M4maths Previous Puzzles

27 Feb 2014 Quadratic Equation

Solve
{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2

OPtion

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

Solution

Put {(3x+1)/(x+1)} = P and reverse 1/P
and solve
we get x = 1 and x = -1/5

Option 3)

Correct Option: 3 Best Solution (54)

Akanksha Mishra   10 years AGO

Given;
{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2

[(3x + 1)^2 + (x + 1)^2]/(3x+1)(x+1) = 5/2

(10x^2 + 8x + 2)/(3x^2 + 4x + 1) = 5/2

20x^2 + 16x +4 = 15x^2 +20x + 5

5x^2 - 4x -1 = 0

5x^2 -(5 - 1)x = 0

5x^2 - 5x + x - 1 = 0

5x(x - 1) + 1(x - 1) = 0

5x + 1 = 0
x = -1/5
or

x -1 = 0
x = 1

so answer will be 1,-1/5(option 3)

Like? Yes (19) | No (13)  

shilpi   10 years AGO

solving the equation
x=1 and -1/5

Like? Yes (1) | No   

NAGENDRA P   10 years AGO

let a=3x+1,b=x+1
a/b+b/a=2.5...
a^2+b^2-2.5ab=0..
a^2+b^2-2ab=.5ab...
(a-b)^2=(ab)/2...
by substituting for a and b..
8x^2=3x^2+4x+1....
5x^2-4x-1=0..
by solving the above quadratic equation..
we get x=1,x=-0.2..(-1/5)
so option (3)...

Like? Yes (1) | No   

TINKU MITTAL   10 years AGO

(3x+1)^2+(x+1)^2=(5/2)*(3x+1)*(x+1)
after expanding and rearranging,we get;
5x^2-4x-1=0
(5x+1)(x-1)=0
x=1,-1/5

Like? Yes (1) | No   

viveka   10 years AGO

solving 2{(3x+1)^2(x+1)^2}=5{(3x+1)(x+1)}
5x^2-4x-1=0
-1/5 ,1

Like? Yes (1) | No   

santhosh   10 years AGO

1,-1/5 satisfy the given equation

Like? Yes (1) | No   

Rituparna Chakraborty   10 years AGO

Let, (3x+1)=a
(x+1)=b
then the given equation becomes=
a/b+b/a=5/2
=>2a^2-5ab+2b^2=0
=>2a^2-4ab-ab+2b^2=0
=>(a-2b)(2a-b)=0
So,(a-2b)=0
putting the value of a,b we obtain,
=>(3x+1)-2(x+1)=0
=>x=1
and (2a-b)=0
putting the value of a,b we obtain,
=>2(3x+1)-(x+1)=0
=>x=-1/5
S0,x=1,-1/5

Like? Yes (1) | No   

Mahalakshmi   10 years AGO

solving this equation i get 5x^2-4x-1=0
solving it
x=1,-1/5

Like? Yes (1) | No   

Vaishali Dhiman   10 years AGO

option is 3
{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2
{(3x+1)^2+(x+1)^2}/{(3x+1)(x+1)}=5/2
{9x^2+1+6x+x^2+1+2x}/{3x^2+4x+1}=5/2
{10x^2+8x+2}/{3x^2+4x+1}=5/2
20x^2+16x+4=15x^2+20x+5
5x^2-4x-1=0
5x^2-5x+x-1=0
5x(x-1)+1(x-1)=0
(5x+1)(x-1)=0
x=(-1/5),1

Like? Yes (1) | No   

neeraj   10 years AGO

3x+1/x+1=x+1/3x+1=5/2
(3x+1)^2+(x+1)^2/x+1*3x+1=5/2
solve the equation x=1 and x=-1/5

Like? Yes (1) | No   

ranu   10 years AGO

3x+1/x+1=x+1/3x+1=5/2
(3x+1)^2+(x+1)^2/x+1*3x+1=5/2
solve the equation x=1 and x=-1/5

Like? Yes (1) | No   

Vijit Mangla   10 years AGO

{(3x+1)2+(x+1)2}/{(3x+1)(x+1)}=5/2
10x2+8x+2/3x2+4x+1=5/2
20x2+16x+4/3x2+4x+1=5/2
5x2-4x-1=0
x= 4+_{(36)1/2}/10
x=1,-1/5

Like? Yes (1) | No   

PAYAL GAUTAM   10 years AGO

let a=3x+1 & b=x+1
2(a^2)+2(b^2)-5ab=0
=> a=b/2 or a=2b
=> 3x+1=(x+1)/2 or 3x+1=2*(x+1)
=> x=-1/5 or x=1

Like? Yes (1) | No   

Vijaya Bharathi   10 years AGO

Solve equation,
We get X=1
X=-1/5

Like? Yes (1) | No   

techknowxpress   10 years AGO

2*((3x+1)^2+(x+1)^2)-5*((x+1)*(3x+1))=0
therefore, 5x^2-4x-1=0
solving we get,
x=1,(-1/5)

Like? Yes (1) | No   

Aditya Verma   10 years AGO

{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2
2{(3x+1)^2+(x+1)^2}=5{(3x+1)(x+1)}
=>5x^2-4x-1=0
=>x=1,-1/5(answer)

Like? Yes (8) | No (9)  

chethan   10 years AGO

by solving the equation,we ll get d answer..

Like? Yes (1) | No (4)  

komathi rajan   10 years AGO

answer is simple. on simplification of the equation by squaring,expanding and furthur calculation. we get the answer as x=1 or x=(-1/5)

Like? Yes (1) | No (4)  

Parthi Smith   10 years AGO

[((3*1)+1)/(1+1)]+[(1+1)/((3*1)+1)]=5/2
[((3*(-1/5)+1)/(1+1)]+[((-1/5)+1)/(3*(-1/5)+1)]=5/2

Like? Yes (1) | No (4)  

karthik hariharan   10 years AGO

{(3x+1)/(x+1)}+{(x+1)/(3x+1)}=5/2

(3x+1)^2+(x+1)^2/(x+1)(3x+1)=5/2

10x^2+8x+2/3x^2+4x+1=5/2

20x^2+16x+4=15x^2+20x+5
5x^2-4x-1=0
x=1,-1/5

Like? Yes (1) | No (4)  

Jigyasa Kumar   10 years AGO

on solving the equation 5x^2-4x-1=0
we get the roots x=1,-1/5

Like? Yes (2) | No (5)  

abhilasha   10 years AGO

on solving the equation 5x^2-4x-1=0
we get the roots x=1,-1/5

Like? Yes (2) | No (5)  

RAKESH   10 years AGO

(3x+1)/(x+1)=y then
y+ 1/y = 5/2
=>(y-2)(2y-1)=0
=> y=2, 1/2
so, (3x+1)/(x+1)= 2, 1/2 => x=1,-1/5

