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Thanks oLahav , actually i was confused between words tens and tenths .. got it !!

Hey Indicar , I m still confused..In 70.1203 ( which one is thousandth digit ) ?

I got it .. Thanks guys !!

I think , CR’s ans is {D} , not so sure though !!

Hi Suresh ,

In the second ( problem solving ) explanation , I think x must divide 210 rather than 120. ( since 840 / 4 = 210 ). If we factorize 210 = 2 * 3 * 5 * 7 . We get 22*22 = 16 factors ( So ans does is 16 ).

Yes Uday , ans is B. Thanks for explaination !!

FIRST STATEMENT SAYS , E on Y axiz , 4 units away from origin.. But in which direction ? It could ( 0,+4 ) or ( 0,-4) .. so A is alone not sufficient.

THE SECOND STATEMENT is also not sufficient ,  since we dont know E.

But as OLahah explain , if we use both the statement , we can cross check E’s distance form C with two points i.e. (0,+4) and (0,-4). Out of these , only ( 0,+4 ) sutisfy the second condition.

So answer is C

How did u come to the conclusion that it was seen continuous till 11:30 am? Question does not mention till what time it was seen continuous. ( It just says that it was seen continuous at 3 hours before it really stopped ) . I agree with oLahav. Answer should be C.

Pretty confusing question, but Suresh solved it with simplistic manner. That is what everybody should learn here ( rather than just focusing on ques’s ans ) , the way to approach the problem. Really good job Suresh , keep it up.

CR :=> C PS :=> B

Here’s one approach: Set S: We know that the largest number is b and the median is 3b/4. Since the median is in the middle of the numbers of set S, we know that the smallest number in set S is 2b/4 You can see that 3b/4 (the median) lies directly in the middle 2b/4 and b (i.e., 2b/4, 3b/4, 4b/4) So, set S ranges from b/2 to b

Set Q: Here the largest number is c and the median is 7c/8. Using similar logic from above, we can conclude that set Q ranges from 6c/8 to 8c/8, with 7c/8 squarely in the middle.

Here’s the important step. 6c/8 is the smallest number in set Q. This value is equivalent to b (the largest value in set S)

So, now we have a relationship between c and b. We know that 6c/8 = b From here, we can find two integer values for b and c that satisfy this equation amd determine how sets S and Q might look.

One solution is b=6 and c=8 For set S, we know that the largest value is 6, so keep adding numbers until we meet the criterion that the median is 3/4 b. We do the same with set S, beginning with c=8. From here we can conclude that set S={3,4,5,6} and set Q={6,7,8} The median of {3,4,5,6,7,8} is 5.5, which means the median is 11/16 of c

We could have chosen other numbers for b and c but the fraction remains 11/16 For example, b=12 and c=16 gives us S={6,7,8,9,10,11,12} and Q={12,13,14,15,16} The median of {6, 7, . . . 15, 16} is 11, making it 11/16 of c

ANS :=> C. 11/16

Prepare the equations like 1) For set S (b-a)/2 = (3/4)b i.e. a = -1/2 b 2) For set Q ( c-b ) / 2 = (7/8) c i.e. b = -6/8 c

Now for set R , c-a/2 = ? put value of a in the form of b , then put value of b in the form of c , Solve it ( ans : 11/16 c )

really helpful stuff !!

is there any setting needs to be done to play the vedio in iexplorer? I am not able to see the vedio!!!!!

CR : (E) DS : (A)

brillient way of solving the complicated maths problems !! Keep it up !!