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Question Number 216284 by issac last updated on 02/Feb/25

if i have 7200 coin  and Each A,B,C are 500 coin  at this time how many Should  i buy each so that i can buy as many  possible???

$$\mathrm{if}\:\mathrm{i}\:\mathrm{have}\:\mathrm{7200}\:\mathrm{coin} \\ $$$$\mathrm{and}\:\mathrm{Each}\:\boldsymbol{\mathrm{A}},\boldsymbol{\mathrm{B}},\boldsymbol{\mathrm{C}}\:\mathrm{are}\:\mathrm{500}\:\mathrm{coin} \\ $$$$\mathrm{at}\:\mathrm{this}\:\mathrm{time}\:\mathrm{how}\:\mathrm{many}\:\mathrm{Should} \\ $$$$\mathrm{i}\:\mathrm{buy}\:\mathrm{each}\:\mathrm{so}\:\mathrm{that}\:\mathrm{i}\:\mathrm{can}\:\mathrm{buy}\:\mathrm{as}\:\mathrm{many} \\ $$$$\mathrm{possible}??? \\ $$

Commented by AntonCWX last updated on 03/Feb/25

Buy as many what as possible?  A B or C

$${Buy}\:{as}\:{many}\:{what}\:{as}\:{possible}? \\ $$$${A}\:{B}\:{or}\:{C} \\ $$

Answered by AntonCWX last updated on 03/Feb/25

((7200)/(500))=14.4  ((14)/3)=4(2/3)  (5,5,4) or (5,4,5) or (4,5,5)

$$\frac{\mathrm{7200}}{\mathrm{500}}=\mathrm{14}.\mathrm{4} \\ $$$$\frac{\mathrm{14}}{\mathrm{3}}=\mathrm{4}\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\mathrm{5},\mathrm{5},\mathrm{4}\right)\:{or}\:\left(\mathrm{5},\mathrm{4},\mathrm{5}\right)\:{or}\:\left(\mathrm{4},\mathrm{5},\mathrm{5}\right) \\ $$

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