3) x = 1, x = -1/5

Like? Yes (4) | No (8)  

saurabh   10 years AGO

3x+1/x+1 + x+1/3x+1 =5/2
by taking LCM
(3x+1)(3x+1)+(x+1)(x+1)/(x+1)(3x+1)=5/2------1
by solving equation 1 and cross multiplication
we get
20x^2+16x+4=15x^2+20x+5
20x^2-15x^2+16x-20x+4-5=0
5x^2-4x-1=0
5x^2-5x+x-1=0
5x(x-1)+(x-1)=0
(x-1) (5x+1)-----------------2
by equating equation 2 with 0
we get
x-1=0 , x=1
and 5x+1=0 , x=-1/5

hence x=1,-1/5

Like? Yes (2) | No (6)  

ASHISH MITTAL   10 years AGO

{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2
let 3x+1=a and x+1=b
so equation become
(a/b)+(b/a)=(5/2)
after solving this equation we get
2a=b and a=2b
so after putting value of a and b we get
x=-1/5 and x=1

Like? Yes (1) | No (5)  

roshani munde   10 years AGO

put 3x+1=p & x+1=q then solve fr p & q

Like? Yes (1) | No (5)  

Snehasis   10 years AGO

Given,
{(3x+1)/(x+1)} + {(x+1)/(3x+1)} = 5/2

Take LCM in left hand side then,
(3x+1)^2 + (x+1)^2/(x+1)(3x+1) = 5/2

Now, using [(a+b)^2 = a^2+b^2+2ab] formula for the LHS(left hand side) numerator,
we get,

9x^2 + 1 + 6x + x^2 + 1 + 2x/(x+1)(3x+1) = 5/2

now, open the brackets of the denominator and solve the numerator, we get,

10x^2 + 8x + 2/3x^2 + 4x + 1 = 5/2

Applying cross-multiplication to the above equation,

20x^2 + 16x +4 = 15x^2 + 20x + 5

Bringing all the right hand side(RHS) term to LHS and solving them,

5x^2 - 4x - 1 = 0

Now, Applying middle term factorization to the above equation then,

5x^2 - 5x + x - 1 = 0

=> 5x(x-1) + 1(x-1) = 0

=> (5x+1)(x-1) = 0

Now solving the equations separately,

5x+1=0 and, x-1=0

=> x = -1/5 => x = 1

The required quadratic root values of x are:

x = 1 and x = -1/5.

Like? Yes (3) | No (7)  

ganesh saicharan rao   10 years AGO

a+1/a=5/2
=2+1/2
->a=2or1/2
3x+1=2(x+1) or 2(3x+1)=x+1
x=1 x=-1/5

Like? Yes (1) | No (5)  

aviral bhargava   10 years AGO

on solving above equation we get,
2(10x^2+8x+2)=5(3x^2+4x+1)
=>5x^2-4x-1=0
Leading to the values of 1 and -1/5.
Hence the option 3rd is correct

Like? Yes (2) | No (6)  

MADHU   10 years AGO

{(3x+1)/(x+1)}+{(x+1)/(3x+1)}=5/2
=> [(3x+1)^2 + (x+1)^2]/[(3x+1)(x+1)]=5/2
=> 5x^2-4x-1=0
=> x=1,-1/5
so ans is 3rd option.

Like? Yes (1) | No (5)  

pari   10 years AGO

let M=(3x+1)/(x+1) n put it in eq.

so we get M+1/M=5/2 -- (I)

solve this eq. u will get M=2,1/2

now put real value of M
like (3x+1)/(x+1)=2 and (3x+1)/(x+1)=1/2

so we wil get x= 1 and -1/5

Like? Yes (1) | No (5)  

Satish Zod   10 years AGO

{(3x+1)/(x+1)} + {(x+1)/(3x+1)} = 5/2
Solving this,
(10x^2 + 8x + 2) / (3x^2 + 4x + 1) = 5/2
5x^2 - 4x – 1 = 0
a= 5, b= - 4, c= - 1

x = [4 ± √(16 + 20)] / 10
x= 1
x= -1/5

Like? Yes (1) | No (5)  

SIVA BALSUBRAMANIAN   10 years AGO

by solving the question we get final quation as
5x^2-4x-1=0
from that
x=1,-1/5

Like? Yes (1) | No (5)  

julee verma   10 years AGO

(3x+1)^2+(x+1)^2/(3x+1)(x+1)=5/2
9x^2+1+6x+x^2+1+2x/(3x^2+3x+x+1)=5/2
2*(10x^2+8x+2)=5*(3x^2+4x+1)
20x^2+16x+4=15x^2+20x+5
5x^2-4x-1=0
5x^2-5x+x-1=0
(5x+1)(x-1)=0
x=-1/5,x=1

Like? Yes (1) | No (5)  

Anand Kumar Saini   10 years AGO

Equation will become :5x^2-4x-1=0
solution of this eq.
5x^2-5x+x-1=0
5x(x-1)+(x-1)=0
(5x+1)(x-1)=0
so result will be:x=1,x=-1/5

Like? Yes (1) | No (5)  

shamma mohammad   10 years AGO

Option cheking:3
By substituting ‘x’ values in the equation we get 5/2
The values satisfy dis eqn is
x=1,x=-1/5

Like? Yes (1) | No (5)  

Prashasti Sharma   10 years AGO

(3x+1)^2 + (x+1)^2=5/2(3x+1)(x+1)
5x^2-4x-1=0
on solving x=1 and x=-1/5

Like? Yes (1) | No (5)  

basha shaik   10 years AGO

Wen we solve dis eqn,we get
5x^2-4x-1=0
roots for abov eqn are
x=1,x=-1/5

Like? Yes (1) | No (5)  

sowmya   10 years AGO

option verification so ans is x=1 & x=-1/5

Like? Yes (1) | No (5)  

Ashish Bhandari   10 years AGO

only x=1 and x=-1/5 is satisfy this equation

Like? Yes (2) | No (6)  

IFFAT IRAM   10 years AGO

{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2
2{(3x+1)^2+(x+1)^2}=5{(3x+1)(x+1)}
2(10x^2+2+8x)=5(3x^2+1+4x)
5x^2-4x-1=0
(5x+1)(x-1)=0
x=1 or -1/5

Like? Yes (1) | No (5)  

Rakesh Kumar Kohli   10 years AGO

{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2
=>2*[{(3x+1)^2+{(x+1)^2] = 5*(3x+1)*(x+1)
=>2*[{(3x+1-x-1)^2] = (3x+1)*(x+1)
=>8x^2 = 3x^2+4x+1
=>5x^2-4x-1=0
=>(5x+1)*(x-1)=0
=> x = -1/5 OR 1 which is option 3).

Like? Yes (1) | No (5)  

Aarush Singh   10 years AGO

on solving the equation 5x^2-4x-1=0
we getthe roots x=1,-1/5

Like? Yes (1) | No (5)  

Shiksha gautam   10 years AGO

{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2
(3x+1)/(x+1) =(5/2)-{(x+1)/3x+1}
(3x+1)/(x+1)=(13x+3)/(6x+2)
5x^2-4x-1=0
5x^2-5x+x-1=0
5x(x-1)+1(x-1)=0
(5x+1)(x-1)=0
x=1,x=-1/5

Like? Yes (1) | No (5)  

Varunraj Reddy   10 years AGO

Simplifying the given equation becomes
5x^2-4x-1=0
By solving we get
x=1,-1/5 option 3.

