आण्विकता तथा अभिक्रिया की कोटि में अन्तर-
Answers
Answer:-
Age of father is 50 years.
Age of daughter is 20 years.
Explanation:-
Given:-
Sum of ages of father and daughter = 70
After 10 years fathers age is twice as old as daughter.
To Find:-
The present ages of father and daughter.
Solution:-
Let the ages of father and daughter be x and y respectively.
So ATQ:-
\begin{gathered} \\ \tt \mapsto x + y = 70...eq(1).\end{gathered}
↦x+y=70...eq(1).
After 10 years:-
Age of father = 10+x.
Age of daughter = 10+y.
\begin{gathered}\\ \tt \mapsto(10 + x) = 2(10 + y).\end{gathered}
↦(10+x)=2(10+y).
\begin{gathered}\\ \tt \mapsto10 + x = 20 + 2y.\end{gathered}
↦10+x=20+2y.
\begin{gathered}\\ \tt \mapsto x - 2y = 20 - 10.\end{gathered}
↦x−2y=20−10.
\begin{gathered}\\ \tt \mapsto x - 2y = 10...eq(2).\end{gathered}
↦x−2y=10...eq(2).
Subtracting eq(2) from eq(1),
\begin{gathered}\\ \tt \mapsto(x + y) - (x - 2y) = 70 - 10.\end{gathered}
↦(x+y)−(x−2y)=70−10.
\begin{gathered}\\ \tt \mapsto \cancel{x} + y \cancel{ - x} + 2y = 60.\end{gathered}
↦
x
+y
−x
+2y=60.
\begin{gathered}\\ \tt \mapsto3y = 60.\end{gathered}
↦3y=60.
\begin{gathered}\\ \tt \mapsto y = \cancel \frac{60}{3} .\end{gathered}
↦y=
3
60
.
\begin{gathered}\\ \large\bf \mapsto \boxed{ \bf y = 20.}\end{gathered}
↦
y=20.
Therefore The Present age of daughter is 20 years.
By putting the value of y in eq(1) :-
\begin{gathered}\\ \tt \mapsto x + y = 70.\end{gathered}
↦x+y=70.
\begAnswer:-
Age of father is 50 years.
Age of daughter is 20 years.
Explanation:-
Given:-
Sum of ages of father and daughter = 70
After 10 years fathers age is twice as old as daughter.
To Find:-
The present ages of father and daughter.
Solution:-
Let the ages of father and daughter be x and y respectively.
So ATQ:-
\begin{gathered} \\ \tt \mapsto x + y = 70...eq(1).\end{gathered}
↦x+y=70...eq(1).
After 10 years:-
Age of father = 10+x.
Age of daughter = 10+y.
\begin{gathered}\\ \tt \mapsto(10 + x) = 2(10 + y).\end{gathered}
↦(10+x)=2(10+y).
\begin{gathered}\\ \tt \mapsto10 + x = 20 + 2y.\end{gathered}
↦10+x=20+2y.
\begin{gathered}\\ \tt \mapsto x - 2y = 20 - 10.\end{gathered}
↦x−2y=20−10.
\begin{gathered}\\ \tt \mapsto x - 2y = 10...eq(2).\end{gathered}
↦x−2y=10...eq(2).
Subtracting eq(2) from eq(1),
\begin{gathered}\\ \tt \mapsto(x + y) - (x - 2y) = 70 - 10.\end{gathered}
↦(x+y)−(x−2y)=70−10.
\begin{gathered}\\ \tt \mapsto \cancel{x} + y \cancel{ - x} + 2y = 60.\end{gathered}
↦
x
+y
−x
+2y=60.
\begin{gathered}\\ \tt \mapsto3y = 60.\end{gathered}
↦3y=60.
\begin{gathered}\\ \tt \mapsto y = \cancel \frac{60}{3} .\end{gathered}
↦y=
3
60
.
\begin{gathered}\\ \large\bf \mapsto \boxed{ \bf y = 20.}\end{gathered}
↦
y=20.
Therefore The Present age of daughter is 20 years.
By putting the value of y in eq(1) :-
\begin{gathered}\\ \tt \mapsto x + y = 70.\end{gathered}
↦x+y=70.
\beg