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Question Number 114863 by mathdave last updated on 21/Sep/20

find four consecutive multiples of 5  such that twice the sum of the two  greatest integer exceed five times the  least by 5

$${find}\:{four}\:{consecutive}\:{multiples}\:{of}\:\mathrm{5} \\ $$$${such}\:{that}\:{twice}\:{the}\:{sum}\:{of}\:{the}\:{two} \\ $$$${greatest}\:{integer}\:{exceed}\:{five}\:{times}\:{the} \\ $$$${least}\:{by}\:\mathrm{5} \\ $$

Answered by PRITHWISH SEN 2 last updated on 21/Sep/20

let the integrs are           (5m−10),(5m−5),5m,(5m+5)  by the problem     2{5m+(5m+5)}=5(5m−10)+5           ⇒m= 11       ∴ the numbers are                             45,50,55,60

$$\mathrm{let}\:\mathrm{the}\:\mathrm{integrs}\:\mathrm{are} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{5}\boldsymbol{\mathrm{m}}−\mathrm{10}\right),\left(\mathrm{5}\boldsymbol{\mathrm{m}}−\mathrm{5}\right),\mathrm{5}\boldsymbol{\mathrm{m}},\left(\mathrm{5}\boldsymbol{\mathrm{m}}+\mathrm{5}\right) \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{problem} \\ $$$$\:\:\:\mathrm{2}\left\{\mathrm{5m}+\left(\mathrm{5m}+\mathrm{5}\right)\right\}=\mathrm{5}\left(\mathrm{5m}−\mathrm{10}\right)+\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\boldsymbol{\mathrm{m}}=\:\mathrm{11} \\ $$$$\:\:\:\:\:\therefore\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{numbers}}\:\boldsymbol{\mathrm{are}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{45},\mathrm{50},\mathrm{55},\mathrm{60} \\ $$

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