Question

Two numbers are in ratio 4 : 5. If the sum of the numbers is 90, find the numbers

Answer 1

Answer:

Let the numbers be 4x and 5x

4x + 5x = 90

9x = 90

x = 9

4x = 40

5x = 50

Therefore, the numbers are 40 and 50.

Step-by-step explanation:

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Answer 2

Answer:

Required two numbers are 40 and 50.

Step-by-step explanation:

Given,two numbers are in ratio 4 : 5.

Let the first number be 4x and second number be 5x.

It is also given,the sum of the numbers is 90.

According to question,

$4x + 5x = 90$

We are adding 4x and 3x,

$9x = 90 \\ x = \frac{90}{9}$

90 is dividing by 9,

$x = 10$

Therefore first number is

$(4 \times 10) = 40 \\ (5 \times 10) = 50$

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

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