# GF's Passtime Puzzles--With New Format

Question asked by Harshad Italiya in #Brainy Puzzles on Jul 2, 2009

Harshad Italiya · Jul 2, 2009

Rank A1 - PRO

**GF's Crazy Puzzles--With New Format**

Hello Ceans,

Here i am going to Start a simple thread for puzzles but with some interesting Rules...

Hop you guys will enjoy it..

**Rules:**

- You will get a points on answering the Puzzles.

- You have
**5 Tries**to answer the Puzzle.

- If you answer Correctly in...

1st Try - 5 Points

2nd Try - 4 Points

3rd Try - 3 Points

4th Try - 2 Points

5th Try - 1 Point

- If you answer Correctly 3 Puzzles continuously you will get 5 Bonus Points.

So... Are you ready for this one !! 😁 Posted in: #Brainy PuzzlesPoint Card:

RajdeepCE............10 Points

pradeep_agrawal...9 Points

shalini_goel14.......5 Points

silverscorpion........5 Points

English-Scared......9 Points

mech_guy.............10 Points

Harshad Italiya · Jul 2, 2009

Rank A1 - PRO

**1.**

There are 50 Balls in a Bowl of Red and Yellow Colors.

All balls are not of same color.

You don't know how many balls are of Red or Yellow Colors.

but, When you pick two balls randomly you get Yellow color ball always.

So How many of them are Red and how many of them are Yellow color?

RajdeepCE · Jul 2, 2009

Rank B3 - LEADER

There are 49 yellow & 1 red ball.

Am I right?

Am I right?

Harshad Italiya · Jul 2, 2009

Rank A1 - PRO

You got it Rajdeep !!

+5 Points added.

+5 Points added.

Harshad Italiya · Jul 2, 2009

Rank A1 - PRO

**2.**

There are two friends they are sharing same

*birthdate*but one of them is 2555 days older then other one...

So Whats the age of both friends? (Assume that both are living till date) 😉

pradeep_agrawal · Jul 2, 2009

Rank C2 - EXPERT

Is it 1939yr and 1932yr?

-Pradeep

-Pradeep

Harshad Italiya · Jul 2, 2009

Rank A1 - PRO

You missed it by Short Pradeep.. Good Try there.

there is 2557 days between 1932 and 1939 😀

there is 2557 days between 1932 and 1939 😀

RajdeepCE · Jul 2, 2009

Rank B3 - LEADER

The birthyear of the both friends should be 1896 and 1903 & the birth month should be after February. The reason behind this is that 2555/365 = 7, but we all know that there is a leap year at every 4 year. Thats why I am confused & it took my time to solve the puzzle. Than I referred calender and I found that year 1900 doesnt have a leap day in it. And I found the solution!!!

Am I right, GF?

Am I right, GF?

pradeep_agrawal · Jul 2, 2009

Rank C2 - EXPERT

By 1939yr and 1932yr i mean the age of the people. That i calculated based on the tribulation period which was of 7years (2555 days).

If we go by other logic, the birth date can be

March 1, 1896 and March 1, 1903 where the difference of age in terms of days is 2555.

There can be many answers by this logic, e.g.,

March 2, 1896 and March 1, 1903

March 1, 1796 and March 1, 1803

.

.

.

can also be answers to this.

-Pradeep

If we go by other logic, the birth date can be

March 1, 1896 and March 1, 1903 where the difference of age in terms of days is 2555.

There can be many answers by this logic, e.g.,

March 2, 1896 and March 1, 1903

March 1, 1796 and March 1, 1803

.

.

.

can also be answers to this.

-Pradeep

Harshad Italiya · Jul 3, 2009

Rank A1 - PRO

You again got the correct answer... and nice explainationRajdeepCEThe birthyear of the both friends should be 1896 and 1903 & the birth month should be after February. The reason behind this is that 2555/365 = 7, but we all know that there is a leap year at every 4 year. Thats why I am confused & it took my time to solve the puzzle. Than I referred calender and I found that year 1900 doesnt have a leap day in it. And I found the solution!!!

Am I right, GF?

**+5 added**

Yes we can get the same result between 100 years interval... You missed it just by one Pradeep Nice effort impressive. 😀pradeep_agrawalBy 1939yr and 1932yr i mean the age of the people. That i calculated based on the tribulation period which was of 7years (2555 days).

If we go by other logic, the birth date can be

March 1, 1896 and March 1, 1903 where the difference of age in terms of days is 2555.

