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Question Number 171989 by Mikenice last updated on 22/Jun/22

solve: A two digit number is such  that 4 times the unit digit is 5 more   than thrice the ten digit. when the  digit are reversed the number is   increase by nine, find the number

$${solve}:\:{A}\:{two}\:{digit}\:{number}\:{is}\:{such} \\ $$$${that}\:\mathrm{4}\:{times}\:{the}\:{unit}\:{digit}\:{is}\:\mathrm{5}\:{more}\: \\ $$$${than}\:{thrice}\:{the}\:{ten}\:{digit}.\:{when}\:{the} \\ $$$${digit}\:{are}\:{reversed}\:{the}\:{number}\:{is}\: \\ $$$${increase}\:{by}\:{nine},\:{find}\:{the}\:{number} \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 23/Jun/22

u: unit digit ,  t: ten-digit  Given:  •4u=3t+5.................(i)  •Reversed=Original+9     10u+t=(10t+u)+9....(ii)  Process:   (ii)⇒ 9u−9t=9                  u−t=1                  u=t+1  (i)⇒4(t+1)=3t+5                t=1              u=t+1=1+1=2  ∴ The number is 12

$${u}:\:{unit}\:{digit}\:,\:\:{t}:\:{ten}-{digit} \\ $$$${Given}: \\ $$$$\bullet\mathrm{4}{u}=\mathrm{3}{t}+\mathrm{5}.................\left({i}\right) \\ $$$$\bullet\mathcal{R}{eversed}=\mathcal{O}{riginal}+\mathrm{9} \\ $$$$\:\:\:\mathrm{10}{u}+{t}=\left(\mathrm{10}{t}+{u}\right)+\mathrm{9}....\left({ii}\right) \\ $$$${Process}: \\ $$$$\:\left({ii}\right)\Rightarrow\:\mathrm{9}{u}−\mathrm{9}{t}=\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}−{t}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}={t}+\mathrm{1} \\ $$$$\left({i}\right)\Rightarrow\mathrm{4}\left({t}+\mathrm{1}\right)=\mathrm{3}{t}+\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{u}={t}+\mathrm{1}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$$\therefore\:\mathcal{T}{he}\:{number}\:{is}\:\mathrm{12} \\ $$

Commented by Mikenice last updated on 23/Jun/22

thanks sir

$${thanks}\:{sir} \\ $$

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