M4maths Previous Puzzles

13 Feb 2014 Solve

How many methods are possible to make 1 rupee by using 10 paisa, 20 paisa and 50 paisa coin. It is not necessary to use all type of coins and any coin can be used more than 1 times.

OPtion

1) 12
2) 9
3) 8
4) 7
5) 6
6) 11
7) 10
8) 13
9) 14
10)None of these

Solution

0.10*10
0.10*8+0.20*1
0.10*6+0.20*2
0.10*4+0.20*3
0.10*2+0.20*4
0.20*5
0.50*1+0.10*5
0.50*1+0.20*1+0.10*3
0.50*1+0.20*2+0.10*1
0.50*2

Option 7) 10

Correct Option: 7 Best Solution (25)

S. Singh   10 years AGO

Possible combinations for 50p,20p,10p are
0,0,10
0,1,8
0,2,6
0,3,4
0,4,2
0,5,0
1,0,5
1,1,3
1,2,1
2,0,0

Like? Yes (13) | No (2)  

Dr.Dipin Singh   10 years AGO

when there is no 50 p coin,
possible number of 20 p coins are 0,1,2,3,4,5
and 10p coins are used accordingly.
when there is one 50 p coin,
possible number of 20 p coins are 0,1,2
and 10p coins are used accordingly.
when there are two 50 p coin,
possible number of 20 p coins is 0
and number of 10p coin is also 0.
So total possible combinations are 6+3+1=10

Like? Yes (5) | No (1)  

madanpal vishwakarma    10 years AGO

the all possible ways to make 1 rupees using 10,20 and 50 paisa coin are 10 (listed below)
the combination of single coins
1>10*10=100
2>20*5=100
3>50*2=100
the combination of 10 and 20 paisa cions
4>20*4 +10*2=100
5>20*3+10*4=100
6>20*2+10*6=100
7>20*1+10*8=100
the combination of 50 and 10 paisa coin
8>50*1+10*5=100
the combination of all three coin
9>50*1+20*1+10*3=100
10>50*1+20*2+10*1=100

so option 7 is correct

Like? Yes (3) | No (1)  

MP    10 years AGO

solving for integer values of x,y,z in
5x+2y+z=10
we get 10 sets from
(0,0,10)
(0,1,8)
..
..
(1,2,1)
(2,0,0)

Like? Yes (4) | No (2)  

neeraj   10 years AGO

50,50
50,10,10,10,10,10
50,10,10,10,20
50,10,20,20
20,20,20,20,20
20,20,20,20,10,10
20,20,20,10,10,10
20,20,10,10,10,10,10,10
20,10,10,10,10,10,10,10,10
10,10,10,10,10,10,10,10,10,10

Like? Yes (2) | No   

NAGENDRA P   10 years AGO

1.10paise---10coins..
2.20paise---5coins..
3.50paise---2coins..
4.50paise 1coin,20paise 2coins,10paise 1 coin..
5.50paise 1coin,20paise 1coin,10paise 3 coins..
6.50paise 1coins,10paise 5coins..
7.20paise 2coins,10paise 6coins..
8.20paise 3coins,10paise 4coins..
9.20paise 4coins,10paise 2coins..
10.20paise 1coins,10paise 8coins..
so there are 10methods to make 1rupee..so option(7)..

Like? Yes (2) | No   

Sutanu Shit   10 years AGO

1.(10,0,0)
2.(8,1,0)
3.(6,2,0)
4.(4,3,0)
5.(2,4,0)
6.(0,5,0)
7.(5,0,1)
8.(3,1,1)
9.(1,2,1)
10.(0,0,2)

Like? Yes (1) | No   

Suman   10 years AGO

methods
10*10
20*5
50*2
1*20+10*8
2*20+10*6
3*20+10*4
4*20+10*2
2*20+10+50
1*20+3*10+50
50+10*5

Like? Yes (2) | No (1)  

rajani   10 years AGO

there are totally 10 combinations for this solution.
10p 20p 50p
--- ---- ---
10 0 0
0 5 0
0 0 2
1 2 1
5 0 1
3 1 1
8 1 0
6 2 0
4 3 0
2 4 0

Like? Yes (1) | No   

Shubham Gupta   10 years AGO

By 10ps coin we can make-->
10*10p
10p+20p*2+50p
2*10p+20p*4
3*10p+20p+50p
4*10p+20p*3
5*10p+50p
6*10p+20p*2
8*10p+20p
2*50p
5*20p

Like? Yes (1) | No   

Shahnawaz Ahmad   10 years AGO

50,50
50,20,20,10,
50,20,10,10,10
50,10,10,10,10,10
20*5
20*4,10*2
20*3,10*4
20*2,10*6
20*1,10*8
10*8

Like? Yes (1) | No   

KRISHNAN   10 years AGO

1) 0.10*10
2) 0.10*8+0.20*1
3) 0.10*6+0.20*2
4) 0.10*4+0.20*3
5) 0.10*2+0.20*4
6) 0.20*5
7) 0.50*1+0.10*5
8) 0.50*1+0.20*1+0.10*3
9) 0.50*1+0.20*2+0.10*1
10)0.50*2
Total 10 methods so option 7 is correct

Like? Yes (1) | No (1)  

Paras   10 years AGO

just make the trail and get the answer

Like? Yes (1) | No (1)  

T G   10 years AGO

1)50 50
2)50 20 20 10
3)50 20 10 10 10
4)50 10 10 10 10 10
5)20 20 10 20 20 10
6)20 20 10 20 10 10 10
7)20 20 10 10 10 10 10 10
8)20 10 10 10 10 10 10 10 10

9)10 *10
10)20*5

Like? Yes (3) | No (3)  

Leelanand Kumar Yalla   10 years AGO

10 Methods are possible...

Like? Yes (1) | No (1)  

j.b.manasa   10 years AGO

the possible ways are
(50,50),(20*5),(10*10),(50,20,20,10),(50,20,10,10,10),(50,10,10,10,10,10),(20,20,20,20,10,10),(20,20,20,10,10,10,10),(20,10*8),(20,20,10*6)

Like? Yes (1) | No (1)  

Dinesh Tripathi   10 years AGO

1)10*10=100
2)10*8+20*1=100
3)10*6+20*2=100
4)10*4+20*3=100
5)10*2+20*4=100
6)20*5=100
7)10*5+50*1=100
8)50*1+10*3+20*1=100
9)50*1+10*1+20*2=100
10)50*2=100

Like? Yes (1) | No (1)  

Srinivasa Vamsi Krishna Dodda   10 years AGO

if 0 50 paise coins are used : then in 6 ways we use 20 or 10 paise combinations to have a rupee

If we use 1 50 paise : then in 3 ways we make use of 20 or 10 paise coins Combinations to have a rupee

and finally we can make 1 rupee with 2 50 paise coins

So... Total ways = 6+3+1 = 10

Like? Yes (1) | No (1)  

K PAIDAM NAIDU   10 years AGO

two 50 coins,five 20,ten 10,one 50 two 20 one 10,one 50 one 20 three 10,one 50 five 10,one 20 eight 10,two 20 six 10,three 20 four 10,four 20 two 10.

Like? Yes (1) | No (1)  

Younus   10 years AGO

10x10
10x8 20x1
10x6 20x2
10x4 20x3
10x2 20x4
20x5
50x2
50x1 10x5
50x1 10x3 20x1
50x1 10x1 20x2

Like? Yes (2) | No (2)  

ATUL KUMAR SINGH   10 years AGO

use of 50paise coin
.50*2=1;
.50*1+.10*5=1;
.50*1+.20*1+.10*3=1;
.50*1+.20*2+.10*1=1;
use of 20paise coin(excluding 50 paise coin)
.20*5=1;
.20*4+.10*2=1;
.20*3+.10*4=1;
.20*2+.10*6=1;
.20*1+.10*8=1;
use of 10paise coin(excluding 50&20 paise coin)
.10*10=1;
//total 10 ways//

Like? Yes (1) | No (1)  

aviral bhargava   10 years AGO

the possible combinations are :-
1)10 coins of 10 paise
2)5 coins of 20 paise
3)2 coins of 50 paise
4)1*10,2*20,1*50
5)3*10,1*20,1*50
6)5*10,1*50
7)1*20,8*10
8)2*20,6*10
9)3*20,4*10
10)4*20,2*10

which sum up to total 10 ways.

Like? Yes (1) | No (2)  

Varunraj Reddy   10 years AGO

These are the Possibilities

10P 0 0 1 2 3 4 5 6 8 10

20P 0 5 2 4 1 3 0 2 1 0

50P 2 0 1 0 1 0 1 0 0 0

So there are total 10 possible ways

So ans is option 7)

Like? Yes (1) | No (2)  

Hanumanth Naik   10 years AGO

10 20 50 total
-- -- --
10 0 0 100
8 1 0 100
6 2 0 100
5 0 1 100
4 3 0 100
3 1 1 100
2 4 0 100
1 2 1 100
0 5 0 100
0 0 2 100

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Deb Das   10 years AGO

With 10 possible ways

Like? Yes (1) | No (2)  

12 Feb 2014 Solve

Total cost of 2 rubbers and 3 pencils is greater than 20 rupee whereas total cost of 3 rubbers and 2 pencils is less than 24 rupee. If cost of pencil is less then 7 rupee then for what prize of rubber and pencil total cost of 4 rubbers and 5 pencils is minimum

OPtion

1) 1, 3
2) 2, 6
3) 3, 6
4) 4, 5
5) 3, 5
6) 1, 6
7) 2, 5
8) 5, 4
9) 3, 4
10)None of these

Solution

2x+3y > 20
3x+2y < 24
y < 7
We have to find a condition such that 4x+5y is minimum

Correct Option: 5 Best Solution (16)

madanpal vishwakarma    10 years AGO

ACCORDING TO THE QUESTION
2X+3Y >20
3X+2Y20 : 2*1+3*3=11 (condition false)

option 2)2,6
2X+3Y >20 : 2*2+3*6=22 (condition true)
3X+2Y20 : 2*3+3*6=24 (condition true)
3X+2Y20 : 2*4+3*5=23 (condition true)
3X+2Y20 : 2*3+3*5=21 (condition true)
3X+2Y20 : 2*1+3*6=20 (condition false)


option 7)2,5
2X+3Y >20 : 2*2+3*5=19 (condition false)

option 8)5,4
2X+3Y >20 : 2*5+3*4=22 (condition true)
3X+2Y20 : 2*3+3*4=18 (condition false)

so the option 5 gives the lowest cast price value

Like? Yes (8) | No (1)  

S.SHAFI   10 years AGO

ans) option 5
rubber cost = r
pencil cost = p

2r + 3p > 20
3r + 2r < 24 and p < 7

Total cost of 4r + 5p is minimum if r = 3, p = 5

Like? Yes (3) | No   

Mahalakshmi   10 years AGO

3,5 satisfy all conditions

Like? Yes (1) | No   

KRISHNAN   10 years AGO

Let cost of rubber be x and pencils be y
then
2x+3y > 20
3x+2y < 24
y < 7
4x+5y is minimum at x = 3 and y = 5 so option 5 is correct

Like? Yes (2) | No (1)  

Kaustubh karekar   10 years AGO

4*3+5*5=37 is minimum
so it is 3,5

Like? Yes (1) | No   

viveka   10 years AGO

2r+3p>20
3r+2p2.4 r

Like? Yes (1) | No   

aviral bhargava   10 years AGO

choosing by options

Like? Yes (2) | No (2)  

Varunraj Reddy   10 years AGO

Given Conditions are
2r+3p>20-----i)
3r+2p

Like? Yes (1) | No (1)  

PINTU KUMAR   10 years AGO

it gives min value

Like? Yes (1) | No (1)  

sravani   10 years AGO

solve for the common values then we get p is < r
p:pencil
r:rubber
then for r=3
p=5 we get min values satisfying the above conditions.

