Question

$\Large\underline{\underline{\pmb{\sf{\bigstar \; Questiion \; ;}}}}$



Find three consecutive whole numbers who's sum is more than 45 but less than 54.





____

The answer ;; 15, 16, 17 OR 16, 17, 18
I wanna know how tf do I solve this one!? :abnormal-crying-noises:



– Well explained answer please! :^)~

Answer 1

Answer:

15, 16, 17 OR 16, 17, 18

Step-by-step explanation:

Consecutive numbers ;; Numbers which are one after another/following continuously.

Whole Numbers ;; Numbers that start with a 0.

• e.g. :: 0, 1, 2, 3, 4, 5, ....∞

Given ;; Three consecutive numbers whose sum is more than 45 but less than 54.

To Find ;; The three consecutive numbers.

Let the three consecutive whole numbers be x, x+1, x+2.

According to the given condition,

Three consecutive whole numbers whose sum is more than 45.

Thiss can be written as–

$\qquad :\implies \sf{x + (x+1) +(x+2) > 45}$

You might obviously know what '>' means.

Lol If not, lemme tell you, '>' shows that the number on the left side is greater than the number on the right side.

p.s. '<' shows the number on the right side is greater than the number on the left side. & '=' means its equal. ( idk why you alive if you dont know this :> )

Solving the equation,

$\implies \sf{x + x+1 + x+2 > 45} \\\\\implies \sf{ 3x + 3 > 45 } \\\\\implies \sf{3x > 45 - 3} \\\\\implies \sf{ 3x > 42 } \\\\\implies \sf{ x > \dfrac{\cancel{42}}{\cancel{3}} } \\\\\implies \boxed{\underline{\purple{\pmb{\sf{x = 14}}}}}$

   

∴ x = 15

• ∵ they said that the sum is greater than 45, therefore 15 > 14

   

Now, they said that the other three consecutive numbers whose sum is less than 54.

Lets do this one too xD

Thiis can be written as ;;

$\qquad :\implies \sf{x + (x+1) +(x+2) > 54}$

   

Lets solve this one too -.-

   

$\implies \sf{x + x+1 + x+2 < 54} \\\\\implies \sf{ 3x + 3 < 54 } \\\\\implies \sf{3x < 54 - 3} \\\\\implies \sf{ 3x < 51 } \\\\\implies \sf{ x < \dfrac{\cancel{51}}{\cancel{3}} } \\\\\implies \boxed{\underline{\purple{\pmb{\sf{x = 17}}}}}$

   

∴ x = 16

• ∵ they said that the sum is greater than 54, therefore 16 < 17

   

Phewwww Finally!

The three consecutive whole numbers are 15, 16, 17 OR 16, 17, 18

   

   

Hope it helps you Anuuuu! <3

Answer 2

Answer:

$\huge \dag \sf \red{question}$

Find three consecutive whole numbers who's sum is more than 45 but less than 54.

$\huge \sf \green{answer} \downarrow$

Let three consecutive numbers be x, (x+1)and (x+2)

According to the condition,

45<x+x+1+x+2<54

$\therefore \: 45 < 3x + 3 < 54 \\ \\ \therefore \: 45 - 3 < 3x + 3 - 3 < 54 - 3 \\ \\ 42 < 3x < 51 \\ \\ \frac{42}{3} < \frac{3x}{3} < \frac{51}{3} \\ \\ 14 < x < 17 \\ \\ \therefore \: x = 15 \: or \: x = 16$

If x=15 then x+1=15+1=16 and

x+2=15+2=17

If x=16 then x+1=16+1=17 and x+2=16+2=18

So,

required three numbers are 15,16,17 or 16,17,18

Hope it helps you from my side

:)