GATE CSE 2014 | Set 1 | Question: 37

Question

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 ___.

Comments

Simple yet elegant! what amazes me more is the best answer selected, by @minal. hatsoff!!

Answer 1

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 $=1\times 3\times 4=12.$

Comments

first of all question is too good and explaination was simply superb

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11+22+44+80+166=323 labels of 11 gm coins are 1,2,3,5 so answer will be 30 why this is wrong pls anyone clarify?? Ohh shit what a silly mistake 11*16=176 na not 166 sorry

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i also thought in the same way  :|

Answer 2

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

Answer 3

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.

Comments

18gm is extra. Why are you making combination of number of coins?

Answer 4

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 5

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

Answer 6

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.