the floor of a room is 3 M wide and 4 long how many square tile of side 50 cm are required to cover the floor of the room
Answers
Step-by-step explanation:
Answer:-
According to the Question
It is given that,
Length of room ,L = 3m = 300cm
Breadth of room ,B = 4m = 400cm
Side of square tiles ,S = 50cm
We have to calculate the no. tiles required to cover the floor of the room.
Firstly we calculate the area of room .
Area of Room = Length × Breadth
Substitute the value we get
→ Area of Room = 300 × 400
→ Area of Room = 120000 cm²
Now, Calculating the area occupied by one titles .
Area of Square tiles = side × side
substitute the value we get
→ Area of Square tiles = 50×50
→ Area of Square tiles = 2500cm²
So, the no. of titles = Area of Room/Area of Square tiles
→ No. of tiles = 120000/2500
→ No. of tiles = 1200/25
→ No. of tiles = 48
Hence, the number of tiles required to cover the floor of the room is 48 .
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Answer:
To find : The distance between the school and his house
Solution :
Speed of a student = 2 ½ km/h = 5/2km/h
★ A student reaches his school 6 minutes late.
Consider the time be x
60min = 1 hour
Time = (x + 6) min = (x + 6/60) = (x + 1/10)h
As we know that
★ Speed = distance/time
Consider the distance be y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ x + \dfrac{1}{10}}⟹25=x+101y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ \dfrac{10x + 1}{10} }⟹25=1010x+1y
\implies \sf \dfrac{5}{2} = y \times \dfrac{10}{10x +1}⟹25=y×10x+110
\implies \sf \dfrac{5}{2} = \dfrac{10y}{10x +1}⟹25=10x+110y
\implies \sf 5(10x +1) = 20y⟹5(10x+1)=20y
\implies \sf 50x +5=20y⟹50x+5=20y
\implies \sf 50x - 20y = - 5 \: \: \: \bf(Equation \: 1)⟹50x−20y=−5(Equation1)
★ Next day starting at the same time, he increases his speed by 1 km/hour and reaches 6 minutes early.
\implies \sf \dfrac{5}{2} + 1 = \dfrac{y}{ x - \dfrac{1}{10}}⟹25+1=x−101y
\implies \sf \dfrac{5 + 2}{2} = \dfrac{y}{\dfrac{10x - 1}{10}}⟹25+2=1010x−1y
\implies \sf \dfrac{7}{2} = \dfrac{10y}{10x - 1}⟹27=10x−110y
\implies \sf 7(10x - 1) = 20y⟹7(10x−1)=20y
\implies \sf 70x - 20y = 7 \: \: \: \bf(Equation \: 2)⟹70x−20y=7(Equation2)
Subtract both the equations
→ 50x - 20y - (70x - 20y) = -5 - 7
→ 50x - 20y - 70x + 20y = - 12
→ - 20x = - 12
→ x = 12/20 = 3/5
Put the value of x in eqⁿ (1)
→ 50x - 20y = - 5
→ 50 × 3/5 - 20y = - 5
→ 30 - 20y = - 5
→ 30 + 5 = 20y
→ 35 = 20y
→ y = 35/20 = 7/4
→ y = 1 3/4 km
•°• The distance between school and house is 1 3/4 km.