I am not sure of what i am doing is absolutely correct, but i m getting 6.

Let alok start with an amount k.
So, after 10 rounds he wld have -> 1024k - 1023.
1024k - 1023x = 1023.
1024k = 1023(x+1).
Now for the min value of x, k = 1023 and (x+1) = 1024.
=> x = 1023.
Hence, SOD = 6.

abe mei kahan galat soch raha hoon



Find all natural numbers n such that the number n(n+1)(n+2)(n+3) has exactly three different prime divisors.

I am not sure of what i am doing is absolutely correct, but i m getting 6.

Let alok start with an amount k.
So, after 10 rounds he wld have -> 1024k - 1023.
1024k - 1023x = 1023.
1024k = 1023(x+1).
Now for the min value of x, k = 1023 and (x+1) = 1024.
=> x = 1023.
Hence, SOD = 6.

Sood bhai ye 1023x kaise aaya, har round mai x gaya toh 10 round mai 10x jaana chahiye na???

Alok went to casino to play a card game. He played 10 rounds of that game. In each round he doubled his amount and then gave Rs x to his friend. After playing 10 rounds, he had Rs 1023. Find the sum of the digits of the least possible value of x.

1.5 2.6 3.7 4.8

I am not sure of what i am doing is absolutely correct, but i m getting 6.

Let alok start with an amount k.
So, after 10 rounds he wld have -> 1024k - 1023.
1024k - 1023x = 1023.
1024k = 1023(x+1).
Now for the min value of x, k = 1023 and (x+1) = 1024.
=> x = 1023.
Hence, SOD = 6.

QHow many real number sets (x, y, z, w) satisfy the equation x^2 + y^2 + z^2 + w^2 = x(y + z +w)?
OPTIONS
1)0....2)1.........3)2........4)3.........5)Infinite

My take 0

(x - y)^2 + (x -z)^2 + (x -w)^2 = 2x^2 - x(y+z+w) ->1

On substituting x(y + z +w) = x^2 + y^2 + z^2 + w^2 in 1

-2x(y + w + z) = 0 which means either x(y + w + z) = 0, subsequently x^2 + y^2 + z^2 + w^2 = 0 which happens only when all x,y,z,and w are 0

QIn a playschool Gods Souls a teacher decided to encourage the kids by distributing prizes to them. Each of the prizes was different from the other. The total number of prize was k, and total number of kids was p. To encourage the kids, the number of prizes was more than the number of kids. But, the teacher imposed a restriction on herself that each kid would receive (k 1) prizes at the most. Thus, the number of ways in which she could distribute the prize is:
OPTIONS
1)p^k 1...2)k^p....3)p^k k......4)None of these

Required numbers = Distribution of K different things among P kids - the cases where each kid gets all the prizes

so,it's p^k k

Alok went to casino to play a card game. He played 10 rounds of that game. In each round he doubled his amount and then gave Rs x to his friend. After playing 10 rounds, he had Rs 1023. Find the sum of the digits of the least possible value of x.

1.5 2.6 3.7 4.8

least possible value of X can be generated when we keep value of P as much close as 1023 after 10 rounds

if we take p=2

so , after 10 rounds amount becomes 1024

1024 - 1023*X = 1023

so , X = 1/1023

i think i misinterpret question !! somebody help

snipped!!!!!!!!!!!!!!

snowdrop Says

Is the answer 30?

u r right refer the link http://www.pagalguy.com/forum/quantitative-questions-and-answers/74662-official-quant-thread-cat-2012-a-480.html for varun bhai solution to this question

In a certain class the number of students who do not study at home or who do not attend classes is one third more than that of those who either study at home or attend classes.Further the number of students who do not study at home but attend classes is 2/5 th more than that of those who study at home but do not attend classes..while the number of students who study at home as well as attend classes is 1/2 that of those who neither study at home nor attend classes..total students is 150,then find number of students who study at home as well as attend classes..?? :banghead::banghead:

sabhi puys ko mera ____________/\___________________

Is the answer 30?

g(x+1) -g(x) =x+2; g(0)=0

find g(20)

my take 230
g(1) -g(0) = 2
g(2) - g(1) =3
.
.
...
g(20) - g(19) = 21
adding we get g(20) - g(0) = 230
or g(20) = 230

