💡First, I recognize that Christopher's age is given as 18 years.Next, I calculate George's age by adding 8 years to Christopher's age, which results in 26 years.Then, I determine Ford's age by subtracting 2 years from Christopher's age, giving me 16 years.Finally, I sum the ages of George, Christopher, and Ford to find the total sum of their ages.

💡Okay, so I've got this problem here where I need to find the value of a certain expression involving two perfect squares, a and b. The problem states that the product of a and b exceeds their sum by 4844. Then, I have to compute this expression:[left(sqrt a + 1right)left(sqrt b + 1right)left(sqrt a - 1right)left(sqrt b - 1right) - left(sqrt{68} + 1right)left(sqrt{63} + 1right)left(sqrt{68} - 1right)left(sqrt{63} - 1right)]Alright, let's break this down step by step.First, since a and b are perfect squares, I can express them as ( a = x^2 ) and ( b = y^2 ) where x and y are integers. The problem says that the product of a and b exceeds their sum by 4844. So, mathematically, that translates to:[ab = a + b + 4844]Substituting ( a = x^2 ) and ( b = y^2 ), we get:[x^2 y^2 = x^2 + y^2 + 4844]Hmm, okay. So, that's our main equation to work with.Now, looking at the expression we need to compute, it's a bit complicated, but I notice that it involves products of terms like ( (sqrt{a} + 1) ) and ( (sqrt{a} - 1) ). Maybe I can simplify this expression first before plugging in the values.Let me focus on the first part of the expression:[left(sqrt a + 1right)left(sqrt b + 1right)left(sqrt a - 1right)left(sqrt b - 1right)]I remember that when you have something like ( (x + y)(x - y) ), it simplifies to ( x^2 - y^2 ). Maybe I can apply that here.Let's pair the terms:First, pair ( (sqrt{a} + 1) ) with ( (sqrt{a} - 1) ):[(sqrt{a} + 1)(sqrt{a} - 1) = (sqrt{a})^2 - (1)^2 = a - 1]Similarly, pair ( (sqrt{b} + 1) ) with ( (sqrt{b} - 1) ):[(sqrt{b} + 1)(sqrt{b} - 1) = (sqrt{b})^2 - (1)^2 = b - 1]So, the entire first part of the expression simplifies to:[(a - 1)(b - 1)]Alright, that's much simpler. So, the expression becomes:[(a - 1)(b - 1) - left(sqrt{68} + 1right)left(sqrt{63} + 1right)left(sqrt{68} - 1right)left(sqrt{63} - 1right)]Now, let's look at the second part of the expression:[left(sqrt{68} + 1right)left(sqrt{63} + 1right)left(sqrt{68} - 1right)left(sqrt{63} - 1right)]I can apply the same pairing technique here. Let's pair ( (sqrt{68} + 1) ) with ( (sqrt{68} - 1) ):[(sqrt{68} + 1)(sqrt{68} - 1) = 68 - 1 = 67]Similarly, pair ( (sqrt{63} + 1) ) with ( (sqrt{63} - 1) ):[(sqrt{63} + 1)(sqrt{63} - 1) = 63 - 1 = 62]So, the second part simplifies to:[67 times 62]Calculating that:[67 times 62 = (60 + 7)(60 + 2) = 60^2 + 60 times 2 + 7 times 60 + 7 times 2 = 3600 + 120 + 420 + 14 = 3600 + 120 = 3720; 3720 + 420 = 4140; 4140 + 14 = 4154]So, the second part is 4154.Now, going back to the first part, which is ( (a - 1)(b - 1) ). Let's see if we can find the value of this expression using the given condition ( ab = a + b + 4844 ).Expanding ( (a - 1)(b - 1) ):[(a - 1)(b - 1) = ab - a - b + 1]From the given condition, we know that ( ab = a + b + 4844 ). So, substituting this into the expression:[ab - a - b + 1 = (a + b + 4844) - a - b + 1 = 4844 + 1 = 4845]So, ( (a - 1)(b - 1) = 4845 ).Now, putting it all together, the expression we need to compute is:[4845 - 4154 = 691]Therefore, the value of the given expression is 691.**Final Answer**boxed{691}

