E - All talls (10) are dolls (5). A - All balls(10) are talls (5). C - Some dolls(5) are balls (5). Option d is the answer.
@vikas_saini sir, in the above triplet (example that you had solved above in your post) in the conclusion, balls has a value of 5, while in the statement 'A' it has a value of 10 - so there's a mismatch, then how could it be the answer? please spare time & explain
@tanveer-malhotra this might need some explanation. (this is commonly called as camel and banana puzzle and is a frequent visitor in interviews for programming roles!) 3000 km from Source to Destination. Elephant need 1 banana per km and it can carry at max 1000 banana at a time. If elephant try to go straight to destination it will eat all the 1000 bananas in 1 km itself). So we need to find an intermediate position where we can store the bananas on the way. So the elephant should shuttle between source and intermediate and this brings in one more constraint that the intermediate should never be more than 500 km away from the source. Else, elephant won't have any banana left to return to source.
With 3000 bananas in total, elephant can make 3 trips from source. So it would be like take 1000 bananas from source ---> travel to intermediate ---> store all the bananas at intermediate after taking enough bananas for return travel --> travel to source --> repeat (3 times) So we can see the path between source --- intermediate is covered 5 times here. Now comes the key part. This stretch of Source -- Intermediate cost us 5 banana per km. So we need to ensure the intermediate is located at a place where the elephant can store max of 2000 bananas. Why ? Because after that, 2000 bananas can be transported to destination (or another intermediate) at a cost of 3 bananas per km (explained below) which is better than spending 5 bananas per km.
take 1000 bananas from intermediate 1 ---> travel to intermediate 2 ---> store all the bananas at intermediate 2 after taking enough bananas for return travel --> travel to intermediate --> repeat (2 times)
So here we can see the stretch intermediate 1 --- intermediate 2 is covered at a cost of 3 banana per km.
So we have to be left with 2000 bananas at intermediate 1. As we are spending 5 bananas per km in this stretch, we can say 5d = 1000 ==> d = 200. So Intermediate 1 is 200 km away from source.
similarly we should ensure we bring 1000 bananas to intermediate 2, (as then the elephant can just take them all to destination in one shot - spending 1 banana per km)
So 3d = 1000 => d = 333
So intermediate 2 is 333 km away from intermediate 1.
Distance left till destination from intermediate 2 = 1000 - (200 + 333) = 467
If elephant takes 1000 bananas from intermediate 2, there will be 1000 - 467 = 533 bananas transported to destination.
Here if you see, what we did is to optimize the price per km (in banana units!) and ensure we place the intermediate points such that it would be a multiple of the maximum carrying capacity. Like source (3000 bananas), Intermediate 1 (2000 bananas) and intermediate 2 (1000 bananas)..
@SHAKHI There is two things which we have to remember in that set
Irfan has to practice all three aspects on a particular day.Three should not be gap of more than two days. Now coming to your doubt.. among 3rd,4th and 5th day is the day where he practices all three aspects. If he does practices on 4th day then the gap between fielding is 3 days. Same case with 3rd day also. 5th day is the day which is satisfying above cases. Hence 5th day is wednesday. I hope it's clear now.
Q3) Let say we had x coins in the beginning
After 1st loot : x - 1 - (x-1)/3 = 2/3(x-1) = (2x-2)/3
After 2nd loot : (2x - 2)/3 - 1 - [(2x - 2)/3 - 1]/3 = 2/3[(2x-2)/3 - 1] = (4x - 10)/9
After 3rd loot : (4x - 10)/9 - 1 - [((4x - 10)/9 - 1]/3 = 2/3 [(4x - 10)/9 - 1] = (8x - 38)/27
Now the final value (after throwing away the fake coin) = (8x - 38)/27 - 1 = (8x - 65)/27 is a multiple of 3.
So we can write, (8x - 65)/27 = 3k
x = (81k + 65)/8
x is an integer, 81K + 65 should be a multiple of 8, means (81K + 65) Mod 8 = 0
We know 81 Mod 8 = 1 and 65 mod 8 = 1. So to get remainder 0, we need to adjust K in a way that 81K mod 8 = 7.
Least value would be K = 7
For K = 7, x = (81 * 7 + 65)/8 = 79, which should (could) be our answer.
Let me know in case of any Logical/Calculation errors.
Looks like your connection to MBAtious was lost, please wait while we try to reconnect.