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

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

I+II+III=145............1

I+2II+3III=390...........2
____________

II+2III=245.................3

We have to minimize III, we have to take I=0 and correspondigly maximize II;

Also II <=150

So, we get, II+III=145

AND II+2III=245;

so, III=100 (minimum);

minimum students who play all III=100

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

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

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 ?

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 ??

29 hai kya????

vinay_sharma Says

there are 8436 steel balls, each with a radius of 1 cm ,stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third and 10 in the fourth, and so on. the number of horizontal layers in the pile ??????????

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.....

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 ??

vinay_sharma Says

there are 8436 steel balls, each with a radius of 1 cm ,stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third and 10 in the fourth, and so on. the number of horizontal layers in the pile ??????????

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

I am getting 14.......

25-7
10-1
5-1
2-5

total=14
N = abc is a three-digit number, a, b and c being their respective digits. N is a perfect square of an even number. To find N which of following informations is/are necessary/sufficient?

A. a, b, c are three consecutive digits but not in order.
B. N is divisible by 18.
C. Units digit of N^2 is c.
(1) Only A and B together are sufficient
(2) A and C together are sufficient
(3) Only B and C together are sufficient
(4) Either A and C together or B and C together are sufficient

Question fir se padiya bs bhai option se kar na padega
if a sum of 200 is to be paid using denominators of Rs2,Rs5,Rs10,Rs25 only,using at least one denomination of each kind,then which could be number of notes used???
11

bs0409 Says

629^ 24 divided by 21, find the remainder.

629^24 mod21=(-1)^24mod 21=1

If a sum of 200 is to be paid using denominators of Rs2,Rs5,Rs10,Rs25 only,using at least one denomination of each kind,then which could be number of notes used???
11
28
84
99

I am getting 14.......

25-7
10-1
5-1
2-5

total=14

let us put conditions and take cases
1.at least one denomination:
take 2,5,25,10 one coin each=>total =42 with 4 coins
we need 158 more we now find a range

for max range i.e max amount of coins
take only Rs 2 now....

158/2 = 79 ....79 number of Rs2 coins
total = 79+4 =83 coins

for min range i.e minimum amount of coins
take only Rs 25 now....

158/25 = 6 ....6 number of Rs25 coins and Rs8 left
for Rs 8. take 4number of Rs 2 coins now.

total = 6+4+4 =14 coins

max = 83
min = 14

14<=n<=83

only 28 satisfies

bs0409 Says

629^ 24 divided by 21, find the remainder.

is it 1 ????

there are 8436 steel balls, each with a radius of 1 cm ,stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third and 10 in the fourth, and so on. the number of horizontal layers in the pile ??????????

If a sum of 200 is to be paid using denominators of Rs2,Rs5,Rs10,Rs25 only,using at least one denomination of each kind,then which could be number of notes used???
11
28
84
99

28 is my take
with
rs25=1
rs5=1
rs2=5
rs10=16
we can range from 23 coins till 38 coins by replacing one rs10 by two rs5

In triangle XYZ,

bs0409 Says

snipeed...................

sumit bhai

consider for a case of 3 repetion

it will be 2(2(2k-x)-x)-x)= 2^3-7x

similarly for 4 terms
it will be (2(2(2(2k-x)-x)-x)-x)= 2^4-15x

so for ten it would be 2^10-1023x

sumit99 Says

Vivek bhai zara explain karna is concept ko, mai bhul gaya ise???????

initially 4096 hai
after a win it will b 25% more or 5/4 times 4096
so after 2nd win it will be 5/4*5/4*4096
similarly after third win

and consequently a loss will reduce the resulatant to 3/4th
so three losses will reduce it to (3/4)^3 of initial

vinay_sharma Says

what is the sum of all positive integers up to 990, which are divisible by 7 and not by 2?

7+21+35+49+.................987

sum = 71*994/2 = 35287

629^ 24 divided by 21, find the remainder.

Right....
but can u explain the bold part???

a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to
(1) 60
(2) 90
(3) 120
(4) None of these

1)
1st round........2p........x........2p-x
2nd round.......2(2p-x)....x.......4p-3x
3rd round.......8p-6x.......x......8p-7x

clear!!

If a sum of 200 is to be paid using denominators of Rs2,Rs5,Rs10,Rs25 only,using at least one denomination of each kind,then which could be number of notes used???
11
28
84
99

I am getting 14.......

25-7
10-1
5-1
2-5

total=14
N = abc is a three-digit number, a, b and c being their respective digits. N is a perfect square of an even number. To find N which of following informations is/are necessary/sufficient?

A. a, b, c are three consecutive digits but not in order.
B. N is divisible by 18.
C. Units digit of N^2 is c.
(1) Only A and B together are sufficient
(2) A and C together are sufficient
(3) Only B and C together are sufficient
(4) Either A and C together or B and C together are sufficient

upudit0 Says

vinay bhai isme kya galti hai

even integers bola hai bhai
how many even integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine.
even integer divisible by 7 = 112,126,140,154,168,182 and 196=7
even integer divisible by 9 = 108,126,144,162,180,198=6
126 is repeated so total (7+6)-1=12
so OA 51-12=39

viveksingh25 Says

4096 x (5/4 )^3 x (3/4)^3 = 3375

Vivek bhai zara explain karna is concept ko, mai bhul gaya ise???????

Right....
but can u explain the bold part???

a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to
(1) 60
(2) 90
(3) 120
(4) None of these

3x^2 + 4x + 5 = 0
it has complex roots.

and complex roots occur in pairs(conjugates).

if ax^2 + bx + c = 0 has one common root (which is complex) with 3x^2 + 4x + 5 = 0 the other root must be same as well.(its conjuate)

hence both equations are same
a=3 ,b= 4, c=5 pythaorean triplet.....right angled triangle

angle C is opposite to side c

C=90 degree