    # What least number must be subtracted from each of 13, 16, 17, and 21 to make them in proportion?

 SSC CGL 2021 Skill Test Dates Announced! The Skill Test will be taking place on 4th and 5th January 2023. The SSC CGL Application Status for all Regions and SSC CGL Admit Card for NER, ER, WR, NWR, CR & MPR Regions is active. Candidates can log in to the regional website of SSC and check their application status. SSC CGL 2022 Tier I Prelims Exam will be conducted from 1st to 13th December 2022. The SSC CGL 2022 Notification was out on 17th September 2022. The SSC CGL Eligibility is a bachelor’s degree in the concerned discipline. This year, SSC has completely changed the exam pattern and for the same, the candidates must refer to SSC CGL New Exam Pattern.  Prev Question 7 Ratio and Proportion Exercise 7.2 Next Answer: Consider x be subtracted from each term 23 – x, 30 – x, 57 – x and 78 – x are proportional It can be written as 23 – x: 30 – x :: 57 – x: 78 – x (23 – x)/ (30 – x) = (57 – x)/ (78 – x) By cross multiplication (23 – x) (78 – x) = (30 – x) (57 – x) By further calculation \begin{array}{l} 1794-23 x-78 x+x^{2}=1710-30 x-57 x+x^{2} \\ x^{2}-101 x+1794-x^{2}+87 x-1710=0 \end{array} So we get -14x + 84 = 0 14x = 84 x = 84/14 = 6 Therefore, 6 is the number to be subtracted from each of the numbers. Was This helpful?  Answer VerifiedHint: To answer this type of problem suppose the least number is x then subtract x from each of the given numbers. At last apply the given conditions that the remainders are in continued proportion then calculate the value of by solving the algebraic equation.Complete step-by-step answer:The given numbers are 7, 17, and 47We have to calculate the least number when subtracted from each of the numbers such that the remainders are in continued proportion.Let x be the number that needs to be subtracted from these numbers to make it proportional.The new numbers will be $7 - x,17 - x,{\text{ }}47 - x$ Since these numbers should be proportional,So their ratios will be in proportion$\dfrac{{17 - x}}{{7 - x}} = \dfrac{{47 - x}}{{17 - x}}$ ​By simplification we get,$\Rightarrow {\left( {17 - x} \right)\left( {17 - x} \right) = \left( {47 - x} \right)\left( {7 - x} \right)} \\ \Rightarrow {{x^2} - 34x + 289 = 329 + {x^2} - 54x} \\ \Rightarrow {20x = 40 \Rightarrow x = 2}$ Now put the value of x in $7 - x,17 - x,{\text{ }}47 - x$ we get the numbers asThe new numbers will be 5,15,45Therefore ′2 ′ is the least number to be subtracted so that the numbers will be proportional.Note: If there are two values of x then we will choose that value of x which satisfy the condition of proportionality as the question says and discard the other one. 