There can be many answers by this logic, e.g.,

March 2, 1896 and March 1, 1903

March 1, 1796 and March 1, 1803

.

.

.

can also be answers to this.

-Pradeep

Harshad Italiya · Jul 3, 2009

Rank A1 - PRO

oh i just missed the time you both has replied at same time... Pradeep way to go you also get

@RajdeepCE.. now you have a Golden chance to get bonus Points. 😁

**4 points**. as you got it in your second reply..@RajdeepCE.. now you have a Golden chance to get bonus Points. 😁

Harshad Italiya · Jul 3, 2009

Rank A1 - PRO

**3.**

Here is one Series for Series Solver..😉

0,2,9,24,50,90,.........

Find the Next 3 figures.

shalini_goel14 · Jul 3, 2009

Rank A3 - PRO

If I am not wrong AGES of the friends was asked in the question, not their BIRTHYEARS right? Can anyone tell me their AGES? 😕RajdeepCEThe birthyear of the both friends should be 1896 and 1903 & the birth month should be after February. The reason behind this is that 2555/365 = 7, but we all know that there is a leap year at every 4 year. Thats why I am confused & it took my time to solve the puzzle. Than I referred calender and I found that year 1900 doesnt have a leap day in it. And I found the solution!!!

Harshad Italiya · Jul 3, 2009

Rank A1 - PRO

One is

**113**Years Old and other is of**106**years old.Harshad Italiya · Jul 4, 2009

Rank A1 - PRO

i wrote wrong series.. i had edited it now please try to solve it.. sorry for your inconvenience..

3.

Here is one Series for Series Solver..:wink:

0,2,9,24,50,90,.........

Find the Next 3 figures.

shalini_goel14 · Jul 4, 2009

Rank A3 - PRO

Is the series like following

0, 2, 9, 24, 50, 90,

0, 2, 9, 24, 50, 90,

**147**,**224**,**324**?Harshad Italiya · Jul 4, 2009

Rank A1 - PRO

You got the right three Shalini 😁 +5 Points added.shalini_goel14Is the series like following

0, 2, 9, 24, 50, 90,147,224,324?

Can you please post a Common Equation of this series?

shalini_goel14 · Jul 4, 2009

Rank A3 - PRO

Well, common series I don't know, I used following way

1[sup]st[/sup] Element = 1 * (0)= 1 * 0 =0

2[sup]nd[/sup] Element = 2 * (0+1) = 2 * 1= 2

3[sup]rd[/sup] Element = 3 * (0+1+2)= 3 * 3 =9

4[sup]th[/sup] Element = 4 * (0+1+2+3) = 4 * 6= 24

5[sup]th[/sup] Element = 5 * (0+1+2+3+4) = 5 * 10 =50

6[sup]th[/sup] Element = 6 * (0+1+2+3+4+5) = 6 * 15 =90

7[sup]th[/sup] Element = 7 * (0+1+2+3+4+5+6) = 7 *21 =147

8[sup]th[/sup] Element = 8 * (0+1+2+3+4+5+6+7) = 8 * 28 =224

9[sup]th[/sup] Element = 9 * (0+1+2+3+4+5+6+7+8) = 9 * 36 =324

So common series can be [not sure ]

N[sup]th[/sup] element = N * [0+1+...(N-1)] where N >=1

1[sup]st[/sup] Element = 1 * (0)= 1 * 0 =0

2[sup]nd[/sup] Element = 2 * (0+1) = 2 * 1= 2

3[sup]rd[/sup] Element = 3 * (0+1+2)= 3 * 3 =9

4[sup]th[/sup] Element = 4 * (0+1+2+3) = 4 * 6= 24

5[sup]th[/sup] Element = 5 * (0+1+2+3+4) = 5 * 10 =50

6[sup]th[/sup] Element = 6 * (0+1+2+3+4+5) = 6 * 15 =90

7[sup]th[/sup] Element = 7 * (0+1+2+3+4+5+6) = 7 *21 =147

8[sup]th[/sup] Element = 8 * (0+1+2+3+4+5+6+7) = 8 * 28 =224

9[sup]th[/sup] Element = 9 * (0+1+2+3+4+5+6+7+8) = 9 * 36 =324

So common series can be [not sure ]

N[sup]th[/sup] element = N * [0+1+...(N-1)] where N >=1

Harshad Italiya · Jul 4, 2009

Rank A1 - PRO

Yep you are in right way..

here is one more Common equation:

here is one more Common equation:

**(N[sup]3[/sup]-N[sup]2[/sup])/2 where N >=1**Harshad Italiya · Jul 5, 2009

Rank A1 - PRO

**4.**

Prof. Crazy is on a long drive with his brand new TATA NANO.

