@ pkaman
Hi Buddy ,
Can u plz paste Similar Links which deals with fundas and short cuts for DI & LR sections.
Thanks
@ pkaman
Hi Buddy ,
Can u plz paste Similar Links which deals with fundas and short cuts for DI & LR sections.
Thanks
Well, am not sure if any such thread exists. May be you could search it in PG itself. Moreover, for DI, its how fast we are in calculations, and about LR, I doubt if there is any formula to solve questions, as the section itself says reasoning .. :P
But, if I come across any of such links, I'll let you know.
Please solve the following question : LOD 2 ,Number Systems
Q 79.A set S is formed by including some of the First one thousand naural numbers.S contains the max no. of numbers such that they satisfy the following questions.
1.No number of the set S is prime.
2.When the numbers of the set S are selected two at a time , we always see co prime numbers.
what is the number of elements in the set S?
a)11 b)12 c)13 d)7
can somebody post the ans for this one...acc to me it shud be 11
ksagarwa Sayscan somebody post the ans for this one...acc to me it shud be 11
Can you please post the detailed solution.
Please solve the following question : LOD 2 ,Number Systems
Q 79.A set S is formed by including some of the First one thousand naural numbers.S contains the max no. of numbers such that they satisfy the following questions.
1.No number of the set S is prime.
2.When the numbers of the set S are selected two at a time , we always see co prime numbers.
what is the number of elements in the set S?
a)11 b)12 c)13 d)7
Since all are composite numbers and all are co-prime to each other, we can say that all the numbers are product of prime numbers.
Also, 32 is greater than 1000, both the prime numbers in the product can not be greater than 31.
So, basically we just have to count number of prime numbers less than or equal to 31, i.e., 11 in numbers.
Also, we can have 1, as its neither prime nor composite.
=> We can have 12 number of elements
Need help with this one from Progressions LOD3:
If x, y, z are in GP and ax, by, cz are equal then a, b, c are in
1. AP
2. GP
3. HP
4. None of these
5. Cannot be determined
Let common ratio of P be r
So y=xr, z=yr
ax=by....ax=bxr....a=br....
Similarly, b=cr....
So, b=a/r, c=a/r^2....
So, a,b,c ar in GP with common ration 1/r....
Hi there ,
Can you plz help me with these questions from Arun Sharma (Numbers/ Progressions):
1)1.2.3+ 2.3.4+ 3.4.5+...................... + 10.11.12?
Find the sum.
2)8+ 88+888+...............................+8888888888
Find the sum.
3)1/1*5+1/5*9+1/9*13+ ............................. +1/221*225
Find the sum.
4)How many 3 digit no's have the property that the digits taken left to right form an AP or GP
5)A 23*23 grid of numbers is such that the nos. in each horizontal row form an AP and each Column form an AP.
7 AND 17 column of 5th row contain digits- 47 & 63 respectively
and 7 AND 17 column of 15th row contain digits- 53 & 77 respectively
Find sum of all the numbers in the grid.
6) 1/+ 1/+1/.......................... 1/
Thanks!!
Hi there ,
Can you plz help me with these questions from Arun Sharma (Numbers/ Progressions):
1. T(n)=n(n+1)(n+2)=n(n^2+2n+2)=n^3+2n^2+2n
Sum=Summation of (n^3)+2*Summation of (n^2)+ 2* Summatio of (n)
=^2+2*n(n+1)(2n+1)/2+2*n(n+1)/2
2. T(n)=8+80+800+....+8*10^(n-1)=8, n being 10
This is a GP with common ratio 10......
So, T(n)=8*/9
Sum=summation of T(n)=8/9*
Put n=10 to get the huge answer.....
3. 1/1*5+1/1*9+1/9*13+........+1/221*225=1/4
=1/4=224/4*225=56/225
4. AP
123,234,345......789......7 numbers.....
135,246......579......5 numbers....
147,258,369.........3 numbers....
159.....1 number.....
These numbers in reverse will also be APs with negative common difference......so total=16*2......
GP
111,222,...999....9 numbers....
124,248.....421,842.....4 numbers.....
369....and 963....2 numbers...
469....964......2 numbers.....
So total=49....
6. S=Summation 1/....multiply top and bottom by
Then, S=sqrt(2)-sqrt(1)+sqrt(3)-sqrt(1)+...........+sqrt(121)-sqrt(120)
After subtracting you will get S=sqrt(121)-sqrt(1)=10
Hello people,
I came across this question of probability in Arun Sharma.
There are 8 orators A,B,C,D,E,F,G and H in how many ways the arrangements be made so that A always comes before B and B always comes before C
The options are as follows :
1.) 8!/3!
2.) 8!/6!
3.) 5!.3!
4.) 8!/(5!.3!)
