givan sale is ₹ 200000 and ₹ 400000 in year 2018 and 2019 respectively . loss is ₹ 10000 in 2018 and profit is ₹ 20000 in 2019 respectively . compute p/ v ratio
Answers
Answer:
Question⇰
The total cost of flooring a room at Rs.8.50 per sq. metre is Rs.510. If the length of the room is 8 metres, find its breadth ?
Given :-
Flooring of a room = ₹ 8.50
Per square Metre = ₹ 510
Length = 8 m
To find :-
Find its breadth
Let's Do...!
\sf \: Cost \: of \: flooring \: per \: square \: meter = Rs.8.50Costofflooringpersquaremeter=Rs.8.50
\sf \implies Total \: cost \: of \: flooring \: of \: room = Rs.510⟹Totalcostofflooringofroom=Rs.510
\sf∴ Area \: of \: the \: room = \frac{510}{8.50} = 60 \: m {}^{2}∴Areaoftheroom=
8.50
510
=60m
2
Formula used :- ( Area of rectangle )
\sf \pink{ Area \: of \: the \: room = l×b}Areaoftheroom=l×b
Calculation...
\sf \purple\implies 60=8×b⟹60=8×b
\sf \purple\implies b = \frac{60}{8} = 7.5 \: m⟹b=
8
60
=7.5m
\red{\boxed{ \sf∴ Breadth \: of \: the \: room \: is \: 7.5 \: m}}
∴Breadthoftheroomis7.5m
Step-by-step explanation:
Consider the left hand side :
\: \: \pink{: \implies}( \sf \dfrac{x {}^{a} }{x {}^{b}}) {}^{a {}^{2} +ab + b {}^{2}} \times( \dfrac{x {}^{b}}{x {}^{c}}) {}^{{b}^{2} + {bc}^{} + {c}^{2}} \times ( \dfrac{x {}^{2} }{x {}^{a}}) {}^{c {}^{2} + ca + a {}^{2} }:⟹(
x
b
x
a
)
a
2
+ab+b
2
×(
x
c
x
b
)
b
2
+bc
+c
2
×(
x
a
x
2
)
c
2
+ca+a
2
\pink{: \implies}\sf\dfrac{x {}^{a( {a}^{2} + ab + {b}^{2})}}{ {x}^{b( {a}^{2} + ab + {b}^{2})}} \times \dfrac{ {x}^{b( {b}^{2} + bc + {c}^{2})}}{ {x}^{c( {b}^{2} + bc + {c}^{2})}} \times \dfrac{ {x}^{c( {c}^{2} + ca + {a}^{2})}}{ {x}^{a( {c}^{2} + ca + {a}^{2})}}:⟹
x
b(a
2
+ab+b
2
)
x
a(a
2
+ab+b
2
)
×
x
c(b
2
+bc+c
2
)
x
b(b
2
+bc+c
2
)
×
x
a(c
2
+ca+a
2
)
x
c(c
2
+ca+a
2
)
Now, Open the brackets :
\pink{: \implies}\sf{x}^{a( {a}^{2} + ab + {b}^{2})} - {b}^{( {a}^{2} + ab + {b}^{2})} \times {x}^{b( {b}^{2} + bc + {c}^{2})} - {c}^{( {b}^{2} + bc + {c}^{2})} \times {x}^{c( {c}^{2} + ca + {a}^{2})} - {a}^{( {c}^{2} + ca + {a}^{2})}:⟹x
a(a
2
+ab+b
2
)
−b
(a
2
+ab+b
2
)
×x
b(b
2
+bc+c
2
)
−c
(b
2
+bc+c
2
)
×x
c(c
2
+ca+a
2
)
−a
(c
2
+ca+a
2
)
\pink{:\implies}\sf{x}^{(a - b)({a}^{2} + ab + {b}^{2})} \times {x}^{(b - c)({b}^{2} + bc + {c}^{2})} \times {x}^{(c - a)({c}^{2} + ca + {a}^{2})}:⟹x
(a−b)(a
2
+ab+b
2
)
×x
(b−c)(b
2
+bc+c
2
)
×x
(c−a)(c
2
+ca+a
2
)
\pink{: \implies}\sf {x}^{({a}^{3} - {b}^{3})} \times x(({{b}^{3} - c{}^{3}))} \times \:{x}^{({c}^{3} - {a}^{3})}:⟹x
(a
3
−b
3
)
×x((b
3
−c
3
))×x
(c
3
−a
3
)
\: \: \: \: \: \: \: \: \: \pink{: \implies}\sf{x}^{({a}^{3} - {b}^{3} + {b}^{3} - {c}^{3} + {c}^{3} - {a}^{3})}:⟹x
(a
3
−b
3
+b
3
−c
3
+c
3
−a
3
)
Final answer :
\begin{gathered} \: \: \boxed{ \sf\pink{: \implies}{x}^{0}} \\ \\ \: \: \: \: \: \: \: \: \: {\boxed {\boxed{{\pink{: \implies}1}}}}\end{gathered}
:⟹x
0
:⟹1
L.H.S = R.H.S
Hence, Verified.