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A snail is at the bottom of a 6 meter well. Each day the snail walks up 2.5 meters but at night it slips down 1.5 meters. On which day will the snail get out of the well?

The following diagram shows the first three patterns of squares in a sequence.

How many squares are there in the 50th pattern?

Twenty lamp posts are placed in a straight line. The distance between any two consecutive posts is 6 meters. What is the distance between the 1st and the 10th lamp post?

Answers

Each day the snail walks up 2.5 meters but at night it slips down 1.5 meters.

Net distance travelled each day = 2.5 – 1.5 = 1m

In 4 days, the snail will climb 4 meters.

On the 5th day, the snail will climb 4 + 2.5 = 6 meters and will be out of the well and will not slide back inside.

Therefore the snail will come out of the well on the 5th day.

The way to solve such questions is to look at the total distance – the distance the snail travels up each day and divide this by the net distance travelled + 1 more day

Total distance – the distance travelled up in one day = 6 – 2.5 = 3.5 meters.

Now the net distance travelled by the snail per day is 1 meters. So it will travel this distance in 3.5 or 4 days. And on the 5th day, it will go out of the well.

As you observe the pattern,

1st figure has 1 square

2nd figure has 3 squares

3rd figure has 5 squares, so on

The pattern they follow is that the number of squares is one less than double the number of the figure.

So the nth figure will have 2n – 1 squares. i.e. (double the no) – 1

The 50th figure will have = (2 x 50) – 1 = 100 – 1 = 99 squares

The distance between any two consecutive posts is 6 meters.

Number of gaps between 1st to 10th lamp post = 10 – 1 = 9

The distance between the 1st and the 10th lamp post = 9 x 6 = 54 meters

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