But Prof. Crazy want to maintain Car Tyre so he use one trick he started replacing the Spare wheel with the main four wheel after traveling a fix distance.For example after traveling some fix distance he replaces the front wheel with spare wheel, again after that traveling that fix distance replaces the another front wheel with spare wheel and this process continues.If Prof. Crazy travels total

**6275 Km**. and the rotation of each wheel is same then calculate how much distance is covered by each wheel?

shalini_goel14 · Jul 5, 2009

Rank A3 - PRO

Do we have to tell the answer without knowledge of "Fix Length" traveled by the car. If so, assuming fixed length is 1km then distance traveled by

Spare wheel = 6273 km

First wheel = 6273km

Second wheel = 6274 km

Third wheel = 6274 km

Fourth wheel = 6274 km

as per me.

Spare wheel = 6273 km

First wheel = 6273km

Second wheel = 6274 km

Third wheel = 6274 km

Fourth wheel = 6274 km

as per me.

silverscorpion · Jul 5, 2009

Rank A3 - PRO

Well, it's not like that.

The question is, after going a distance X, the front tyre is changed with a spare tyre.

After another X, the other front tyre is changed with the first front tyre. Another X, and one of the back tyres is changed. Like this, it goes..

If this is so, then my solution is as follows.

Total distance travelled by the car is 6275 Km

So, total distance travelled by all the tyres is, 6275 * 4 = 25100 Km

This distance is covered by 5 tyres. So each tyre would have covered,

25100 / 5 = 5020 Km.

So, ans: Distance covered by each tyre =

Is it correct?? 😀😀

The question is, after going a distance X, the front tyre is changed with a spare tyre.

After another X, the other front tyre is changed with the first front tyre. Another X, and one of the back tyres is changed. Like this, it goes..

If this is so, then my solution is as follows.

Total distance travelled by the car is 6275 Km

So, total distance travelled by all the tyres is, 6275 * 4 = 25100 Km

This distance is covered by 5 tyres. So each tyre would have covered,

25100 / 5 = 5020 Km.

So, ans: Distance covered by each tyre =

**5020 Km**.Is it correct?? 😀😀

Harshad Italiya · Jul 5, 2009

Rank A1 - PRO

Good try shalini but in your case the distance traveled by each tyre is not same..shalini_goel14Do we have to tell the answer without knowledge of "Fix Length" traveled by the car. If so, assuming fixed length is 1km then distance traveled by

Spare wheel = 6273 km

First wheel = 6273km

Second wheel = 6274 km

Third wheel = 6274 km

Fourth wheel = 6274 km

as per me.

Yes Scorpion you are absolutely correct and very good explanation there now its time to add +5 points to your account 😁silverscorpionWell, it's not like that.

The question is, after going a distance X, the front tyre is changed with a spare tyre.

After another X, the other front tyre is changed with the first front tyre. Another X, and one of the back tyres is changed. Like this, it goes..

If this is so, then my solution is as follows.

Total distance travelled by the car is 6275 Km

So, total distance travelled by all the tyres is, 6275 * 4 = 25100 Km

This distance is covered by 5 tyres. So each tyre would have covered,

25100 / 5 = 5020 Km.

So, ans: Distance covered by each tyre =5020 Km.

Is it correct?? 😀😀

shalini_goel14 · Jul 5, 2009

Rank A3 - PRO

Okies, may be my concept about distance traveled by a car and its wheels was wrong but I just want to get clear on this thing now. Let us suppose a car travels 1 km from some point A to some point B then that means wheels traveled 4 km[as per Scorpion 1*4 =4] from point A to point B in total? Please make itgodfatherGood try shalini but in your case the distance traveled by each tyre is not same..

**CLEAR**-how ? 😕

pradeep_agrawal · Jul 5, 2009

Rank C2 - EXPERT

Shalini, at a time car will be running on 4 tyres and the if the car is traveling 1km then each of the tyre will also be covering 1km. So combined distance traveled by 4tyres = 1km * 4 = 4km.shalini_goel14OLet us suppose a car travels 1 km from some point A to some point B then that means wheels traveled 4 km[as per Scorpion 1*4 =4] from point A to point B in total? Please make itCLEAR-how ?

-Pradeep

silverscorpion · Jul 5, 2009

Rank A3 - PRO

Yep. Pradeep got it right.

@Shalini : Are you clear now??

