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ruslelena [56]
3 years ago
9

Two tourists went on a hike at dawn. One went from a to b and another one went from b to a. They met at noon but did not stop an

d continued walking maintaining same speed for the whole trip. One finished his hike at 4pm in B and another one came to A at 9 pm. At what hour was dawn that day?
Mathematics
1 answer:
Alecsey [184]3 years ago
8 0
<span>Dawn was at 6 am. Variables a = distance from a to passing point b = distance from b to passing point c = speed of hiker 1 d = speed of hiker 2 x = number of hours prior to noon when dawn is The first hiker travels for x hours to cover distance a, and the 2nd hiker then takes 9 hours to cover that same distance. This can be expressed as a = cx = 9d cx = 9d x = 9d/c The second hiker travels for x hours to cover distance b, and the 1st hiker then takes 4 hours to cover than same distance. Expressed as b = dx = 4c dx = 4c x = 4c/d We now have two expressions for x, set them equal to each other. 9d/c = 4c/d Multiply both sides by d 9d^2/c = 4c Divide both sides by c 9d^2/c^2 = 4 Interesting... Both sides are exact squares. Take the square root of both sides 3d/c = 2 d/c = 2/3 We now know the ratio of the speeds of the two hikers. Let's see what X is now. x = 9d/c = 9*2/3 = 18/3 = 6 x = 4c/d = 4*3/2 = 12/2 = 6 Both expressions for x, claim x to be 6 hours. And 6 hours prior to noon is 6am. We don't know the actual speeds of the two hikers, nor how far they actually walked. But we do know their relative speeds. And that's enough to figure out when dawn was.</span>
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(a) Find the three-digit number that increases by 30% when each of its
Elden [556K]

Step-by-step explanation:

So a three digit number can be expressed as: 100a + 10b + c where a is the third digit, b is the second digit, and c is the first digit. Or in other words a is the hundreds place, b is the tens place, and c is the ones place. When something "increases" by 30%, it is 130% it's original value, and to calculate how much that is, you simply convert 130% to a decimal by dividing by 100, which gives you 1.30. And since all the digits are increased by 1, you have the equation:

100(a+1) + 10(b+1) + 1(c+1) = 1.30(100a+10b+c)

Distribute the multiplication on the left side:

100a+100+10b+10+c+1=1.30(100a+10b+c)

Distribute the multiplication on the right side:

100a+100+10b+10+c+1=130a+13b+1.3c

Add like terms on the left side:

100a+10b+c+111=130a+13b+1.3c

Subtract 100a, 10b, and c from both sides

111=30a+3b+0.3c

So "technically" you can just plug in any two values, and then solve for the last value, but since you have a 3 digit number, you have the restriction of a < 10, b < 10, and c < 10, and also a, b, and c, should only be integers and all have the same sign or it wouldn't be a 3 digit number.

So let's start with a, since it has the highest coefficient, well you can fit 30 into 111, 3 times without going over so that's the first value

111 = 30(3) + 3b + 0.3c

111 = 90 + 3b + 0.3c

Now subtract the 90 from both sides

21=3b+0.3c

Well 3 can fit into 21, 7 times!

21 = 3(7) + 0.3c

21 = 21 + 0.3c

subtract 21 from both sides

0 = 0.3c

and now obviously c is 0, if you want you can divide both sides by 0.3 but it's a bit redundant

c = 0

This gives you the three values, a=3, b=7, c=0. which is the number 370. Now let's double check. Adding 1 to each digit would give you 481 and 481/370 = 1.3, so it is correct!

part b:

So to prove there is no three digit number, is to realize there is no solution, given the restriction or integers, greater than or equal to 0, and less than 10, and all of them must have the same sign.

So let's start with the same equation except this time instead of 1.3 it's 1.4

100(a+1) + 10(b+1) + 1(c+1) = 1.40(100a+10b+c)

Distribute on the left side;

100a+100+10b+10+c+1=1.40(100a+10b+c)

Distribute on the right side:

100a+100+10b+10+c+1=140a + 14b + 1.4c

Add like terms on left side:

100a + 10b + c + 111 = 140a + 14b + 1.4c

Subtract 100a, 10b, and c from both sides:

111 = 40a + 4b + 0.4c

Now to do the same process, let's start by finding how many times we can fit 40 into 111, and if you're wondering why we start with 40, it's because let's say for example I just say, I can fit another 40 into it, but I decide not to, and let b do that, well even if it's just 40, b will have be at least 10, which does not fit our restrictions, so you have to fit as many 40's into the number first then go the other numbers.

So only 2 40's can fit in 111 without going over the value

111 = 40(2) + 4b + 0.4c

Subtract 80 from both sides

31=4b+0.4c

4 can fit into 31, 7 times

31 = 4(7) + 0.4c

31 = 28 + 0.4c

subtract 28 from both sides

3 = 0.4c

divide both sides by 0.4

7.5 = c.

Since c is not an integer there is no 3 digit number that exists that increases by 40% whenever you increase it by 1.

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zhuklara [117]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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