Like? Yes (1) | No (1)  

Suman   10 years AGO

ans will be (3,5)
Using hit and trial method, option five satisfy given conditions and price of 4 rubber and 5 pencil will be minimum =3*4+5*5=12+25=37rupees

Like? Yes (1) | No (1)  

T G   10 years AGO

ans is 5
x=price of rubber
y=price of pencil
2x+3y>20...(1)
3x+2y

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Sushma Reddy   10 years AGO

By options we get minimum amount at 3,5 as
4*3+5*5= 37

Like? Yes (1) | No (1)  

Praneeth   10 years AGO

By options we get minimum value 37 for 3,5

Like? Yes (1) | No (1)  

R.RAVIKUMAR   10 years AGO

By substitution and comparing with original equations, we will get minimum cost for 3,5

Like? Yes (1) | No (1)  

Srinath   10 years AGO

Check all options for given data:
2r+3p>20
3r+2p

Like? Yes (1) | No (2)  

11 Feb 2014 Quadratic equation

For what value of k, equation: 2x^2+kx-5 = 0 & x^2 - 3x - 4 = 0 may have one root common.

OPtion

1) -4, 4/39
2) -2/9, 3/4
3) -3, -27/4
4) -3, -3/4
5) -3, -2/5
6) -3, 27/4
7) 3, -27/4
8) 3, -2/5
9) 3, 2/5
10)None of these

Solution

Let p be the common root.
2p^2+kp-5 = 0 (i)
p^2-3p-4 = 0 (ii)
solve
k=-3, k = -27/4
option 3)

Correct Option: 3 Best Solution (28)

Saidinesh   10 years AGO

2x^2+kx-5=0.........(1)
x^2-3x-4=0..........(2)
solving equation 2
x^2-3x+x-4=0;
(x+1)(x-4)=0;
x=-1;x=4;
put x value in first equation
x=-1;
2-k-5=0;
k=-3;
x=4;
32+4k-5=0
k=-27/4

Like? Yes (16) | No (5)  

Dr.Dipin Singh   10 years AGO

x^2 - 3x - 4 = 0
(x-4)(x+1)=0 means two roots are 4, and -1.
when common root is -1,
Value of k in 2x^2+kx-5 = 0 is -3.
when common root is 4,
value of k in 2x^2+kx-5 = 0 is -27/4.

Like? Yes (13) | No (7)  

Mahalakshmi   10 years AGO

By solving equation x^2-3x-4=0 will get x=4,-1
by substituting x value in equation 2x^2+kx-5 =0 will get k= -3,-27/4

Like? Yes (5) | No (1)  

abhilasha   10 years AGO

Answer -3, -27/4
on keeping the values -3, -27/4 we get the roots -1,4 respectively.And from the second equation we got the roots 4,-1 so the roots are matching and thus this is the answer.

Like? Yes (2) | No   

S.SHAFI   10 years AGO

The roots of the equation x^2-3-4=0 are 4,-1...
It can be calculated by using

x = {=-b+-sqrt(b^2-4*a*c)}/2a

Substitute k=-3 and -27/4 in the equation 2x^2+kx-5=0 to have one root in common.

When k=-3, the roots are 5/2, -1

When k=-27/4, the roots are -5/8, 4

So, ans is option 3...

Like? Yes (4) | No (3)  

T G   10 years AGO

ans is 3
the root of the 2nd equation is 4 and -1.
now put the value of x in the 1 st equation respectively....get the result
-3 and -27/4

Like? Yes (2) | No (2)  

R.RAVIKUMAR   10 years AGO

for second eqn we get roots as 4 and -1.
by substituting -3,-27/4 in first eqn, we get roots as -1 and 4 respectively...

Like? Yes (2) | No (2)  

K Venkata Subba Rao   10 years AGO

2x^2+kx-5 =0
-2*(x^2-3x-4 )=0
(+)---------------
x= -3/(k+3)
(sub x in eq)
u'll get k=-3 or -27/4

Like? Yes (1) | No (1)  

SID PRASANTH KUMAR   10 years AGO

%alpha ={ca1 -ac1 } over {ab1 - ba1 }={bc1 - cb1 } over {ca1 - ac1 }

Like? Yes (1) | No (1)  

Varunraj Reddy   10 years AGO

From the equation 2 we get the roots are 4,-1

For the value of k=-3,-27/4 we get the same roots
from the equation 1

So option 3) is answer.

Like? Yes (1) | No (1)  

IFFAT IRAM   10 years AGO

let 'a' be the common root
2a^2+ka-5=0........(1)
a^2-3a-4=0.........(2)
by cross multiplication,


a^2/(-4k-15)=a/(8-5)=1/(-6-k)
now,
a^2=(4k+15)/(6+k)
a=-3/(6+k)

again,

9/(6+k)^2=(4k+15)/(6+k)
(4k+15)(6+k)=9
4k^2+39k+81=0
(4k+27)(k+3)=0
k=-27/4 or -3

Like? Yes (2) | No (2)  

j.b.manasa   10 years AGO

x^2-3x-4=0
=(x-4)*(x+1)=0
=x=-1,4.
Common roots means when these x values obtained are substituted in the equation1 then it must be 0
2(-1)^2+k(-1)-5=0 and 2(4)^2+K(4)-5=0
=2-k-5=0 and 32+4k-5=0
k=-3,-27/4

Like? Yes (1) | No (1)  

kothakota sai priya   10 years AGO

x^2-3x-4=0 solve this equation we get x values as
x=4 and x=-1
substitute these x values in other equation i.e
for x=4

2(4)^2+k(4)-5=0
32+4k-5=0
k=-27/4

for x=-1
2(-1)^2+k(-1)-5=0
2-k-5=0
k=-3

Like? Yes (1) | No (1)  

MP    10 years AGO

two roots of x^2 - 3x - 4 = 0 are 4 and -1.
consider one of these ( one by one) as common root with 2x^2+kx-5=0,
we get k = -3,-27/4

Like? Yes (3) | No (3)  

Sourav Das   10 years AGO

2nd eqn has two roots x=4,-1.
put these value in the 1st eqn and solve for the value of k.
k=-27/4 for x=4
k=-3 for x=-1.

Like? Yes (2) | No (2)  

Nitesh Jadhav   10 years AGO

solve using [-b (+/-)root(b^2-4ac)]/2a formula

Like? Yes (1) | No (1)  

sai sarma   10 years AGO

x^2-3x-4 can be divided as
x^2-4x+x-4
=>x(x-4)+1(x-4)
=>x=4,x=-1

we have a condition may be one root matches
take x=4
then sub in eqn1 then k=-27/4
take x=-1 and k=-3

so option (3) is correct

Like? Yes (2) | No (2)  

hema   10 years AGO

roots of second equations are 4 and -1.
substitute in 1st equ

Like? Yes (1) | No (1)  

susmit   10 years AGO

By solving 2nd equation we get x = -1 or x = 4
Substituting both the values in the first equztion, we get k = -3 or k = -27/4

Like? Yes (2) | No (2)  

Udhaya kumar   10 years AGO

For solving 2nd eqn we get
x=4,-1
sub x in first eqn we get
k = -3,-6.75(-27/4).......

Like? Yes (1) | No (1)  

Sandeep Kumar   10 years AGO

x^2-3x-4 =0
it gives.....
x= 4,-1

putting these two values in eq 1 we get,
k= -27/4 , -3

Like? Yes (1) | No (1)  

Vikram Saini   10 years AGO

x^2-3x-4=0
solve it nd get x=4,-1
nd put the valu in othr eqn
nd get k=-27/4,-3
so option 3 is corrct answr.
thnxxxx

Like? Yes (1) | No (2)  

Jigyasa Kumar   10 years AGO

Answer -3, -27/4
on keeping the values -3, -27/4 we get the roots -1,4 respectively.And from the second equation we got the roots 4,-1 so the roots are matching and thus this is the answer.

Like? Yes (1) | No (2)  

Aarush Singh   10 years AGO

Answer -3, -27/4
on keeping the values -3, -27/4 we get the roots -1,4 respectively.And from the second equation we got the roots 4,-1 so the roots are matching and thus this is the answer.

Like? Yes (1) | No (2)  

ASHISH MITTAL   10 years AGO

x^2 - 3x - 4 = 0
root of this equation after solving
x=-1,4
And after putting value of x in 2x^2+kx-5 = 0
we get x=-3, -27/4

Like? Yes (1) | No (2)  

Krishna Chaitanya   10 years AGO

Given equation: 2x^2+kx-5 = 0 & x^2 - 3x - 4 = 0 may have one root common.
On solving x^2 - 3x - 4 = 0 we get roots -1,4
By substituting x values in 2x^2+kx-5 = 0 we get k values which are -3,-27/4

Like? Yes (1) | No (2)  

Devendra Marghade   10 years AGO

2x^2+kx-5 = 0 ---(i) &
x^2-3x-4 =0 ---(ii)
Solving (ii),x^2+x-4x-4=0, (x+1)(x-4)=0,x=-1,4
For value of k,
Substituting x=-1 in eqn.(i),2(-1)^2+k(-1)-5=0
k=-3 &
with x=4 in eqn.(i),2(4)^2+k(4)-5=0,
k=-27/4

Like? Yes (1) | No (2)  

Srinath   10 years AGO

solve x^2 - 3x - 4 = 0, X=-1,4
pui it on 2x^2+kx-5 = 0 ,
we get k=-3,-27/4 respectively...

Like? Yes (1) | No (2)  

10 Feb 2014 Sequence

What is the next number of the following sequence
1,6,31,156,781,.....

