The sum of a series of 5 consecutive odd numbers is 225 The second num

1.	The sum of a series of 5 consecutive odd numbers is 225 The second number of this series is 15 less than the second lowest number of another series of 5 consecutive ever numbers. What is 60% of the highest number of this series of consecutive even numbers -1). 36.02). 34.63). 38.44). 40.8
Answer» Solution Let the five consecutive odd numbers in increasing order = $(x-4) , (x-2) , (x) , (x+2) , (x+4)$ SUM of these numbers = $(x-4) + (x-2) + (x) + (x+2) + (x+4) = 225$ => $5x = 225$ => $x = \frac{225}{5} = 45$ Thus, the odd numbers are = 41 , 43 , 45 , 47 , 49 Let another series of even numbers in increasing order = $(y-4) , (y-2) , (y) , (y+2) , (y+4)$ Also, $43 = (y - 2) - 15$ => $y = 43 + 15 + 2 = 60$ Thus, HIGHEST number of the even series = 60 + 4 = 64 $\therefore$ 60% of 64 = $\frac{60}{100} \TIMES 64 = 38.4$

The sum of a series of 5 consecutive odd numbers is 225 The second number of this series is 15 less than the second lowest number of another series of 5 consecutive ever numbers. What is 60% of the highest number of this series of consecutive even numbers -1). 36.02). 34.63). 38.44). 40.8

Answer»

Solution

Let the five consecutive odd numbers in increasing order = $(x-4) , (x-2) , (x) , (x+2) , (x+4)$

SUM of these numbers = $(x-4) + (x-2) + (x) + (x+2) + (x+4) = 225$

=> $5x = 225$

=> $x = \frac{225}{5} = 45$

Thus, the odd numbers are = 41 , 43 , 45 , 47 , 49

Let another series of even numbers in increasing order = $(y-4) , (y-2) , (y) , (y+2) , (y+4)$

Also, $43 = (y - 2) - 15$

=> $y = 43 + 15 + 2 = 60$

Thus, HIGHEST number of the even series = 60 + 4 = 64

$\therefore$ 60% of 64 = $\frac{60}{100} \TIMES 64 = 38.4$