💡Alright, so I have this problem here where I need to find the largest possible 3-digit sum for the following addition:``` XXZ+ YXZ+ Z------```And X, Y, Z are different digits. The options given are (A) 981, (B) 987, (C) 976, and (D) 982. I need to figure out which one is the correct answer.First, let me understand the structure of the problem. We have three numbers being added together:1. The first number is a 3-digit number where the hundreds and tens place are both X, and the units place is Z. So, it's like XXZ.2. The second number is also a 3-digit number where the hundreds place is Y, the tens place is X, and the units place is Z. So, it's YXZ.3. The third number is just a single digit Z.When we add these three numbers together, we get a sum that's a 3-digit number. Our goal is to maximize this sum, given that X, Y, and Z are different digits.Let me break down each number into its decimal components to make it easier to handle:1. XXZ can be written as 100*X + 10*X + Z = 110*X + Z.2. YXZ can be written as 100*Y + 10*X + Z.3. Z is just Z.So, when we add them together:(110*X + Z) + (100*Y + 10*X + Z) + ZLet me simplify this expression:110*X + Z + 100*Y + 10*X + Z + ZCombine like terms:(110*X + 10*X) + 100*Y + (Z + Z + Z)Which simplifies to:120*X + 100*Y + 3*ZSo, the total sum is 120*X + 100*Y + 3*Z.Now, since we want the largest possible 3-digit sum, we need to maximize this expression. However, we have constraints:1. X, Y, Z must be different digits (0-9).2. The sum must be a 3-digit number, so it must be less than 1000.Given that 120*X and 100*Y are the largest contributors to the sum, we should try to maximize X and Y first, and then Z.Let me start by trying to assign the largest possible values to X and Y.Let's consider X = 9. Then, 120*X = 120*9 = 1080. But 1080 is already a 4-digit number, which is too big. So, X can't be 9 because even without adding Y and Z, it's already over 1000.Next, try X = 8. Then, 120*X = 120*8 = 960. That's a good start because 960 is a 3-digit number, and we can add to it.Now, let's assign the next largest value to Y. Since X is 8, Y can be 9 (as long as Y is different from X). So, Y = 9. Then, 100*Y = 100*9 = 900.So, adding that to 960, we get 960 + 900 = 1860. But wait, that's already 1860, which is a 4-digit number. That's too big. So, we need to adjust.Hmm, maybe Y can't be 9 if X is 8 because the sum becomes too large. Let's try Y = 7. Then, 100*Y = 700. Adding that to 960 gives us 960 + 700 = 1660. Still too big.Wait, this approach isn't working. Maybe I need to consider that the total sum must be less than 1000. So, 120*X + 100*Y + 3*Z < 1000.Given that, let's set up the inequality:120*X + 100*Y + 3*Z < 1000We need to find the maximum value of 120*X + 100*Y + 3*Z such that X, Y, Z are distinct digits (0-9) and the sum is less than 1000.Let me try to find the maximum possible values for X and Y such that 120*X + 100*Y is as large as possible without exceeding 1000 when adding 3*Z.Assuming Z is at most 9, 3*Z is at most 27. So, 120*X + 100*Y must be less than 1000 - 27 = 973.So, 120*X + 100*Y < 973Let me try X = 8:120*8 = 960Then, 100*Y < 973 - 960 = 13So, Y must be less than 13/100, which is 0.13. But Y has to be a digit, so Y can only be 0.But if Y = 0, then Z has to be different from X and Y, so Z can be 1-9 except 8 and 0.But let's check the total sum:120*8 + 100*0 + 3*Z = 960 + 0 + 3*ZTo make it as large as possible, Z should be 9:960 + 27 = 987That's a 3-digit number, and all digits are different: X=8, Y=0, Z=9.Wait, but Y=0 might be an issue because in the second number YXZ, Y is in the hundreds place. If Y=0, then YXZ would effectively be a 2-digit number, which is allowed in arithmetic, but I need to confirm if leading zeros are permitted. In standard arithmetic, leading zeros are not written, but since the problem didn't specify, I think it's acceptable.So, the sum would be 987, which is one of the options (B). But let me check if I can get a higher sum with different values.Let me try X=7:120*7 = 840Then, 100*Y < 973 - 840 = 133So, Y can be up to 1 (since 100*1 = 100). Let's try