QIn a playschool Gods Souls a teacher decided to encourage the kids by distributing prizes to them. Each of the prizes was different from the other. The total number of prize was k, and total number of kids was p. To encourage the kids, the number of prizes was more than the number of kids. But, the teacher imposed a restriction on herself that each kid would receive (k 1) prizes at the most. Thus, the number of ways in which she could distribute the prize is:
OPTIONS
1)p^k 1...2)k^p....3)p^k k......4)None of these

normally it would have been

p^k
but since there is a restriction that a kid can get k-1 prize
total no. of cases that defies the restriction is when a single kid gets k or all prizes
so p cases

so total =p^k-p

QHow many real number sets (x, y, z, w) satisfy the equation x^2 + y^2 + z^2 + w^2 = x(y + z +w)?
OPTIONS
1)0....2)1.........3)2........4)3.........5)Infinite


QIn a playschool Gods Souls a teacher decided to encourage the kids by distributing prizes to them. Each of the prizes was different from the other. The total number of prize was k, and total number of kids was p. To encourage the kids, the number of prizes was more than the number of kids. But, the teacher imposed a restriction on herself that each kid would receive (k 1) prizes at the most. Thus, the number of ways in which she could distribute the prize is:
OPTIONS
1)p^k 1...2)k^p....3)p^k k......4)None of these

For any 2 natural numbers,x and y ,if x2+y2 is divisible by 21, what is the remainder when x,y are divided by 21 respectively

a) 3,4 b) 2,5 c) 0,0 d) 1,6 e) none of these

x^2+y^2 mod 21 =0

21 =3*7
x^2+y^2 mod 3 or 7=0


any perfect square is of form 3k and 3k+1 ,or 3k-1 so leave remainder 0,1,2 respectively therefore sum of sqaures will give either but sum is divisible by 3 so x,y must be of form 3k i.e. divisible by 3


square of any number will leave remainder by 7
0,1,2 or 4 becoz sum of squares is divisible by 7 so number must leave remainder 0 becoz any other possible sombination will not leave reminder 0by 7

see x=21a & y=21b so divisible by 21 hence leave remainder 0,0

samaj nahi aa raha...what if x=3 (<6) and y = 0 (y>-2)...
x+y <4????

yaar aise to y ki har value ke liye bahut saare x honge jiske liye condition satisfy nai karega
but question is asking for what value of y this inequalities would hold true (atleast for one value of x)
so for y less than or equal to -2 ke liye ek bhi x aisa nai hoga jiske liye expression would be true.
so y>-2

In a certain class the number of students who do not study at home or who do not attend classes is one third more than that of those who either study at home or attend classes.Further the number of students who do not study at home but attend classes is 2/5 th more than that of those who study at home but do not attend classes..while the number of students who study at home as well as attend classes is 1/2 that of those who neither study at home nor attend classes..total students is 150,then find number of students who study at home as well as attend classes..?? :banghead::banghead:

sabhi puys ko mera ____________/\___________________

nothing yaar just draw x=6 area on left side would be x<6
now draw x+y=4 area on the upper side would be x+y>6
now the intersection these two areas would be the solution set and this common area starts from y=-2 so y>-2

samaj nahi aa raha...what if x=3 (<6) and y = 0 (y>-2)...
x+y <4????

upudit0 Says

how can u be sure that x=21k1 & y=21k2

There will be other values of x,y of course
If the above expression is true for the set of natural nos (x,y) for which the above expression holds good
then x,y of the form 21k1,21k2 would be a subset of that set
So, if true for set then should be true for a subset too right
to make things easier I took a special case of 21k1,21k2

option 3
just take x any y 21k1 and 21k2

how can u be sure that x=21k1 & y=21k2

awesomeusername Says

thoda explain kar do yaar..kaise kiya ?

nothing yaar just draw x=6 area on left side would be x<6
now draw x+y=4 area on the upper side would be x+y>6
now the intersection these two areas would be the solution set and this common area starts from y=-2 so y>-2

wouldnt it be violated for x=0 and value of y<-2???

