solve these story sums.
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5. The dimensions of one tile are given as 5cm and 8cm. Hence area of 1 tile= 5*8=40 sq. cms. Hence total tiles needed to tile the floor= 80000/40=2000.
6. Cost of levelling the badminton court = 7656 m² × Rs. 80. Cost of levelling the badminton court = Rs. 612480.
7.∴ The breadth of the rectangular playing field is 20m.
8.Now let us consider the formula for the area of the rectangle.
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×b
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×bAs we are given that area of the class room is equal to 594 square meters, let us substitute it in the above equation. Then we get,
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×bAs we are given that area of the class room is equal to 594 square meters, let us substitute it in the above equation. Then we get,⇒594=27×b
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×bAs we are given that area of the class room is equal to 594 square meters, let us substitute it in the above equation. Then we get,⇒594=27×bTaking 27 to other side we get,
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×bAs we are given that area of the class room is equal to 594 square meters, let us substitute it in the above equation. Then we get,⇒594=27×bTaking 27 to other side we get,⇒b=59427
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×bAs we are given that area of the class room is equal to 594 square meters, let us substitute it in the above equation. Then we get,⇒594=27×bTaking 27 to other side we get,⇒b=59427Calculating the above value, we get the value of b as,
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×bAs we are given that area of the class room is equal to 594 square meters, let us substitute it in the above equation. Then we get,⇒594=27×bTaking 27 to other side we get,⇒b=59427Calculating the above value, we get the value of b as,⇒b=22
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×bAs we are given that area of the class room is equal to 594 square meters, let us substitute it in the above equation. Then we get,⇒594=27×bTaking 27 to other side we get,⇒b=59427Calculating the above value, we get the value of b as,⇒b=22So, the breadth of the class room is 22 meters.
let us consider the formula for the area of the rectangle.Area=(length)×(breadth)Using this formula, we can substitute the given value of length and breadth in it. Then we get,⇒Area=27×bAs we are given that area of the class room is equal to 594 square meters, let us substitute it in the above equation. Then we get,⇒594=27×bTaking 27 to other side we get,⇒b=59427Calculating the above value, we get the value of b as,⇒b=22So, the breadth of the class room is 22 meters.Hence, the answer is 22 meters.
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