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Question Number 167828 by otchereabdullai@gmail.com last updated on 26/Mar/22

 Houses on one side of a particular    street are assigned odd numbers,    starting from 11. If the sum of the    numbers is 551, how many houses   are there?

$$\:\mathrm{Houses}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particular}\: \\ $$$$\:\mathrm{street}\:\mathrm{are}\:\mathrm{assigned}\:\mathrm{odd}\:\mathrm{numbers},\: \\ $$$$\:\mathrm{starting}\:\mathrm{from}\:\mathrm{11}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{551},\:\mathrm{how}\:\mathrm{many}\:\mathrm{houses} \\ $$$$\:\mathrm{are}\:\mathrm{there}? \\ $$

Answered by mr W last updated on 27/Mar/22

say there are n houses.  the first house′s number =11  the last house′s number =11+2(n−1)  sum of numbers of all houses:  ((n(11+11+2(n−1)))/2)=551  n^2 +10n−551=0  (n−19)(n+29)=0  ⇒n=19  ⇒n=−29 (reject)

$${say}\:{there}\:{are}\:{n}\:{houses}. \\ $$$${the}\:{first}\:{house}'{s}\:{number}\:=\mathrm{11} \\ $$$${the}\:{last}\:{house}'{s}\:{number}\:=\mathrm{11}+\mathrm{2}\left({n}−\mathrm{1}\right) \\ $$$${sum}\:{of}\:{numbers}\:{of}\:{all}\:{houses}: \\ $$$$\frac{{n}\left(\mathrm{11}+\mathrm{11}+\mathrm{2}\left({n}−\mathrm{1}\right)\right)}{\mathrm{2}}=\mathrm{551} \\ $$$${n}^{\mathrm{2}} +\mathrm{10}{n}−\mathrm{551}=\mathrm{0} \\ $$$$\left({n}−\mathrm{19}\right)\left({n}+\mathrm{29}\right)=\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{19} \\ $$$$\Rightarrow{n}=−\mathrm{29}\:\left({reject}\right) \\ $$

Commented by otchereabdullai@gmail.com last updated on 27/Mar/22

supper prof W God bless you!

$$\mathrm{supper}\:\mathrm{prof}\:\mathrm{W}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}! \\ $$

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