twice of a number with thrice of another number gives 23.four times the first number and five times the second number when added gives 41. find the number?
Answers
Answer:
⋆
Solution:−
\begin{lgathered}\\ \\\end{lgathered}
\orange{\boxed{\sf{Relative\: Density\:of\:a\: substance = \dfrac{Density\:of\:the\: substance}{Density\:of\:water} }}}
RelativeDensityofasubstance=
Densityofwater
Densityofthesubstance
\begin{lgathered}\\\end{lgathered}
\begin{lgathered}\sf\red{Here,} \\ \\ \sf{Relative\: density\:of\: silver = 10.8} \\ \\ \sf{Density\:of\: silver =?} \: \: \: \sf\purple{(To\:be\: calculated)} \\ \\ \sf{and,} \\ \\ \sf{Density\:of\:water = 1.0 \times 10^3\: kg \:m^-3 } \\ \\ \sf\red{Now,} \\ \\ \sf{Putting\: these\: Values :-} \\ \\\end{lgathered}
Here,
Relativedensityofsilver=10.8
Densityofsilver=?(Tobecalculated)
and,
Densityofwater=1.0×10
3
kgm
−
3
Now,
PuttingtheseValues:−
\begin{lgathered}{\sf{10.8 = \dfrac{Density\:of\: silver}{1.0 \times 10^3} }} \\ \\ \sf{Now,} \\ \\ \sf{ Density\:of\: silver = 10.8 \times 1.0 \times 10^3 \:kg \: m^-3 } \\ \\ \implies{\boxed{\sf{10.8 \times 10^3 kg\:m^-3}}}\end{lgathered}
10.8=
1.0×10
3
Densityofsilver
Now,
Densityofsilver=10.8×1.0×10
3
kgm
−
3
⟹
10.8×10
3
kgm
−
3
\begin{lgathered}\\\end{lgathered}
\begin{lgathered}\sf{Thus,} \\ \\ \sf\blue{The\: density\:of\:silver\:in\:SI\:units\:is\:10.8 \times 10^3 \: kg\: m^-3.} \\ \\\end{lgathered}
Thus,
ThedensityofsilverinSIunitsis10.8×10
3
kgm
−
3.
\sf\underline\pink{Thanks}
Thanks