Question

solve 4x+3 < 6x+7 , XER​

Answer 1

♣GIVEN :

Length of the floor of the room (l) = 4.84 m = 484 cm

Breadth of the floor of the room (b) = 3.1 m = 310 cm

Length of each tile (l) = 22 cm

Breadth of each tile (b) = 10 cm

Rate of paving (r) = ₹ 6.50/tile

♣TO FIND :

Cost of the tiles (c)

♣SOLUTION :

❶ First we need to find the area of the floor:-

❐ USING FORMULAE :

$\huge \boxed{ \sf{A = \bigg(l \: \times b \bigg) unit}}$

A=(l×b)unit

✯ WHERE :-

A = Area

l = length

b = breadth

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➲ Putting all values :-

➨$\:\bf{A = \bigg( 484 \times 310 \bigg)} \: cm^{2}$

➠$\: \underline{\boxed{\bf{A = 150040} \: {cm}^{2} }}$

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❷ Now we have to find the area of each tile:-

❐ USING FORMULAE :

$\huge \boxed{ \sf{A = \bigg(l \: \times b \bigg) unit}}$

A=(l×b)unit

✯ WHERE :-

A = Area

l = length

b = breadth

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➲ Putting all values :-

➨$\:\bf{A = \bigg( 22\times 0 \bigg)} \: cm^{2}$

➠$\: \underline{\boxed{\bf{A = 220} \: {cm}^{2} }}$

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❸ Now let us find the number of tiles required for paving the floor:-

❐ USING FORMULAE :

$\huge \boxed{ \sf{ Tiles \: Required = \bigg( \dfrac{ Area \: of \: the \: floor }{Area \: of \: each \: tile} \bigg)}}$

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➲ Putting all values :-

➨ $\:\bf{Tiles \: required= \dfrac{15004 \cancel0}{22 \cancel0} }$

➨$\:\bf{Tiles \: required= \dfrac{ \cancel{15004} ^{682}}{ \cancel{22}_{1} }}$

➠$\: \underline{\boxed{\bf{Tiles \: required= 682} }}$

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❹ Finally we find the cost of the tiles:-

We have :-

❍ Rate per tile = ₹ 6.50

❐ USING FORMULAE :

$\huge \boxed{ \sf{ Cost \: of \: tiles= ₹\bigg( Rate × \: tiles \: required \bigg)} }$

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➲ Putting all values :-

➨$\:\bf{ Cost \: of \: tiles \: = ₹\bigg(6.50 \times 682 \bigg)}$

➠ $\: \underline{ \boxed{\bf{ Cost \: of \: tiles \: = ₹4433 }}}$

♣ANSWER :

Our required answer is ₹4433