Y=1.Then, 840 + 100 = 940Now, 3*Z can be up to 27, so total sum would be 940 + 27 = 967, which is less than 987.Alternatively, if Y=2:100*2 = 200840 + 200 = 1040, which is already over 1000, so that's invalid.So, with X=7, the maximum sum is 967, which is less than 987.What about X=6:120*6 = 720100*Y < 973 - 720 = 253So, Y can be up to 2 (200). Let's try Y=2.720 + 200 = 920Then, 3*Z can be up to 27, so total sum is 920 + 27 = 947, which is still less than 987.Similarly, higher Y would exceed the limit.So, X=8 seems to give the highest possible sum.Wait, but earlier when I tried X=8 and Y=9, the sum was 1860, which is too big. But when I set Y=0, I got 987, which is valid.Is there a way to have Y=1 with X=8?Let's check:X=8, Y=1120*8 + 100*1 = 960 + 100 = 1060Adding 3*Z, which is at most 27, gives 1060 + 27 = 1087, which is too big.So, Y=1 with X=8 is invalid.What about Y=0? As before, that gives 960 + 0 + 27 = 987.Is there a way to have Y=0 and Z=9, which gives 987, and all digits are different: X=8, Y=0, Z=9.Yes, that works.But let me check if there's another combination where X=8, Y=0, and Z=9 gives 987, which is option B.Alternatively, could I have X=8, Y=0, Z=9, which is 987.Is there a way to get a higher sum?Wait, if I set X=8, Y=0, Z=9, sum is 987.If I set X=8, Y=0, Z=8, but Z must be different from X, so Z can't be 8.Similarly, Z=7 would give 960 + 0 + 21 = 981, which is option A, but 987 is higher.So, 987 is higher.Is there a way to have X=8, Y=1, but as we saw earlier, that would make the sum exceed 1000.Alternatively, what if X=7, Y=9?Let me check:X=7, Y=9120*7 = 840100*9 = 900Total so far: 840 + 900 = 1740Adding 3*Z, which is at most 27, gives 1740 + 27 = 1767, which is way too big.So, that's invalid.What if X=7, Y=8?120*7 = 840100*8 = 800Total: 840 + 800 = 1640Adding 3*Z: 1640 + 27 = 1667, still too big.So, no.What about X=7, Y=7? But Y must be different from X, so Y can't be 7.So, X=7, Y=6:120*7 = 840100*6 = 600Total: 840 + 600 = 1440Adding 3*Z: 1440 + 27 = 1467, still too big.So, no.It seems that the only way to get a valid 3-digit sum with X=8 is when Y=0, which gives us 987.Wait, but let me check if there's another combination where X=8, Y=0, Z=9, which gives 987, but maybe with different digits.Alternatively, could I have X=8, Y=0, Z=9, which is 987.Is there a way to have a higher sum?Wait, if I set X=8, Y=0, Z=9, sum is 987.If I set X=8, Y=0, Z=8, but Z must be different from X, so Z can't be 8.Similarly, Z=7 would give 960 + 0 + 21 = 981, which is option A, but 987 is higher.So, 987 is higher.Is there a way to have X=8, Y=1, but as we saw earlier, that would make the sum exceed 1000.Alternatively, what if X=8, Y=0, Z=9, which is 987.Wait, but let me check the actual addition to make sure.So, XXZ is 889.YXZ is 089, which is 89.Z is 9.Adding them together:889+ 89+ 9Let's add 889 + 89 first.889 + 89:889 + 80 = 969969 + 9 = 978Then, add 9:978 + 9 = 987.Yes, that works.So, the sum is 987, which is a 3-digit number, and all digits X=8, Y=0, Z=9 are different.Is there a way to get a higher sum?Let me think.If I set X=8, Y=0, Z=9, sum is 987.What if I set X=8, Y=0, Z=9, which is the same.Alternatively, could I set X=8, Y=0, Z=9, which is 987.Is there a way to have X=8, Y=0, Z=9, which is 987.Wait, but what if I set X=8, Y=0, Z=9, which is 987.Is there a way to have a higher sum?Wait, if I set X=8, Y=0, Z=9, sum is 987.If I set X=8, Y=0, Z=9, which is 987.I think that's the maximum.Alternatively, could I set X=8, Y=0, Z=9, which is 987.Yes, that seems to be the highest possible.Wait, but let me check if there's another combination where X=8, Y=0, Z=9, which is 987.Alternatively, could I set X=8, Y=0, Z=9, which is 987.I think that's the maximum.So, the answer should be 987, which is option B.

