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47 votes
47 votes
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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Best answer
125 votes
125 votes

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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49 votes
49 votes
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
11 votes
11 votes
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.
10 votes
10 votes

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)

Answer:

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