Question

Two numbers A and B are in the ratio 3:5 and their LCM is 120. Find the values of A and B.​

Answer 1

Answer:

Let the two numbers be 3x and 4x and HCF be x

We know that

Product of two number=Product of their LCM and HCF

So,

4x×3x=180×x12x

2=180x

12x=180

x=15

3x=3×15=45

4x=4×15=60

Therefore the numbers are 45 and 60.

Answer 2

Answer:

Required value of A is 24 and value of B is 40

Step-by-step explanation:

Given,Two numbers A and B are in the ratio 3:5.

Let value of A be 3x and value of B be 5x.

We know,LCM means Lowest Common Multiples.

By prime factorisation,

$3x = 3 \times x \\ 5x = 5 \times x$

So, lowest common multiple of 3x and 5x is $(3 \times 5 \times x) = 15x$

So,LCM of 3x and 5x is 15x.

It is also given LCM of A and B is 120.

So according to question,

$15x = 120 \\ x = \frac{120}{15} \\ x = 8$

So, value of A is (3×8) = 24 and value of B is (5×8) = 40

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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