OPtion

1) 1275
2) 2250
3) 3225
4) 3906
5) 3750
6) 975
7) 1575
8) 3450
9) 2525
10)None of these

Solution

1 = 1
1+5 = 6
1+5+5^2 = 1+5+25 = 31
1+5+5^2+5^3 = 1+5+25+125 = 156
1+5+5^2+5^3+5^4 = 1+5+25+125+625 = 781
Similarly
1+5+5^2+5^3+5^4+5^5 = 1+5+25+125+625+3125 = 3906
Option 4)

Correct Option: 4 Best Solution (118)

Akanksha Mishra   10 years AGO

Given Series:
1,6,31,156,781,.....

1 * 5 + 1 = 6
6 * 5 + 1 = 156
156 * 5 + 1 = 781

similarly

781 * 5 + 1 = 3906
so answer will be option 4

Like? Yes (26) | No (10)  

Varunraj Reddy   10 years AGO

Given sequence 1,6,31,156,781,...
The differences between numbers are 5,25,125,625,...
i.e., 5^1,5^2,5^3,5^4,...
So next number with difference of 5^5=3125
So 781+3125=3906
Ans is option 4)

Like? Yes (4) | No (1)  

Vikram Saini   10 years AGO

1,1*5+1,6*5+1,31*5+1,156*5+1,781*5+1=3906

Like? Yes (3) | No   

shivam tiwari   10 years AGO

FROM THE SERIES.....1,6,31,156,781 (1X5)+1=6 (6X5)+1=31 (31X5)+1=156 (156X5)+1=781 (781X5)+1=3906

Like? Yes (3) | No (1)  

NAGENDRA P   10 years AGO

(5^n-1)/4
put n=1,4/4=1
put n=2,(25-1)/4=6
put n=3,(125-1)/4=31
put n=4,(625-1)/4=156
put n=5,(3125-1)/4=781
put n=6,(15625-1)/4=3906..So option (4)

Like? Yes (2) | No   

saranya sankar   10 years AGO

5^0=1
5^1+1=6
5^2+6=31
5^3+31=156
5^4+156=781

5^5+781=3906

Like? Yes (2) | No   

VENKATA SIVA AVINASH A   10 years AGO

ANS: 3906

6=5^1+1
31=5^2+6
156=5^3+31
781=5^4+156
so next term is 5^5+781=3906

Like? Yes (2) | No   

OM PRAKASH JHA   10 years AGO

FROM THE SERIES.....1,6,31,156,781
(1X5)+1=6
(6X5)+1=31
(31X5)+1=156
(156X5)+1=781
(781X5)+1=3906

Like? Yes (2) | No (1)  

OM PRAKASH JHA   10 years AGO

WE CAN SEE FROM THE GIVEN SERIES THAT.....1,6,31,156,781
(1X5)+1=6
(6X5)+1=31
(31X5)+1=156
(156X5)+1=781
(781X5)+1=3906

Like? Yes (3) | No (2)  

abhilasha   10 years AGO

fibonnacci series is followed...so the next number will be 3906

Like? Yes (1) | No   

Chirag Ranjan Lahiri   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
so ans will be - 781*5+1=3906

Like? Yes (1) | No   

Mallikharjuna   10 years AGO

follwoing series

1,
(1*5)+1 = 6
(6*5)+1 =31
(31*5)+1 =156
(156*5)+1 =781
(781*5 )+1 =3906.

Like? Yes (1) | No   

Vijay Chauhan   10 years AGO

1 * 5 = 5 +1 = 6
6 * 5 = 30 +1 = 31
31 * 5 = 155 +1 = 156
156* 5 = 780 +1 = 781
781* 5 = 3905+1 = 3906
so, ans is 4) 3906

Like? Yes (3) | No (2)  

sumasindhu   10 years AGO

ans is previous number multiplied with 5 and add 1
like: 1*5=5+1==6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906
so ans is 3906 option 4

Like? Yes (1) | No   

Ranjeet k gupta   10 years AGO

next number is multiple of 5 and previous number and after multiplication addition of 1.

Like? Yes (1) | No   

nalini singh   10 years AGO

1,6/2
=3
second digit/2 first digit=31

Like? Yes (1) | No   

saketh reddy   10 years AGO

1,6...der is gap of 5 ....and 6*5=30+1=31
31*5=155+1=156
similarly 781*5=3905+1=3906

Like? Yes (1) | No   

saragadam gopi krishna   10 years AGO

1-6-31-156-781
1,1*5+1,6*5+1,31*5+1,156*5+1,781*5+1=3906

Like? Yes (1) | No   

Sana   10 years AGO

1*5=5 +1=6
6*5=30 +1=31
31*5=155 +1=156
156*5=780 +1=781
781*5=3905 +1=3906

Like? Yes (1) | No   

Manikumar R   10 years AGO

1
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No   

viveka   10 years AGO

1,6,31,156,781,.....
5. 25 125 625 3125+781=3906

Like? Yes (1) | No   

jyoti   10 years AGO

1,6=(1*5)+1,31=(6*5)+1,156=(31*5)+1,781=(156*5)+1,so 3096=(781*5)+1

Like? Yes (1) | No   

nitesh   10 years AGO

n=5;
y(0)=1;
y(n)={y(n-1)*5}+1;
hence y(5)={y(4)*5}+1=3906;

Like? Yes (1) | No   

sushil gupta   10 years AGO

difference b/w the number is 5 25 125 625 and next is 3125 so the next number is 3125+781 = 3906 then ans is 4

Like? Yes (1) | No   

diwakar kumar singh   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No   

kumar gaurav   10 years AGO

1 mulitiply with 5 then add 1 more. same format

Like? Yes (1) | No   

kiran kumar reddy yanamala   10 years AGO

1*5+1,6*5+1,31*5+1,....

Like? Yes (1) | No   

Sukanya   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No   

C. Veda Kumar   10 years AGO

the diff is 5,25, 125, 625, so next one is 3125...add 3125+781=3906

Like? Yes (1) | No   

Dipak Shaw   10 years AGO

1
1×5+1=6
6×5+1=31
.......
781×5+1=3906

Like? Yes (1) | No   

D.HARINATH REDDY   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No   

Rajasekhar Reddy   10 years AGO

781*5 + 1 = 3096

Like? Yes (1) | No   

Mazeed   10 years AGO

here the reason is the difference between two numbers is multiple with 6 , then we add the previous known number with our product ..
6-1=5, 5*6=30 then 30+1=31
31-6=25 , 25*6=150 then 150+6=156
156-31=125 , 125*6=750 then 750+31=781
again the same logic can obtain the next number ..
781-156=625 , 625*6=3750 then 3750+156=3906

Like? Yes (1) | No   

S.SHAFI   10 years AGO

Given series is in the form of (5^n-1)/4
1,6,31,156,781,..
for n=1, (5^1-1)/4 = 4/4 = 1
for n=2, (5^2-1)/4 = 24/4 = 6
for n=3, (5^3-1)/4 = 124/4 = 31
for n=4, (5^4-1)/4 = 624/4 = 156
....
for n=6, (5^6-1)/4 = 15624/4 = 3906

Like? Yes (2) | No (1)  

ROHIT RAJ   10 years AGO

1, 1*5+1=6, 6*5+1=31, 31*5+1=156, 156*5+1=781, 781*5+1=3904

Like? Yes (1) | No   

S Prakash   10 years AGO

Solution:
1,6,31,156,781,3906
Reason
1=1
1+5=6
6+(5^2)=31
31+(5^3)=156
156+(5^4)=781
781+(5^5)=3906

Like? Yes (1) | No   

RAKESH   10 years AGO

781+5*625=3906

Like? Yes (1) | No   

Anirudh Maheswari   10 years AGO

(0*5)+1=1; 1*5+1=6; 6*5+1=31 so, 781*5+1=3906

Like? Yes (1) | No   

VISHNUVARDHANA   10 years AGO

1*5+1 = 6
6*5+1 = 31
31*5+1 = 156
156*5+1 = 781
781*5+1 = 3906

Like? Yes (1) | No   

rahul   10 years AGO

series is
(1*5)+1 = 6
(6*5)+1 = 31
(31*5)+1 = 156
..
(781*5)+1 = 3906

Like? Yes (1) | No   

Harshit Daga   10 years AGO

0*5+1, 1*5+1, 6*5+1, 31*5+1, 781*5+1...
so the next number will be..
781*5+1 = 3906
that is option 4

Like? Yes (1) | No   

Praveen Tenneti   10 years AGO

1,1*5+1=6,6*5+1=31,31*5+1=156,156*5+1=781,781*5+1=3906

Like? Yes (1) | No   

hemlata ninama   10 years AGO

1*5+1=6 ,6*5+1=31 ,31*5+1=156 ,156*5+1=781 ,781*5+1=3906

Like? Yes (1) | No   

Gaurav Garg   10 years AGO

t(n)=t(n-1)+5^(n-1)

Like? Yes (1) | No   

IFFAT IRAM   10 years AGO

781*5+1=3905+1
=3906

Like? Yes (1) | No   

Suman   10 years AGO

1*5+1=6
5*6+1=31....so on 781*5+1=3906

Like? Yes (1) | No   

yuva   10 years AGO

781*5+1

Like? Yes (1) | No   

Kaustubh karekar   10 years AGO

781*5+1=3906

Like? Yes (1) | No   

sravani   10 years AGO

1,6,31,156,781,..
5,25,125,625,3125

3125+781=3906

Like? Yes (1) | No   

arun prakash singh   10 years AGO

multiplication of 5 with add 1

Like? Yes (1) | No   

SONU GUPTA   10 years AGO

1; 1+5=6; 6+5^2=31; 31+5^3=156; 156+5^4=781; 781+5^5=3906

Like? Yes (1) | No   

hemavathi   10 years AGO

1*5=5, 5+1=6
6*5=30, 30+1=31
31*5=155, 155+1=156
156*5=780, 780+1=781
781*5=3905, 3905+1=3906

Like? Yes (1) | No   

Gourab Dutta   10 years AGO

6=1*5+1;;; 31=6*5+1;; 156=31*5+1;; 781=156*5+1; 3906=781*5+1

Like? Yes (1) | No   

Gudivada sunitha   10 years AGO

1*5= 5.5+1=6
6*5=30,30+1=31
31*5=155,155+1=156
156*5=780,780+1=781
781*5=3905,3905+1=3906

Like? Yes (1) | No   

arunkumar   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No   

K Venkata Subba Rao   10 years AGO

1, 6, 31, 156, 781
6-1 =5 (5^1)
31-6 =25 (5^2)
156-31 =125 (5^3)
781-156 =625 (5^4)
? -781 =3125 ==> ans is 3906

Like? Yes (2) | No (1)  

saratht   10 years AGO

6-1=5
31-6=25
156-31=125
781-156=625
hence the number must be 781+(625*5)=3906

Like? Yes (1) | No   

YUVAN SATHAIAH   10 years AGO

1=1
5^1+1=5+1=6
5^2+6=25+6=31
5^3+31=125+31=156
5^4+156=625+156=781
then
5^5+781=3125+781=3906

Like? Yes (1) | No   

neeraj   10 years AGO

6=1*5+1
31=6*5+1
156=31*5+1
781=156*5+1
3906=781*5+1

Like? Yes (1) | No   

kundan kumar singh   10 years AGO

follow the sequence of (number*5+1)

Like? Yes (1) | No   

Aman Seth   10 years AGO

it is series and every number is multiply by 5 and after that add +1 in to this series

Like? Yes (1) | No   

Sourav Das   10 years AGO

(1*5)+1=6
(6*5)+1=31
(31*5)+1=156
(156*5)+1=781
(781*5)+1=3906
so option (4) is correct

Like? Yes (1) | No   

sangeetha   10 years AGO

(1*5)+1=6
(6*5)+1=31
.
.
.
(781*5)+1=3906

Like? Yes (1) | No   

roushan   10 years AGO

series of 5 ^1 , 5^2 , 5^3 ....5^5.

