@DeAdLy said:Q:Find the sum of all possible distinct remainders which are obtained when squares of a prime number are divided by 6a.7, b.8, c.9, d.10
8
@techgeek2050 said:100 students of a class appeared in monthly test of at least one of the three subjects; Physics, Chemistry or Mathematics. Number of students who appeared in Mathematics test is more than who appeared in Chemistry test which is more than who appeared in Physics test which is more than who appeared in exactly two subjects which is more than who appeared in all three. At least one student appeared in test of all three subjects. What is the minimum number of students who appeared in test of Chemistry?
@chillfactor sir plz see this one.
@Tusharrr said:Suppose there are 10 coins laid out in front of you. All of the coins are fair (i.e. have an equal chance of heads or tails) except one, which flips to heads every time. You draw one coin at random and flip it 5 times. If each of the 5 flips results in heads, then the probability that this coin is fair can be written as a/b, where a and bare coprime positive integers. What is the value of a+b?
probability of getting 5 heads = (1/10)(9*1/32 + 1) = 41/320
Probability that a fair coin is selected and we get 5 heads = (1/10){9*(1/32)} = 9/320
required prob = (9/320)/(41/320) = 9/41
@Buck.up There are 25 points on a plane of which 7 are collinear , how many quadrilaterals can be formed from these points?a)5206
b)2603
c)13015
d)none of these
quadrilaterals can be formed with 7 collinear points and 18 non collinear point
18c4+7c2*18c2+7c1*18c3
b)2603
c)13015
d)none of these
quadrilaterals can be formed with 7 collinear points and 18 non collinear point
18c4+7c2*18c2+7c1*18c3
@Buck.up What is the probability of getting a word with all 'T's together for the word ATTEMPT ?
total cases of arrangement--7!/3!
all three T together-- 5!
5!*3!/7!=1/7
total cases of arrangement--7!/3!
all three T together-- 5!
5!*3!/7!=1/7
100 students of a class appeared in monthly test of at least one of the three subjects; Physics, Chemistry or Mathematics. Number of students who appeared in Mathematics test is more than who appeared in Chemistry test which is more than who appeared in Physics test which is more than who appeared in exactly two subjects which is more than who appeared in all three. At least one student appeared in test of all three subjects.What is the minimum number of students who appeared in test of Chemistry?
@techgeek2050 said:@chillfactor sir plz see this one.
I don;t have any proper approach to solve this one, but intuitively all three should be 1, then make physics and maths only 2
So, now all three = 1
and exactly two = 2
Physics = 3
So, chemistry has to be atleast 4
(put 3 in only chemistry and rest all in only maths)
:HELP:
A)The integer sequence x1,x2,x3.....is such that x(n+2)=x(n+1)-x(n) for n>=1. Sum of first 1492 terms is 1985 and of frst 1985 terms is 1492. Find sum of first 2007 terms.
1)3477 2)986 3)493 4)1491 5)573
B)Find no. of terms common in series:
S1=1,3,6,10,15.......200 terms
S2=3,6,9,12,15..........200 terms
1)21 2)22 3)132 4)133 5)None of these
@AsihekAdhvaryu said:I think max value will be infinite
the max value will be 392. But solving it will be very cumbersome. If the two quadratic polynomials are plotted separately, they touch each other at x=36, which is also the common root. After x=36, they both are increasing, and one of them attains the maxima at x=50 after which it starts decreasing while the other one keeps on increasing. so the maxima must lie close to x=50. Although the exact value of x at which maxima is attained is at x=36 + 14 rt(2) = 55 approx , I didn't solve any furthur and just put x=50 at which the value obtained is approx 340. If we increase x from x=50 to a little higher value, the answer will get closer to the actual max
@meow14 said:
S1=1,3,6,9,10,15.......200 terms
There is a typo in this series i guess 
. please check out whether 9 is there in Q.P.
. please check out whether 9 is there in Q.P. @Buck.up said:What is the probability of getting a word with all 'T's together for the word ATTEMPT ?
Another approach: We only need to know what is the probability of the Ts being together. other letters are irrelevant.
Number of ways of choosing 3 positions out of 7 = 7C3 = 35.
Number of ways of choosing 3 adjacent positions out of 7 = 5 (123, 234, 345, 456, 567)
Hence required probability is 5/35 = 1/7
@REVEALED
regards
scrabbler
Number of ways of choosing 3 positions out of 7 = 7C3 = 35.
Number of ways of choosing 3 adjacent positions out of 7 = 5 (123, 234, 345, 456, 567)
Hence required probability is 5/35 = 1/7
@REVEALED
regards
scrabbler
@albiesriram said:There is a typo in this series i guess . please check out whether 9 is there in Q.P.
Eh..sorry. Its 1,3,6,10,15.....
ABC is an acute triangle with ∠BCA=35∘. Denote the circumcenter of ABC as O and the orthocenter of ABC as H. If AO=AH, what is the value of ∠ABC(in degrees)?
Details and assumptions
The circumcenter of a triangle is the center of a circle which passes through all three vertices of a triangle. The orthocenter of a triangle is the intersection of the 3 altitudes (perpendicular from vertices to opposite side).
Use extended sine rule if needed
@meow14 said::HELP:A)The integer sequence x1,x2,x3.....is such that x(n+2)=x(n+1)-x(n) for n>=1. Sum of first 1492 terms is 1985 and of frst 1985 terms is 1492. Find sum of first 2007 terms.1)3477 2)986 3)493 4)1491 5)573
986...?
My logic: the series repeats every 6 terms. Let us assume the first two terms as a and b, then the series will have terms
a
b
b-a
-a
-b
a-b
a
b
and so on. So sum of every set of 6 terms is 0.
So sum of 1492 (ie 6k+4) terms = sum of 4 terms and sum of 1985 (i.e. 6m+5) terms = sum of 5 terms so the 5th term is 1985 - 1492 = 493. This is nothing but b.
Now we could go on to find a, but we need not do so. Since we need to find sum of 2007 (i.e. 6p+3) terms which is the sum of the first 3 terms = a + b + (b - a) = 2b = 2 * 493 = 986.
regards
scrabbler
@meow14 said::HELP:
B)Find no. of terms common in series:S1=1,3,6,10,15.......200 termsS2=3,6,9,12,15..........200 terms1)21 2)22 3)132 4)133 5)None of these
4)133?
the only non-multiples of 3 in S1 are 1, 1+ 9X1, 1+ 9X(1+2), and so on
the last such term = S1(199) = 19900 = 1 + [9 X (1+2+3+...+n)]
n =66
so there are 67 such terms
balance 133 terms are common terms
@scrabbler said:986...?My logic: the series repeats every 6 terms. Let us assume the first two terms as a and b, then the series will have termsabb-a-a-ba-baband so on. So sum of every set of 6 terms is 0.So sum of 1492 (ie 6k+4) terms = sum of 4 terms and sum of 1985 (i.e. 6m+5) terms = sum of 5 terms so the 5th term is 1985 - 1492 = 493. This is nothing but b. Now we could go on to find a, but we need not do so. Since we need to find sum of 2007 (i.e. 6p+3) terms which is the sum of the first 3 terms = a + b + (b - a) = 2b = 2 * 493 = 986.regardsscrabbler
Thanx. 
But OA not available.
But OA not available.@RDN said:4)133?the only non-multiples of 3 in S1 are 1, 1+ 9X1, 1+ 9X(1+2), and so on...there are 67 such terms
how did you get 67 terms?
In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.