The book says the correct answer is option 1 :shocked:, but I think it is option 3 which is correct.
My approach is since A,B,C are always going to be in a particular order only we can consider them as 1 letter so in all we have 6 i.e. D,E,F,G, H and ABC together as 1. The number of possible arrangements for 6 letters is
6! which is equal to 5!.3! (option c)
I doubt Arun Sharma's answer sheet as many times the answers metioned in them have been found wrong.
Correct me if I am wrong
nopes ,actually there might be letters b4/after/between A,B,C too but in any case A should come before B and B before C
now in the total 8! arrangements of the numbers A,B,C have been arranged in 6! ways but out of every 6! ways of arranging them , only 1 is favorable where they occur in the mentioned order.
no of desired permutations = 8!/6!
nopes ,actually there might be letters b4/after/between A,B,C too but in any case A should come before B and B before C
now in the total 8! arrangements of the numbers A,B,C have been arranged in 6! ways but out of every 6! ways of arranging them , only 1 is favorable where they occur in the mentioned order.
no of desired permutations = 8!/6!
Well you almost got it, just that ABC can be arranged amongst themselves in 3! ways not 6!. So answer is 8!/3!
This question was discussed some 10-15 pages back as well;)
oops my bad type o 😁 ya 8!/3! 😁
and ya so many pages back , just felt like replying 😛
Two men start walking on a road which diverges by 120 degrees.
Find the distance between them after 4 hours , if they walk at speeds of 3km/hr and 2 km/hr respectively.
ans. 4*(19)^1/2
correct or not?
Two men start walking on a road which diverges by 120 degrees.
Find the distance between them after 4 hours , if they walk at speeds of 3km/hr and 2 km/hr respectively.
ans. 4*(19)^1/2
correct or not?
applying the cosine formula, am getting the same ans...
Product of n consecutive integers is always divisible by n!. So for n = 16, 5, 20 and 25, product should be divisible by 16!, 5!, 20!, 25! respectively. But 1000 is not divisible by any of these.
So, I think its a misprint in the book. The correct question should be like:-
If x is positive integer and
x + (x + 1) + (x + 2) + (x + 3) + .... + (x + (n - 1)) = 1000 then which of the following cannot be true about number of terms n
a)The number of terms can be 16
b)The number of terms can be 5
c)The number of terms can be 25
d)The number of terms can be 20
x + (x + 1) + (x + 2) + (x + 3) + .... + (x + (n - 1)) = n(n + 2x - 1)/2 = 1000
=> n(n + 2x - 1) = 2000
Now, we can notice that one of n and (n + 2x - 1) is odd and other will be even and n
2000 = 1*2000 = 5*400 = 16*125 = 25*80
=> n can be 5, 16 and 25
So, 20 should be the answer.
Chill bhai tussi gr8 ho.
Could anyone help me with this. Its from Arun Sharma's Probability LOD 3:banghead:
Probabilities that Rajesh passes in Maths, Physics and Chemistry are m, p and c respectively. Of theses subjects, Rajesh has a 75% chance of passing , in at least two and 40% chance of passing in exactly two. Find which of the following is true?
a) p+m+c =19/20
b)p+m+c = 27/20
c)pmc = 1/20
d)pmc =1/8
I did not get even using the hint
The answer is "b"
I dont understand what "p+m+c" exactly represents:splat:
thanx in advance
anyone
A 23*23 grid of numbers is such that the nos. in each horizontal row form an AP and each Column form an AP.
7 AND 17 column of 5th row contain digits- 47 & 63 respectively
and 7 AND 17 column of 15th row contain digits- 53 & 77 respectively
Find sum of all the numbers in the grid
anyone with dis one ?
Well you almost got it, just that ABC can be arranged amongst themselves in 3! ways not 6!. So answer is 8!/3!
This question was discussed some 10-15 pages back as well;)
If A, B & C would be arranged among themselves, then they won't be following the mentioned constraint, i.e., A before B before C.
I don't think if this is the answer ... 😞 Please clarify
Plz answer this:
a dealer offers a cash discount of 20% and still makes a profit of 20% ,when he further allows 160 articles when the customer buys 120 .how much percent above the c.p were his wares listed.
Plz answer this:
a dealer offers a cash discount of 20% and still makes a profit of 20% ,when he further allows 160 articles when the customer buys 120 .how much percent above the c.p were his wares listed.
Is the answer 50 % ?
If A, B & C would be arranged among themselves, then they won't be following the mentioned constraint, i.e., A before B before C.
I don't think if this is the answer ... 😞 Please clarify
Yaar its like this, 8 orators can be arranged in 8! ways.
also ABC can be arranged in 3! ways.
But out of these 6 combinations only one combination will suit our requirement of ABC.
So 8!/6 i.e. 8!/3!
:D