Like? Yes (1) | No (5)  

Hemalatha   10 years AGO

on solving,
5x^2-4x-1=0;x=1,-1/5

Like? Yes (1) | No (5)  

vasudha   10 years AGO

{(3x+1)/(x+1)}+{(x+1)/(3x+1)} = 5/2
by taking L.C.M,
{(3x+1)^2+(x+1)^2}/{(3x+1)*(x+1)}=5/2
by using (a+b)^2=a^2+b^2+2ab
we get (9x^2+1+6x+x^2+2x+1)/(3x^2+4x+1) = 5/2
i.e.,
{(10x^2+8x+2)/(3x^2+4x+1)}=5/2
by cross multiplication,
20x^2+16x+4=15x^2+20x+5
by simplifying above equation we get,
5x^2-5x-1=0
by finding the roots of the equation we get x=1,x=-1/5

Like? Yes (1) | No (5)  

tejaswini   10 years AGO

by considering (3x+1)/(x+1)=x
so x+(1/x)=5/2
by solving this equation we get x=2,1/2
again by substituting in x value
we get x=1,-1/5

Like? Yes (1) | No (5)  

SIBNATH HALDER   10 years AGO

{(3x+1)^2+(x+1)^2}/(x+1)(3x+1)=5/2,
(10x^2+8x+2)/(3x^2+4x+1)=5/2,
20x^2+16x+4=15x^2+20x+5,
5x^2-4x-1=0,
(5x+1)(x-1)=0,
x=1,x=-1/5;

Like? Yes (1) | No (5)  

sravan   10 years AGO

by solving above eqation you will get
5x^2-4x-1=0
5x^2-5x+x-1=0
5x(x-1)+1(x-1)=0
so the roots are x=1,x=-1/5

Like? Yes (1) | No (5)  

VIKASH RAY   10 years AGO

by,solving the given equation we get the following answer..

Like? Yes (1) | No (5)  

ARJYA HALDER   10 years AGO

Let, [(x+1)*(3*x+1)]=a
Then the equation becomes, a+(1/a)=(5/2)
Solving the equation we get, a=2 & a=(1/2)
a=2 gives x=(-1/5) &
a=(1/2) gives x=1
So, option 3 is correct.

Like? Yes (1) | No (5)  

Gudivada sunitha   10 years AGO

substitute the options. by substituting 3 option
(3*1+1)/(1+1)+(1+1)/(3*1+1)=4/2+2/4= 2+1/2= 5/2

Like? Yes (1) | No (5)  

SONU GUPTA   10 years AGO

answer will be x=1, x=-1/5

Like? Yes (1) | No (6)  

26 Feb 2014 Solve

Radius of earth is 6400 km. If mass of earth remains conserved and radius of earth would be 1cm and then its density would be (approximately) :-

OPtion

1) 10^26 times the present value of density of earth
2) 10^24 times the present value of density of earth
3) 10^22 times the present value of density of earth
4) 10^20 times the present value of density of earth
5) 10^28 times the present value of density of earth
6) 10^30 times the present value of density of earth
7) 10^8 times the present value of density of earth
8) 10^16 times the present value of density of earth
9) 10^9 times the present value of density of earth
10)None of these

Solution

Since mass is conserved hence
d2/d1 = V1/V2
= R13/R23
= 6400km*6400km*6400km/ 1cm*1cm*1cm
= 6400000m*6400000m*6400000m/0.01m*0.01m*0.01m
= 6.4*6.4*6.4*10^18/10^-6
= 262.144*10^24
approx 2.6*10^26
appro 10^26

Correct Option: 1 Best Solution (9)

RAKESH   10 years AGO

(6400*1000*100)^3= 10^26

Like? Yes (14) | No (5)  

aviral bhargava   10 years AGO

density of earth with radius being 6400 km=
d=m/v
=m/(4/3)pie (r^3)
=m/m/(4/3)pie(6400*10^5)^3
density of earth with radius being 1 cm
d=m/v
=m/(4/3)pie(1^3)
density2/density2 =
2.6*(10^26)
Hence the first option is correct

Like? Yes (5) | No (1)  

S.SHAFI   10 years AGO

Density = mass/volume
volume = 4/3*pi*r^3
density of earth when radius is 6400km is (3/4)*(m/pi)*(1/6400km^3)
density of earth when radius is 1cm is (3/4)*(m/pi)*(1/1cm^3)
Calculating the value..
it is 10^26 times the present value of density of earth..
so answer is option 1.

Like? Yes (5) | No (2)  

rakesh   10 years AGO

density=mass/volume
since, mass is constant density is inversily proportion to volume.
d1*v1=d2*v2
volume=PI*r^3
v1=PI*(64*10^7)^3
v2=PI
d2=d1*2^18*10^21
d2 approx 10^26 times of d1

Like? Yes (3) | No (1)  

shilpi   10 years AGO

d= present density * 10^26*2.6

Like? Yes (1) | No   

Harsh Poddar   10 years AGO

mass remains constant
mass =x in both cases
case 1
radius=6400km=640000000cm
density=mass/ volume
x/(640000000)^3
case 2
new radius=1cm
density=x/1^3
so density of case 1=x/262144000000000000000000000
mass will be cancelled
after 2 there are 20 digits so earth will be 10^20 times density of current earth
option 4

Like? Yes (2) | No (1)  

anubhav verma   10 years AGO

(100*1000*100)^3=10^21
64^3=262144=2.6*10^5
total 10^26

Like? Yes (1) | No (1)  

ASHISH MITTAL   10 years AGO

(r1)^3/(r2)^3

Like? Yes (1) | No (1)  

ashish kumar   10 years AGO

D=M/V
Volume of sphere= (4*3.14*R^3)/3
if mass is conserved
so Dnew*Vnew =Dearth*volume of earth
Dnew=(6.4)^3*10^24*Dearth
Dnew=2.62*10^26 *Dearth

Like? Yes (1) | No (1)  

25 Feb 2014 Age Problem

After 5 years a man's age will be three times his son's age and 5 years ago he was seven times as old as his son. Find their present ages.

