Question

Ahola Brainliacs ❤❤✌

Solev the following Maths question :-

★ Find three consecutive whole numbers whose sum is more than 45 but less than 54.

______________________

Spammers be away ✔️

Answer 1

$\huge\bold{heyy\: SISTA}$

$<b><u>here's thy solution$

==>>let those no's be,,, x-1, x , x+1==(1)

the given condition is,,,,

there sum,, i.e,

x - 1 + x + x +1 = 3x..

45< 3x,,,

SIMILARLY ,,,

3x<54..

according to the equation..

45<3x.

x > 15,,

x = 15..

then the consecutive no's be,, put x value in eq 1.

14 , 15 ,16..

if the condition is ,,,

54> 3x

x< 18,,

then consecutive members are,,,,

17, 16 ,15...

$\huge{Bye}$

$<b><u>RGRDS@STYLG...$

Answer 2

Answer:

The numbers are 15, 16 and 17 or 16, 17 and 18.

Step-by-step explanation:

Let the whole numbers be x - 1, x and x + 1

We have been given that the sum of the numbers is more than 45 but less than 54.

According to question:-

( x - 1 ) + (x) + ( x + 1 ) > 45

3x > 45

x > $\frac{45}{3}$

x > 15                                       .... (1)

( x - 1 ) + (x) + ( x + 1 ) > 45

3x > 54

x < $\frac{54}{3}$        

x < 18                                         ...(2)

From (1) and (2) we can say that x is either 16 or 17 :-

If x is 16

Then the numbers would be 15, 16 and 17

Their sum is 48, which satisfies the conditions.

If x is 17 :-

Then the numbers would be 16, 17 and 18

Their sum is 51, which satisfies the conditions too !

$\large\boxed{\large\boxed{\large\boxed{Solved !}}}}$