For what values of y would the following inequalities be consistent

x + y > 4 and x < 6

y>-2
since these are easy equations did my graph method

For what values of y would the following inequalities be consistent

x + y > 4 and x < 6

g(x+1) -g(x) =x+2; g(0)=0

find g(20)

g(20) = (19 + 2) + (18 + 2) + (17 + 2) + ....+ (1 + 2) + (0 + 2)

=> 2*20 + 19*20 / 2 = 40 + 190 = 230

choosen04 Says

Find the largest positive integer such that n^3 + 100 is divisible by (n+10)?

is it 890 ?????????????/

g(x+1) -g(x) =x+2; g(0)=0

find g(20)

I am also getting 230

g(20)-g(0)=20(0+2)+(1+2+3.......+19)

g20=230

g(x+1) -g(x) =x+2; g(0)=0

find g(20)

answer is 230 !!

g(20) = 21+20+19+18+......+2+g(0) !!

am i rite ??

for any 2 natural numbers,x and y ,if x2+y2 is divisible by 21, what is the remainder when x,y are divided by 21 respectively

a) 3,4 b) 2,5 c) 0,0 d) 1,6 e) none of these

0,0.........

Find the largest positive integer such that n^3 + 100 is divisible by (n+10)?

g(x+1) -g(x) =x+2; g(0)=0

find g(20)

For any 2 natural numbers,x and y ,if x2+y2 is divisible by 21, what is the remainder when x,y are divided by 21 respectively

a) 3,4 b) 2,5 c) 0,0 d) 1,6 e) none of these

option 3
just take x any y 21k1 and 21k2

For any 2 natural numbers,x and y ,if x2+y2 is divisible by 21, what is the remainder when x,y are divided by 21 respectively

a) 3,4 b) 2,5 c) 0,0 d) 1,6 e) none of these

123456789123456789.......upto 180 digits when divided by 11 will leave remainder of:
a)0 b)4 c)5 d)6

zero !!
Sum of odd terms-sum of even terms is zero !!
hence,no is div by 11 !!

123456789123456789.......upto 180 digits when divided by 11 will leave remainder of:
a)0 b)4 c)5 d)6

sum of the digits at even places - sum of the digits at odd places mod 11
=2+4+6+8+1+3+5+7+9 ....(repeat it 10 times) - 1+3+5+7+9+2+4+6+8 .... 10 times mod 11
= 450 - 450 mod 11 = 0

123456789123456789.......upto 180 digits when divided by 11 will leave remainder of:
a)0 b)4 c)5 d)6

N is made of 10 repeating sets where each set is

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9


By rule of divisbility for 11, add nos which are alternate and the remaining in to another
1+3+5+7+9+2+4+6+8 = 45
2+4+6+8+1+3+5+7+9 = 45
Both the totals are 45. Difference between two = 0

Multiply this for 10 sets. total = 0

Hence the no is divisible by 11.
So the remainder is 0.
Option A.

123456789123456789.......upto 180 digits when divided by 11 will leave remainder of:
a)0 b)4 c)5 d)6

Is it 1/12.......??

Given expression=x^4(1-x)+y^4(1-y)=x^4*y+x*y^4=xy(x^3+y^3)
=xy(x+y)(x^2-xy+y^2)
=xy((x+y)^2-3xy)=xy(1-3xy)

Now by AM-GM inequality, 0<=1/4
and 1-3xy>=1/4

NOW,

(3xy+(1-3xy))/2>=sqrt(3xy*(1-3xy))

so , 3xy*(1-3xy)<=1/4
or, xy*(1-3xy)<=1/12......

yeh kaise aya..?

vinay_sharma Says

suppose n is an integer such that the sum of the digits of n is 2, and 10^10< n <10^11. the number of different values of n is ?

my take 11 such numbers..

10^10 + (10^0 to 10^10)
so total 11 values.

vinay_sharma Says

there are 150 students out of which 125 play cricket , 130 play hockey , 135 play volley ball , 5 don't play any game. How many (minimum)students play all 3 games

Is it 100??

