A no. when divided by11 gives remainder 3,when divided by 16 gived remainder 4 but when divided by 11 gives no remainder.What is d least such no?
pls show d process
dude correct the question
dude correct the question
Asfakul Saysdude correct the question
how is it poss........on one side the ques is saying it gave rem 3 wen divided by 11.........and in the nxt statement it is saying it gave no rem wen divide by 11.......may b i didn't understand the ques....or there will be some new concepts....plz discusss this ques puys.....
thnks
Arun Sharma P&C; LOD3 Pg. 531 Q.32
Six white and six black balls of the same size are distributed among ten urns so that there is at least one ball in each urn. What is the number of different distributions of the balls?
a) 25000
b) 26250
c) 28250
d) 13125
It's eating my head please solve it someone!
I have few queries on remainders : pg . 91..Q 11,15,16
1.What is the remainder when 123412341234..upto 400 digits is divided by 909?
2.What is the remainder when 971 (30^99+61^100)*1148^56 is divided by 31.
3.Remainder when 2^100 is divided by 101.
Please help 
Ten books are arranged in a row on a bookshelf. A student has to select three out of these ten books
in such a way that no two books selected by him must have been lying adjacently. In how many
ways can he make the selection?
....please someone explain it in detail.
its same like placing of three books 7 books such no 2 from that 3 will come together
so, _b_b_b_b_b_b_b_
8 blank spaces 3 books to place
its 8C3=56
Arun Sharma P&C; LOD3 Pg. 531 Q.32
Six white and six black balls of the same size are distributed among ten urns so that there is at least one ball in each urn. What is the number of different distributions of the balls?
a) 25000
b) 26250
c) 28250
d) 13125
It's eating my head please solve it someone!
http://www.pagalguy.com/forum/quantitative-questions-and-answers/70678-official-quant-thread-cat-2011-a-686.html#post2894010
go by this
I have few queries on remainders : pg . 91..Q 11,15,16
1.What is the remainder when 123412341234..upto 400 digits is divided by 909?
2.What is the remainder when 971 (30^99+61^100)*1148^56 is divided by 31.
3.Remainder when 2^100 is divided by 101.
Please help
2. i think it will be zero as (30^99+61^100) will be divisible by 31.
3.is the answer 1??
will post the solution for 1 in a while...
2. i think it will be zero as (30^99+61^100) will be divisible by 31.
3.is the answer 1??
will post the solution for 1 in a while...
yup..answer is 1 ..how did u get it ?
AishR Saysyup..answer is 1 ..how did u get it ?
101 is prime
so E(101) = 100
and using Fermant's Law we have the remainder of M^(N-1) when divided by N, when N being prime is 1...
so 2^100/101 = 1
For question number 1...
12341234.. 400digits
each digit will come for 100 times.. (400/4)
so all we have to do is 100(1+2+3+4) = 100*10 = 1000/909 = 91
pls confirm
Two people, A & B , start from P & Q (distance = D) at the same time towards eachother. They meet at point R, which is at a dist of 0.4D from P. They continue to move to and fro between the two points. Find the distance from P at which the fourth meeting takes place.
a)0.8D
b)0.6D
c)0.3D
d)0.4D
Two people, A & B , start from P & Q (distance = D) at the same time towards eachother. They meet at point R, which is at a dist of 0.4D from P. They continue to move to and fro between the two points. Find the distance from P at which the fourth meeting takes place.
a)0.8D
b)0.6D
c)0.3D
d)0.4D
0.4D oly i guess!!!!
chandrakant.k Says0.4D oly i guess!!!!
Answer is givrn as option a)
Two people, A & B , start from P & Q (distance = D) at the same time towards eachother. They meet at point R, which is at a dist of 0.4D from P. They continue to move to and fro between the two points. Find the distance from P at which the fourth meeting takes place.
a)0.8D
b)0.6D
c)0.3D
d)0.4D
Itz 0.8D...You can do it manual calculation
Ganu02 SaysItz 0.8D...You can do it manual calculation
could u explain??
my method..
since they meet 0.4D at their first meet, the 1st person will travel 0.4D and 2nd 0.6D.. so ratio of their speed is 2:3... when they meet for 4th time, they together cover a distance of 4D.. so 1st person must have travelled 1.6D and 2nd person 2.4D.. so it is 0.4D from P.. where am i gng wrong???:shocked:
In an exam, 80% passed in Physics,70% in chemistry while 15% failed in both
subjects.325 passed in both the subjects. Find total nos of students who appeared
in the exam.
a) 500 b) 400 c) 300 d) 600
Plz give me a bit of explanation.
In an exam, 80% passed in Physics,70% in chemistry while 15% failed in both
subjects.325 passed in both the subjects. Find total nos of students who appeared
in the exam.
a) 500 b) 400 c) 300 d) 600
Please give me a bit of explanation.
In an exam, 80% passed in Physics,70% in chemistry while 15% failed in both
subjects.325 passed in both the subjects. Find total nos of students who appeared
in the exam.
a) 500 b) 400 c) 300 d) 600
Please give me a bit of explanation.
500??? i did back tracking which is advised in these type of questions!!!
could u explain??
my method..
since they meet 0.4D at their first meet, the 1st person will travel 0.4D and 2nd 0.6D.. so ratio of their speed is 2:3... when they meet for 4th time, they together cover a distance of 4D.. so 1st person must have travelled 1.6D and 2nd person 2.4D.. so it is 0.4D from P.. where am i gng wrong???:shocked:
You are correct bro.The only thing is that just you have to calculate reverse...
At first meet,
Time---------A----------B
1st min------0.2D-------0.7D
2ND min-----0.4D-------0.4D-----1st meet
3rd min------0.6D-------0.1D
4th min------0.8D-------0.2D
5th min------1.0D-------0.5D
6th min------0.8D-------0.8D-----2nd meet
7th min------0.6D-------0.9D
8th min------0.4D-------0.6D
9th min------0.2D-------0.3D
10th min-----0.0D-------0.0D-----3rd meet
11th min-----0.8D-------0.3D
12th min-----0.8D-------0.6D
13th min-----0.8D-------0.9D
14th min-----0.8D-------0.8D-----4th meet
there is another method i ll update that too....
In an exam, 80% passed in Physics,70% in chemistry while 15% failed in both
subjects.325 passed in both the subjects. Find total nos of students who appeared
in the exam.
a) 500 b) 400 c) 300 d) 600
Please give me a bit of explanation.
I am getting a)500
let total students be 100x.
then, 80x+70x-k = 100x-15x (k - Students who passed in both)
=> k =65x=325
=> x= 5
Use Venn Diagrams
In an exam, 80% passed in Physics,70% in chemistry while 15% failed in both
subjects.325 passed in both the subjects. Find total nos of students who appeared
in the exam.
a) 500 b) 400 c) 300 d) 600
Please give me a bit of explanation.
solve it using venn diagrams
let students passing in only phy be a and only chem be b
and the total no of students be x
so .15x+(.8x-325)+(.7x-325)+325=x
solving you will get x=500