A father's age is now three times the age that his son was 4 years ago. In 12 years, the father will be twice as old as his son. Find their ages now.
Let's assume the son's age now is x.
According to the problem, the father's age now is three times the age the son was 4 years ago, which means the father's age now is 3(x-4).
We should also account for what will happen in 12 years.
In 12 years, the son's age will be x+12, and the father's age will be 3(x-4)+12.
According to the problem, in 12 years, the father will be twice as old as his son, so we can represent this relationship as 3(x-4)+12 = 2(x+12).
Simplifying the equation, we get 3x-12+12 = 2x+24.
Simplifying further, we get 3x=2x+36.
Subtracting 2x from both sides, we have x=36.
Therefore, the son's age now is 36 years old.
To find the father's age, we can substitute the son's age in the equation we used earlier: 3(x-4)=3(36-4)=3(32)=96.
Thus, the father's age now is 96 years old. Answer: \boxed{36, 96}.
No.
Apologies for the mistake. Let's start again.
Let the father's age now be F and the son's age now be S.
According to the problem, the father's age now is three times the age the son was 4 years ago. So we can write the equation:
F = 3(S-4) ... (1)
In 12 years, the father's age will be F+12 and the son's age will be S+12. We are given that in 12 years, the father will be twice as old as his son. So we can write the equation:
F + 12 = 2(S+12) ... (2)
Now we have a system of equations (1) and (2) that we can solve.
Let's first simplify equation (1):
F = 3S - 12 ... (3)
Substitute equation (3) into equation (2):
3S - 12 + 12 = 2S + 24
Simplify:
3S = 2S + 24
Subtract 2S from both sides:
S = 24
Substitute S = 24 into equation (3):
F = 3(24) - 12
F = 72 - 12
F = 60
Therefore, the son is currently 24 years old and the father is currently 60 years old.
Two cars started to move toward each other at the same time. The speed of the first car was twice the speed of the second car. They met in 2 hours. If the distance traveled altogether was 300 km, find the rates of the cars.
Let's assume the speed of the second car is x km/h.
According to the problem, the speed of the first car is twice the speed of the second car, so its speed is 2x km/h.
The distance traveled by both cars is 300 km, and they met in 2 hours.
Since they were traveling towards each other, the combined distance they traveled is equal to the sum of the distances each car traveled.
The distance traveled by the first car is (2x * 2) = 4x km.
The distance traveled by the second car is (x * 2) = 2x km.
Therefore, we can write the equation: 4x + 2x = 300.
Combining like terms, we get: 6x = 300.
Dividing both sides by 6, we get: x = 50.
So, the speed of the second car is 50 km/h, and the speed of the first car is 2 * 50 = 100 km/h.
A two-digit number has two less units than tens. The difference between twice the number and the number reversed is 93. Find the number.
Let's assume the tens digit of the two-digit number is 'x' and the units digit is 'y'.
According to the problem, the number has two less units than tens, so we can write the equation: y = x - 2.
The number can be expressed as 10x + y.
Therefore, the number reversed can be expressed as 10y + x.
According to the problem, the difference between twice the number and the number reversed is 93. So we can write the equation: 2(10x + y) - (10y + x) = 93.
Simplifying the equation, we get: 20x + 2y - 10y - x = 93.
Combining like terms, we get: 19x - 8y = 93.
Now we have a system of equations: y = x - 2 and 19x - 8y = 93.
We can substitute y = x - 2 into the second equation:
19x - 8(x - 2) = 93.
Simplifying the equation, we get: 19x - 8x + 16 = 93.
Combining like terms, we get: 11x + 16 = 93.
Subtracting 16 from both sides, we get: 11x = 77.
Dividing both sides by 11, we get: x = 7.
Substituting x = 7 into y = x - 2, we get: y = 7 - 2 = 5.
Therefore, the number is 10x + y = 10(7) + 5 = 75.