Like? Yes (1) | No   

Himanshu Bisht   10 years AGO

difference between numbers are multiple of 5.
ex
6 - 1 = 5;
31 - 6= 25;
156-31 = 125;
781-156 = 625;
3906-781 = 3125

Like? Yes (1) | No   

j.b.manasa   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No   

SADINENI RAGHAVENDRA BABU   10 years AGO

1 *5=5 & 5 +1 =6
6 *5=30 & 30 +1 =31
31 *5=155 & 155 +1 =156
156 *5=780 & 780 +1 =781
781 *5=3905 & 3905+1 =3906 is answer

Like? Yes (1) | No   

rahul kumar   10 years AGO

each consecutive no. is the progressive difference of 5, 5^2 , 5^3.....upto 5^5

Like? Yes (1) | No   

Gokul Krishnan   10 years AGO

1,6,31,156,781...
difference is (5*1)+1,(5*5)+6,(25*5)+31,(125*5)+156,
(625*5)+781 .
So,the answer is 3906.

Like? Yes (1) | No   

julee verma   10 years AGO

1+5=6
6+25=31
31+125=156
156+625=781
781+3125=3906

Like? Yes (1) | No   

Anantharaman   10 years AGO

Sequence follows n*5+1 formula
1
6 = 1 * 5 + 1 = 6
31 = 6 * 5 + 1 = 31
156 = 31 * 5 + 1 = 156
781 = 156* 5 + 1 = 781

Similarly 781 * 5 + 1 = 3906
Option 4

Like? Yes (1) | No   

hena samanta   10 years AGO

6-1=5
31-6=25
156-31=125
781-156=625
2nd element=1*5+1=6
3rd element=5*5+6=31
4th element=25*5+31=156
5th element=125*5+156=781
therefor 6th element=625*5+781=3906

Like? Yes (1) | No   

ananthavalli   10 years AGO

6-1=5
31-6=25
156-31=125
781-156=625
The difference between each number is multiply by 5 that is 5*5=25;25*5=125;125*5=625;
Then 625*5=3125
3125+781=3906
option 4)

Like? Yes (1) | No   

Udhaya kumar   10 years AGO

Difference bw numbers are 5,25,125,625,3125 respectively...

Like? Yes (1) | No   

a.k.gayathridevi   10 years AGO

1+(1*5)=6
1+(6*5)=31
1+(31*5)=156
1+(156*5)=781
so
1+(781*5)=3906

Like? Yes (1) | No   

Tarun Sai   10 years AGO

Differences between each number are increasing powers of 5

Like? Yes (1) | No   

Sai Kiran B   10 years AGO

1*5=5, 5+1=6
6*5=30,30+1=31
31*5=155, 155+1=156
156*5=780,780+1=781
781*5=3905,3905+1=3906

Like? Yes (1) | No   

kundan kumar   10 years AGO

the difference is multiplied by 5

Like? Yes (1) | No   

mohini gupta   10 years AGO

next no.= previous no. *5+1

Like? Yes (1) | No   

Srinivasa Vamsi Krishna Dodda   10 years AGO

6-1 = 5
31-6 = 25 (5^2)
156-31 =125 (5^3)
781-156 = 625(5^4)
3906-781 = 3125 (5^5)

So ans : 3906

Like? Yes (1) | No   

Krishna Chaitanya   10 years AGO

Given 1,6,31,156,781 this can be interpreted as
(1*5)+1=6
(6*5)+1=31
(31*5)+1=156
(156*5)+1=781
(781*5)+1=3906

Like? Yes (1) | No   

MD JAVED SALIM   10 years AGO

multiply by 5 and add 1 in each step.

Like? Yes (1) | No   

satya   10 years AGO

1,1*5+1,6*5+1,31*5+1,156*5+1,781*5+1(3905+1)

Like? Yes (1) | No   

Leelanand Kumar Yalla   10 years AGO

1*5=5+1=6;
6*5=30+1=31;
31*5=155+1=156;
156*5=780+1=781;
781*5=3905+1=3906; this is the answer.....

Like? Yes (1) | No   

Greeshma Varghese   10 years AGO

1+(5^1)=6
6+(5^2)=31
156+(5^4)=781
781+(5^5)=3906

Like? Yes (1) | No   

pritipriya   10 years AGO

increasing numbers in sequence by
5,25,125,625,3125

Like? Yes (1) | No   

SID PRASANTH KUMAR   10 years AGO

(1*5)+1,(6*5)+1,(31*5)+1....

Like? Yes (1) | No   

Keshav Kishlay   10 years AGO

6-1=5,31-6=25,156-31=125,781-156=625,
so,625*5=3125+781=3906;

Like? Yes (1) | No   

Shahnawaz Ahmad   10 years AGO

1+1*5=6
6+5*5=31
31+25*5=156
156+125*5=781
781+625*5=3906

Like? Yes (1) | No   

sandy   10 years AGO

1
5^1+1
5^2+6
5^3+31
5^4+156
5^5+781=3906

Like? Yes (1) | No   

Rajat Goyal   10 years AGO

781*5 + 1= 3906
So, option 4 is correct.

Like? Yes (1) | No   

Dinesh Tripathi   10 years AGO

1
1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No   

prerna   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No   

susmit   10 years AGO

0*5+1=1
1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906
option no.4

Like? Yes (1) | No (1)  

Neha Rani   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
so,781*5+1=3906

Like? Yes (1) | No (1)  

Jigyasa Kumar   10 years AGO

fibonacci series is used...so the next number will be 3906

Like? Yes (1) | No (1)  

sudipta sadhukhan   10 years AGO

(5^n-1)/4

Like? Yes (1) | No (1)  

Venkatesh KK   10 years AGO

Difference between successive digits are 5,25(i.e, 5^2),125(i.e, 5^3),625(i.e, 5^4),....

ie.,
1+5=6
6+25=31
31+125=156
156+625=781
Similarly,
781+3125=3906 (option 4)

Like? Yes (2) | No (2)  

Aarush Singh   10 years AGO

fibonnacci rule is followed...so the next number is
3906

Like? Yes (1) | No (1)  

madanpal vishwakarma    10 years AGO

the series is as follows
first term =1
second term =previous term *5+ 1 =1*5+1=6
third term =previous term *5+ 1 =6*5+1=31
fourth term =previous term *5+ 1 =31*5+1=156
fifth term =previous term *5+ 1 =156*5+1=781
sixth term =previous term *5+ 1 =781*5+1=3906

so option 4 is correct

Like? Yes (2) | No (2)  

Srinath   10 years AGO

According 2 antidiagonals....

Like? Yes (1) | No (1)  

T G   10 years AGO

ans 3906
1 6 31 156 781
5 25 125 625
5*5=25
25*5=125
125*5=625
625*5=3125
3125+781=3906

Like? Yes (2) | No (2)  

ASHISH MITTAL   10 years AGO

previous term*5+1

Like? Yes (1) | No (1)  

Kanhaiya Jha   10 years AGO

1,6,31,156,781,...
Here difference between two consecutive terms=
5,25,125,625,..
Therefore all the differences are multiple of 5.
Hence next difference will be 625*5=3125
Therefore next term=3125+781=3906
Therefore option 4)

Like? Yes (1) | No (1)  

Lavankumar Reddy   10 years AGO

Here First 1
then =1+1*5=6
next =1+6*5=31
next =1+31*5=156
next =1+156*5=781

And Finally like these =1+781*5= 3906

Like? Yes (1) | No (1)  

Praneeth   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
156*5+1=781
781*5+1=3906

Like? Yes (1) | No (1)  

Paras   10 years AGO

difference is multiple of 5
5,25,125,625,3125
1+5=6
6+25=31
31+125=156
156+625=781
781+3125=3906

Like? Yes (1) | No (1)  

Grishma Shah   10 years AGO

1
1*5 + 1 = 6
6*5 + 1 = 31
31*5+ 1 = 156
156*5+1 = 781
781*5+1 = 3906

Like? Yes (1) | No (1)  

Saidinesh   10 years AGO

1*5+1=6;
6*5+1=31;
31*5+1=156;
156*5+1=781;
781*5+1=3906

Like? Yes (1) | No (1)  

Devendra Marghade   10 years AGO

1+5^1=6
6+5^2=31
31+5^3=156
156+5^4=781
781+5^5=3906

Like? Yes (1) | No (1)  

Suyog   10 years AGO

1*5+1=6,
6*5+1=31,
31*5+1=156,
156*5+1=781,
781*5+1=3906
So its 4

Like? Yes (1) | No (1)  

Sudipta Mukherjee   10 years AGO

1,6,(6-1)*6+1=31,(6-1)*31+1=156,(6-1)*156+1=781,(6-1)*781+1=3906

Like? Yes (1) | No (1)  

Dinesh thatti   10 years AGO

the series is follows as *5 + 1
1*5+1=6

6*5+1=31

31*5+1=156

156*5+1=781

781*5+1=3906

so answer is 3906 which is 4h option

Like? Yes (1) | No (1)  

Hitesh Kulkarni   10 years AGO

1*5+1=6
6*5+1=31
31*5+1=156
and so on
781*5+1=3906

Like? Yes (1) | No (1)  

Hanumanth Naik   10 years AGO

Given series is 1,6,31,156,781
the difference between first two numbers is 5
then 5*5=25,25+6=31
next two numbers difference is 25,25*5=125,125+31=156
next two numbers difference is 625,625*5=3125,3125+781=3906.

Like? Yes (1) | No (1)  

Sourish Barman   10 years AGO

1*5=5+1=6
6*5=30+1=31
31*5=155+1=156
156*5=780+1=781
781*5=3905+1=3906... ANS..