150-5=145

Ans wd be (125+130+135)-(2*145)=100

upudit0 Says

ans is correct aap approach dal dena m feeling sleepy

the absolute sequence has to be
0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10

Now leaving aside 0, for each x,x we can put -x or x.
So lets say for 5,5 we can have -5,5 or 5,-5

So 2 possibilities for each integer between 1 to 10.

So ans=2^10......

bs0409 Says

Is it 1024............???

ans is correct aap approach dal dena m feeling sleepy

there are 150 students out of which 125 play cricket , 130 play hockey , 135 play volley ball , 5 don't play any game. How many (minimum)students play all 3 games

QHow many ways can the integers from -10 to 10 be arranged in a sequence such that absolute value of the numbers in the sequence is non-decreasing?
OPTIONS
1)1023.....2)1024.......3)2048........4)2047.........5)512

Is it 1024............???

there are 127 coins each of 1 rupee. u have to put the coins in the minimum no. of bags such that u can give any amount from 1 to 127 by just selecting the bags u have.

1) tell the no. of bags .

2) number of coins each bag has inside it.

quetion is based on power of 2's it is the property of power of 2's that it can give any sum till the power it raised for eg:

2^3=8
now any amount <8 i can give you if want three i will give 2bags having 2^0+2^1 coins=3

so i need 2^6=64<128 7 bags

coins in each bag will be 1,2,4,8,16,32,64
GN to all puyzz

there are 127 coins each of 1 rupee. u have to put the coins in the minimum no. of bags such that u can give any amount from 1 to 127 by just selecting the bags u have.

1) tell the no. of bags .

2) number of coins each bag has inside it.

1) 7 bags

2)1,2,4,8,16,32,64

there are 127 coins each of 1 rupee. u have to put the coins in the minimum no. of bags such that u can give any amount from 1 to 127 by just selecting the bags u have.

1) tell the no. of bags .

2) number of coins each bag has inside it.

vinay_sharma Says

suppose n is an integer such that the sum of the digits of n is 2, and 10^10< n <10^11. the number of different values of n is ?

10^10=10000...10times

1100000..9times 0s can arrange -->10 ways

2000000-->1

total 11 ways

Q If the positive real numbers x and y satisfy the equation x + y = 1, what is the maximum possible value of S = x^4 + y^4 (x^5 + y^5)?


OPTIONS..1)1/2..... 2)1/3 ........3)1/4 ......4)1/12...... 5)1/16

snipped......

vinay_sharma Says

suppose n is an integer such that the sum of the digits of n is 2, and 10^10< n <10^11. the number of different values of n is ?

9 Is the Answer ??

Q If the positive real numbers x and y satisfy the equation x + y = 1, what is the maximum possible value of S = x^4 + y^4 (x^5 + y^5)?


OPTIONS..1)1/2..... 2)1/3 ........3)1/4 ......4)1/12...... 5)1/16

Is it 1/12.......??

Given expression=x^4(1-x)+y^4(1-y)=x^4*y+x*y^4=xy(x^3+y^3)
=xy(x+y)(x^2-xy+y^2)
=xy((x+y)^2-3xy)=xy(1-3xy)

Now by AM-GM inequality, 0<=1/4
and 1-3xy>=1/4

NOW,

(3xy+(1-3xy))/2>=sqrt(3xy*(1-3xy))

so , 3xy*(1-3xy)<=1/4
or, xy*(1-3xy)<=1/12......

vinay_sharma Says

suppose n is an integer such that the sum of the digits of n is 2, and 10^10< n <10^11. the number of different values of n is ?

my take -->11

suppose n is an integer such that the sum of the digits of n is 2, and 10^10< n <10^11. the number of different values of n is ?

vinay_sharma Says

in a certain examination paper, there are n questions. for j = 1,2,3.....n, there are 2^(n-j) students who answered j or more questions wrongly. if the total number of wrong answers is 4095, then the value of n is ?

