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Question Number 78979 by jagoll last updated on 22/Jan/20

the first n even  numbers are known. if one number  is deleted then the average  of remaining numbers is ((2582)/(50)).  find the value of deleted number.

$$\mathrm{the}\:\mathrm{first}\:\mathrm{n}\:\mathrm{even} \\ $$$$\mathrm{numbers}\:\mathrm{are}\:\mathrm{known}.\:\mathrm{if}\:\mathrm{one}\:\mathrm{number} \\ $$$$\mathrm{is}\:\mathrm{deleted}\:\mathrm{then}\:\mathrm{the}\:\mathrm{average} \\ $$$$\mathrm{of}\:\mathrm{remaining}\:\mathrm{numbers}\:\mathrm{is}\:\frac{\mathrm{2582}}{\mathrm{50}}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{deleted}\:\mathrm{number}. \\ $$

Commented by mr W last updated on 22/Jan/20

the 51 numbers are 2,4,6,...102  the deletete number is 70

$${the}\:\mathrm{51}\:{numbers}\:{are}\:\mathrm{2},\mathrm{4},\mathrm{6},...\mathrm{102} \\ $$$${the}\:{deletete}\:{number}\:{is}\:\mathrm{70} \\ $$

Commented by jagoll last updated on 22/Jan/20

sir please explain me

$$\mathrm{sir}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{me} \\ $$

Commented by mr W last updated on 22/Jan/20

n even numbers: 2(1+2+...+n)=n(n+1)  deleted number: 2m with m≤n  ((n(n+1)−2m)/(n−1))=((2582)/(50))=((1291k)/(25k))  n(n+1)−((1291)/(25))(n−1)=2m≤2n  n^2 −((1316)/(25))n+((1291)/(25))≤0  ⇒1≤n≤51  ⇒1≤k≤2  k=1: ⇒n=26⇒no integer m  k=2: ⇒n=51⇒m=35 ⇒solution

$${n}\:{even}\:{numbers}:\:\mathrm{2}\left(\mathrm{1}+\mathrm{2}+...+{n}\right)={n}\left({n}+\mathrm{1}\right) \\ $$$${deleted}\:{number}:\:\mathrm{2}{m}\:{with}\:{m}\leqslant{n} \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)−\mathrm{2}{m}}{{n}−\mathrm{1}}=\frac{\mathrm{2582}}{\mathrm{50}}=\frac{\mathrm{1291}{k}}{\mathrm{25}{k}} \\ $$$${n}\left({n}+\mathrm{1}\right)−\frac{\mathrm{1291}}{\mathrm{25}}\left({n}−\mathrm{1}\right)=\mathrm{2}{m}\leqslant\mathrm{2}{n} \\ $$$${n}^{\mathrm{2}} −\frac{\mathrm{1316}}{\mathrm{25}}{n}+\frac{\mathrm{1291}}{\mathrm{25}}\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}\leqslant{n}\leqslant\mathrm{51} \\ $$$$\Rightarrow\mathrm{1}\leqslant{k}\leqslant\mathrm{2} \\ $$$${k}=\mathrm{1}:\:\Rightarrow{n}=\mathrm{26}\Rightarrow{no}\:{integer}\:{m} \\ $$$${k}=\mathrm{2}:\:\Rightarrow{n}=\mathrm{51}\Rightarrow{m}=\mathrm{35}\:\Rightarrow{solution} \\ $$

Commented by jagoll last updated on 22/Jan/20

thanks you lot sir

$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{lot}\:\mathrm{sir} \\ $$

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