Answer Posted / abhishek gupta

[REQUEST- PLEASE READ IT ATLEAST TWICE]

4 ITERATIONS OF WEIGHING....
DIVIDE THE 9 COINS INTO 3 BATCHES (4, 4 AND 1).

1) FIRST WEIGH THE FIRST BATCH WITH THE SECOND ONE. IF THE
BALANCE DOESN'T SHOW ANY MOVEMENT, THEN THE LEFT OUT COIN IS
THE ODD COIN.

2) IF THERE IS ANY MOVEMENT, THEN TEST THE 2 BATCHES
INDIVIDUALLY, i.e., 2 ON EACH SIDE OF THE BALANCE FOR BOTH
THE BATCHES. OUT OF THE 2 BATCHES, 1 WOULD SHOW MOVEMENT.
THEN IT IS CLEAR THAT THE ODD COIN IS FROM EITHER OF THE 4
COINS (SAY, THE SECOND BATCH).

3) [A] THEN, BALANCE ANY 2 COINS FROM THE SECOND BATCH
(HAVING THE ODD COIN) WITH 2 FROM THE FIRST BATCH (NOT
HAVING ANY ODD COIN).

4) IF THE BALANCE SHOWS ANY MOVEMENT, THEN AGAIN BALANCE 2
COINS (1 OF SECOND BATCH FROM THE BALANCE + 1 FROM THE FIRST
BATCH) WITH 2 MORE COINS FROM THE FIRST BATCH. IF THE
BALANCE DOESN'T SHOW ANY MOVEMENT, THEN THE LEFT OUT COIN ON
THE BALANCE [A] OF SECOND BATCH IS THE ODD ONE. BUT IF THE
BALANCE SHOWS SOME MOVEMENT THEN THE COIN WHICH WE TOOK OF
SECOND BATCH FROM THE BALANCE [A] IS THE ODD ONE OUT.

NOTE--- IN ALL OTHER ANSWERS, YOU ARE NOT ABLE TO FIND "THE
1 ODD COIN" TILL THE STEP. BUT HERE YOU GET IT CORRECT IN
ANY WAY, IN MAX-4 STEPS AND NOT 9 STEPS.

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1264

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1631

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I HAVE APPLIED FOR RRB-CHENNAI(ECRC,GOODS GUARD).SO I NEED ECRC ALL PREVIOUS YEAR QUESTION SOLVED PAPER..PLS SEND IT TO ramya.raj@hotmail.com

4123

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1294

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2578

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1340

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1261

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1406

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1193

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1606

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2274

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The square of a two digit number is divided by half the number. After 36 is added to the quotient, this sum is then divided by 2. The digits of the resulting number are the same as those in the original number, but they are in reverse order. The ten's place of the original number is equal to twice the difference between its digits. What is the number?

1422

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