OPtion

1) 60, 15 years
2) 48, 12 years
3) 52, 13 years
4) 44, 11 years
5) 48, 15 years
6) 60, 20 years
7) 40, 10 years
8) 55, 11 years
9) 40, 12 years
10)None of these

Solution

Let present age of father be x and son be y
After 5 years, father's age = 5+x
After 5 years, son's age = 5+y
By the question
(5+x) = 3(5+y) (i)
(x-5) = 7(y-5) (ii)
Solve (i) and (ii)
we get
father's present age = 40 years
and son's present age = 10 years
opiton 7)

Correct Option: 7 Best Solution (75)

Akanksha Mishra   10 years AGO

Consider the age of father in present = x year
Consider the age of father in present = y year

after five year age of father will be = (x+5) year
after five year age of father will be = (y+5) year

five year ago age of father = (x - 5)year
five year ago age of son = (y - 5)year

according to question;
5 years a man's age will be three times his son's age

(x + 5) = 3(y + 5)
you will get
x - 3y -10 = 0 ---------(1)

5 years ago he was seven times as old as his son
(x - 5) = 7(y - 5)
you will get equation;
x - 7y + 30 = 0--------------(2)

according to equation (1) and (2)
x = 40
y = 10
so
the age of father in present = x = 40 year
the age of father in present = y = 10 year
option 7

Like? Yes (30) | No (5)  

Harsh Poddar   10 years AGO

let son age =x
five years after=x+5
after 5 years father's age=3(x+5)
before five years son age=x-5
before five years fathers age =7(x-5)
present father age=3(x+5)-5 and 7(x-5)+5
equation
3(x+5)-5=7(x-5)+5
solve
x=10
present son age =10years
father age =3(10+5)-5
=45-5=40years
so present ages=40,10 years
option 7

Like? Yes (15) | No (1)  

SAM   10 years AGO

Let Man\'s age=x;son\'s age=y.
x+5=3(y+5)
x-5=5(y-5)
now solve it....

Like? Yes (7) | No   

saurabh   10 years AGO

The dad(d) is 40, and the son(s) is 10.
d=7s (the dad and son five years ago)
and
d+10=3(s+10) (the dad and son five years from now)
d+10=3s+30
substitute in the value of d from the first equation
7s+10=3s+30
rearrange
7s-3s=30-10
4s=20
s=5
Note this is the son five years ago...
so the son is now 10
the dad five years ago = 7s = 35, so the dad is now 40

((((((((now if you want to check your answer,
then
five years ago, the son was 5 and dad was 35
5*7=35 Check :)
Five years from now, the don is 15, and the dad is 45
15*3=45)))))))))

Like? Yes (4) | No   

shilpi   10 years AGO

suppose the age of father is =x
and the age of son is =y

acc to equation 1
x+5=3(y+5)
and acc to equation 2
x-5=7(y-5)

on solving the above equation
x(age of father)=40
y(age of son)=10

Like? Yes (4) | No   

Kaustubh karekar   10 years AGO

let their current ages be x,y
so, after 5 yrs, it will be x+5,y+5
now, x+5=3(y+5), then x+5=3y+15
also, 5 yrs ago the age will be x-5, y-5
now, x-5=7(y-5), then x-5=7y-35
solving the equations we get,
x=40 and y=10
so, option 7 is correct

Like? Yes (4) | No   

S.SHAFI   10 years AGO

After 5 years,
Lets son's age = x, then
Mans's age is three times his son's age = 3x

5 years ago,
son's age will be x-5-5 = x-10
Man's age will be 3x-5-5 = 3x-10
Man's age was seven times his son = 7(x-10)
so,
3x-10 = 7(x-10)
3x-10 = 7x-70
4x = 60
x = 15

Son's age after 5 years is 15 years and Man's age is 3x = 45 years.
Their Present Ages are 10 and 40.

Like? Yes (4) | No (1)  

Hariom   10 years AGO

let, man's present age is X
and son is y.
After 5 years their age will be x+5 and y+5.

x+5=3(y+5)...eqn 1
neo 5 years ago, their age was x-5 and y-5.
therefor acc. to the question,
x-5=7(y-5)...eqn 2
on solving 1 and 2, we get x=40 and y=10.

Like? Yes (4) | No (1)  

Leelanand Kumar Yalla   10 years AGO

Let the present Ages of Man and his son is M,S;
So according to statement M+5=3(s+5);------>(1)
M-5=7(5-5);------->(2)
M-3S-10=0---->(1)
M-7S+30=0---->(2)
By Solving These two Equations we will get Son's age is 10 years;
and Man's Age is 40 Years.... This is The answer...

Like? Yes (4) | No (1)  

shamma mohammad   10 years AGO

son=x
man=7x after 5years
option cheking:7
40,10
40-5=35
10-5=5
7x=7(5)=35

Like? Yes (5) | No (2)  

ANUP KUMAR   10 years AGO

x=present age of father;
y=present age of son;
given,
x+5=3*(y+3)
x-3y=10---(1)
&
x-5=7*(y-5)
x-7y=-30----(2)

solve (1)-(2)=>y=10;put the value of y in eqn (1)
then x=40;
present age of father is 40
& present age of son is 10

Like? Yes (3) | No   

Sajal Garg   10 years AGO

Let father's age = x; son's =y
Condition 1) (x+5)=3(y+5) =>x-3y=10...(1)
condition 2) (x-5)=7(y-5) =>x-7y=-30...(2)
solving 1 & 2) x=40 & y=10..
Answer=7) 40,10 years

Like? Yes (2) | No   

Devendra Marghade   10 years AGO

If present age of man=x, son=y
3(y+5)=(x+5), x-3y=10 ---(i) &
7(y-5)=(x-5), x-7y=-30 ---(ii)
Solving (i) & (ii), x=40, y=10

Like? Yes (2) | No   

Jigyasa Kumar   10 years AGO

the option 40, 10 is correct.because,
5 years ago, man will be 35 years old and son's age will be 5 years.So, man's age is 5 times his son's age.

Like? Yes (2) | No   

Sukanya   10 years AGO

after 5 years the ages will be
father - 45 son - 15(15*3=45)
before 5 years the ages will be
father-35 son-5(5*7=35)

Like? Yes (2) | No   

R.RAVIKUMAR   10 years AGO

Let's assume that father's present age is 'x' yrs and son's present age is 'y' yrs..
As per the given instructions,
x+5=3(y+5) => x-3y=10
x-5=7(y-5) => x-7y=-30
by solving those eqns, we get x=40 and y=10..

Like? Yes (2) | No   

Venkatesh KK   10 years AGO

Let the present ages of father and son be F and S respectively. After 5 years their age will be (F+5) and (S+5) respectively...
Therefore,
(F+5) = 3(S+5) ...(1)

5 years ago,
(F-5) = 7(S-5) ...(2)

Solving equations (1) and (2) we get S = 10 and F=40... i.e., father's age = 40 and son's age = 10.. (Option 7)

Like? Yes (3) | No (1)  

Aarush Singh   10 years AGO

Answer is 40, 1o.Because,
5 years ago, man will be 35 years old and son\'s age will be 5 years.So, man\'s age is 5 times his son\'s age.

Like? Yes (2) | No   

Rajasekhar Reddy   10 years AGO

M+5 = 3*(S+5)
M-3S = 10 -------> 1
M-5 = 7*(S-5)
M-7S = -30 ---------> 2
SOLVING ABOVE TWO EQUATIONS
WE GET
S = 10 YEARS
M = 40 YEARS

Like? Yes (2) | No   

Subhash chandra pal   10 years AGO

Let present ages are as follows:
Father=x
Son=y
After 5 year:
Father's age=x+5
Son's age=y+5
According to problem:
x+5=3(y+5)..............(i)
5 years ago:
Father's age=x-5
Son's age=y-5
According to problem:
x-5=7(y-5)...................(ii)
Now solving equation (i)and (ii)
We have x=40 and y=10.

