what must be substracted from each of the numbers 23, 40, 57 and 108. so that the remainders are in proportional ?
ritvikjain2090ow0ydl:
i think the question is wrong
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hello,
let the no. be subtracted from each no. be x
(23-x):(40-x)::(57-x):(108-x)
product of means=product of extremes
(23-x)(108-x)=(40-x)(57-x)
⇒23(108-x)-x(108-x)=40(57-x)-x(57-x)
⇒2484-23x-108x-x²=2280-40x-57x-x²
⇒2484-135x=2280-97x
⇒2484-2280=97x+135x
⇒204=38x
⇒x=5.36...... ∴5.36.........should be subtracted so that the remainders will in proportion
hope this helps,if u like it please mark it as brainliest
let the no. be subtracted from each no. be x
(23-x):(40-x)::(57-x):(108-x)
product of means=product of extremes
(23-x)(108-x)=(40-x)(57-x)
⇒23(108-x)-x(108-x)=40(57-x)-x(57-x)
⇒2484-23x-108x-x²=2280-40x-57x-x²
⇒2484-135x=2280-97x
⇒2484-2280=97x+135x
⇒204=38x
⇒x=5.36...... ∴5.36.........should be subtracted so that the remainders will in proportion
hope this helps,if u like it please mark it as brainliest
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