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1-3a-1-4b-1-6c-and-a-b-c-27-find-a-c-

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Question Number 179140 by Shrinava last updated on 25/Oct/22
(1/(3a)) = (1/(4b)) = (1/(6c)) and a+b+c=27 find a−c=?
$$\frac{\mathrm{1}}{\mathrm{3a}}\:=\:\frac{\mathrm{1}}{\mathrm{4b}}\:=\:\frac{\mathrm{1}}{\mathrm{6c}}\:\:\:\mathrm{and}\:\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{27} \\ $$$$\mathrm{find}\:\:\:\mathrm{a}−\mathrm{c}=? \\ $$
Answered by Frix last updated on 25/Oct/22
a=4t b=3t c=2t a+b+c=27 9t=27 t=3 a=12 b=9 c=6 a−c=6
$${a}=\mathrm{4}{t} \\ $$$${b}=\mathrm{3}{t} \\ $$$${c}=\mathrm{2}{t} \\ $$$${a}+{b}+{c}=\mathrm{27} \\ $$$$\mathrm{9}{t}=\mathrm{27} \\ $$$${t}=\mathrm{3} \\ $$$${a}=\mathrm{12} \\ $$$${b}=\mathrm{9} \\ $$$${c}=\mathrm{6} \\ $$$${a}−{c}=\mathrm{6} \\ $$
Commented by Frix last updated on 25/Oct/22
yes thank you
$$\mathrm{yes}\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Oct/22
3a=4b=6c ∧ a+b+c=27; a−c=? [ _( _( ) ) 3a=6c⇒a−2c=0⇒a−c=c_(Hence we need value of c) ] ▶Eleminating b from a+b+c=27 3a=4b=6c⇒ { ((b=(3/2)c)),((b=(3/4)a)) :} •a+b+c=27⇒a+(3/4)a+c=27 ⇒7a+4c=108....(i) •a+b+c=27⇒a+(3/2)c+c=27 ⇒2a+5c=54....(ii) ▶Eleminating a from (i) & (ii) { ((2×(i)⇒14a+8c=216)),((7×(ii)⇒14a+35c=378)) :}⇒27c=162 ⇒c=((162)/(27))=6 •3a=6c⇒a−2c=0⇒a−c=c=6
$$\mathrm{3}{a}=\mathrm{4}{b}=\mathrm{6}{c}\:\wedge\:{a}+{b}+{c}=\mathrm{27};\:{a}−{c}=? \\ $$$$\left[\:_{\underset{\:} {\:\:}} \underset{{Hence}\:\:\:{we}\:{need}\:\:\:{value}\:{of}\:\:{c}} {\mathrm{3}{a}=\mathrm{6}{c}\Rightarrow{a}−\mathrm{2}{c}=\mathrm{0}\Rightarrow{a}−{c}={c}}\:\:\right] \\ $$$$\blacktriangleright\mathcal{E}{leminating}\:\boldsymbol{{b}}\:{from}\:{a}+{b}+{c}=\mathrm{27} \\ $$$$\mathrm{3}{a}=\mathrm{4}{b}=\mathrm{6}{c}\Rightarrow\begin{cases}{{b}=\frac{\mathrm{3}}{\mathrm{2}}{c}}\\{{b}=\frac{\mathrm{3}}{\mathrm{4}}{a}}\end{cases}\: \\ $$$$\bullet{a}+{b}+{c}=\mathrm{27}\Rightarrow{a}+\frac{\mathrm{3}}{\mathrm{4}}{a}+{c}=\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{7}{a}+\mathrm{4}{c}=\mathrm{108}….\left({i}\right) \\ $$$$\bullet{a}+{b}+{c}=\mathrm{27}\Rightarrow{a}+\frac{\mathrm{3}}{\mathrm{2}}{c}+{c}=\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{2}{a}+\mathrm{5}{c}=\mathrm{54}….\left({ii}\right) \\ $$$$\blacktriangleright\mathcal{E}{leminating}\:\boldsymbol{{a}}\:{from}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\begin{cases}{\mathrm{2}×\left({i}\right)\Rightarrow\mathrm{14}{a}+\mathrm{8}{c}=\mathrm{216}}\\{\mathrm{7}×\left({ii}\right)\Rightarrow\mathrm{14}{a}+\mathrm{35}{c}=\mathrm{378}}\end{cases}\Rightarrow\mathrm{27}{c}=\mathrm{162} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{c}=\frac{\mathrm{162}}{\mathrm{27}}=\mathrm{6} \\ $$$$\bullet\mathrm{3}{a}=\mathrm{6}{c}\Rightarrow{a}−\mathrm{2}{c}=\mathrm{0}\Rightarrow{a}−{c}={c}=\mathrm{6} \\ $$