Like? Yes (2) | No   

abhilasha   10 years AGO

the option 40, 10 is correct.because,
5 years ago, man will be 35 years old and son\'s age will be 5 years.So, man\'s age is 5 times his son\'s age.

Like? Yes (2) | No   

aviral bhargava   10 years AGO

let father's age be x and son's age be y
first equation would be formed as,
(x+5)=3(y+5)
Second equation would be formed as,
(x-5)=7(y-5)
On solving the above equations we get,
x=40 years and y=10 years
father's age is 40 years and son's age is 10 years .

Like? Yes (2) | No   

Mahalakshmi   10 years AGO

from above question we can form equations
x-5=7(y-5)
x+5=3(x+5)

solving this equation will get
x=40,
y=10
so fathers age 40 and sons age 10

Like? Yes (2) | No   

RAKESH   10 years AGO

x+5=3(y+5)
x-5=7(y-5)
=> x=40,y=10

Like? Yes (3) | No (1)  

Saidinesh   10 years AGO

five years back son age =5;
father's age =35;
i.e 7 times
now their age is 10,40
after five years
their age is 15, 45
i.e son age is trice as father age

Like? Yes (2) | No   

Pallekonda Rajesh   10 years AGO

today both years lets a and b
after 5 years b+5=3*(a+5)
before 5 years b-5=7*(a-5)
if we solve above two equations we will get
the ans:a=10 b=40

Like? Yes (2) | No   

TRIMURTULU   10 years AGO

man+5=3(son+5)
man-5=7(son-5)
2 equation 2 unknowns solve
man=40
son=10

Like? Yes (2) | No   

ranu   10 years AGO

let assume sons age is x year and father age is y year
5 year ago
7(x-5)=y-5
7x-35=y-5
7x-y=30 equation1
after 5 year
3(x+5)=y+5
3x+15=y+5
3x-y=-10 equation2
solve eq1 and eq2
x=10 and y=40

Like? Yes (2) | No   

SIVA BALSUBRAMANIAN   10 years AGO

x present age of father, y present age of mother
so after 5years
x+5=3(y+5)
before 5 years
x-5=7(y-5)
by solving this equation
x=40,y=10

Like? Yes (2) | No   

Sanjay Bhatt   10 years AGO

Let the Father's Age be X years.
Let the Son's Age be Y years.


After 5 years...
X+5=3(Y+5)...(1)
Before 5 years...
X-5=7(Y-5)...(2)

Solving (1) and (2) simultaneously we get,
X=40
Y=10

Like? Yes (2) | No   

ASHISH MITTAL   10 years AGO

Let father present age is f and son age is s so
f+5=3(s+5)------1
f-5=7(s-5)------2
by solving equation 1 and 2 we get
f=40
s=10

Like? Yes (2) | No (1)  

Harshit Daga   10 years AGO

Let man current age = x
and let son current age = y

After 5 years,
x+5 = 3*(y+5)

and,
before 5 years..

(x-5) = 7*(y-5)

so from above equations..
x=40,
and
y=10

which means..
Man current age = 40 years.
and, son current age = 10 years..

i.e. option 7 is correct.

Like? Yes (1) | No   

NAGENDRA P   10 years AGO

let the age of man be M,let the age of his son be S..
M+5=3(S+5)
M=3S+10....(1)
M-5=7(S-5)
3S+5=7S-35...
40=4S
S=10..
M=40..
SO OPTION (7)

Like? Yes (1) | No   

magamutha   10 years AGO

5 yrs ago the age of the son and father is x and y respectively. The eqn is y=7x
after 5 yrs y+10=3(x+10)
x=5,y=35 present ages are 10 and 40

Like? Yes (1) | No   

Sudipta Mukherjee   10 years AGO

from options ..7 is r8....40+5=3(10+5)....(i)
7(10-5)=40-5...(ii)

Like? Yes (1) | No   

rajat joshi   10 years AGO

only this option fits best

Like? Yes (1) | No   

TANUSREE MAHAPATRA   10 years AGO

let,x=son's present age
y=father's present age
3(x+5)=y+5
7(x-5)=y-5
after solving we get,x=10,y=40

Like? Yes (1) | No   

Gudivada sunitha   10 years AGO

after 5 years 40 is 45 son age is a5 and 15*3 =45.
before 5 years father age is 35 , son age is 5, 5*7=35

Like? Yes (1) | No   

Lavankumar Reddy   10 years AGO

40,10 years that are after 5 years = 40+5 = 45 and 10+5 = 15 so age was 45/15 = 3 times
5 years ago age is = 40-5=35 and 10-5 = 5
S0 different is =35/5= 7 times

Like? Yes (1) | No   

Dhwan Lalwani   10 years AGO

Present man n son age be M & S
After 5yr, M+5=3(S+5)
Before 5yr, M-5=7(S-5)
solving them we get,
M=40yrs & S= 10yrs

Like? Yes (1) | No   

chandrappa   10 years AGO

x+5=3(y+5)

Like? Yes (1) | No   

pari   10 years AGO

father son
5yr ago 7x x
present age 7x+5 x+5
5yr after 7x+5+5 = 3(x+5+5) solve this eq.

we will get x=5
so ans is 40 and 10

Like? Yes (1) | No   

Rajat Goyal   10 years AGO

Best Option

Like? Yes (1) | No   

Swathi   10 years AGO

Ans:7)40,10
F+5=3(X+5)
F-5=7(X-5)
by solving these two equations we will get F=40 and X=10

Like? Yes (1) | No   

rakesh   10 years AGO

son age father age
present x
after 5yrs x+5 3(x+5)
before 5yrs x-5 7(x-5)

3(x+5)-7(x-5)=10
3x+15-7x+35-=10
4x=40
x=10


man present age=40
son present age=10

Like? Yes (1) | No   

sravani   10 years AGO

let man's age =x
let son's age =y
x+5=3(y+5)
x-5=7(y-5)
are given
we get x+10=5y by solving the above eq
so option 7 only satisfies it.

Like? Yes (1) | No   

Sapan Suman   10 years AGO

two equation
M+5=3(S+5)
M-5=7(S-7)

Like? Yes (1) | No   

vijayanirmala   10 years AGO

x=father age y=sons age
(x-5)=7(y-5)so (x+5)=3(y+5)
(x-5) = 7(y-5)

(x+5) = 3(y+5)
--> x+5 = 3y+15
--> x-3y = 10

Like? Yes (1) | No   

Parthi Smith   10 years AGO

man's age = x
son's age = y
1.x+5=3(y+5) => x=3y+10
2.x-5=7(y-5) => x=7y-30
Therefore,y=10,x=40

Like? Yes (1) | No   

ARJYA HALDER   10 years AGO

Let, present father\'s age=x
present son\'s age=y
So, from question we can get,
x+5=3(y+5) &
x-5=7(y-5)

Solving the above equations we get, x=40 & y=10

Like? Yes (1) | No   

VIKASH RAY   10 years AGO

Let,the present age of the man is- 40 years.
& the age of his son is- 10 years.
so, 5 years ago, he was seven times as old as his son.
and after5 years,he will be three times his son's age

Like? Yes (1) | No   

MD NAZIR   10 years AGO

Ans: 7) 40, 10
Eliminating Answers will take lesser time.
(40+5)=3 (10+5)
(40-5)=7 (10-5)

Like? Yes (1) | No   

MADHU   10 years AGO

let father's age be F
son's age be S
after 5 yrs F+5=3(S+5)=> F=3S+10
ago 5 yrs F-5=7(S-5)=> F=7S-30
by equating we get F=40
S=10

Like? Yes (1) | No   

susmit   10 years AGO

f+5 = 3(s+5)
f-5 = 7(s-5)

Solving this f = 40
s = 10
hence option no. 7

Like? Yes (1) | No   

Sumit Kumar Dharmani   10 years AGO

let us assume that the present age of man be M and of his son be S
then according to given condition
(M+5) = 3(S+5)
and
(M-5) = 7(S-5)
so solving these 2 equations;
we get M = 40 and S = 10
hence the answer is option 7.

