A two digits number is such that four times the unit digit is five times greater than the tens digit. When the digits are reversed, the number is increased by nine. Find the number?
4u=5t
10u+t = 10t+u + 9
now just solve for t and u
For the original number let the unit digit be x and the tens digit by y
so the number is 10y + x
so the number reversed will be 10x + y , which is now 9 more than the original, so
10x + y - (10y + x) = 9
9x - 9y = 9
x - y = 1
also 4x = 5y ---> x = 5y/4
using substitution:
5y/4 - y = 1
5y - 4y = 4
y =4, then x = 5(4)/4 = 5
So the original number is 45
check: number reversed is 54, is this greater by 9? YES
4 times the unit digit ----> 20
5 times the tens digit ----> 20
ok then!! 1/2,1/3,1/4
2nd problem:
2 of the 3 events to happen--> 1st and 2nd, 2nd and 3rd, 1st and 3rd
prob = (1/2)(1/3) + (1/3)(1/4) + (1/2)(1/4)
= 1/6 + 1/12 + 1/8
= 3/8
The probability of three independent events are 1/2,1/3,1/4 respectively. What's the probability that two of them will occur
To find the two-digit number, let's assign variables to the digits.
Let:
- x be the tens digit
- y be the units digit
From the given information, we can create two equations.
Equation 1: "Four times the unit digit is five times greater than the tens digit."
4y = 5x
Equation 2: "When the digits are reversed, the number is increased by nine."
10y + x = 10x + y + 9
Now, we can solve these equations simultaneously to find the values of x and y.
From Equation 1, we can express x in terms of y:
4y = 5x
x = (4y)/5
Substituting x in Equation 2, we have:
10y + (4y)/5 = 10((4y)/5) + y + 9
Let's simplify the equation by multiplying through by 5:
50y + 4y = 40y + 5y + 45
Combining like terms:
54y = 45y + 45
Simplifying further by subtracting 45y from both sides:
9y = 45
Dividing by 9:
y = 5
Now, substitute the value of y back into Equation 1:
4(5) = 5x
Simplifying:
20 = 5x
Divide by 5:
x = 4
So, the tens digit (x) is 4, and the units digit (y) is 5.
Therefore, the number is 45.