Answered by som(math1967) last updated on 25/Oct/22
(1/(3a))=(1/(4b))=(1/(6c)) 3a=4b=6c ((3a)/(12))=((4b)/(12))=((6c)/(12)) (a/4)=(b/3)=(c/2)=k (let) ∴a=4k, b=3k,c=2k a+b+c=27 ⇒4k+3k+2k=27 ⇒9k=27⇒k=3 a−c=4k−2k=2k=2×3=6
$$\frac{\mathrm{1}}{\mathrm{3}{a}}=\frac{\mathrm{1}}{\mathrm{4}{b}}=\frac{\mathrm{1}}{\mathrm{6}{c}} \\ $$$$\mathrm{3}{a}=\mathrm{4}{b}=\mathrm{6}{c} \\ $$$$\frac{\mathrm{3}{a}}{\mathrm{12}}=\frac{\mathrm{4}{b}}{\mathrm{12}}=\frac{\mathrm{6}{c}}{\mathrm{12}} \\ $$$$\frac{{a}}{\mathrm{4}}=\frac{{b}}{\mathrm{3}}=\frac{{c}}{\mathrm{2}}={k}\:\left({let}\right) \\ $$$$\therefore{a}=\mathrm{4}{k},\:{b}=\mathrm{3}{k},{c}=\mathrm{2}{k} \\ $$$${a}+{b}+{c}=\mathrm{27} \\ $$$$\Rightarrow\mathrm{4}{k}+\mathrm{3}{k}+\mathrm{2}{k}=\mathrm{27} \\ $$$$\Rightarrow\mathrm{9}{k}=\mathrm{27}\Rightarrow{k}=\mathrm{3} \\ $$$$\:{a}−{c}=\mathrm{4}{k}−\mathrm{2}{k}=\mathrm{2}{k}=\mathrm{2}×\mathrm{3}=\mathrm{6} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Oct/22
(1/(3a)) = (1/(4b)) = (1/(6c)) ∧ a+b+c=27; a−c=? (1/(3a)) = (1/(4b)) = (1/(6c))⇒((4×1)/(4(3a))) = ((3×1)/(3(4b))) = ((2×1)/(2(6c))) =(9/(12(a+b+c)))=(3/(4×27))=(1/(36)) 3a=36 ∧ 6c=36 3a=36 ∧ 3c=18 ⇒3a−3c=18 ⇒a−c=6
$$\frac{\mathrm{1}}{\mathrm{3a}}\:=\:\frac{\mathrm{1}}{\mathrm{4b}}\:=\:\frac{\mathrm{1}}{\mathrm{6c}}\:\:\wedge\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{27};\:\mathrm{a}−\mathrm{c}=? \\ $$$$\frac{\mathrm{1}}{\mathrm{3a}}\:=\:\frac{\mathrm{1}}{\mathrm{4b}}\:=\:\frac{\mathrm{1}}{\mathrm{6c}}\Rightarrow\frac{\mathrm{4}×\mathrm{1}}{\mathrm{4}\left(\mathrm{3a}\right)}\:=\:\frac{\mathrm{3}×\mathrm{1}}{\mathrm{3}\left(\mathrm{4b}\right)}\:=\:\frac{\mathrm{2}×\mathrm{1}}{\mathrm{2}\left(\mathrm{6c}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{9}}{\mathrm{12}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}=\frac{\mathrm{3}}{\mathrm{4}×\mathrm{27}}=\frac{\mathrm{1}}{\mathrm{36}} \\ $$$$\:\:\:\:\:\:\mathrm{3a}=\mathrm{36}\:\wedge\:\mathrm{6c}=\mathrm{36} \\ $$$$\:\:\:\:\:\:\mathrm{3a}=\mathrm{36}\:\wedge\:\mathrm{3c}=\mathrm{18} \\ $$$$\:\:\:\:\:\:\Rightarrow\mathrm{3a}−\mathrm{3c}=\mathrm{18} \\ $$$$\:\:\:\:\:\:\Rightarrow\mathrm{a}−\mathrm{c}=\mathrm{6} \\ $$

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