@Shalini : Are you clear now??

shalini_goel14 · Jul 5, 2009

Rank A3 - PRO

@Pradeep and Scorpion Thanks for trying to clarifying but I am still not clear. The thing you explained is quite obvious and now I have even realised what all things I messed up in my answer. Anyways leave that, my doubt is this "Divide by 5 rule" works only for cases when the car travels distance in a multiple of 5 right? [Please correct me if wrong ] . What if distance travelled is not a multiple of 5. Suppose car travels distance of 12 km then

Distance travelled by car = 12 km

Distance travelled by 4 wheels in total = 12 *4 =48 km

But distance is equally covered by 5 wheels [Note: Assumption made which is true only when distance travelled is a multple of 5]

so distance travelled by each wheel = 48 /5 =

Distance travelled by car = 12 km

Now if I assume 2km is the fixed distanec travelled by the car then there will be total 6 [12/2 =6 ] intervals of changing the tyres right ?

1st tyre covers =2km

2nd tyre covers =2km

3rd tyre covers =2km

4th tyre covers = 2km

Spare tyre covers =Nil

1st tyre covers =Nil

2nd tyre covers =2km

3rd tyre covers =2km

4th tyre covers = 2km

Spare tyre covers =2km

1st tyre covers =2km

2nd tyre covers =Nil

3rd tyre covers =2km

4th tyre covers = 2km

Spare tyre covers =2km

1st tyre covers =2km

2nd tyre covers =2km

3rd tyre covers =Nil

4th tyre covers = 2km

Spare tyre covers =2km

1st tyre covers =2km

2nd tyre covers =2km

3rd tyre covers = 2km

4th tyre covers = Nil

Spare tyre covers =2km

1st tyre covers =2km

2nd tyre covers =2km

3rd tyre covers = 2km

4th tyre covers = 2km

Spare tyre covers =Nil

So now if I total distance covered by each tyre in all intervals

1st tyre covers =10km

2nd tyre covers =10km

3rd tyre covers =10 km

4th tyre covers = 10km

Spare tyre covers =8km

And if you sum up the distance covered by my approach =10+10+10+10+8=48 km

Now doubt is where I am going wrong in my approach. Please make it CLEAR .😕

**Case 1: According to YOUR approach of solving**Distance travelled by car = 12 km

Distance travelled by 4 wheels in total = 12 *4 =48 km

But distance is equally covered by 5 wheels [Note: Assumption made which is true only when distance travelled is a multple of 5]

so distance travelled by each wheel = 48 /5 =

**9600 m or 9.6 km****Case 2: According to MY approach of solving**Distance travelled by car = 12 km

Now if I assume 2km is the fixed distanec travelled by the car then there will be total 6 [12/2 =6 ] intervals of changing the tyres right ?

**1st Interval covered**1st tyre covers =2km

2nd tyre covers =2km

3rd tyre covers =2km

4th tyre covers = 2km

Spare tyre covers =Nil

**2nd Interval covered**1st tyre covers =Nil

2nd tyre covers =2km

3rd tyre covers =2km

4th tyre covers = 2km

Spare tyre covers =2km

**3rd Interval covered**1st tyre covers =2km

2nd tyre covers =Nil

3rd tyre covers =2km

4th tyre covers = 2km

Spare tyre covers =2km

**4th Interval covered**1st tyre covers =2km

2nd tyre covers =2km

3rd tyre covers =Nil

4th tyre covers = 2km

Spare tyre covers =2km

**5th Interval covered**1st tyre covers =2km

2nd tyre covers =2km

3rd tyre covers = 2km

4th tyre covers = Nil

Spare tyre covers =2km

**6th Interval covered**1st tyre covers =2km

2nd tyre covers =2km

3rd tyre covers = 2km

4th tyre covers = 2km

Spare tyre covers =Nil

So now if I total distance covered by each tyre in all intervals

1st tyre covers =10km

2nd tyre covers =10km

3rd tyre covers =10 km

4th tyre covers = 10km

Spare tyre covers =8km

*Proof:*Distance covered by 4 wheels in total = 12 *4 =48 kmAnd if you sum up the distance covered by my approach =10+10+10+10+8=48 km

Now doubt is where I am going wrong in my approach. Please make it CLEAR .😕

silverscorpion · Jul 5, 2009

Rank A3 - PRO

First of all, 48/5 gives

The thing here is, you cannot assume the fixed distance, because that's what you have to find. So, the way to go would be, there are 5 tyres totally and so, you'd change tyres 4 times.

