{4} = $$     $$\frac{{{y_1} + {y_2} + {y_3}}}{3} - 6$$
$$ \Rightarrow \frac{{{x_1} + {x_2} + {x_3} + {x_4}}}{4} = $$     $$\frac{{{y_1} + {y_2} + {y_3} - 18}}{3}$$
$$ \Rightarrow 3\left( {{x_1} + {x_2} + {x_3} + {x_4}} \right) = $$     $$4\left( {{y_1} + {y_2} + {y_3}} \right) - 72$$
Also,
$$ \Rightarrow {y_1} + {y_2} + {y_3} = $$     $${x_1} + {x_2} + {x_3} + {x_4} - 16$$
$$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} = $$     $${y_1} + {y_2} + {y_3} + 16$$   .....(i)
So we have,
$$ \Rightarrow 3\left( {{y_1} + {y_2} + {y_3} + 16} \right) = $$     $$4\left( {{y_1} + {y_2} + {y_3}} \right) - 72$$
$$ \Rightarrow 3{y_1} + 3{y_2} + 3{y_3} + 48 = $$     $$4{y_1} + 4{y_2} + 4{y_3} - 72$$
$$ \Rightarrow 4{y_1} + 4{y_2} + 4{y_3}$$   $$ - 3{y_1} - 3{y_2} - 3{y_3}$$     = 48 + 72
$$ \Rightarrow {y_1} + {y_2} + {y_3} = 120$$
$$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4}$$     = 120 + 16 = 136 [From (i)]
∴ Average of four odd numbers :
$$\eqalign{ & = \frac{{{x_1} + {x_2} + {x_3} + {x_4}}}{4} \cr & = \frac{{136}}{4} \cr & = 34 \cr} $$