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1 vote

$\text{Let the first number is x, the second number is y & the third number is z}$

$\text{According to the question}$

$x+y=45 ...(1)$

$y+z=55 ...(2)$

$z+3x=90...(3)$

from (1) & (2)

$x-z=-10...(4)$

from (3) & (4):

$x=20$

from equation (1):

$y=25$

from equation (3):

$z=30$

So option $(C)$ is correct.

$\text{According to the question}$

$x+y=45 ...(1)$

$y+z=55 ...(2)$

$z+3x=90...(3)$

from (1) & (2)

$x-z=-10...(4)$

from (3) & (4):

$x=20$

from equation (1):

$y=25$

from equation (3):

$z=30$

So option $(C)$ is correct.