Now you redo the problem in your method and you'll get the answer. Now if we assume x is the fixed distance after which tyres are changed, then as per your method, we have,

1st tyre covers = x km

2nd tyre covers = x km

3rd tyre covers = x km

4th tyre covers = x km

Spare tyre covers =Nil

1st tyre covers =Nil

2nd tyre covers = x km

3rd tyre covers = x km

4th tyre covers = x km

Spare tyre covers = x km

1st tyre covers = x km

2nd tyre covers =Nil

3rd tyre covers = x km

4th tyre covers = x km

Spare tyre covers = x km

1st tyre covers = x km

2nd tyre covers = x km

3rd tyre covers =Nil

4th tyre covers = x km

Spare tyre covers = x km

1st tyre covers = x km

2nd tyre covers = x km

3rd tyre covers = x km

4th tyre covers = Nil

Spare tyre covers = x km

So, if you total everything, you have each tyre travelling 4x Km.

Each tyre travels 4x Km and so, the total distance travelled is 20x. Now, this 20x is equal to 12*4=48.

If 20x is 48, then 4x is 9.6. There you have the answer..

Where you went wrong is, you assumed the fixed distance to be something.

Assume that distance to be 3 instead of 2, and you'll see that you get a different answer. For each different assumption, you get a different answer. For 3, you get 9, 9, 9, 9 and 12 whose total also comes to 48. But the distances of all the tyres must be equal..

Clear??

**9.6 Km**and not**960m**The thing here is, you cannot assume the fixed distance, because that's what you have to find. So, the way to go would be, there are 5 tyres totally and so, you'd change tyres 4 times.

Now you redo the problem in your method and you'll get the answer. Now if we assume x is the fixed distance after which tyres are changed, then as per your method, we have,

**1st Interval covered**1st tyre covers = x km

2nd tyre covers = x km

3rd tyre covers = x km

4th tyre covers = x km

Spare tyre covers =Nil

**2nd Interval covered**1st tyre covers =Nil

2nd tyre covers = x km

3rd tyre covers = x km

4th tyre covers = x km

Spare tyre covers = x km

**3rd Interval covered**1st tyre covers = x km

2nd tyre covers =Nil

3rd tyre covers = x km

4th tyre covers = x km

Spare tyre covers = x km

**4th Interval covered**1st tyre covers = x km

2nd tyre covers = x km

3rd tyre covers =Nil

4th tyre covers = x km

Spare tyre covers = x km

**5th Interval covered**1st tyre covers = x km

2nd tyre covers = x km

3rd tyre covers = x km

4th tyre covers = Nil

Spare tyre covers = x km

So, if you total everything, you have each tyre travelling 4x Km.

Each tyre travels 4x Km and so, the total distance travelled is 20x. Now, this 20x is equal to 12*4=48.

If 20x is 48, then 4x is 9.6. There you have the answer..

Where you went wrong is, you assumed the fixed distance to be something.

Assume that distance to be 3 instead of 2, and you'll see that you get a different answer. For each different assumption, you get a different answer. For 3, you get 9, 9, 9, 9 and 12 whose total also comes to 48. But the distances of all the tyres must be equal..

Clear??

shalini_goel14 · Jul 6, 2009

Rank A3 - PRO

Sorry Dear Scorpion, again troubling you but my doubt is still not cleared , may be you people are not cleared what I am trying to say - my bad 😔

@GF Please grant an excuse for spamming your thread again and again.

Now here goes my doubts 😕

Now as I am already talking about the case of distance travelled by car which is

Possible values of x would be 1,2, 3,4,6,12 giving no of intervals as

12,6,4,3,2,1 respectively . Am I right or wrong? 😕

Thanks !

@GF Please grant an excuse for spamming your thread again and again.

Now here goes my doubts 😕

9.6km was a typo- edited. Sorry for that. Why we cannot assume fixed length? If we were supposed to find out that an you tell em what was that length in GF's question ? Also what was the total no. of intervals in his question ?silverscorpionFirst of all, 48/5 gives9.6 Kmand not960m

The thing here is, you cannot assume the fixed distance, because that's what you have to find. So, the way to go would be, there are 5 tyres totally and so, you'd change tyres 4 times.

