RIDDLES -- Dimag ki batti jalao

Raam and shyam ki century to ho gayi.
try this one...

In a calculator, 3 buttons to be press to get result 100.
condition: zero not to be used.
 
My answer complies with the rules of cricket. there are times when i have seen 5 runs awarded to batsman if the ball hits the helmet(kept behind keeper) or the cap of the fielder.

This question demanded both players to score century without getting out(last pair on crease). i simply did that! :)

PS : Still having doubts then do revert back. i will tell you how to take advantage of a "short run" ;)

When ball hits the helmet behing keeper...its a dead ball and 5 panelty runs are given to batting side.
 
5 runs penalty ke run team ko add hote hai player ko nahi.


If this is the case then why are overthrows added to batsman's account if he has hit the ball with his bat??
(seen players scoring centuries with overthrows)

Your answer applies to the situation when batsman has not hit the ball with the bat! :)
 
Try This....

The five school children in couples(Two at a time) weigh 129 pounds, 125 pounds, 124 pounds, 123 pounds, 122 pounds, 121 pounds, 120 pound, 118 pounds, 116 pounds and 114 pounds on a weighing machine. What was the weight of each one of the five little girls if taken separately?
 
Try This....

The five school children in couples(Two at a time) weigh 129 pounds, 125 pounds, 124 pounds, 123 pounds, 122 pounds, 121 pounds, 120 pound, 118 pounds, 116 pounds and 114 pounds on a weighing machine. What was the weight of each one of the five little girls if taken separately?

65, 64, 60, 58, 56 kg
ya dear, i missed that last pair word in previous ques. thanx
 
१२ गेंदों में से एक गेंद बाकियों से भारी या हलकी है बाकि ११ गेंद एक समान वजन की है आपके पास एक तराजू है कम से कम कितनी तोलो में विषम गेंद को ढूंड निकल पावोगे
 
१२ गेंदों में से एक गेंद बाकियों से भारी या हलकी है बाकि ११ गेंद एक समान वजन की है आपके पास एक तराजू है कम से कम कितनी तोलो में विषम गेंद को ढूंड निकल पावोगे

Ans. Three...min. 3 balls weigh karni padengi
 
१२ गेंदों में से एक गेंद बाकियों से भारी या हलकी है बाकि ११ गेंद एक समान वजन की है आपके पास एक तराजू है कम से कम कितनी तोलो में विषम गेंद को ढूंड निकल पावोगे
teen baar tol k bata sakte hain k , kon c ball halki ya bhari hai,,,,

first tolna
pehle 6-6 ko tarazu k paldo me dalo.....( ek palda bhari ya halka hoga jisme asamanya wazan ki ball hai)
second tolna

fir us palde ki balls ko 3-3 karke tarazu k palde me dalo...( ek palda bhari ya halka hoga jisme asamanya wazan ki ball hai)
teesra tol
fir un 3 balls me se 1-1 balls tarazu k palde me dalo,,agar palda barabar hai to bachi hui ball asamanya wazan ki hai...otherwise palde se hi pata lag jayega ki kon c ball hai.
 
bhai 1 palda to bhari and dusra halka hoga hi..par ye naa pta ki odd ball halki hai ya bhaari. ho to jayega 3 baar mai hi.par ye solution theek naa.

teen baar tol k bata sakte hain k , kon c ball halki ya bhari hai,,,,

first tolna
pehle 6-6 ko tarazu k paldo me dalo.....( ek palda bhari ya halka hoga jisme asamanya wazan ki ball hai)
second tolna

fir us palde ki balls ko 3-3 karke tarazu k palde me dalo...( ek palda bhari ya halka hoga jisme asamanya wazan ki ball hai)
teesra tol
fir un 3 balls me se 1-1 balls tarazu k palde me dalo,,agar palda barabar hai to bachi hui ball asamanya wazan ki hai...otherwise palde se hi pata lag jayega ki kon c ball hai.
 
Try This....

The five school children in couples(Two at a time) weigh 129 pounds, 125 pounds, 124 pounds, 123 pounds, 122 pounds, 121 pounds, 120 pound, 118 pounds, 116 pounds and 114 pounds on a weighing machine. What was the weight of each one of the five little girls if taken separately?

lets suppose that five kids A,B,C,D and E with their weights in descending order.
now equations :


  1. A+B=129
  2. A+C=125
  3. A+D=124
  4. A+E=123
  5. B+C=122
  6. B+D=121
  7. B+E=120
  8. C+D=118
  9. C+E=116
  10. D+E=114
------------------
SOLVING equations 8,9,10,

we get sure values of C,D and E i.e 60,58 and 56 respectively.

using these values we get three values of A, B each as follows :

  1. A=67
  2. A=66
  3. A=65

  • B= 62
  • B= 63
  • B= 64
suiting the requirements A=65 and B=64 and the rest of the values make all combinations..!!
 
bhai 1 palda to bhari and dusra halka hoga hi..par ye naa pta ki odd ball halki hai ya bhaari. ho to jayega 3 baar mai hi.par ye solution theek naa.

ya, if it is sure that culprit ball is heavier or lighter than only ravinder explanation is correct

try again ravinder
 
A+B is highest=129,A+C second highest=125
D+E is lowest=114,C+E second lowest=116.
Now, (A+B)+(A+C)+(A+D)+(A+E)+(B+C)+(B+D)+(B+E)+(C+D)+(C+E)+(D+E)=129+125+124+123+122+121+120+118+116+114
4(A+B+C+D+E)=1212
A+B+C+D+E=303
Put the value of A+B and D+E
C=303-129-114=60
A+C=125 so A=65
A+B=129 so B=64
C+E=116 so E=56
D+E=114 so D=58

lets suppose that five kids A,B,C,D and E with their weights in descending order.
now equations :



  1. A+B=129
  2. A+C=125
  3. A+D=124
  4. A+E=123
  5. B+C=122
  6. B+D=121
  7. B+E=120
  8. C+D=118
  9. C+E=116
  10. D+E=114

------------------
SOLVING equations 8,9,10,

we get sure values of C,D and E i.e 60,58 and 56 respectively.

using these values we get three values of A, B each as follows :


  1. A=67
  2. A=66
  3. A=65



  • B= 62
  • B= 63
  • B= 64

suiting the requirements A=65 and B=64 and the rest of the values make all combinations..!!
 
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