Like? Yes (1) | No   

Rituparna Chakraborty   10 years AGO

Let
Man's present age=x
Son's present age=y
After 5 years
Man's age=(x+5)
Son's age=(y+5)
Given,(x+5)=3(y+5)
=>x-3y=10 ---------
5 years ago
Man's age=(x-5)
Son's age=(y-5)
Given,(x-5)=7(y-5)
=>x-7y=-30 -----------
Solving & we obtain,
x=40
y=10
So,Man's present age=40
Son's present age=10

Like? Yes (1) | No   

sudipta sadhukhan   10 years AGO

suppose their age after 5 yrs X & 3X.
present age X-5 & 3X-5
age before 5yrs X-10 3X-10

A/Q 7X-70=3X-10

=> X=15

present age 10& 40

Like? Yes (1) | No   

suraj jha   10 years AGO

mans age=x;
son age=y;
after 5 years
x+5=3(y+5);
x-3y=10.....(1)
5 years ago;
x-5=7(y-5);
x-7y=-30...(2)
eq1-eq2
x=40 & y=10
mans age=40;
son age=10;

Like? Yes (1) | No   

Chithra   10 years AGO

Present ages 40,10
After 5 yrs
45=15
5 yrs ago
35=5

Like? Yes (1) | No   

AKASH KUMAR   10 years AGO

LET MAN\'S PRESENT AGE =X AND SON\'S AGE=Y
THEN X+5=3(Y+5)---1
X-5=7(Y-5)---2
SOLVE WE GET X=40 AND Y=10

Like? Yes (1) | No   

ramkumar   10 years AGO

a+5=3(b+5);-----1
a-5=7(b-5);-----2
a=7(b-5)+5;
a=7b-35+5;
a=7b-30;------3
put equ(3) in (1)
7b-30+5=3(b+5);
7b-25=3b+15;
7b-3b=+15+25;
4b=40;
b=10;
put b=10 in equ (3);
a=7*10-30;
a=40;
so ans is a=10
b=40

Like? Yes (1) | No   

ayush   10 years AGO

3(x+5)=y+5
7(x-5)=y-5

* x & y are present age of child and father

Like? Yes (1) | No   

techknowxpress   10 years AGO

m+5=3*(s+5)
m-5=7*(s-5)
Solving these 2 eqns , we get
m(man's age)=40yrs
s(son's age)=10yrs

Like? Yes (1) | No   

anubhav verma   10 years AGO

x+5=3(y+5)
x-5=7(y-5)

Like? Yes (1) | No   

karthik hariharan   10 years AGO

Father age is taken as X;
Son age is taken as Y;
X+5=3(Y+5);----1
X-5=7(Y-5);-----2
X=3(Y+5)-5;
X=3Y+15-5;
X=3Y+10;-----3
PUT EQU 3 IN 2
3Y+10-5=7Y-35;
3Y+5=7Y-35;
3Y-7Y=-35-5;
-4Y=-40;
Y=10;
PUT Y=10 IN EQU 1
X+5=3(Y+5);
X+5=3(10+5);
X+5=45;
X=40;

Like? Yes (1) | No   

DHAKAD NEETENDRA SINGH   10 years AGO

only this option satisfy the condition

Like? Yes (2) | No (1)  

madanpal vishwakarma    10 years AGO

assume that men current age is x year and his son is y year old

then
according to the question

After 5 years a man's age will be three times his son's age

x+5=3(y+5).....equation (i)
II conditon
5 years ago he was seven times as old as his son

x-5=7(y-5)... equation(II)

on solving
we get x=40 and y=10.
so option 7 is correct

Like? Yes (4) | No (4)  

D.HARINATH REDDY   10 years AGO

x+5=3(y+5)
x-5=7(y-5)
then x=40,y=10

Like? Yes (1) | No (1)  

SID PRASANTH KUMAR   10 years AGO

40+5=3*(10+5)

Like? Yes (2) | No (2)  

IFFAT IRAM   10 years AGO

let the present age of son be x and father be y.


after 5 years,
son's age= x+5
father's age=y+5
y+5=3(x+5)
y=3x+10.........(1)

5 years ago,
son's age=x-5
father's age=y-5
y-5=7(x-5)
y=7x-30...........(2)

3x+10=7x-30
4x=40
x=10

y=3*10+10
y=40
son's present age=10yrs
father's present age=40 yrs.

Like? Yes (1) | No (1)  

Prashasti Sharma   10 years AGO

f=father and s=son
f(+5):(s+5)=3:1---(1)
f(-5):s(-5)=7:1--(2)
ratio is unequal hence..
eq1 * 6 and eq2*2
f(+5):s(+5)=18:6
f(-5):s(-5)=14:2
diff is 4
s0 if 4--10
then 1-- 10/4 which is 2.5
present age of father= (18*2.5-5)=40 years
present age of son=(6*2.5-5)=10 years

or by eqn method
3x-10=7(x-10)
x=15
father's age= 3*15-5=40 yrs
son's age= 15-5=10 yrs

Like? Yes (1) | No (1)  

Chaitanya Singh   10 years AGO

x -> age of father
y -> age of son

2 equations are
(x-5) = 7(y-5)
(x+5) = 3(y+5)

on solving
x = 40
y = 10

Like? Yes (1) | No (1)  

Vikram Saini   10 years AGO

40,10

Like? Yes (1) | No (1)  

Dhruv Patel   10 years AGO

x+5=3(y+5)
x-5=7(y-5)

Like? Yes (1) | No (1)  

Shiv Kumar   10 years AGO

40*10 because before 5 year man age 35 son 7 or after 5 year 45 son 15

Like? Yes (1) | No (1)  

24 Feb 2014 Sequence

What is the missing number of the following sequence
1, 3, 9, 81, ? , 43046721

OPtion

1) 5929
2) 6241
3) 6400
4) 6561
5) 6724
6) 6889
7) 7056
8) 7227
9) 7369
10)None of these

Solution

1 = 1
1*3 = 3
3*3 = 9
9*9 = 81

81*81 = 6561 (Option 4)

6561*6561 = 43046721

Correct Option: 4 Best Solution (86)