Now you redo the problem in your method and you'll get the answer. Now if we assume x is the fixed distance after which tyres are changed, then as per your method, we have,What thing again makes you to give

**5 intervals ?**😕 5 intervals thing will be valid only if "Distance traveled by car is a multiple of 5 and x is a factor of 5." Am I right or wrong?Now as I am already talking about the case of distance travelled by car which is

**not a multiple of 5**[12 km at present] then**x**should always be a factor of distance travelled by car[12 km at present] right? then only we can get no. of interval as a whole number right?Possible values of x would be 1,2, 3,4,6,12 giving no of intervals as

12,6,4,3,2,1 respectively . Am I right or wrong? 😕

Assume that distance to be 3 instead of 2, and you'll see that you get a different answer. For each different assumption, you get a different answer.Yes , I know and its reality also we would definitely get different answers if the car changes tyres after a specific fixed interval of time. That's why I asked for that length in the question.

For 3, you get 9, 9, 9, 9 and 12 whose total also comes to 48. But the distances of all the tyres must be equal..You are saying it would be 9,9,9,9,12 and at the same time saying "distances of all the tyres

**must be equal" -**what makes you to think that it**must**be equal ? 😕 Any solid reason behind this ?Thanks !

Harshad Italiya · Jul 6, 2009

Rank A1 - PRO

@Shalini.. No Problem you can carry on!!

your question is right if we have not given the fix distance but here we give the Total distance covered by Car and also given that distance covered by all wheels are same so here we have to keep that distance in mind..and have to calculate according to that.

your question is right if we have not given the fix distance but here we give the Total distance covered by Car and also given that distance covered by all wheels are same so here we have to keep that distance in mind..and have to calculate according to that.

silverscorpion · Jul 6, 2009

Rank A3 - PRO

That was the question, Shalini.

Question says that

And that is the sole reason why assuming an initial distance won't work.;

If you assume so, then whatever is that distance, in the end, 4 of the tyres will have one distance and the other tyre will have a different distance. To make all of them equal distance, we follow my method. (ie, multiply by 4 and divide by 5.)

Now are you clear? Shall we go to the next question? 😀😀

Question says that

**the distances covered by all the five tyres are equal.**And that is the sole reason why assuming an initial distance won't work.;

If you assume so, then whatever is that distance, in the end, 4 of the tyres will have one distance and the other tyre will have a different distance. To make all of them equal distance, we follow my method. (ie, multiply by 4 and divide by 5.)

Now are you clear? Shall we go to the next question? 😀😀

shalini_goel14 · Jul 6, 2009

Rank A3 - PRO

Question asked was following right? Well, I could not see something similar to this "

**the distances covered by all the five tyres are equal." .**Are you talking about "**the rotation of each wheel is same**" then sorry for troubling you guys. I was having concept that rotation is different from distance travelled in actual that's why I haven't taken into account.Prof. Crazy is on a long drive with his brand new TATA NANO.

But Prof. Crazy want to maintain Car Tyre so he use one trick he started replacing the Spare wheel with the main four wheel after traveling a fix distance.For example after traveling some fix distance he replaces the front wheel with spare wheel, again after that traveling that fix distance replaces the another front wheel with spare wheel and this process continues.If Prof. Crazy travels total6275 Km. and the rotation of each wheel is same then calculate how much distance is covered by each wheel?

Thanks a lot for your patience. Sure, go ahead. !silverscorpionNow are you clear? Shall we go to the next question? 😀😀

Harshad Italiya · Jul 7, 2009

Rank A1 - PRO

**5.**

Fill in the gaps with the number 1 to 9. you can not use any figure twise.

**_ _ * _ _ _=_ _ _ _**

There are multiple answer of this question so if you write them there is 2 points extra for each answer.

Saandeep Sreerambatla · Jul 8, 2009

Rank A2 - PRO

The answers i have got are

138 * 42 = 5796

157 * 28 = 4396

159 * 48 = 7632

138 * 42 = 5796

157 * 28 = 4396

159 * 48 = 7632

Harshad Italiya · Jul 8, 2009

Rank A1 - PRO

ES you got 5 points+ 4 Points..😁English-ScaredThe answers i have got are

138 * 42 = 5796

157 * 28 = 4396

159 * 48 = 7632

NIMISH PRABHU · Jul 8, 2009

Rank D3 - MASTER

hmmm nice one 😀

RajdeepCE · Jul 8, 2009

Rank B3 - LEADER

39 x 186 = 7254,Congrats ES, I show this puzzle little bit late otherwise...😁

18 x 297 = 5346,

27 x 198 = 5346,

12 x 483 = 5796.

Anyway all the best for the next puzzle.

