Question

The average of 7 consecutive numbers which are positive integers is 10 the average of lowest and highest such number

Answer 1

Given :

The average of 7 consecutive numbers which are positive integers = 10

To Find :

The average of lowest and highest such number

Solution :

Let The 7 consecutive numbers = x, (x + 1),(x + 2) ,(x + 3), (x + 4), (x + 5), (x +6)

According to question

The average of 7 consecutive numbers which are positive integers = 10

So,

$\dfrac{x + (x + 1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)}{7}$ = 10

i.e 7 x + 21 = 10 × 7

Or,  7 x + 21 = 70

Or,  7 x = 70 - 21

Or,  7 x = 49

∴        x = $\dfrac{49}{7}$

i.e       x = 7

So,

The 7 consecutive numbers = x, (x + 1),(x + 2) ,(x + 3), (x + 4), (x + 5), (x +6)

                                               = 7, (7 + 1),(7 + 2) ,(7 + 3), (7 + 4), (7 + 5), (7 +6)

                                               = 7 , 8 , 9 , 10 , 11 , 12 , 13

Now,

The lowest number = 7

The highest number = 13

So, The  average of lowest and highest number = $\dfrac{7+13}{2}$

                                                                                 = $\dfrac{20}{2}$ = 10

Hence,  The average of lowest and highest such number is 10 . Answer