Like? Yes (2) | No (2)  

Balakumaran N.A   10 years AGO

differences: 5, 25, 125, 625, ...

Like? Yes (1) | No (2)  

Avishek Pattanaik   10 years AGO

Add the current term with increasing power of 5 to get the next term:
1+5^1=6
6+5^2=31
31+5^3=156
156+5^4=781
781+5^5=3906

Like? Yes (1) | No (2)  

07 Feb 2014 Coded language

If in coded langauge
RAJESH = 3380
MUKESH = 1877
SURESH = 5658
Then
NAVEEN = ?

OPtion

1) 1150
2) 1310
3) 1350
4) 1480
5) 1470
6) 1550
7) 1580
8) 1680
9) 1710
10)None of these

Solution

A(1), B(2), C(3), D(4), E(5), F(6), G(7), H(8), I(9), J(10), K(11), L(12), M(13), N(14), O(15), P(16), Q(17), R(18), S(19), T(20), U(21), V(22), W(23), X(24), Y(25), Z(26)

RAJESH = (R*J*S)-(A*E*H) = (18*10*19)-(1*5*8) = 3380
MUKESH = (M*K*S)-(U*E*H) = (13*11*19)-(21*5*8)= 1877
SURESH = (S*R*S)-(U*E*H) = (19*18*19)-(21*5*8)= 5658
Similarly
NAVEEN = (N*V*E)-(A*E*N) = (14*22*5)-(1*5*14) = 1470

Option 5)

Correct Option: 5 Best Solution (2)

Devendra Marghade   10 years AGO

Assigning numbers to alphabets as A=1,B=2,...,Z=26 & taking difference of product of alternate numbers.
RAJESH=(18)(1)(10)(5)(19)(8)=(18*10*19)-(1*5*8)= 3380
MUKESH=(13)(21)(11)(5)(19)(8)=(13*11*19)-(21*5*8)= 1877
SURESH=(19)(21)(18)(5)(19)(8)=(19*18*19)-(21*5*8)=5658
Then
NAVEEN=(14)(1)(22)(5)(5)(14)=(14*22*5)-(1*5*14)=1470

Like? Yes (13) | No (1)  

KRISHNAN   10 years AGO

Each alphabets are replaced with there corresponding number like a-1,b-2,c-3....z-26

so R A J E S H = 18 1 10 5 19 8
NOW the odd places are multiplied and even places are multiplied seperately

SO R*J*S = 18*10*19 = 3420 and A*E*H= 1*5*8 =40
now 40 is subtracted from 3420 so 3380

Like wise MUKESH is (M*K*S)-(U*E*H)
or (13*11*19)-(21*5*8)= 2717-840=1877

SURESH is (S*R*S)-(U*E*H) OR
(19*18*19)-(21*5*8)= 6498-840= 5658

Similarly NAVEEN is (N*V*E)-(A*E*N) or
(14*22*5)-(1*5*14) = 1540-70 or 1470
So option 5 is the correct answer

Like? Yes (11) | No   

06 Feb 2014 Solve

Number of methods to post 4 letters in 4 envelopes such that only 1 letter goes to the correct envelope is:-

OPtion

1) 3
2) 4
3) 5
4) 6
5) 7
6) 8
7) 9
8) 10
9) 12
10) None of these

Solution

a b c d
A C D B
A D B C
Similarly with B, C, D

Correct Option: 6 Best Solution (15)

nitesh   10 years AGO

ABCD(envelopes)
ACDB(letters)
ADBC(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8

Like? Yes (13) | No (3)  

KRISHNAN   10 years AGO

Choosing right letter-right envelope can be done in 4C1 ways.
In each of the ways, for the remaining 3 letters and 3 envelopes, there are only 2 ways of putting all three in wrong envelopes as follows:
First one can be put in 2 wrong envelopes. Second and third can only be put in 1 remaining wrong envelope. So 2 ways.
Total 4C1*2 ways.
or 8 ways thus option 6 is correct

Like? Yes (3) | No   

saranya sankar   10 years AGO

Total # of ways of choosing envelopes=4!=24.
Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct)
ABCD(envelopes)
ACDB(letters)
ADBC(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8

hence the ans is 8...

Like? Yes (3) | No   

ASHISH MITTAL   10 years AGO

one letter out of 4 that must be put in its correct envelope. We can do this in 4 ways. In how many ways can we put the rest of the 3 letters into 3 envelopes such that all 3 letters go into incorrect envelopes? We found that 3 letters can be put into 3 envelopes such that each letter is put in incorrectly in 2 ways. So total number of ways such that only one letters go into the envelopes with their correct addresses = 4*2 = 8 ways

Like? Yes (2) | No   

Manikumar R   10 years AGO

one letter out of 4 that must be put in its correct envelope. We can do this in 4 ways.
rest of the 3 letters into 3 envelopes such that all 3 letters go into incorrect envelopes
so all ways is 4*2=8

Like? Yes (2) | No   

Vijay Chauhan   10 years AGO

1,2,3,4 and we have to set only 1 letter to correct envelope...so...
suppose 1 is in correct envelope...so remains are different.... brackets are in correct envelope
(1),(2),(3),(4) -> wrong
(1),(2), 4 , 3 -> wrong
(1), 3 , 2 ,(4) -> wrong
(1), 3 , 4 , 2 -> right .......
(1), 4 ,(3), 2 -> wrong
(1), 4 , 2 , 3 -> right .......
...
For 1 set there are 2 possibility are right...but we have 4 set ...
2*4 = 8
so, ans is 6)8

Like? Yes (3) | No (1)  

Perambadur Jyothikrishna   10 years AGO

A B C D submit in a four 'a b c d' envelopes
if you fix A-----a
the other possibility in another envelopes is
b c d
----- ---- ----
C D B
D B C
so there is two possiblities if you fix the A
if you B,C,D totally 2*4==8.

Like? Yes (2) | No   

Surbhi Verma   10 years AGO

Ans is 8

Like? Yes (1) | No   

D.HARINATH REDDY   10 years AGO

In 8 ways we can post letters to envolop such that only 1 letter goes to the correct envelope.

Like? Yes (1) | No   

Praneeth   10 years AGO

Consider => L1-e1 is right => then L2 has 2 possible combinations such as e3 or e4 (we consider L2-e3); it cannot take e2 bcoz then there will be a combination with 2 letters in right envelop. Now the obvious choice for L3 to take e4 and not e3 (same logic for L2) and thus L4 is paired bydefault with e2.....Now plz refer below sequence with 2 possibilities

Hence => [L1-e1 | L2-e3 | L3-e4 | L4-e2] OR [L1-e1 | L2-e4 | L3-e2 | L4-e3]

Now this sequence is only for L1-e1 to be the only right and all others are wrong; similarly we have 4 pairs such as (L1-e1 | L2-e2 | L3-e3 | L4-e4 ) with 2 possibilities for each

Hence Ans is 4 x 2 = 8 ways

Like? Yes (1) | No   

Sudipta Mukherjee   10 years AGO

Let’s first select the one letter out of 4 that must be put in its correct envelope. We can do this in 4 ways. In how many ways can we put the rest of the 3 letters into 3 envelopes such that all 3 letters go into incorrect envelopes? We found that 3 letters can be put into 3 envelopes such that each letter is put in incorrectly in 2 ways. So total number of ways such that only two letters go into the envelopes with their correct addresses = 4*2 = 8 ways

Like? Yes (1) | No   

Dr.Dipin Singh   10 years AGO

Correct letter can go to any one of the 4 envelopes.
Balance three letters can be placed in 2 ways such that none of these 3 letters go in correct envelope.
so total possible methods to post 4 letters in 4 envelopes such that only 1 letter goes to the correct envelope = 4*2=8

Like? Yes (1) | No   

Surya Narayana K   10 years AGO

Those possibilites are...
ADBC
ACDB
CBDA
DBAC
BDCA
DACB
BCAD
CABD

So,there are 8 possibilities(Option 6).

Like? Yes (2) | No (1)  

santhosh   10 years AGO

let 1 letter and 1 envelope make together
then remaining arrangement has only 2 chances for each envelope so 2*2*2=8

Like? Yes (1) | No   

viveka   10 years AGO

l1 l2 l3. l4

l1 l3 l4 l2
l1 l4 l2 l3. 2 ways for l1
2 ways for l2 ,l3, l4 each
so 8

Like? Yes (1) | No   

05 Feb 2014 Solve

The sum of all 2 digit divisors of the number obtained by the difference of a 3 digit number of successive digits in decreasing order and a 3 digit number having same digits in reciprocal order. Only 1, 2, 3, 4, 5, 6, 7, 8, 9 can be used as digits

OPtion

1) 250
2) 31
3) 448
4) 29
5) 19
6) 16
7) 249
8) 198
9) 150
10)None of these

Solution

321-123=198

2 digit divisors of 198 are 11, 18, 22, 33, 66, 99.

Correct Option: 7 Best Solution (12)

Venkatesh KK   10 years AGO

The three digit numbers having successive digits in decreasing order are,
987
876
765
...
...

Then difference between three digit number and reciprocal order of the same number is,

987-789=198
876-678=198
765-567=198
...
...

The difference is 198. The divisors of 198 are 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 198.

Now the sum of two digit divisors of 198 is,

11+18+22+33+66+99 = 249 (Option 7)

Like? Yes (11) | No (2)  

Praneeth   10 years AGO

321-123=198
432-234=198
543-345=198
so all 2digit factors of 198 are
11+18+22+33+66+99=249

Like? Yes (4) | No   

Srinath   10 years AGO

Take any 3 successive digits in giver series.
I take 456, the reciprocal for this num is 654
The difference is 654-456=198
The 2 digit divisors for the number 198 are: 11,18,22,33,66,99
Now finally the sum is 11+18+22+33+66+99=249

Like? Yes (4) | No   

nitesh   10 years AGO

always diff=198, 11,18,22,33,66,99 are divisor ,,sum=249

Like? Yes (3) | No   

madanpal vishwakarma    10 years AGO

the 3 digit numbers in

decreasingorder -reciprocal order
123 - 321=198
234-432=198
345-543=198
456-654=198
567-765=198
678-876=198
789-987=198

so two digit divisors of 198
99+66+33+22+18+11 =249
so option 7 is correct

Like? Yes (3) | No   

KUMAR ASHUTOSH   10 years AGO

ans:249
diff=198
divisors:99+66+33+22+11+18=249

Like? Yes (3) | No   

SADINENI RAGHAVENDRA BABU   10 years AGO

the difference of a any 3 digit number of successive digits in decreasing order and a 3 digit number having same digits in reciprocal order is 198.
and its 2 digit numbers sum is 11+18+22+33+66+99=249

Like? Yes (3) | No   

S.SHAFI   10 years AGO

The digits are 1,2,3,4,5,6,7,8,9
3 digit number of successive digits in decreasing order = 987, 876, 765, 654, 543, 432, 321
3 digit number having same digits in reciprocal order = 789, 678, 567, 456, 345, 234, 123

So, the difference is 198 for all the values..
and the 2 digit divisors of 198 are 11,18,22,33,66,99 and the sum is 249

Like? Yes (3) | No (1)  

Niraj   10 years AGO

Difference of a 3 digit number of successive digits in decreasing order and a 3 digit number having same digits in reciprocal order will be 198

These are 2 digits divisors of 198 11, 18, 22, 33, 66, 99 and sum of these will be 249.

