Rectangular courtyard is 57m 50cm and 22m broad. Determine the least possible number of square stones of the same size required to pave it.
6 answers
math question
Length=5750 cm
Width= 2200cm
AREA = length of largest tile=H.C.F of (5750cm and 2200cm)
=2x5x5x4=200cm
Required number of tiles= Area of floor/Area of tile
=5750x2200 / 200x200
=316.25
so, 316.25 is the least possible number of square stone s of the same size to pave it.
Width= 2200cm
AREA = length of largest tile=H.C.F of (5750cm and 2200cm)
=2x5x5x4=200cm
Required number of tiles= Area of floor/Area of tile
=5750x2200 / 200x200
=316.25
so, 316.25 is the least possible number of square stone s of the same size to pave it.
Good
Determine the largest number of four digit which is divisible by 18 , 25 and 35
I don't know
To find the largest number that is divisible by 18, 25, and 35, we need to find the least common multiple (LCM) of these three numbers.
The LCM of 18, 25, and 35 is the smallest multiple that is divisible by all three numbers. To find the LCM, we can prime factorize each number and take the highest power of each prime factor.
Prime factorization of 18 = 2 * 3^2
Prime factorization of 25 = 5^2
Prime factorization of 35 = 5 * 7
Taking the highest power of each prime factor, we have:
2^1 * 3^2 * 5^2 * 7^1 = 2 * 9 * 25 * 7 = 3150
So, the largest four-digit number divisible by 18, 25, and 35 is 3150.
The LCM of 18, 25, and 35 is the smallest multiple that is divisible by all three numbers. To find the LCM, we can prime factorize each number and take the highest power of each prime factor.
Prime factorization of 18 = 2 * 3^2
Prime factorization of 25 = 5^2
Prime factorization of 35 = 5 * 7
Taking the highest power of each prime factor, we have:
2^1 * 3^2 * 5^2 * 7^1 = 2 * 9 * 25 * 7 = 3150
So, the largest four-digit number divisible by 18, 25, and 35 is 3150.
13 answers