Question Number 49036 by Tinkutara last updated on 01/Dec/18
Commented by prakash jain last updated on 02/Dec/18
$$\mathrm{2}^{{n}−\mathrm{1}} \:{gave}\:{at}\:{least}\:\mathrm{1}\:{wrong}\:{answer} \\ $$$$\left({include}\:{student}\:{with}\:\mathrm{2}\:{wrong}\:{answers}\right) \\ $$$$\mathrm{2}^{{n}−\mathrm{2}} \:{gave}\:{at}\:{least}\:\mathrm{2}\:{wrong}\:{amswers}\: \\ $$$$\left({only}\:{one}\:{additional}\:{wrong}\:{answer}\right. \\ $$$$\left.{should}\:{be}\:{counted}\right) \\ $$$$… \\ $$$$\mathrm{2}^{{n}−{k}} \:{gave}\:{k}\:{wrong}\:{answer} \\ $$$${S}=\mathrm{2}^{{n}−\mathrm{1}} +\mathrm{2}^{{n}−\mathrm{2}} +…+\mathrm{2}^{{n}−{k}} \\ $$$$\mathrm{2}^{{n}−\mathrm{1}} \left(\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)=\mathrm{2}^{{n}} ×\frac{\mathrm{2}^{{k}} −\mathrm{1}}{\mathrm{2}^{{k}} } \\ $$$${example} \\ $$$${n}=\mathrm{6},\:{k}=\mathrm{3} \\ $$$$\mathrm{32}\:−\:\mathrm{1}\:{or}\:{more}\:{wrong} \\ $$$$\mathrm{16}\:−\:\mathrm{2}\:{of}\:{more}\:{wrong}\: \\ $$$$\mathrm{8}\:\:\:\:−\:\mathrm{3}\:{or}\:{more}\:{wrong} \\ $$$${total}\:{wrong}\:{answer}\:\mathrm{32}+\mathrm{16}+\mathrm{8}=\mathrm{56} \\ $$$$ \\ $$
Commented by prakash jain last updated on 02/Dec/18
$$\mathrm{Book}\:\mathrm{answer}\:\mathrm{will}\:\mathrm{be}\:\mathrm{correct}\:\mathrm{if}\:{n}={k}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Dec/18
$${thank}\:{you}\:{sir}… \\ $$
Commented by Tinkutara last updated on 03/Dec/18
Thank you very much Sir! I got the answer. ��������
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18
$${put}\:{i}=\mathrm{1} \\ $$$$\mathrm{2}^{{n}−\mathrm{1}} {students}\:{give}\:\mathrm{1}{wrong}\:{answer} \\ $$$$\mathrm{2}^{{n}−\mathrm{2}} \:{students}\:\:{give}\:\mathrm{2}{wrong}\:{ansrwer} \\ $$$$\mathrm{2}^{{n}−\mathrm{3}} \:{students}\:\:{give}\:\mathrm{3}\:{wrong}\:{answer} \\ $$$$… \\ $$$$… \\ $$$$\mathrm{2}^{{n}−{k}} \:{students}\:{give}\:{k}\:{wrong}\:{answer} \\ $$$${total}\:{wrong}\:{answer} \\ $$$$\:\:\:\:\:\:{S}=\mathrm{2}^{{n}−\mathrm{1}} ×\mathrm{1}+\mathrm{2}^{{n}−\mathrm{2}} ×\mathrm{2}+\mathrm{2}^{{n}−\mathrm{3}} ×\mathrm{3}…+\mathrm{2}^{{n}−{k}} ×{k} \\ $$$${S}×\mathrm{2}^{−\mathrm{1}} =\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{{n}−\mathrm{2}} ×\mathrm{1}\:\:\:\:+\mathrm{2}^{{n}−\mathrm{3}} ×\mathrm{2}..\:\:+\mathrm{2}^{{n}−{k}} ×\left({k}−\mathrm{1}\right)+\mathrm{2}^{{n}−{k}−\mathrm{1}} ×{k} \\ $$$${substruct} \\ $$$$\left({S}−\frac{{S}}{\mathrm{2}}\right)=\left(\mathrm{2}^{{n}−\mathrm{1}} +\mathrm{2}^{{n}−\mathrm{1}} +…+\mathrm{2}^{{n}−\mathrm{1}} \:{n}\:{times}\right)−\mathrm{2}^{{n}−{k}−\mathrm{1}} ×{k} \\ $$$$\frac{{S}}{\mathrm{2}}={n}\mathrm{2}^{{n}−\mathrm{1}} −\mathrm{2}^{{n}−{k}−\mathrm{1}} ×{k} \\ $$$${S}={n}\mathrm{2}^{{n}} −\mathrm{2}^{{n}−{k}} ×{k} \\ $$
Commented by Tinkutara last updated on 02/Dec/18
But answer is independent of k.
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18
$$\mathrm{0}{k}\:{let}\:{others}\:{try}\:{or}\:{you}\:{post}\:{answer}\:{in}\:{details} \\ $$
Commented by Tinkutara last updated on 02/Dec/18
$${Ans}\:{is}\:\mathrm{2}^{{n}} −\mathrm{1} \\ $$$${I}\:{don}'{t}\:{understand}\:{the}\:{solution}\:{given} \\ $$