Like? Yes (2) | No   

Anantharaman   10 years AGO

Number formed by three digit consecutive numbers in decreasing order = 321, 432, 543, 654, 765, 876, 987

Number formed by reversing the above numbers = 123, 234, 345, 456, 567, 678, 789 respectively

Difference between any of the above pairs is 198

2 digit divisors of 198 = 11, 18, 22, 33, 66, 99

Sum of all 2 digit divisors = 249

Option 7

Like? Yes (2) | No   

D.HARINATH REDDY   10 years AGO

987-789=198
876-678=198
similarly for all numbers.
2 digit divisors of 198 =11,18,22,33,66,99
sum= 11+18+22+33+66+99 = 249

Like? Yes (1) | No   

Manikumar R   10 years AGO

432-234=198
543-345=198
654-456=198
.
.
.
.
198 two digit divisors is 11, 18, 22, 33, 66, 99
sum of divisors is 249

Like? Yes (2) | No (1)  

04 Feb 2014 Solve

Sum of all 120 numbers formed by using 1, 2, 3, 3, 3, and 4 is?

OPtion

1) 35555516
2) 35555520
3) 45555516
4) 34555606
5) 45555520
6) 45555506
7) 34562816
8) 48355520
9) 48355516
10) None of these

Solution

Unit place =20(1+2+3+3+3+4)
Tens place=20(1+2+3+3+3+4) +32 ;( 32 is carry from units place)
Hundreds place=20(1+2+3+3+3+4) +35(carry from previous)
Similarly others

Correct Option: 2 Best Solution (14)

ROHIT UPADHYAY   10 years AGO

total numbers =120
there are 6 digit 1,2,3,3,3,4 each digit appear 120/6 =20 times in each position
for example.. 1 appear--- 1,2,3,3,3,4
1,2,3,3,4,3
1,2,3,4,3,3
1,2,4,3,3,3
1,4,2,3,3,3
1,4,3,2,3,3
1,4,3,3,2,3
. .
. .
. .
. .
Now 1+2+3+3+3+4 = 16
thus the sum of all 120 numbers is----->16*20(1+10+100+1000+10000+100000)=35555520

Like? Yes (34) | No (8)  

Srinath   10 years AGO

6n=120, n=20.
20*1=20
20*2=40
20*3=60
20*3=60
20*3=60
20*4=80
sum these numbers: 320

Now 320(100000+10000+1000+100+10+1)
=320(111111)
=35555520.

Like? Yes (8) | No (1)  

madanpal vishwakarma    10 years AGO

if you were to list them all in a column to add, you'd have

if consider all 3's as different
-------


Among all the 120 units digits, 1/6 or 20 of them are 1's,
1/6 or 20of them are 2's, ... , 1/6 or 20of them are 3's and so on

20 of them are 1's, so their sum is 20x1 = 20
20 of them are 2's, so their sum is 20x2 = 40
20 of them are 3's, so their sum is 20X3 = 60
20 of them are 3's, so their sum is 20x3 = 60
20 of them are 3's, so their sum is 20x3 = 60
20 of them are 4's, so their sum is 20x4= 80
----------------------------------------------
total of digits in the units column = 320

So the sum of the digits in the units column is 320. The other four
columns also have the same digits as the units digits.

So the sum of each column is 320. So the sum is

320*10000 + 320*1000 + 320*100 + 320*10 + 320

320(10000 + 1000 + 100 + 10 + 1)

320(11111)

3555520

Like? Yes (3) | No   

Venkatesh KK   10 years AGO

if you were to list them all in a column to add, you'd have

123334
123343
123433
124333
.....
.....
.....
433132
433312
433321
-------


Among all the 120 units digits, 1/6 or 20 of them are 1's,
1/6 or 20 of them are 2's, 1/6 or 20 of them are 3's, ... , 1/6 or 20 of them are 4's

20 of them are 1's, so their sum is 20x1 = 20
20 of them are 2's, so their sum is 20x2 = 40
20 of them are 3's, so their sum is 20X3 = 60
20 of them are 3's, so their sum is 20x3 = 60
20 of them are 3's, so their sum is 20x3 = 60
20 of them are 4's, so their sum is 20x4 = 80
----------------------------------------------
total of digits in the units column = 320

So the sum of the digits in the units column is 320. The other five
columns also have the same digits as the units digits.

So the sum of each column is 320. So the sum is

320*100000 + 320*10000 + 320*1000 + 320*100 + 320*10 + 320

320(100000 + 10000 + 1000 + 100 + 10 + 1)

320(111111)

35555520 (option 2)

Like? Yes (3) | No   

Suyog   10 years AGO

111111*20*16=35555520

Like? Yes (3) | No   

T G   10 years AGO

(6-1)!/3! *111111*16

Like? Yes (4) | No (2)  

S.SHAFI   10 years AGO

There are 6P6/3! = 120 ways of forming 6-digit numbers.
Consider first the one's place. In those 120 numbers, 1,2,3,3,3,4 appear exactly the same number of times, that is, each digit appears 120/6 = 20 times.
For the tens place, same thing: 1,2,3,3,3 and 4 appear equally the same number of times.
Each digit appears 20 times. The argument for the hundreds and thousands places are the same.

Thus the sum of all those distinct 120 numbers is
10^5*20*(1+2+3+3+3+4) + 10^4*20*(1+2+3+3+3+4) + 10^3*20*(1+2+3+3+3+4) + 10^2*20*(1+2+3+3+3+4) +
10^1*20*(1+2+3+3+3+4) + 1*20*(1+2+3+3+3+4) + =

1111110*2*(1+2+3+4+5) = 1111110*2*15 = 35555520

Like? Yes (2) | No   

Sudipta Mukherjee   10 years AGO

total numbers=120

if you were to list them all in a column to add, you'd have

12345
12354
12435
12453
13245
.....
.....
.....
42133
42313
43123
43213
-------


Among all the 120 units digits, 1/6 or 20 of them are 1's,
1/6 or 20 of them are 2's, ... , 1/6 or 20 of them are 4's

20 of them are 1's, so their sum is 20x1 = 20
20 of them are 2's, so their sum is 20x2 = 40
20 of them are 3's, so their sum is 20X3 = 60
20 of them are 3's, so their sum is 20x3 = 60
20 of them are 3's, so their sum is 20x3 = 60
20 of them are 4's, so their sum is 20x4 = 80
----------------------------------------------
total of digits in the units column = 320

So the sum of the digits in the units column is 320. The other four
columns also have the same digits as the units digits.

So the sum of each column is 320. So the sum is

320*100000 + 320*10000 + 320*1000 + 320*100 + 320

300(100000 + 10000 + 1000 + 100 + 10+1)

35555520

Like? Yes (2) | No   

Praneeth   10 years AGO

1,2,3,3,3,4 can be arranged in 6!/3! = 120 ways

Since each of the 6 digits have an equal chance of
appearing, then they will appear 120/6 = 20 ways

sum of all the digits is 16

so 20*16*(111111)= 35555520

Like? Yes (2) | No   

Manikumar R   10 years AGO

Total numbers that can be formed from 1,2,3,3,3&4 is 120

So there are in total 6 x 120 = 720 numerals equally placed with 1,2,3,3,3&4;

so in each place each of the numerals occurring for 120/60=20 times.

[Ten thousands place 20 ones, 20 Twos, 60 Threes, 20 Fours; this gives total 120; similalrly every other place is filled]


Thus the sum is = 20(111111 + 222222 + 3*333333 + 444444)

= 1777776×20 = 35555520

Like? Yes (1) | No   

Hanumanth Naik   10 years AGO

using the probability get those numbers in one's ,10,100,1000 places

Like? Yes (1) | No   

Mallikharjuna   10 years AGO

each digit appears at each palce is 120/6=20 times
unit place 1*(1+2+3+3+3+4)*20=320
tens palce 10*(1+2+3+3+3+4)*20=3200
similarly 100*16*20
1000*16*20
10000*16*20
100000*16*20

TOTAL SUM OF NUMBERS IS=35555520

Like? Yes (1) | No   

D.HARINATH REDDY   10 years AGO

Sum of digits which are in same position of all numbers is =320
Then form a number with all these position digits sum. 00000320
+0000320
+000320
+00320
+0320
+320
=35555520
We get 35555520.
This is the required answer.

Like? Yes (1) | No   

sravani   10 years AGO

(1+100+1000+10+10000+1000000+100000)=111111
(1+2+4)*5!/3!*111111+3*5!/2!*11111
35555520

Like? Yes (1) | No   

03 Feb 2014 Sequence

What is the next number of the following sequence
4,36,100,196,......

OPtion

1) 407
2) 132
3) 306
4) 324
5) 380
6) 270
7) 483
8) 575
9) 506
10)None of these

Solution

2^2 = 4
6^2 = 36
10^2 = 100
14^2 = 196
similarly
18^2 = 324
Or
(1*3)+1 = 3+1 = 4
(5*7)+1 = 35+1 = 36
(9*11)+1 = 99+1 = 100
(13*15)+1 = 195+1 = 196
Similarly
(17*19)+1 = 323+1 = 324
Option 4)

Correct Option: 4 Best Solution (91)

Manikumar R   10 years AGO

2^2=4
6^2=36
10^2=100
14^2=196
18^2=324

Like? Yes (8) | No (1)  

VENKATA SIVA AVINASH A   10 years AGO

2^2=4
6^2=36
10^2=100
14^2=196
so there is an increase of
4 in every stmt
so next no is 18
18^2=324

Like? Yes (3) | No   

Grishma Shah   10 years AGO

The series is as follows:

x^2 ,(x+4)^2 ,(x+4+4)^2 ,(x+4+4+4)^2 ,(x+4+4+4)^2

2^2, 6^2, 10^2, 14^2, 18^2

Hence answer is 18^2=324

Like? Yes (3) | No   

Sivanjali   10 years AGO

4,36,100,196
4= 2^2 +1^2
36= 2^2 +3^2
100=2^2+5^2
196=2^2+7^2
2^2+9^2=324

Like? Yes (2) | No   

Srinivasa Vamsi Krishna Dodda   10 years AGO

2^2=4
6^2=36 (2+4)
10^2=100 (6+4)
14^2=196 (10+4)
18^2=324 (14+4)

Like? Yes (2) | No   

Vijay Chauhan   10 years AGO

2^2 =4
6^2 =36
10^2=100
14^2=196...
every time add 4 and square that number
so, 18^2 = 324, ans 4)324

Like? Yes (2) | No (1)  

Ratnesh Pandey   10 years AGO

the series can be written as
2^2,6^2,10^2,14^2
so the next number will be 18^2
i.e. 324
ans is 4)324

Like? Yes (2) | No (1)  

sowjanya   10 years AGO

4,36,100,196...
2^2=4
6^2=36
10^2=100
14^2=196
the diffrence is 4..(6-2=4,10-6=4,14-4=10,18-4=14)
sooo..18^2=324
answer is 4th option 324

Like? Yes (1) | No   

kalyani   10 years AGO

4,36,100,196....
2^2=4 (2+4=6, so next comes 6^2)
6^2=36 (6+4=10,so next comes 10^2)
10^2=100 (10+4=14, so next comes 14^2)
14^2=196 (14+4=18, so next comes 18^2)
18^2=324
answer is 324.

