Question

I added 4 to one-third of a number and obtained an answer which was half of 44. What was the number?

Answer 1

Answer:

answer is 54

Step-by-step explanation:

let the required no be x

so, ATP= 4+1/3x = 22

1/3x= 22-4

1/3x= 18

x= 18÷ 1/3

x= 18× 3

x= 54. Ans.

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Answer 2

Let's break this question into 2 parts:

• Adding 4 to One-Third of a number
• Obtaining the number which is half of 44

Solution:

First solve the first part which is the LHS.

Let the unknown number be x.

So, the expression becomes:

$\tt{ \leadsto \: 4 + \dfrac{x}{3} }$

Now, simplify the second part which becomes RHS.

We know that half of 44 is 22.

Equating the above 2 expressions,

$\tt{4 + \dfrac{x}{3} = 22 }$

$\implies \tt{ \dfrac{x}{3} = 22 - 4}$

$\implies \tt{ \dfrac{x}{3} = 18 }$

$\implies \tt{x = 18 \times 3}$

$\therefore \boxed{ \bf{x = 54}}$

So, the number was 54

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebraic\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$