GATE2014-1-37

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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 ___.

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Suppose $X$ is the number of coins of $11$ gm and $Y$ is the number of $10$ gm coins.
According to question,

$\qquad 11X+10Y= 323 \qquad \to (1)$
$\qquad X+Y=31\qquad \qquad \to (2)$

Solving $(1),(2),$ we get $X=13$ and $Y=18.$ So, here number of coins of $11$ gm is $13$ and the only possible combination for $13$ coins is

• $1$ coin from bag$1$
• $4$ coins from bag$3$
• $8$ coins from bag$4$

So, product of label of bags will be $=1\times 3\times 4=12.$

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How you got the 2nd equation X+Y=31 ?
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:P
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Just perfect way of thinking :) . Many say solving as many problems as you can build this intuition but even after solving so many questions, I took this paper as test today. It was disaster :(

@minal @Ayush Upadhyaya how to deal with this??

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its kind of aptitude ..it will come :) no worry (y) keep doing
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Elegant solution
There are 5 bags , I assumed initially all bags are having 10 gm coins, and picked them as per the given condition

1,2,4,8,16   of all bags have 10 gm coins then total weight will come to

10 + 20 + 40 + 80+ 160 = 310  but total weight should be 323, but 13 is less, i divided 13 into 1 + 4 + 8
11 + 20 + 44+ 88 + 160, means 1st, 3rd and 4th bags have 11 gm coins. so product of labels will be 1*3*4 = 12

better explanation

let xi ------> coin weight in bag i
hence xi can be 10 or 11
basic crux of problem is

odd +even=odd ---(0)
even +even =even--(1)

x1 + 2*x2 + 4*x3 +8*x4 + 16*x5 =323 --(2)

from (0), we get
x1 is odd since other terms are of form 2*x.
hence x1 can only be 11.

2*x2 + 4*x3 +8*x4 + 16*x5 =323 -11=312 --(3)

dividing (3) by 2
x2 + 2*x3 +4*x4 + 8*x5 =156 --(4)

here 156 is even and 2*x3 +4*x4 + 8*x5  is even,hence from (1), x2 must be even
i.e      x2 = 10
so,
2*x3 +4*x4 + 8*x5 =156 -x2 =156-10=146 --(5)

dividing (5) by 2

x3 +2*x4 + 4*x5 =73 ---(6)

similarly continuing
x3 is odd hence x3=11

After taking (eq(6)-x3)/2

x4 + 2*x5 =31
x4 is odd hence x4 =11

implies x5=10

xi={11,10,11,11,10}

Answer is 1*3*4=12 . correct is (B)

Simplest logic according to me,
Total coins= 1+2+4+8+16=31
Now, if we put 11 coins in all the bags(maximum situation) then maximum weight= 31*11=341.
Now, we want only 323.
So, 341-323= 18 gm weight is extra for us.
Now, we try to remove this extra 18 gm weight by checking any of pairs.
We get 2+16=18 these two bags doesn't need 11 gm coins
So answer is 1st bag*3rd bag*4th bag
12
Hope you get it!
P.S. With this approach we can solve for anything amount from 0 to 341.
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18gm is extra. Why are you making combination of number of coins?

The total number of coins taken=31

The minimum possible weight is 310gm , but given total weight of the coins is 323gm.

We need 13gm, which is possible with only 1st bag, 3rd bag and 4th bag.

Ans: 1×3×4=12

1 vote
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.
1 vote
Total weight given is 323. We can get 3 at the end only by multiple of 11. If we take 3 coins we get weight distribution as-

290+33= 323.

We can have bag 1 and bag 2 for having 3 coins of weight 11 but remaining 4,8,16 coins of weight 10 account to only 280.

So our only choice left is to have 13 coins of weight 11.

11*13=143

323-143=180. The remaining coins 16+2=18*10=180.

Hence bag 1,3,4 contains coins of weight 11. So 12 is the answer.

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