Like? Yes (1) | No   

Rajasekhar Reddy   10 years AGO

the sequence follows as the square of
2,6,10,14,18
the next number is 18^2 =324

Like? Yes (1) | No   

Komala.G   10 years AGO

Ans 4
4 36 100 196 324
+32*1 +32*2 +32*3 +32*4
First see the difference between each number in the series
4+(32*1=32) = 36
36+(32*2=64) = 100
100+(32*3=96) = 196
196+(32*4=128) = 324

Like? Yes (1) | No   

Devendra Marghade   10 years AGO

2^2=4
6^2=36
10^2=100
14^2=196
18^2=324

Like? Yes (1) | No   

rajdeep pal   10 years AGO

324 ...dif 32*1=32+4=36

32*2=64+36=100
32*3=96+100=196
32*4=128+196=324

Like? Yes (1) | No   

D.HARINATH REDDY   10 years AGO

above sequence is squares of 2,6,10,14
next number in sequence is 18 square =324

Like? Yes (1) | No   

rishav   10 years AGO

4+16*2=36
4+16*4=100
4+16*6=196
so ,4+16*8=324

Like? Yes (1) | No   

keerthika   10 years AGO

2^2=4,(2+4)=6 6^2=36, (6+4)=10 10^2=100 lik wise series goes so 18^2=324..

Like? Yes (1) | No   

Jigyasa Kumar   10 years AGO

next number will be the square of 18 i.e. 324

Like? Yes (1) | No   

Aarush Singh   10 years AGO

next number will be the square of 18 = 324

Like? Yes (1) | No   

Rajat Goyal   10 years AGO

square of 2 = 4
square of 6 =36
square of 10=100
square of 14=196
square of 18=324

Like? Yes (1) | No   

abhilasha   10 years AGO

square of 18 will be the next number...

Like? Yes (1) | No   

Keshav Kishlay   10 years AGO

36-4=32,
100-36=64 and
196-100=96 so,
96+32=128+196=324

Like? Yes (1) | No   

kishor Mali   10 years AGO

2x2=4
6x6=36
10x10=100
14x14=196
then 18x18=324 every time increment of 4 .

Like? Yes (1) | No   

santhosh   10 years AGO

4(1^2),4(3^2),4(5^2),4(7^2) so next 4(9^2)
4(81)=324

Like? Yes (1) | No   

sravani   10 years AGO

2^2,6^2,10^2,14^2,18^2

Like? Yes (1) | No   

satyen nabar   10 years AGO

2^2, 6^2, 10^2, 14^2, 18^2 = 324

Like? Yes (1) | No   

NAGENDRA P   10 years AGO

4=2^2..
36=6^2..
100=10^2..
196=14^2..
in the same way..18^2=324..
so option (4)

Like? Yes (1) | No   

OM PRAKASH JHA   10 years AGO

From the series we can see that...
2^2=4
6^2=36
10^2=100
14^2=196
All the series is in sequence as 2,6,10,14. Hence next will be 18.
thus answer is 18^2=324

Like? Yes (1) | No   

OM PRAKASH JHA   10 years AGO

Given series is the square of 2,6,10,14. Here the difference between every number is 4.
2^2=4
6^2=36
10^2=100
14^2=196
Hence next will be 14+4=18.
Thus answer will be 18^2=324

Like? Yes (1) | No   

Chirag Ranjan Lahiri   10 years AGO

as 2x2=4,then 2+4=6 so 6x6=36, then6+4=10 so 10x10=100, then 10+4=14 so 14x14=196, hence ans will be 14+4=18 i.e 18x18=324...

Like? Yes (1) | No   

Harshit Daga   10 years AGO

2^2, 6^2, 10^2, 14^2..
so next number would be 18^2,
means 324..
so option 4 is correct

Like? Yes (1) | No   

Naveen Bhatta   10 years AGO

Diff of 36-4 = 32
100-36 = 64 ( 32*2 )
196-100 = 96 ( 32*3 )
Let us say nextdiff is x then diff should be x=32*4, i.e x=128
therefore ans should be 196+128=324

Like? Yes (1) | No   

udhayakumar   10 years AGO

2*2=4
6*6=36
10*10=100
14*14=196
18*18=324
2,6,10,14,18, Difference is 4 .

Like? Yes (1) | No   

hena samanta   10 years AGO

2nd digit=36
36-4=32
3rd digit=100
100-36=64,64=32*2
4th digit=196
196-100=96,96=32*3
32*4=128
=>5th digit(196+128)=324

Like? Yes (1) | No   

g shriniwas   10 years AGO

4+32*0=4
4+32*1=36
36+32*2=100
100+32*3=196
196+32*4=324

Like? Yes (1) | No   

divya   10 years AGO

36-4=32;32+4
32*2=64;64+36=100
32*3=96;100+96=196
32*4=128;196+128=324

Like? Yes (1) | No   

Sudipta Mukherjee   10 years AGO

2*2=4
(2+4)*(2+4)=36
(2+8)*(2+8)=100
(2+12)*(2+12)=196
(2+16)*(2+16)=324

Like? Yes (1) | No   

DIPANKAT BHATTACHARYYA   10 years AGO

2^2=4,6^2=36,10^2=100,14^2=196,18^2=324
The difference between 6 & 2,10 & 6,14 & 10 is 4.
So the next number is (14+4)^2=18^2=324

Like? Yes (1) | No   

Dinesh thatti   10 years AGO

4=2^2
36=6^2
100=10^2
196=14^2
324=18^2

so answer is 324

we get this by squaring alternative numbers
2,6,10,14,18

Like? Yes (1) | No   

hemavathi   10 years AGO

4+32=36
36+(32+32)=100
100+(32+32+32)=196
196+(32+32+32+32)=324

Like? Yes (1) | No   

Kaleem Pasha   10 years AGO

Answer is 324 i.e,
=>4
=>4+(32*1)=36
=>36+(32*2)=100
=>100+(32*3)=196
=>196+(32*4)=324

Like? Yes (1) | No   

ASHISH MITTAL   10 years AGO

4*1^2=4
4*3^2=36
4*5^2=100
4*7^2=196
4*9^2=324

Like? Yes (1) | No   

Naveen Pandey   10 years AGO

18sqr

Like? Yes (1) | No   

Amarjit kumar   10 years AGO

According to the given question,
4+(32*1)=36
36+(32*2)=100
100+(32*3)=196
196+(32*4)=324
So,option 4 is correct

Like? Yes (1) | No   

Sukanya   10 years AGO

2^2=4
6^6=36
10^10=100
14^14=196
18^18=324

Like? Yes (1) | No   

Vaishali Dhiman   10 years AGO

the series is 2^2,6^2,10^2,14^2........
so,
6-2=4,
10-6=4,
so...
14+4=18
& 18^2=324

Like? Yes (1) | No   

Bholanath    10 years AGO

2^2=4
6^2=36
10^2=100
14^2=196
so, 18^2=324

Like? Yes (1) | No   

Vikram Saini   10 years AGO

18^2
324

Like? Yes (1) | No   

abhay gupta   10 years AGO

2*2=4
+4
6*6=36
+4
10*10=100
+4
14*14=196
+4
18*18=324

Like? Yes (1) | No   

prem kumar   10 years AGO

look at series
the diff. of 4
2^2=4
6^2=36
10^2=100
14^2=169
so
18^2=324

Like? Yes (5) | No (4)  

akhilesh kumar   10 years AGO

4/4=1
36/4=9
100/4=25
196/4=49
1,9,25,49,squre of odd no.
1,3,5,7,so next is 9
324/4=81
so ans is 324

Like? Yes (1) | No   

abilash   10 years AGO

196 +32+96=324

Like? Yes (1) | No   

S Prakash   10 years AGO

Solution:
4,36,100,196,324
Explanation
2,2+4,6+4,10+4,14+4 square this numbers
2*2=4
6*6=36
10*10=100
14*14=196
18*18=324

Like? Yes (2) | No (1)  

mounika   10 years AGO

Ans is 324.
36-4=32------>(32*1)
100-36=64----->(32*2)
196-100=96----->(32*3)
Add multiple of 32 in sequence
(32*4)=128
196+128=324.
so option is 4.

Like? Yes (1) | No   

SONU GUPTA   10 years AGO

incressing 32,64,96 so next 128 so the next number is 324

Like? Yes (1) | No   

Deb Das   10 years AGO

36-4=32
4+32=36
36+32*2=100
100+32*3=196
196+32*4=324

Like? Yes (1) | No   

Chaitanya Singh   10 years AGO

2^2, 6^2, 10^2, 14^2, 18^2

18^2 = 324

Like? Yes (1) | No   

Komal Uttamrao Patil   10 years AGO

(2)^2, (2+4)^2, (2+8)^2, (2+12)^2, (2+16)^2

Like? Yes (1) | No   

Bojedla Durgabhavani   10 years AGO

2*2=4
6*6=36
10*10=100
14*14=196
18*18=324

Like? Yes (1) | No   

Hanumanth Naik   10 years AGO

square of 18

Like? Yes (1) | No   

Srinath   10 years AGO

according 2: (4n+2)^2.

