Afterall they were right.......
In earlier post, i had mentioned i am not convinced about the usage of plug-in methods. Seems like someone overheard this.
If a sequence of consecutive integers of increasing value has a sum of 63 and a first term of 6, how many integers are in the sequence?
11
10
9
8
7
As i forgot the sum on n terms formula in AP i found this difficult to solve. The key didn't involve AP formulae. Instead they found out by working out the sum themselves starting from 6.
In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?
a) 9/10
b) 16/20
c) 2/5
d) 3/5
e) ½
I read about a type of probability problem where atleast one is used. I forgot what was the trick involved there and i cudn't think more into the problem also. i was stuck. (During the morning trip in the van i tried this) I know for sure this problem wud involve 5 and 4 in denominator and the numerator involved 3. Hence, a shud be the answer. Bingo!!! the guess is right. Though i understood the problem later, i now fully convinced that guessing and plug-in r effecient methods especially if u r stuck with the problem.
If a sequence of consecutive integers of increasing value has a sum of 63 and a first term of 6, how many integers are in the sequence?
11
10
9
8
7
As i forgot the sum on n terms formula in AP i found this difficult to solve. The key didn't involve AP formulae. Instead they found out by working out the sum themselves starting from 6.
In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?
a) 9/10
b) 16/20
c) 2/5
d) 3/5
e) ½
I read about a type of probability problem where atleast one is used. I forgot what was the trick involved there and i cudn't think more into the problem also. i was stuck. (During the morning trip in the van i tried this) I know for sure this problem wud involve 5 and 4 in denominator and the numerator involved 3. Hence, a shud be the answer. Bingo!!! the guess is right. Though i understood the problem later, i now fully convinced that guessing and plug-in r effecient methods especially if u r stuck with the problem.
posted by Ramkumar @ 1:46 AM
1 Comments:
At 1:15 PM ,
Iday said...
I accept. But they are not dependable methods u see. The first problem is a no brainer - keep adding consecutive numbers to 6 and u'll have ur count. Dont sweat - the AP method is actually longer than this one :)
The second problem is a pure guess that u made. Teh trick, as u must be knowing now, is to convert this into a problem of "1 - P(zero red balls)". Even if u missed that, u shud have tried the longer method of "P(one red ball) + P(two red balls)". This is basically RB+BR+RR i.e. ((3/5)*(2/4)) + ((2/5)*(3/4)) + ((3/5)*(2/4)) = 9/10.
Take some time da - this is not CAT. No need for short cuts. And no takinf risks - u need to be as sure as possible abt the answer choices u make.
Post a Comment
Subscribe to Post Comments [Atom]
<< Home