GF's Passtime Puzzles--With New Format

Harshad Italiya

@harshad-ukH5ww • Oct 26, 2024

Oct 26, 2024

1.4K

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.

Point Card:
RajdeepCE............10 Points
pradeep_agrawal...9 Points
shalini_goel14.......5 Points
silverscorpion........5 Points
English-Scared......9 Points
mech_guy.............10 Points

So... Are you ready for this one !! 😁

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Harshad Italiya

@harshad-ukH5ww • Jul 1, 2009

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?
0
RajdeepCE

@rajdeepce-7UdrG8 • Jul 1, 2009

There are 49 yellow & 1 red ball.
Am I right?
0
Harshad Italiya

@harshad-ukH5ww • Jul 1, 2009

You got it Rajdeep !!

+5 Points added.
0
Harshad Italiya

@harshad-ukH5ww • Jul 1, 2009

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) 😉
0
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 2, 2009

Is it 1939yr and 1932yr?

-Pradeep
0
Harshad Italiya

@harshad-ukH5ww • Jul 2, 2009

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

there is 2557 days between 1932 and 1939 😀
0
RajdeepCE

@rajdeepce-7UdrG8 • Jul 2, 2009

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?

0
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 2, 2009

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
0
Harshad Italiya

@harshad-ukH5ww • Jul 2, 2009

RajdeepCE
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?

You again got the correct answer... and nice explaination +5 added

pradeep_agrawal
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
Yes we can get the same result between 100 years interval... You missed it just by one Pradeep Nice effort impressive. 😀
0
Harshad Italiya

@harshad-ukH5ww • Jul 2, 2009

oh i just missed the time you both has replied at same time... Pradeep way to go you also get 4 points. as you got it in your second reply..

@RajdeepCE.. now you have a Golden chance to get bonus Points. 😁
0
Harshad Italiya

@harshad-ukH5ww • Jul 2, 2009

3.

Here is one Series for Series Solver..😉
0,2,9,24,50,90,.........
Find the Next 3 figures.
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 2, 2009

RajdeepCE
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!!!
If I am not wrong AGES of the friends was asked in the question, not their BIRTHYEARS right? Can anyone tell me their AGES? 😕
0
Harshad Italiya

@harshad-ukH5ww • Jul 2, 2009

One is 113 Years Old and other is of 106 years old.
0
Harshad Italiya

@harshad-ukH5ww • Jul 4, 2009

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.

0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 4, 2009

Is the series like following

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

@harshad-ukH5ww • Jul 4, 2009

shalini_goel14
Is the series like following

0, 2, 9, 24, 50, 90, 147, 224, 324 ?
You got the right three Shalini 😁 +5 Points added.
Can you please post a Common Equation of this series?
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 4, 2009

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
0
Harshad Italiya

@harshad-ukH5ww • Jul 4, 2009

Yep you are in right way..
here is one more Common equation: (N[sup]3[/sup]-N[sup]2[/sup])/2 where N >=1
0
Harshad Italiya

@harshad-ukH5ww • Jul 4, 2009

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?
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 5, 2009

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.
0
silverscorpion

@silverscorpion-iJKtdQ • Jul 5, 2009

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 = 5020 Km.

Is it correct?? 😀😀
0
Harshad Italiya

@harshad-ukH5ww • Jul 5, 2009

shalini_goel14
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.
Good try shalini but in your case the distance traveled by each tyre is not same..

silverscorpion
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 = 5020 Km.

Is it correct?? 😀😀
Yes Scorpion you are absolutely correct and very good explanation there now its time to add +5 points to your account 😁
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 5, 2009

godfather
Good try shalini but in your case the distance traveled by each tyre is not same..
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 it CLEAR -how ? 😕
0
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 5, 2009

shalini_goel14
OLet 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 it CLEAR -how ?
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.

-Pradeep
0
silverscorpion

@silverscorpion-iJKtdQ • Jul 5, 2009

Yep. Pradeep got it right.

@Shalini : Are you clear now??
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 5, 2009

@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

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 km

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 .😕
0
silverscorpion

@silverscorpion-iJKtdQ • Jul 5, 2009

First of all, 48/5 gives 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??
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 5, 2009

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 😕

silverscorpion
First of all, 48/5 gives 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.
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 ?

