Here given E x E = E and A x A =A.
So E = 1 or 5, A = 5 or 1.
Let’s put E = 1 and A = 5.
5B1 = 1 x B1
Not possible.
Let’s put E = 5 and A = 1.
1B5 = 5 x B5
So B = 2.
BBA + CCD = EEE.
221 + CCD = 555.
1 + D = 5.
D = 4.
2 + C = 5.
C = 3.
AB x CE = 12 x 35 = 420.
Let assume length of the train is ‘L’ and speed of train = a.
Speed of man 1 = b.
Speed of man 2 = c.
20 = L / a – b.
18 = L / a + c.
18a + 18c = 20a – 20b.
a = 10b + 9c.
Distance of two men = 600 x (a + c).
Time = 600(a+c) – 600(b+c) / (b+c)
= 600 ( a – b) / (b + c)
= 600 (10b + 10c – b) / (b + c)
= 5400 seconds
= 90 min (ans)
@amar-rajput ;-we only have two choice for B and E and that is 300 and 500,now if we take E=500 the net amount of E would become more than that of A.While in question they have mentioned that A must be greater than E.
@surya_prakash There are 200 students in a class. 20% students are there who don’t like any subject. 50% students like physics, 60% students like Chemistry and 70% students like Mathematics.
I + II + III = 100 - 20 = 80......(1)
I + 2II + 3III = 50+60+70 = 180.....(2)
Now if we put I = 0 then
II + III = 80....(3)
2II + 3III = 180....(4)
II should be maximum so that III cold be minimum.
Now equation (4) - 2 x equation (3)
III = 20.
This 20 is percent actually.
20% of 200 = 40.
I hope it's clear now.
@vikas_saini said in Data Sufficiency - Vikas Saini - Part 2:
Q4. X & y are positive integers such that x = 8y + 12, what is the greatest common divisor of x & y ?I. x = 12 u, where u is an integer.II. y = 12 z, where z is an integer.
why is statement 1 not sufficient?we come to conclusion y = 3(u-1) similar to x = 12(8z + 1)
@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.
Source ---------------------------- Intermediate ------------------------- Destination
With 3000 bananas in total, elephant can make 3 trips from source. So it would be liketake 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.(adsbygoogle = window.adsbygoogle || []).push({});
Source ---------------------------- Intermediate1 -------------------- Intermediate 2 ------------------------- Destination
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)..(adsbygoogle = window.adsbygoogle || []).push({});
Got it ?
That was lot of typing... Phew!!!
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
@vikas_saini a few doubts, please clarify them -
In cases where using either of the two given statements, we don't get a unique answer, shouldn't we mark the answer as 'can't be solved using either of the two statements' ?For example, take the question -When three times of the unit's digit of a number is subtracted from the number,28 is obtained.What is the number ?A. The digit at the ten's place is greater than the digit at the unit's place.B. The digit at the ten's place is less than the digit at the unit's place.
from statement A, we got answer as 31, from statement B it is 46, so isn't this conflicting?
Second DoubtIn a class of 200 students, the highest and the lowest scores in a test are 98 and 18 respectively. Is 50 the average score of the class in the test ?A. 100 students score above 50 and the remaining 100 students score below 50 in the test.B. If the highest score and the lowest score in the test are excluded, the sum of the top 99 scores is exactly double of the sum of the bottom 99 scores.
here we are to only ascertain if 50 is class average or not? So, doesn't statement B lead to a conclusion that the average is beyond 50?