2^(n-j) student answered j or more questions wrongly.
For j = 1, 2^(n-1) students answered 1 or more questions wrongly.
For j = 2, 2^(n-2) students answered 2 or more questions wrongly.
It means the number of students who answered exactly 1 question wrong is 2^(n-1)-2^(n-2).
Similarly, the number of students who answered exactly 2 questions wrong = 2^(n-2)-2^(n-3) and so on.
Now total wrong answered question is 4095 which we get due to the contribution of exactly 1 wrong, exactly 2 wrong ... and so on.
It means
1*(2^(n-1)-2^(n-2))+2*(2^(n-2)-2^(n-3))+....+(n-1)*(2^(n-(n-1))-2^(n-n))+n*2^(n-n) = 4095
2^(n-1)+2^(n-2)+....+2^1+1=4095
2^n-1=4095
2^n=4096
n=12.

QHow many ways can the integers from -10 to 10 be arranged in a sequence such that absolute value of the numbers in the sequence is non-decreasing?
OPTIONS
1)1023.....2)1024.......3)2048........4)2047.........5)512

vinay_sharma Says

in a certain examination paper, there are n questions. for j = 1,2,3.....n, there are 2^(n-j) students who answered j or more questions wrongly. if the total number of wrong answers is 4095, then the value of n is ?

2^n-1 students answered 1 or more questions wrongly

2^n-2 students answered 2 or more questions wrongly.

=> 2^n-1 - 2^n-2 students answered 1 question wrongly => no of wrong answers=2^n-2

similarly, 2^n-2 - 2^n-3 students answered 2 quetions wrongly => no of wrong answers =2* 2^n-3.;

similarly no. of students answered 3 quetions wrongly = 3*2^n-3...

.......so on

no of students who answered n questions wrongly = n

no of students who answered n-1 questions wrongly = n-1 (2^n-(n-1)) =(n-1)2...

=> 2^n-2 + 2.2^n-3 +..........2(n-1) + n =4095

=>2^n -1 =4095

=>2^n=4096

=>n=12

36 layer

s = 1+3+6+10+....
tn = n(n+1)/2

sigma tn = n(n+1)(2n+1)/12- (n)(n+1)/4 = 8436

on simplification

n(n+1)(2n+1)/6= 8436
or n(n+1)(2n+1) = 50616

now 40^3 = 64000> 50616 > 30^3 = 27000

so it has to b 36*37*38

vinay_sharma Says

if comeone do not know the formula of triangular number how is he going to solve this one ?

vivek bhai wale method se karna hai.....Series ka value nikal ke.....

Q If the positive real numbers x and y satisfy the equation x + y = 1, what is the maximum possible value of S = x^4 + y^4 (x^5 + y^5)?


OPTIONS..1)1/2..... 2)1/3 ........3)1/4 ......4)1/12...... 5)1/16
QThe minimum value of 17x^2 + 5y^2 + 12xy + 10x - 8y + 40 for real x, y is :
OPTIONS
1)40....2)27........3)31.........4)0...........5)16

vinay_sharma Says

let t be the set of integers{3, 11 , 19 ,27 , ....... 451 , 459 ,467} and s be a subset of t such that the sum of no two elements of s is 470. the maximum possible number of elements in s is ??

total number of terms = 59

if we take 467 we can not take 3
if we take 459 we can not take 11

and so on........

and 30 th term is 235

we can take terms from
3,8......,235
or
235,243.......,459,467

either way 30 terms.

vinay_sharma Says

let t be the set of integers{3, 11 , 19 ,27 , ....... 451 , 459 ,467} and s be a subset of t such that the sum of no two elements of s is 470. the maximum possible number of elements in s is ??

Is it 30???

There are 59 elements.......

(59+1)/2=30

1
3=1+2
6=1+2+3
10=1+2+3+4...all are triangular numbers

sum of triangular numbers is given by
n(n+1)(n+2)/6

n(n+1)(n+2)/6 =8436
=>we can check options here...bt since they are not given..

n=36.....

if comeone do not know the formula of triangular number how is he going to solve this one ?

vinay_sharma Says

let t be the set of integers{3, 11 , 19 ,27 , ....... 451 , 459 ,467} and s be a subset of t such that the sum of no two elements of s is 470. the maximum possible number of elements in s is ??

total numbers=(467-3)/8+1=58+1=59

3+467=470
11+459=470
go like this
so total element 59/2 approx29 will have 470 as sum

so sum of no two elements of s is 470=59-29=30