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 !
0
Harshad Italiya

@harshad-ukH5ww • Jul 5, 2009

@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.
0
silverscorpion

@silverscorpion-iJKtdQ • Jul 6, 2009

That was the question, Shalini.

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? 😀😀
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 6, 2009

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 total 6275 Km. and the rotation of each wheel is same then calculate how much distance is covered by each wheel?
silverscorpion
Now are you clear? Shall we go to the next question? 😀😀
Thanks a lot for your patience. Sure, go ahead. !
0
Harshad Italiya

@harshad-ukH5ww • Jul 7, 2009

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.
0
Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • Jul 7, 2009

The answers i have got are

138 * 42 = 5796

157 * 28 = 4396

159 * 48 = 7632
0
Harshad Italiya

@harshad-ukH5ww • Jul 7, 2009

English-Scared
The answers i have got are

138 * 42 = 5796

157 * 28 = 4396

159 * 48 = 7632
ES you got 5 points+ 4 Points..😁
0
NIMISH PRABHU

@nimish-prabhu-sOwQMf • Jul 7, 2009

hmmm nice one 😀
0
RajdeepCE

@rajdeepce-7UdrG8 • Jul 7, 2009

39 x 186 = 7254,
18 x 297 = 5346,
27 x 198 = 5346,
12 x 483 = 5796.
Congrats ES, I show this puzzle little bit late otherwise...😁
Anyway all the best for the next puzzle.
0
Harshad Italiya

@harshad-ukH5ww • Jul 7, 2009

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?
0
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 7, 2009

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
0
Harshad Italiya

@harshad-ukH5ww • Jul 7, 2009

You are High-speed Puzzle solver Pradeep.

5 points added!
0
silverscorpion

@silverscorpion-iJKtdQ • Jul 10, 2009

Hey GF, where's the next puzzle?? 😀
0
Harshad Italiya

@harshad-ukH5ww • Jul 10, 2009

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

<a href="https://iharshad.wordpress.com/2009/07/11/sharp-your-brain/" target="_blank" rel="nofollow noopener noreferrer">Sharp Your Brain!! | MadhaV's Magic Blog</a>

0
Harshad Italiya

@harshad-ukH5ww • Jul 11, 2009

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?
0
mech_guy

@mech-guy-TMwuj2 • Jul 11, 2009

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
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 11, 2009

godfather
7.

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

Bob = 20 years
John = 15 years

Am I right ?
0
mech_guy

@mech-guy-TMwuj2 • Jul 11, 2009

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
0
shalini_goel14

@shalini-goel14-ASmC2J • Jul 11, 2009

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.
0
mech_guy

@mech-guy-TMwuj2 • Jul 11, 2009

You are welcome Shalini 😀

Regards
0
Harshad Italiya

@harshad-ukH5ww • Jul 11, 2009

mech_guy
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
Sorry for that Type mistake your answer is Right Mech_Guy 😉
0
Harshad Italiya

@harshad-ukH5ww • Jul 11, 2009

godfather
Till then try to Solve this one 😉

<a href="https://iharshad.wordpress.com/2009/07/11/sharp-your-brain/" target="_blank" rel="nofollow noopener noreferrer">Sharp Your Brain!! | MadhaV's Magic Blog</a>
anyone had tried this one? 😛

#-Link-Snipped-#
#-Link-Snipped-#

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shalini_goel14

@shalini-goel14-ASmC2J • Jul 11, 2009

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 !
0
Harshad Italiya

@harshad-ukH5ww • Jul 11, 2009

shalini_goel14
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 !
Sorry Shalini But i didn't get you.

mech_guy gives the Right answer before you.. i accept that there is my typo Mistake.
0
Harshad Italiya

@harshad-ukH5ww • Jul 11, 2009

English-Scared
20 levels done...
How many times you use Google? 😉
now what about 21,22,23?
0
Harshad Italiya

@harshad-ukH5ww • Jul 11, 2009

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?
0
Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • Jul 11, 2009

godfather
How many times you use Google? 😉
now what about 21,22,23?

Few times i used google 😀

i got only 20 levels in that game 😀
0
mech_guy

@mech-guy-TMwuj2 • Jul 11, 2009

godfather
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?

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
0
Harshad Italiya

@harshad-ukH5ww • Jul 12, 2009

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
You Rocking man 😁

Check out your Points 😉
0

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