# Question 22

## Question 22

Question 22 :

Tintin unlocks the chest, and plugs the key into the main door. A dial emerges on the door. Beginning with the dial set at zero, the dial must be turned counter-clockwise to the first combination number, (then clockwise back to zero), and clockwise to the second combination number, (then counter-clockwise back to zero), and counter-clockwise again to the third and final number, where upon the door shall immediately spring open. There are 40 numbers on the dial, including the zero. Without knowing the combination numbers, what is the maximum number of trials required to open the safe (one trial equals one attempt to dial a full three-number combination)?

(a) 64000 (b) 1600 (c) 63999 (d) 1599

Tintin unlocks the chest, and plugs the key into the main door. A dial emerges on the door. Beginning with the dial set at zero, the dial must be turned counter-clockwise to the first combination number, (then clockwise back to zero), and clockwise to the second combination number, (then counter-clockwise back to zero), and counter-clockwise again to the third and final number, where upon the door shall immediately spring open. There are 40 numbers on the dial, including the zero. Without knowing the combination numbers, what is the maximum number of trials required to open the safe (one trial equals one attempt to dial a full three-number combination)?

(a) 64000 (b) 1600 (c) 63999 (d) 1599

**TECHNOTHLON**- Admin
- Posts : 39

Join date : 2016-07-19

## Re: Question 22

Now in this question i hav huge doubt.. Is the combination "000" allowed as a password? If yes then the ans shud be (a), n if no thn (b).. I think we shud count 000 as it is a possible combination.. What abt others?

**ABC**- Guest

## Re: Question 22

The answer should be 64000.

This can be achieved by permutation and combination.

Three places each can be filled by 40 numbers .So, 40*40*40 =64000.☺☺☺

This can be achieved by permutation and combination.

Three places each can be filled by 40 numbers .So, 40*40*40 =64000.☺☺☺

**Peeyush Sahu**- Guest

## The Options don't seem to be correct.

Using P&C,

we get 40*40*40 = 64,000.

But 000 won't be counted

=> 64,000-1 = 63,999.

But even 63,999 is WRONG.

Consider a case with 0.

Ex-

504

Here, when he turns the Knob from 5 back to 0, 0 is already counted.

Same goes for- 540, 054 or any no. with 0.

504, 054 & 540 are already counted while doing 541 or 542 or 54... .

This means any no. with 0 already gets counted.

So,

It will be 39*39*39 by P&C.

=> 59319

we get 40*40*40 = 64,000.

But 000 won't be counted

=> 64,000-1 = 63,999.

But even 63,999 is WRONG.

Consider a case with 0.

Ex-

504

Here, when he turns the Knob from 5 back to 0, 0 is already counted.

Same goes for- 540, 054 or any no. with 0.

504, 054 & 540 are already counted while doing 541 or 542 or 54... .

This means any no. with 0 already gets counted.

So,

It will be 39*39*39 by P&C.

=> 59319

**Chinmay Hebbar**- Guest

## Answer is 1600

it is so because.. total possible combinations are 40x40x40.. but in the question it is written that the lock will spring open in the 3rd rotation... so we can eliminate the trials of the 3rd rotation.. so the possible combinations are.. 40x40 =1600

**Guest**- Guest

## Correction to my solution

Here,

The solution (which is (d)) is obtained by -

Using Permutation&Combinations,

we get 40*40*1 (Since the last round, it opens immediately) - 1 (Case of 000) = 1599

However,

This solution is wrong because-

If Consider a case with 0

For example,

504

Here, when he turns the Knob from 5 back to 0, 0 is already counted.

As stated in the passage "(then counter-clockwise back to zero)".

Same goes for- 540, 054 or any no. with 0.

504, 054 & 540 are already counted while doing 541 or 542 or 54... .

This means any no. with 0 already gets counted.

So,

It will be 39*39*1 by P&C.

=> 1521

The solution (which is (d)) is obtained by -

Using Permutation&Combinations,

we get 40*40*1 (Since the last round, it opens immediately) - 1 (Case of 000) = 1599

However,

This solution is wrong because-

If Consider a case with 0

For example,

504

Here, when he turns the Knob from 5 back to 0, 0 is already counted.

As stated in the passage "(then counter-clockwise back to zero)".

Same goes for- 540, 054 or any no. with 0.

504, 054 & 540 are already counted while doing 541 or 542 or 54... .

This means any no. with 0 already gets counted.

So,

It will be 39*39*1 by P&C.

=> 1521

**Chinmay Hebbar**- Guest

## Re: Question 22

Answer keys are available on Technothlon downloads page.Type your score here accordingly.

**All the best**- Guest

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