Answer:
Rider 1 does one round in 15 min, and will complete another in each consecutive multiple of 15 min
Rider 2 does one round in 18 min, and will complete another in each consecutive multiple of 18 min
Assuming that they start together, they will complete another round together in a time that is both multiples of 15min and 18 min.
Then we need to find the smallest common multiple between 15 and 18.
To smallest common multiple between two numbers, a and b, is equal to:
a*b/(greatest common factor between a and b).
Now, the greatest common factor between 15 and 18 can be found if we write those numbers as a product of prime numbers, such as:
15 = 3*5
18 = 2*3*3
The greatest common factor is 3.
Then the smallest common multiple will be:
(15*18)/3 = 90
This means that after 90 mins, they will meet again at the starting place.
Multiply the GCF of the numerical part 3 and the GCF of the variable part x^2y to get
3x^2y.
Answer:
The number is 91
Step-by-step explanation:
Let x be the ones place digit and y be the tens place digit,
Then the number would be 10y + x,
We have,
y - x = 8
Possible values of y and x = { (8, 0), (9, 1) }
∵ 0 is not the digit of the number,
Hence, y = 9 and x = 1
Therefore, required number = 10(9) + 1 = 90 + 1 = 91
3/8 = 0,375
5/12 = 0,416
5/12 is bigger
