Train A and train B stops at Swindon at 10:30 . Train A stops every twelve minutes and train B stops every 14 Mins , when do they both stop at Swindon again ?

at 11:54

Step-by-step explanation:

Let's define the 10:30 as our t = 0 min.

We know that Train A stops every 12 mins, and Train B stops every 14 mins, they will stop at the same time in the least common multiple of 12 and 14.

To find the least common multiple of two numbers, we must do:

LCM(a,b) = a*b/GCD(a,b)

Where GCD(a, b) is the greatest common divisor of a and b.

In this case the only common divisior of 12 and 14 is 2.

So we have:

LCM(12, 14) = 12*14/2 = 84.

Then the both trains will stop 84 minutes after 10:30

one hour has 60 mins, so we can write 84 minutes as:

1 hour and 24 minutes = 1:24

Then they will stop at the same time at 10:30 + 1:24 = 11:54

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)