Like? Yes (1) | No   

Paras   10 years AGO

the difference in each no is 32
so
32,64,96,128
128+196=324

Like? Yes (1) | No   

Mallela Srilakshmi   10 years AGO

4,36,100,196....
4=2*2
next number=(2+4)*(2+4)
36=6*6
next number=(6+4)*(6+4)
100=10*10
next number=(10+4)*(10+4)
196=14*14
next number=(14+4)*(14+4)

Like? Yes (1) | No   

surendhar   10 years AGO

2*2=4,
18*2=36,
50*2=100,
98*2=196
like wise 162*2=324

then 18-2=16
50-18=32
98-50=48
next will be x-98=64 which follows 16 th table
so x would be 162
162*2=324

Like? Yes (1) | No   

cute honey   10 years AGO

the series is of squares with difference 4

Like? Yes (1) | No   

haritha   10 years AGO

take the difference 36-4=32 100-36=64, 196-100=96
now 64-32=32 ,96-64=32,now 96+32=128 and 196+128=324

Like? Yes (1) | No   

jyoti   10 years AGO

as 36-4=32,100-36=64(32*2),196-100=96(32*3).so 324-196=128(32*4)

Like? Yes (1) | No   

Leelanand Kumar Yalla   10 years AGO

Here Seies is 4,36,100,196..
4*9= 4*3^2=36;
4*25=4*5^2=100;
4*49=4*7^2=196;
4*81=4*9^2=324; This is the answer...

Like? Yes (2) | No (1)  

Charanraj k   10 years AGO

18*18=324

Like? Yes (1) | No   

Kaustubh karekar   10 years AGO

4=2*2
36=6*6
100=10*10
196=14*14
common difference between their square roots is 4
so next is 18*18=324

Like? Yes (1) | No   

gowthamigupta   10 years AGO

4*9=36,2*25=100,4*49-196,,next 4*81=324..9,25,49,81...are perfect squares.

Like? Yes (1) | No   

rohit mishra   10 years AGO

in series
4,36,100,196 the difference b/w each number is 4 so

2^2=4,6^6=36,10^10=100,14^14=196,18^18=324

Like? Yes (1) | No   

gaurav sningh   10 years AGO

2^2 = 4
6^2 = 36
10^2= 100
14^2 = 196
therefore, 18^2 = 324

Like? Yes (1) | No   

Naresh   10 years AGO

squaring of 2 is 4,squaring of 6 is 36,squaring of 10 is 100,squaring of 14 is 196,squaring of 18 is 324(Answer). That means the next number is increased by 4 and square of that number is the result.

Like? Yes (1) | No   

AVIJIT   10 years AGO

4 is multipled with square of odd consequtive nums.
1,3,5,7,9 square of these are 1,9,25,49,81
and ans will be 81*4=324

Like? Yes (1) | No   

ekta rani   10 years AGO

4=2^2 the difference is always 4
36=6^6 (6-2=4)
100=10^10 (10-6)=4
196=14^14 (14-10=4)
324=18^18 (18-14=4)

Like? Yes (1) | No   

Anucia Mary A   10 years AGO

2 square is 4
2+4=6 square is 36
6+4=10 square is 100
10+4=14 square is 196
14+4=18 square is 324
so the answer is 4)324

Like? Yes (1) | No   

madanpal vishwakarma    10 years AGO

given series 4 ,36,100,196 ...
the pattern is 2^2,6^2,10^2,14^2,since the difference is 4 so next is 18^2
so option 4 is correct 18^=324

Like? Yes (1) | No (1)  

VIKASH RAY   10 years AGO

According to the given sequence,
4+(32*1)=36
36+(32*2)=100
100+(32*3)=196
196+(32*4)=324
so, option 4 is correct..

Like? Yes (1) | No (1)  

sudipta sadhukhan   10 years AGO

2²,(2+4)²,(6+4)²,(10+4)²,(14+4)²,...........

Like? Yes (1) | No (1)  

sumasindhu   10 years AGO

given numbers are 4= 2^2 ; 36=6^2; 100=10^2; 196= 14^2; the difference between 2,6,10,14 is 4. so next number is square of 14+4 i.e: 18
ans is : 18^2= 324

Like? Yes (1) | No (1)  

Surya Narayana K   10 years AGO

Given sequence is 4,36,100,196,...

Divided with 4 then becomes 1,9,25,49,..

Difference between successive number is 8,16,24..

So,next one is 49+32=81..

So, next number in the sequence is 81*4=324(option 4).

Like? Yes (1) | No (1)  

Varunraj Reddy   10 years AGO

Given series 4,36,100,196,....
Clearly the difference between numbers are 32,64,96,....i.e. multiples of 32
So next number with a difference of 32*4=128
So 196+128=324
Ans is option 4)

Like? Yes (1) | No (1)  

YUVAN SATHAIAH   10 years AGO

4-36 =32(1*32)
36-100 =64(2*32)
100-196=96(3*32)
then
196-324=128(4*32)

Like? Yes (2) | No (2)  

KRISHNAN   10 years AGO

The sequence is formed by (4*n+2)^2
where n starts from 0
so (4*0+2)^2 = 4
(4*1+2)^2 = 36
(4*2+2)^2 = 100
(4*3+2)^2 = 196
(4*4+2)^2 = 324 SO Option 4 is correct answer

Like? Yes (1) | No (1)  

DHAKAD NEETENDRA SINGH   10 years AGO

2^2 6^2 10^2 14^2

there is a gap of 4 in between them
so next number is = 18^2 = 324

Like? Yes (1) | No (1)  

Priya Dharshini   10 years AGO

the answer is 2^2,6^2,10^2,14^2,18^2
add 4 then square the number;

Like? Yes (1) | No (1)  

karthik hariharan   10 years AGO

2*2=4
6*6=36
10*10=100
14*14=196
18*18=324

Like? Yes (1) | No (1)  

Anirudh Maheswari   10 years AGO

2^2=4; 6^2=36; 10^2=100; 14^2=196;
Here, series can be seen as 2^2, 6^2, 10^2, 14^2
we can see The bases are in difference of 4:- 2+4=6, 6+4=10, 10+4=14.
So next number must be 14+4=18
and 18^2=324
So option 4.

Like? Yes (1) | No (1)  

RAKESH   10 years AGO

4,36,100,196,....
2^2,6^2,10^2,14^2,18^2

Like? Yes (1) | No (2)  

Vinay Sipani    10 years AGO

2^2 , 6^2 , 10^2 , 14^2 ... (4n - 2)^2

=> next term = 5 th term = 18^2 = 324

Like? Yes (1) | No (2)  

nitesh   10 years AGO

2,6,10,14,18 square diff 4

Like? Yes (1) | No (2)  

31 Jan 2014 Coded language

Out of these 10 codes, different 1 is

OPtion

1) TSRQPONMLK
2) SYEKQWCIOU
3) JMPSVYBEHK
4) GECAYWUSQO
5) LUDMVENWFO
6) AFKPUZFKPU
7) XCHMRWBGLQ
8) BWRMHCXSNI
9) KOSWAEIMQU
10)None of these

Solution

code has incorrect sequence.

Correct Option: 6 Best Solution (21)

ROHIT UPADHYAY   10 years AGO

A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z

see the sequence between alphabets....

in 6) AFKPUZFKPU
space should be 4 alphabets but in this code, not constant in "ZF" ----->
A----F
F----K
K----P
P----U
U----Z
Z----E
sequence wrong......

Like? Yes (45) | No (2)  

jyoti   10 years AGO

sequence is not constant in 6th code

Like? Yes (5) | No   

Komala.G   10 years AGO

OPTION 6
AFKPUZFKPU correct code is AFKPUZEJOT
Their is a difference of place between each letter
- means REVERSE + means FOWARD
1) TSRQPONMLK - 1 place (-1)
2) SYEKQWCIOU - 6 place (+6)
3) JMPSVYBEHK - 3 place (+3)
4) GECAYWUSQO - 2 place (-2)
5) LUDMVENWFO - 9 place (+9)
6) AFKPUZFKPU - AFKPUZ - (+5) ZFKPO - (+6)
7) XCHMRWBGLQ - 5 place (+5)
8) BWRMHCXSNI - 5 place (-5)
9) KOSWAEIMQU - 4 place (+4)

Like? Yes (5) | No (1)  

AVIJIT   10 years AGO

A F K P U Z FKPU
the diffrnce between each word is 4.but after z changed the sequence. i will be EJOT

Like? Yes (2) | No   

S.SHAFI   10 years AGO

ans) option: 6
AFKPUZFKPU

Write numbers to the alphabets correspondingly..

Every letter in the word is +5 to the previous letter.
A = 1
F = 6
K = 11
P = 16
U = 21
Z = 26
E = 31 = 26+5 = 5 but given F..
After 26 the letters are repeated..

Like? Yes (2) | No   

VENKATA SIVA AVINASH A   10 years AGO

there is an increase in 5 numbers between two alphabets but between z and f there is an increase of 6

Like? Yes (2) | No   

Udhaya kumar   10 years AGO

difference b/w letters varied.....

Like? Yes (2) | No   

Ratnesh Pandey   10 years AGO

answer is 6
bcoz in any other series the diff between two alphabet is unique but in 6th option
you can see U+5=z, Z+5=F( it should be E)

Like? Yes (1) | No   

Manikumar R   10 years AGO

difference b/w letters was different

Like? Yes (1) | No   

mounika   10 years AGO

Becoz this code starts with vowel letter 'A'.
Except this all are starts with consonants.
so aption is 6.

Like? Yes (1) | No   

Varunraj Reddy   10 years AGO

All the options contain equal differences between their adjacent alphabet but in option 6)it is having difference of 5 between adjacent alphabet except Z&F.
So ans is option 6)

Like? Yes (1) | No   

S Prakash   10 years AGO

Solution:
AFKPUZFKPU
A+5th letter= F
F+5th letter= K
and so on......
In place of F,E has to come

Like? Yes (2) | No (1)  

G.Madhumitha   10 years AGO

In 6th code the difference between each letter is 4... but the difference between 'Z' and 'F' is 5. So option 6 is odd.

Like? Yes (1) | No   

Rajat Goyal   10 years AGO

In Rest all the words all letters are at fixed gap.

Like? Yes (1) | No   

Mayuri Rangari   10 years AGO

In the given 10 code all code having same difference between two letter
only 6th code option having no same dofference
hence ans is option 6

Like? Yes (1) | No   

Subham Kumar Jena   10 years AGO

Every other options are in sorted way of letters.
But in Option-6, "AFKPUZFKPU"...After Z; the letters are not in sorted way...

Like? Yes (1) | No   

santhosh   10 years AGO

option 6 has not followed it's sequence

Like? Yes (1) | No   

yadagiri reddy   10 years AGO

A-F=4
F-K=4
K-P=4
U-Z=4
Z-F=5
F-K=4
K-P=4
P-U=4
difference at Z-F=5
it is different in all

Like? Yes (1) | No   

Hemalatha   10 years AGO

it differs in the pattern..

Like? Yes (1) | No   

hena samanta   10 years AGO

instead of option no 6 every codes contains ten different letters.but code 6 contain two P,two U,two K,and two F.

Like? Yes (1) | No   

Dayashankar Rajhans   10 years AGO

there should be a in place of z

Like? Yes (1) | No