GATE | GATE-CS-2014-(Set-1) | Question 65
There are 5 bags labeled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Then the product of the labels of the bags having 11 gm coins is ___.
There are 5 bags numbered 1 to 5. We don't know how many bags contain 10 gm and 11 gm coins. We only know that the total weights of coins is 323. Now the idea here is to get 3 in the place of total sum's unit digit. Mark no 1 bag as having 11 gm coins. Mark no 2 bag as having 10 gm coins. Mark no 3 bag as having 11 gm coins. Mark no 4 bag as having 11 gm coins. Mark no 5 bag as having 10 gm coins. Note: The above marking is done after getting false results for some different permutations, the permutations which were giving 3 in the unit place of the total sum. Now, we have picked 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Hence total sum coming from each bag from 1 to 5 is 11, 20, 44, 88, 160 gm respectively. For the above combination we are getting 3 as unit digit in sum. Lets find out the total sum, it's 11 + 20 + 44 + 88 + 160 = 323. So it's coming right. Now 11 gm coins containing bags are 1, 3 and 4. Hence, the product is : 1 x 3 x 4 = 12.
Alternative solution :
There are bags with coins of weight 11g and 10g. The total weight of 1,2,4,8,16 coins picked from bags 1,2,3,4,5 is given as 323.
Now we need to find no.of coins of 10g and 11g that give a sum of 323g.
Let x be the no.of coins of weight 10g and y be the no.of coins of weight 11g
Therefore 10x + 11y = 323………1
Now we now the total no.of coins picked i.e 1+2+4+8+16=31.
Therefore x + y = 31…………2
Solving equations 1 and 2 we get x=18, y=13.
Therefore bags 2 and 5(2+16=x=18) have 10g coins and bags 1, 3 and 4(1+4+8=y=13) have 11g coins.
Therefore the product of the labels of the bags having 11 gm coins is 1 x 3 x 4 = 12.
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