💡Okay, so I've got this math problem here: 300 × 2 + (12 + 4) × 1 / 8. Hmm, I need to figure out what this equals. Alright, let's break it down step by step. First, I remember that when solving equations, I should follow the order of operations, which is often remembered by the acronym PEMDAS—Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).Looking at the equation, I see there are parentheses: (12 + 4). According to PEMDAS, I should solve what's inside the parentheses first. So, 12 plus 4 equals 16. Now, the equation simplifies to 300 × 2 + 16 × 1 / 8.Next, I need to handle the multiplication and division from left to right. Let's start with 300 multiplied by 2. 300 times 2 is 600. So now, the equation is 600 + 16 × 1 / 8.Moving on, I have 16 multiplied by 1. Well, any number multiplied by 1 is itself, so 16 times 1 is still 16. Now the equation is 600 + 16 / 8.Now, I need to perform the division: 16 divided by 8. 16 divided by 8 equals 2. So, the equation simplifies to 600 + 2.Finally, I just need to add those two numbers together: 600 plus 2 equals 602. So, the result of the equation is 602.Wait, let me double-check to make sure I didn't make any mistakes. Starting over, 300 times 2 is definitely 600. Then, 12 plus 4 is 16. 16 times 1 is 16, and 16 divided by 8 is 2. Adding 600 and 2 gives me 602. Yep, that seems right. I think I've got it!

💡Okay, so I have this problem about a freight train that was delayed by 12 minutes and then made up for the lost time by increasing its speed over a distance of 60 km. I need to find the original speed of the train. Hmm, let me try to break this down step by step.First, let's understand what's happening. The train was supposed to travel a certain distance at its original speed, but it got delayed by 12 minutes. To make up for that delay, it increased its speed by 15 km/h for the next 60 km. So, by going faster for those 60 km, it managed to save enough time to compensate for the 12-minute delay.Alright, so I think I can model this with some equations. Let me denote the original speed of the train as ( x ) km/h. Then, the increased speed would be ( x + 15 ) km/h.Now, the train was delayed by 12 minutes. I should convert that into hours because the speed is in km/h. So, 12 minutes is ( frac{12}{60} = 0.2 ) hours.The key here is that the time saved by increasing the speed over the 60 km equals the 0.2 hours delay. So, the time it would have taken at the original speed minus the time it took at the increased speed equals 0.2 hours.Let me write that as an equation. The time taken to travel 60 km at the original speed is ( frac{60}{x} ) hours, and the time taken at the increased speed is ( frac{60}{x + 15} ) hours. The difference between these two times is 0.2 hours.So, the equation is:[frac{60}{x} - frac{60}{x + 15} = 0.2]Now, I need to solve this equation for ( x ). Let me see. To solve this, I can first find a common denominator for the fractions on the left side. The common denominator would be ( x(x + 15) ).So, rewriting the equation:[frac{60(x + 15) - 60x}{x(x + 15)} = 0.2]Simplifying the numerator:[60x + 900 - 60x = 900]So, the equation becomes:[frac{900}{x(x + 15)} = 0.2]Now, I can cross-multiply to get rid of the fraction:[900 = 0.2 times x(x + 15)]Let me compute the right side:[0.2 times x(x + 15) = 0.2x^2 + 3x]So, the equation is:[900 = 0.2x^2 + 3x]To make it easier, I can multiply both sides by 5 to eliminate the decimal:[5 times 900 = 5 times (0.2x^2 + 3x)][4500 = x^2 + 15x]Now, let's rearrange this into a standard quadratic equation:[x^2 + 15x - 4500 = 0]Okay, now I have a quadratic equation. I can solve this using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Where ( a = 1 ), ( b = 15 ), and ( c = -4500 ).Plugging in the values:[x = frac{-15 pm sqrt{15^2 - 4 times 1 times (-4500)}}{2 times 1}][x = frac{-15 pm sqrt{225 + 18000}}{2}][x = frac{-15 