Question

Question:-


1) Find a number whose double is 45 greater than its half.


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Answer 1

$\: \: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Let's \: understand \: the \: concept \: 1^{st}}}}}}}$

➛ We have to find a number whose double' is 45' greater than its half'. In short there is a number ( To find ) whose double is greater ( 45 ) than it's half.

$\: \: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Full \: solution}}}}}}$

~ Let's assume the original number as a

~ Now let's work according to the question,

➨ 2a = 45 + a/2

➨ 2a = ( 45 × 2 + a ) / 2

[ Multiplying both sides by 2 ]

➨ 2 × 2a = 90 + a

➨ 4a = 90 + a

[ Subtracting a from both sides ]

➨ 4a - a = 90

➨ 4a - 1a = 90

➨ 3a = 90

➨ a = 90/3

➨ a = 30

$\: \: \: \: \: \: \: \: \:{\bold{\sf{No \: need \: to \: be \: confused !}}}$

$\: \: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Let's \: verify \: it !}}}}}}$

➨ 2a = 45 + a/2

➨ 2(30) = 45 + 30/2

➨ 60 = 45 + 15

➨ 60 = 60

➨ LHS = RHS

➨ Henceforth, 30 = 30 and 30 is the original number.

$\: \: \: \: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Knowledge \: drink}}}}}}$

➛ This question is from mathematics very nice and intersecting topic named, Linear Equation. Let's learn something about linear equation !

❶ Linear equation = Here we will have to deal with linear expressions in just one variable. Such equations are known to be “linear equation in one variable”

❷ An algebraic equation in an equality involving variable. It has an equality sign. The expression on the left of equality sign is LHS. The expression on the right of equality sign is RHS like in expression 2x - 3 = 7

↦ 2x is variable

↦ = is the sign of equality

↦ 7 is equation

↦ 2x - 3 is LHS

↦ 7 is RHS

❸ In an equation the values, of the expression on LHS and RHS are equal. This happen to be true ! for certain values of that variable. The values are the solution of that equation like,

↦ x = 5 is the solution of the equation

↦ 2x - 3 = 7 [ x = 5 ]

↦ LHS = 2 × 5 - 3 = 7 = RHS

↦ On the other hand, x = 10 is nor a solution of the equation [ x = 10 ]

↦ LHS = 2 × 10 - 3 = 17

↦ This isn't equal to the RHS.

❹ How to find solution for equation?

We have to assume that the 2 sides of the equation are in a balance. We have to perform the same mathematical operation on both sides of the equation so that the balance isn't disturbed. A few such steps give you your solution always...!

Answer 2

Solution :

1) Find a number whose double is 45 greater than its half.

Let the number be "y"

Now according to question :

$: \implies\sf 2y = \dfrac{y}{2} + 45 \\\\ \\ : \implies\sf 2y = \dfrac{y + 90}{2} \\\\ \\ :\implies\sf 2y \times 2 = y + 90 \\\\ \\ : \implies\sf 4y = y + 90 \\\\ \\ : \implies\sf 4y - y = 90 \\\\ \\ : \implies\sf 3y = 90 \\\\ \\ : \implies\sf y = \dfrac{90}{3} \\\\ \\ : \implies{\underline{\boxed{\bold{y = 30}}}} \ \star$

$\therefore\underline{\textsf{Required number = {\textbf{30}}}}$

$\rule{200}{1}$

Verification :

$\longmapsto\sf 2 \times 30 = \dfrac{30}{2} + 45 \\\\ \\ \longmapsto\sf 60 = 15 + 45 \\\\ \\ \longmapsto\sf 60 = 60 \\\\ \\ \longmapsto\bold{LHS = RHS}$

Therefore, required number = 30 .