## Solved Distance Puzzles For HCL Placement Tests

Below are three problems based on distance and time calculations.

**Question 1**

A boy drops a bouncing toy from 1st floor to ground floor. Each time, it bounces vertically to a height of 1/2 of the preceding bounce's height. Find the distance the toy will cross if it totally takes 6 bounces and the distance between each floor is 10m.

a)39 m b)30 m c)40 m d)15 m

**Answer :**b)30 m

Solution :

The distance between two floors is 10m.

The boy dropped the bouncing toy from a height of 10 m.

As the toy is initially dropped, the toy falls 10 meters. Then, the toy rebounds to 10 x 1/2 = 5 meters and falls back down 5 meters. This process will repeat for 5 more times.

The boy dropped the bouncing toy from a height of 10 m.

As the toy is initially dropped, the toy falls 10 meters. Then, the toy rebounds to 10 x 1/2 = 5 meters and falls back down 5 meters. This process will repeat for 5 more times.

Then:

Distance that the toy travels down = 10 + 10x1/2 + 10x 1/2 x 1/2 + ...+ 10x(1/2)

= 10 + 10x(1/2)

^{6}= 10 + 10x(1/2)

^{1}+ 10(1/2)^{2}+ 10(1/2)^{3}+ 10x(1/2)^{4}+ 10(1/2)^{5}+ 10(1/2)^{6}which is a g.p series with 7 terms and first term = 10 common term = 1/2.
we know that, sum of n terms in a g.p series is

sum = a(1 - r

Here, distance = 10 x [1 - (1/2)

= 20x127/128 = 19.84 meters.

Distance that the toy travels up = 10x(1/2)

= 19.84 - 10 = 9.84 meters.

sum = a(1 - r

^{n})/1 - r when r is less than oneHere, distance = 10 x [1 - (1/2)

^{7}]/(1-1/2) = 10 x 127/128 / 1/2= 20x127/128 = 19.84 meters.

Distance that the toy travels up = 10x(1/2)

^{1}+ 10(1/2)^{2}+ 10(1/2)^{3}+ 10x(1/2)^{4}+ 10(1/2)^{5}+ 10(1/2)^{6}= 19.84 - 10 = 9.84 meters.

Thus, the total distance is 19.84 + 9.84 = 29.68 meters = 30 meters (approximately)

**Question 2**

A spider starts jumping up from the bottom of a ladder with 32 steps. Each minute, it jumps 4 steps and slips back 2 steps. How much time would the spider take to reach the top of the ladder?

a)15 minutes b)14 minutes c)16 minutes d)18 minutes

**Answer :**a)15minutes.

Solution :

Think of it like a number line: In the first minute, the spider goes 4 steps up, and slips back 2 steps.

0...1...2...3...4...5...6 |---|---|---|---|---|---| ----------------> ........<-------

(Imagine that's going up instead of horizontal.)

The spider effectively climbs 4 steps - 2 steps = 2 steps higher after the first minute.

In the second minute, the spider goes up 4 more steps, and slips back 2 steps. So he's at 2 steps + 4 steps - 2 steps = 4 steps.

Every minute, it gains 4 steps and loses 2 steps, for a total gain of 2 steps. So after the 14th minute, it is at 28 steps height.

Now, here's the twist: in the 15th minute, it jumps up 4 steps. Since 28 steps + 4 steps = 32 steps, it is at the top of the ladder.

Hence the answer is 15 minutes.

In the second minute, the spider goes up 4 more steps, and slips back 2 steps. So he's at 2 steps + 4 steps - 2 steps = 4 steps.

Every minute, it gains 4 steps and loses 2 steps, for a total gain of 2 steps. So after the 14th minute, it is at 28 steps height.

Now, here's the twist: in the 15th minute, it jumps up 4 steps. Since 28 steps + 4 steps = 32 steps, it is at the top of the ladder.

Hence the answer is 15 minutes.

**Question 3**

A man throws a ball from certain height to the ground and it covers a distance of X meters on the ground. Each time it's bounce covers 2/3 of previous distance. The ball stops bouncing after 4 bounces. Then the total distance it covers on the ground is:

a)191X/43 meters b)250X/21 meters c)172X/75 meters d)211X/81 meters

**Answer :**d)211X/81 meters

Solution :

This problem is slightly different from the 1st one.

Here, we have to find the distance which covered by the ball on the ground.

Given that, it covers X meters by the through.

And it takes four bounce, each bounce covers 2/3 of previous distance.

Then the distance covered in 1st bounce = (2/3)X

Here, we have to find the distance which covered by the ball on the ground.

Given that, it covers X meters by the through.

And it takes four bounce, each bounce covers 2/3 of previous distance.

Then the distance covered in 1st bounce = (2/3)X

The distance covered in 2nd bounce = (2/3) of (2/3)X = X(2/3)

The distance covered in 3rd bounce = (2/3) of X(2/3)

And the distance covered in 4th bounce = (2/3) of X(2/3)

^{2}The distance covered in 3rd bounce = (2/3) of X(2/3)

^{2}= X(2/3)^{3}And the distance covered in 4th bounce = (2/3) of X(2/3)

^{3}= X(2/3)^{4}
Required distance = X + (2/3)X + X(2/3)

The above series is a G.P series of 5 terms with a = X and r = 2/3

Then sum = a(1 - r

= X[1 - (2/3)

= X[1 - 32/243]/(1/3)

= 3X(211/243) = 211X/81

Hence the required answer is 211X/81 meters.

^{2}+ X(2/3)^{3}+ X(2/3)^{4}The above series is a G.P series of 5 terms with a = X and r = 2/3

Then sum = a(1 - r

^{n})/(1 - r)= X[1 - (2/3)

^{5}]/(1 - 2/3)= X[1 - 32/243]/(1/3)

= 3X(211/243) = 211X/81

Hence the required answer is 211X/81 meters.