Question

Three number are in the ratio 2:3:4;
if the sum of their square is 2399, then the
members are ?​

Answer 1

Question:-

• Three number are in the ratio 2 : 3 : 4 ; if the sum of their square is 2394 , then the number are ?

To Find:-

• Find the numbers.

Solution:-

Let the numbers be 2x , 3x and 4x

Given ,

• Sum of squares is 1856

Now ,

We have to find the numbers:-

$\longrightarrow\sf \: { ( 2x ) }^{ 2 } + { ( 3x ) }^{ 2 } + { ( 4x ) }^{ 2 } = 1856$

$\longrightarrow\sf \: 4{x }^{ 2 } + 9{ x }^{ 2 } + 16{ x }^{ 2 } = 1856$

$\longrightarrow\sf \: 29{ x }^{ 2 } = 1856$

$\longrightarrow\sf \: { x }^{ 2 } = \cancel\dfrac { 1856 } { 29 }$

$\longrightarrow\sf \: x = \sqrt { 64 }$

$\longrightarrow\sf \: x = 8$

Hence ,

• First number is 2x = 16

• Second number is 3x = 24

• Third number is 4x = 32

Answer 2

Question:-

Three number are in the ratio 2 : 3 : 4 ; if the sum of their square is 2394 , then the number are ?

To Find:-

Find the numbers.

Solution:-

Let the numbers be 2x , 3x and 4x

Given ,

Sum of squares is 1856

Now ,

We have to find the numbers:-

$\longrightarrow\sf \: { ( 2x ) }^{ 2 } + { ( 3x ) }^{ 2 } + { ( 4x ) }^{ 2 } = 1856$

$\longrightarrow\sf \: 4{x }^{ 2 } + 9{ x }^{ 2 } + 16{ x }^{ 2 } = 1856$

$\longrightarrow\sf \: 29{ x }^{ 2 } = 1856$

$\longrightarrow\sf \: { x }^{ 2 } = \cancel\dfrac { 1856 } { 29 }$

$\longrightarrow\sf \: x = \sqrt { 64 }$

$\longrightarrow\sf \: x = 8$

Hence ,

First number is 2x = 16

Second number is 3x = 24

Third number is 4x = 32