Akanksha Mishra   10 years AGO

Look upon the series carefully
1, 3, 9, 81, ? , 43046721

3 * 3 = 9
9 * 9 = 81

similarly
81* 81 = 6561 (answer)

6561*6561 = 43046721

so answer will 6561

Like? Yes (31) | No (6)  

Dr.Dipin Singh   10 years AGO

series is
3^0,3^1,3^2,3^4,3^8,3^16
so missing number is 3^8= 81*81= 6561

Like? Yes (11) | No (7)  

nazar   10 years AGO

(3)2=9
(9)2=81
(81)2=6561
(6561)2=43046721

Like? Yes (2) | No   

karthik hariharan   10 years AGO

81*81=6561
81*(3*3*3*3)=81

Like? Yes (2) | No   

Rajasekhar Reddy   10 years AGO

the sequence follows as
3^0 = 1
3^1 = 3
3^2 = 9
3^4 = 81
3^8 = 6561
3^16 = 43046721
so, the missing number is 6561

Like? Yes (8) | No (7)  

S.SHAFI   10 years AGO

Given series, the number is in n^2 form of previous number.
1,3
3^2 = 9
9^2 = 81
81^2 = 6561
6561^2 = 43046721

Like? Yes (3) | No (2)  

YUVAN SATHAIAH   10 years AGO

square of the previous number

Like? Yes (2) | No (1)  

K Venkata Subba Rao   10 years AGO

n*n

Like? Yes (3) | No (2)  

NAGENDRA P   10 years AGO

squaring each no. of series we will get next subsequent no. of the series..
1^2=1..
3^2=9..
9^2=81..
81^2=6561..
6561^2=43046721..
so option(4) is the answer..

Like? Yes (1) | No   

Mahalakshmi   10 years AGO

3^2= 9
9^2= 81
81^2 is 6561
6561^2 = 43046721

Like? Yes (1) | No   

Aarush Singh   10 years AGO

1. This is because nine multiplied by nine equals 81. 81 multiplied by 81 equals 6561, and 6561 multiplied by 6561 equals 43046721

Like? Yes (2) | No (1)  

Chithra   10 years AGO

3^2=9
9^2=81
81^2=6561
6561^2=43046721

Like? Yes (2) | No (1)  

lovely verma   10 years AGO

1
1*3=3
3*3=9
9*9=81
81*81=6561
6561*6561=43046721
............................................
so answer is 6561
................................................

Like? Yes (2) | No (1)  

R.RAVIKUMAR   10 years AGO

1,3,9,81,?,43046721

(1^2)*3=3
(3^2)*9=81
(9^2)*81=6561
(81^2)*6561=43046721

Like? Yes (3) | No (2)  

TRIMURTULU   10 years AGO

squares of the numbers obtained from the previous number starting from 1,3
1*1=1
3*3=9
9*9=81
81*81=6561
6561*6561=43046721

Like? Yes (1) | No   

madanpal vishwakarma    10 years AGO

THE GIVEN SERIES
1, 3, 9, 81, ? , 43046721
3^0,3^1,3^2,3^4,3^8,3^16
SO OPTION 4>6561 IS CORRECT

Like? Yes (3) | No (2)  

shilpi   10 years AGO

we are squaring each previous digit to get the next digit. 81 * 81 = 6561, 6561 * 6561 = 43046721

Like? Yes (1) | No   

Pallekonda Rajesh   10 years AGO

next number is the square of the number
3^2=9
9^2=81
81^2=6561
6561^2=43046721

Like? Yes (1) | No   

Vaishali Dhiman   10 years AGO

ans is 6561
3^2=9
9^2=81
81^2=6561
6561^2=43046721

Like? Yes (1) | No   

devendra singh   10 years AGO

ALL ARE SQUARE OF PREVIOUS TERM...

Like? Yes (1) | No   

SID PRASANTH KUMAR   10 years AGO

3^2=9
9^2=81
81^2=6561

Like? Yes (2) | No (1)  

ranu   10 years AGO

1*3=3
3*3=9
9*9=81
81*81=6561
6561*6561=43046721

Like? Yes (1) | No   

MD SOHAIL AKHTER   10 years AGO

3^2=9
9^2=81
81^2=6561
6561^2=43046721

Like? Yes (2) | No (1)  

Sudipta Mukherjee   10 years AGO

81*81=6561

Like? Yes (1) | No   

Shivani Gupta   10 years AGO

1*3=3,3*3=9,9*9=81,81*81=6561,6561*6561=43046721.

Like? Yes (2) | No (1)  

AVIJIT   10 years AGO

81*81=6561

Like? Yes (1) | No   

santhosh   10 years AGO

81*81=6561

Like? Yes (1) | No   

neeraj   10 years AGO

1*3=3
3*3=3
9*9=81
81*81=6561
6561*6561=43046721

Like? Yes (1) | No   

jayabharathy   10 years AGO

3*3=9
9*9=81
81*81=6561

Like? Yes (1) | No   

Vijaya Bharathi   10 years AGO

81*81

Like? Yes (1) | No   

astha   10 years AGO

81*81

Like? Yes (1) | No   

Snehasis   10 years AGO

It follows the square of the given no:
3^2 = 9
9^2 = 81
81^2 = 6561-------->this is the required answer.
6561^2 = 43046721

Like? Yes (1) | No   

haritadara   10 years AGO

3*3=9
9*9=81
81*81=6561
6561*6561=43046721

Like? Yes (1) | No   

HOOR FARHEEN   10 years AGO

3*3=9
9*9=81
81*81=6561
6561*6561=43046721

Like? Yes (1) | No   

Kaleem Pasha   10 years AGO

1*1=1
1*3=3
3*3=9
9*9=81
81*81=6561
6561*6561=43046721

Like? Yes (1) | No   

Manikumar R   10 years AGO

81*81

Like? Yes (1) | No   

vitoba   10 years AGO

81^2=6561,
similarly ....

Like? Yes (1) | No   

FAISAL RIZWAN   10 years AGO

Solution are as follows:-
1*3=3
3*3=9
9*9=81
81*81=6561
6561*6561=43046721

So option 4 is correct

Like? Yes (1) | No   

ajay jha   10 years AGO

as in given series next no is square of prev no like 3*3=9;
9*9=81;
similarly 81*81=6561

Like? Yes (1) | No   

Subhash chandra pal   10 years AGO

except first and second terms all terms are square of the previous one.
ex:
9=3^2
81=9^2
6561=81^2
43046721=6561^2

