#### A hundred stones are placed, in a straight line, a yard distant from each other. How many yards must a person walk, who undertakes to pick them up, and place them in a basket stationed one yard from the first stone?

**Answer: ** In solving this question it is clear that to pick up the first stone and put it into the basket, the person must walk two yards, one in going for the stone and another in returning with it; that for the second stone he must walk four yards, and so on increasing by two as far as the hundredth, when he must walk two hundred yards, so that the sum total will be the product of 202 multiplied by 50, or 10,100 yards. If any one does not see why we multiply 202 by 50 in getting the answer, we refer him to his arithmetic.

**Solution: **

## Show Answer

In solving this question it is clear that to pick up the first stone and put it into the basket, the person must walk two yards, one in going for the stone and another in returning with it; that for the second stone he must walk four yards, and so on increasing by two as far as the hundredth, when he must walk two hundred yards, so that the sum total will be the product of 202 multiplied by 50, or 10,100 yards. If any one does not see why we multiply 202 by 50 in getting the answer, we refer him to his arithmetic.

## Show Answer

In solving this question it is clear that to pick up the first stone and put it into the basket, the person must walk two yards, one in going for the stone and another in returning with it; that for the second stone he must walk four yards, and so on increasing by two as far as the hundredth, when he must walk two hundred yards, so that the sum total will be the product of 202 multiplied by 50, or 10,100 yards. If any one does not see why we multiply 202 by 50 in getting the answer, we refer him to his arithmetic.

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#### My only timepiece is a wall clock. One day I forgot to wind it and it stopped. I went to visit a friend whos watch is always correct, stayed awhile, and then went home. There I made a simple calculation and set the clock right. How did I do this even though I had no watch on me to tell how long it took me to return from my friend’s house?

**Answer: ** Before I left, I wound the wall clock. When I returned, the change in time equaled how long it took to go to my friends house and return, plus the time I spent there. But I knew the latter because I looked at my friends watch when I arrived and left.
Subtracting the time of the visit from the time I was absent from my house, and dividing by 2, I obtained the time it took me to return home. I added this time to what my friend watch showed when I left, and set the sum on my wall clock.

**Solution: **

## Show Answer

Before I left, I wound the wall clock. When I returned, the change in time equaled how long it took to go to my friends house and return, plus the time I spent there. But I knew the latter because I looked at my friends watch when I arrived and left.

Subtracting the time of the visit from the time I was absent from my house, and dividing by 2, I obtained the time it took me to return home. I added this time to what my friend watch showed when I left, and set the sum on my wall clock.

## Show Answer

Before I left, I wound the wall clock. When I returned, the change in time equaled how long it took to go to my friends house and return, plus the time I spent there. But I knew the latter because I looked at my friends watch when I arrived and left.

Subtracting the time of the visit from the time I was absent from my house, and dividing by 2, I obtained the time it took me to return home. I added this time to what my friend watch showed when I left, and set the sum on my wall clock.

Subtracting the time of the visit from the time I was absent from my house, and dividing by 2, I obtained the time it took me to return home. I added this time to what my friend watch showed when I left, and set the sum on my wall clock.

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