Harshad Italiya · Jul 8, 2009

Rank A1 - PRO

**6.**

Once upon a time, four friends went in to the forest. In the night, they need to cross a bridge. But a problem is that the bridge can bear only two people. They have last candle remained which can stay lighten upto

**15 minutes**, so they need to cross bridge in 15 minutes. The timings to cross bridge by each friend is given below,

Friend A = 1 min

Friend B = 2 min

Friend C = 5 min

Friend D = 8 min

So would you like to help them to cross the bridge?

pradeep_agrawal · Jul 8, 2009

Rank C2 - EXPERT

Here is my solution:

Friend A = 1 min

Friend B = 2 min

Friend C = 5 min

Friend D = 8 min

Step 1: A + B cross bridge, Time = 2min.

Total time = 2min.

Step 2: A comes back, Time = 1min.

Total time = 3min.

Step 3: C + D cross bridge, Time = 8min.

Total time = 11min.

Step 4: B comes back, Time = 2min.

Total time = 13min.

Step 1: A + B cross bridge, Time = 2min.

Total time = 15min.

-Pradeep

Friend A = 1 min

Friend B = 2 min

Friend C = 5 min

Friend D = 8 min

Step 1: A + B cross bridge, Time = 2min.

Total time = 2min.

Step 2: A comes back, Time = 1min.

Total time = 3min.

Step 3: C + D cross bridge, Time = 8min.

Total time = 11min.

Step 4: B comes back, Time = 2min.

Total time = 13min.

Step 1: A + B cross bridge, Time = 2min.

Total time = 15min.

-Pradeep

Harshad Italiya · Jul 8, 2009

Rank A1 - PRO

You are High-speed Puzzle solver Pradeep.

5 points added!

5 points added!

silverscorpion · Jul 10, 2009

Rank A3 - PRO

Hey GF, where's the next puzzle?? 😀

Harshad Italiya · Jul 11, 2009

Rank A1 - PRO

Till then try to Solve this one 😉silverscorpionHey GF, where's the next puzzle?? 😀

**Sharp your Brain!!**

Harshad Italiya · Jul 11, 2009

Rank A1 - PRO

**7.**

Bob and John form a team together. Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages. John is as old as Bob was when John was half as old as he will become over ten years.

How old are Bob and John?

mech_guy · Jul 11, 2009

Rank D2 - MASTER

Hi godfather,

Well don't know about the ages of Tom and Jerry, but yeah the ages of Bob and John now should be 40 and 30 years respectively.

Tom and Jerry?? May be i missed something.

Regards

Well don't know about the ages of Tom and Jerry, but yeah the ages of Bob and John now should be 40 and 30 years respectively.

Tom and Jerry?? May be i missed something.

Regards

shalini_goel14 · Jul 11, 2009

Rank A3 - PRO

I don't know the ages of Tom and Jerry but assuming you would have done some typos, according to me answers aregodfather7.

Tom and Jerry form a team together. Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages. John is as old as Bob was when John was half as old as he will become over ten years.

How old are Tom and Jerry?

Bob = 20 years

John = 15 years

Am I right ?

mech_guy · Jul 11, 2009

Rank D2 - MASTER

Hi Shalini,

I think if you cross-check your answer you will know that ages you have computed are wrong.

Last condition says,

"John is as old as Bob was when John was half as old as he will become over ten years"

If, J = 15 years and B = 20 years

Then in 10 years time John will be 25 years and half of this is 12.5 yeras.

When John was 12.5 years, Bob was 17.5 years old

So John's age now should be = 17.5 years but your solution says he is 15 years.

So i guess your solution is wrong here.

Regards

I think if you cross-check your answer you will know that ages you have computed are wrong.

Last condition says,

"John is as old as Bob was when John was half as old as he will become over ten years"

If, J = 15 years and B = 20 years

Then in 10 years time John will be 25 years and half of this is 12.5 yeras.

When John was 12.5 years, Bob was 17.5 years old

So John's age now should be = 17.5 years but your solution says he is 15 years.

So i guess your solution is wrong here.

Regards

shalini_goel14 · Jul 11, 2009

Rank A3 - PRO

Oops, mistake once again. Thanks mech_guy for correcting me. I found out problem in my equations formed.

Yes answer will be Bob =40 years and John = 30 years.

Yes answer will be Bob =40 years and John = 30 years.

mech_guy · Jul 11, 2009

Rank D2 - MASTER

You are welcome Shalini 😀

Regards

Regards

Harshad Italiya · Jul 12, 2009

Rank A1 - PRO

Sorry for that Type mistake your answer is Right Mech_Guy 😉mech_guyHi Shalini,

I think if you cross-check your answer you will know that ages you have computed are wrong.