pm sqrt{18225}}{2}][x = frac{-15 pm 135}{2}]So, there are two possible solutions:1. ( x = frac{-15 + 135}{2} = frac{120}{2} = 60 )2. ( x = frac{-15 - 135}{2} = frac{-150}{2} = -75 )Since speed can't be negative, we discard the negative solution. Therefore, the original speed of the train is 60 km/h.Let me just double-check to make sure this makes sense. If the original speed is 60 km/h, then the increased speed is 75 km/h. The time to travel 60 km at 60 km/h is 1 hour. The time to travel 60 km at 75 km/h is ( frac{60}{75} = 0.8 ) hours. The difference is ( 1 - 0.8 = 0.2 ) hours, which is 12 minutes. That matches the delay, so it checks out.Yeah, I think that's correct. The original speed of the train was 60 km/h.

💡Okay, so I have this problem where there are four smaller circles, each with a radius of 2. They're all externally tangent to each other and internally tangent to a larger circle. I need to find the radius of this larger circle. Hmm, let me try to visualize this. There's a big circle, and inside it, four smaller circles are arranged in such a way that each is touching the other three and also touching the big circle from the inside.First, I should probably draw a diagram to help me understand. Since I can't draw right now, I'll imagine it. The four small circles are placed symmetrically inside the larger one. Each small circle touches two others, and all of them touch the larger circle. So, their centers must form some sort of regular shape inside the larger circle.Wait, if each small circle is externally tangent to the others, that means the distance between any two centers of the small circles should be equal to the sum of their radii. Since each has a radius of 2, the distance between any two centers should be 2 + 2 = 4 units. So, the centers of the four small circles form a square where each side is 4 units long.Okay, so the centers of the four small circles are at the vertices of a square with side length 4. Now, I need to find the radius of the larger circle that circumscribes this square and also has each small circle touching it internally. That means the distance from the center of the larger circle to each center of the small circles plus the radius of the small circle should equal the radius of the larger circle.Let me denote the radius of the larger circle as R. The centers of the small circles are each at a distance of R - 2 from the center of the larger circle because the small circles are internally tangent. So, if I can find the distance from the center of the larger circle to any of the small circle centers, I can add 2 to it to get R.Since the centers of the small circles form a square, the center of the larger circle should coincide with the center of this square. In a square, the distance from the center to any vertex is equal to half the length of the diagonal. The diagonal of a square with side length s is s√2. So, in this case, the diagonal would be 4√2, and half of that is 2√2. Therefore, the distance from the center of the larger circle to any small circle center is 2√2.Adding the radius of the small circle, which is 2, to this distance gives the radius of the larger circle. So, R = 2√2 + 2. I can factor out the 2 to make it look nicer: R = 2(√2 + 1).Wait, let me double-check that. The centers form a square with side length 4, so the diagonal is 4√2. Half of that is 2√2, which is the distance from the center of the square to any vertex. Since each small circle has a radius of 2, the radius of the larger circle must be 2√2 + 2. Yeah, that seems right.I think I got it. The radius of the larger circle is 2 times (√2 + 1), which is approximately 4.828 units. That makes sense because the centers are 2√2 away, and adding the 2 radius gives the total radius needed to encompass all the small circles.**Final Answer**The radius of the large circle is boxed{2(sqrt{2} + 1)}.