Last condition says,

"John is as old as Bob was when John was half as old as he will become over ten years"

If, J = 15 years and B = 20 years

Then in 10 years time John will be 25 years and half of this is 12.5 yeras.

When John was 12.5 years, Bob was 17.5 years old

So John's age now should be = 17.5 years but your solution says he is 15 years.

So i guess your solution is wrong here.

Regards

Harshad Italiya · Jul 12, 2009

Rank A1 - PRO

anyone had tried this one? 😛godfatherTill then try to Solve this one 😉

Sharp your Brain!!

https://www.freestuffhotdeals.com/hacker/1.html

shalini_goel14 · Jul 12, 2009

Rank A3 - PRO

Hey man, THIS IS NOT FAIR. 😐 You mentioned that person can give 5 tries and one those giving 2nd try will get 4 points added in their account. If this rule is not getting applied then better remove it from your list so that people don't get misguided. My answer in 2[sup]nd[/sup] try was correct. Your wish, your thread ignore it or give points. Who cares. 😐

Thanks !

Thanks !

Harshad Italiya · Jul 12, 2009

Rank A1 - PRO

Sorry Shalini But i didn't get you.shalini_goel14Hey man, THIS IS NOT FAIR. 😐 You mentioned that person can give 5 tries and one those giving 2nd try will get 4 points added in their account. If this rule is not getting applied then better remove it from your list so that people don't get misguided. My answer in 2[sup]nd[/sup] try was correct. Your wish, your thread ignore it or give points. Who cares. 😐

Thanks !

**mech_guy**gives the Right answer before you.. i accept that there is my typo Mistake.

Harshad Italiya · Jul 12, 2009

Rank A1 - PRO

How many times you use Google? 😉English-Scared20 levels done...

now what about 21,22,23?

Harshad Italiya · Jul 12, 2009

Rank A1 - PRO

**8.**

One newspaper vendor sells three types of newspaper The Crazy Times, The Crazy Post and The Crazy News. And the copy sold in one day is 70, 60 and 50 respectively. Now something about his client…

17 clients buy The Crazy Times and The Crazy Post both,

Fifteen clients buy The Crazy Post and The Crazy News both,

16 clients buy The Crazy Times and The Crazy News both, while three clients buy All three newspapers daily.

So the question is that what is the total number of Clients that vendor have?

Saandeep Sreerambatla · Jul 12, 2009

Rank A2 - PRO

godfatherHow many times you use Google? 😉

now what about 21,22,23?

Few times i used google 😀

i got only 20 levels in that game 😀

mech_guy · Jul 12, 2009

Rank D2 - MASTER

godfather8.

One newspaper vendor sells three types of newspaper The Crazy Times, The Crazy Post and The Crazy News. And the copy sold in one day is 70, 60 and 50 respectively. Now something about his client…

17 clients buy The Crazy Times and The Crazy Post both,

Fifteen clients buy The Crazy Post and The Crazy News both,

16 clients buy The Crazy Times and The Crazy News both, while three clients buy All three newspapers daily.

So the question is that what is the total number of Clients that vendor have?

From above Venn diagram, where,

CT = Crazy Times

CP = Crazy Post

CN = Crazy News

Answer is 135 Clients (Total Union of above Venn Diagram)

PS: posting an image for the first time, if it doesn't show-up, explanation is:

No. Of Clients buying ONLY CT = 40

No. Of Clients buying ONLY CP = 31

No. Of Clients buying ONLY CN = 22

No. of Clients buying CT and CP = 14

No. of Clients buying CT and CN = 13

No. of Clients buying CN and CP = 12

No. of clients buying all 3 = 3

Total comes out as 135.

Hope its clear.

Regards

Harshad Italiya · Jul 12, 2009

Rank A1 - PRO

You Rocking man 😁mech_guy

From above Venn diagram, where,

CT = Crazy Times

CP = Crazy Post

CN = Crazy News

Answer is 135 Clients (Total Union of above Venn Diagram)

PS: posting an image for the first time, if it doesn't show-up, explanation is:

No. Of Clients buying ONLY CT = 40

No. Of Clients buying ONLY CP = 31

No. Of Clients buying ONLY CN = 22

No. of Clients buying CT and CP = 14

No. of Clients buying CT and CN = 13

No. of Clients buying CN and CP = 12

No. of clients buying all 3 = 3

Total comes out as 135.

Hope its clear.

Regards

Check out your Points 😉