Like? Yes (1) | No   

SADINENI RAGHAVENDRA BABU   10 years AGO

81*81=6561

Like? Yes (1) | No   

saurabh   10 years AGO

81*81

Like? Yes (1) | No   

Neha Rani   10 years AGO

1*3=3
3*3=9
9*9=81
81*81=6561 Ans
6561*6561=43046721

Like? Yes (1) | No   

TANUSREE MAHAPATRA   10 years AGO

81*9*9=6561

Like? Yes (1) | No   

rana chowdhury   10 years AGO

3^2=9
9^2=81
81^2=6561
as well as (43046721)^(1/2)=6561

Like? Yes (1) | No   

pari   10 years AGO

(3)^0+(3)^1+(3)^2+(3)^4 + ? +(3)^16

so ans will be (3)^8= 6561

Like? Yes (1) | No   

Tejaswee Karani   10 years AGO

3*3=9
9*9=81
81*81=6561
6561*6561=43046721

so,6561 is the correct answer

Like? Yes (1) | No   

ASHISH MITTAL   10 years AGO

3^0, 3^1, 3^2, 3^4, 3^8, 3^16

Like? Yes (1) | No   

Rituparna Chakraborty   10 years AGO

1st term=3^0=1
2nd term=3^1=3
3rd term=3^(1*2)=3^2=9
4th term=3^(2*2)=3^4=81
5th term=3^(4*2)=3^8=6561
6th term=3^(8*2)=3^16=43046721
So Ans=6561

Like? Yes (1) | No   

vishwanath   10 years AGO

81 square

Like? Yes (1) | No   

shyam   10 years AGO

3 to the power 8

Like? Yes (1) | No   

PONRAJ.B   10 years AGO

3*3=9
9*9=81
81*81=6561

Like? Yes (1) | No   

rakesh   10 years AGO

3^0,3^1,3^2,3^4,3^8,3^16

Like? Yes (1) | No   

Sukanya   10 years AGO

3^0=1
3^1=3
3^2=9
3^4=81
3^8=6561
3^16=43046721

Like? Yes (1) | No   

vasudha   10 years AGO

1,3,3^2,9^2,81^2,6561^2
1,3,9, 81, 6561.....

Like? Yes (1) | No   

Soumya   10 years AGO

in sequence 3^2=9,9^2=81,likewise 81^2=6561,6561^2=43046721.

Like? Yes (1) | No   

kiran kumar reddy yanamala   10 years AGO

square of the numbers

Like? Yes (1) | No   

DHIRENDRA    10 years AGO

81*81= 6561

Like? Yes (1) | No   

Kaustubh karekar   10 years AGO

the sequence is square of previous number

Like? Yes (1) | No   

Suman Saurabh   10 years AGO

since in this series next number is square of previous number.
so missing number in this series is 81*81=6561

Like? Yes (1) | No   

haritha   10 years AGO

81 ^2=6561

Like? Yes (1) | No   

Hemalatha   10 years AGO

81^2=6561

Like? Yes (1) | No   

srinu   10 years AGO

81*81=6561
6561*6561=43046721

Like? Yes (1) | No   

Leelanand Kumar Yalla   10 years AGO

Here 3^0= 1;
3^1= 3;
3^2= 9;
3^4= 81;
3^8= 6561;
3^16= 43046721;
Like the way it follows... so answer is 6561...

Like? Yes (1) | No   

kiruthika   10 years AGO

The series is
1, 3, 9, 81, ? , 43046721

3^2=9
81^2=6561
6561^2=43046721

Like? Yes (1) | No   

vijayakumar   10 years AGO

formula=n^2
values=1,3,9,......

1)n=1
1^2=1

2)n=3
3^2=9

3)n=9
9^2=81

ans:6561
n=81
81^2=6561

Like? Yes (1) | No   

Keshav Kishlay   10 years AGO

81*81=6561

Like? Yes (1) | No   

Parthi Smith   10 years AGO

6561^2=43046721

Like? Yes (1) | No   

indu   10 years AGO

it is a sqare root of 81 or 81*81=6561

Like? Yes (1) | No   

JYOTIRMAYA MOHAPATRA   10 years AGO

81*81=6561

Like? Yes (1) | No   

KRISHNAN   10 years AGO

Each number is previous numbers square after 1

3^2=9 , 9^2= 81, 81^2= 6561, 6561^2=43046721

So option 4 is correct answer

Like? Yes (1) | No (1)  

Madhvi Soni   10 years AGO

6561

3^2=9
9^2=81
81^2=6561
6561^2=43046721

Ans 81*81

Like? Yes (1) | No (1)  

aviral bhargava   10 years AGO

a series is being formed here of,
3^0,3^1,3^2,3^4,3^?,3^16
Hence the ? mark would be replaced by 8
because the powers are forming again a series of,
0,1,2,4,8,16
every term is the double of its previous term.
Hence the option 4 is correct which is (3^8).

Like? Yes (1) | No (1)  

Balakumaran N.A   10 years AGO

81*81 = 6561.

Like? Yes (1) | No (1)  

susmit   10 years AGO

1*3=3
3*3=9
9*9=81
81*81=6561
6561*6561=43046721
option no. 4

Like? Yes (1) | No (1)  

Jigyasa Kumar   10 years AGO

1. This is because nine multiplied by nine equals 81. 81 multiplied by 81 equals 6561, and 6561 multiplied by 6561 equals 43046721

Like? Yes (1) | No (1)  

achyuth   10 years AGO

in this sequence 1,3,9,81,?,43046721..
81*81=6561

ans;6561
opt(4)

Like? Yes (1) | No (1)  

saisudha sarva   10 years AGO

3^0=1
3^1=3
3^2=9
3^4=81
3^8=6561 & 3^16 =43046721

Like? Yes (1) | No (1)  

gourav saini   10 years AGO

1,3,9,81,?,43046721
sq of 1 is 1
sq of 3 is 9 so we get the next term
sq of 9 is 81 again we get the next term
then sq of 81 is 6561
if we sq the 6561 we get 43046721
so ans 1s 4

Like? Yes (1) | No (1)  

Devendra Marghade   10 years AGO

3^0=1
3^1=3
3^2=9
3^4=81
3^8=6561
3^16=43046721

Like? Yes (1) | No (1)  

Murugan D   10 years AGO

answer is 4
because 81X81=6561
6561X6561=43046721

Like? Yes (1) | No (1)  

Prashasti Sharma   10 years AGO

3^2=9
9^2=81
81^2=6561
6561^2=43046721

Like? Yes (1) | No (1)  

Harshit Daga   10 years AGO

1, 3, 3^2, 9^2..
and next number would be...
81^2 = 6561.

means, option 4 is correct..

Like? Yes (1) | No (1)  

akash   10 years AGO

1^2=1
3^2=9
9^2=81
81^2=6561
6561^2=43046721

Like? Yes (1) | No (1)  

shamma mohammad   10 years AGO

1*3^0=1
1*3^1=3
1*3^2=9
1*3^4=81
1*3^8=6561
1*3^16=43046721

Like? Yes (3) | No (3)  

abhilasha   10 years AGO

1. This is because nine multiplied by nine equals 81. 81 multiplied by 81 equals 6561, and 6561 multiplied by 6561 equals 43046721

Like? Yes (1) | No (2)  

21 Feb 2014 Coded language

If in coded language
ARUN = 62
MOHIT = 75
UDITA = 65
Then
NIRAJ = ?

OPtion

1) 52
2) 53
3) 57
4) 60
5) 62
6) 64
7) 68
8) 72
9) 76
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).
ARUN = CTWP = 3+20+23+16 = 62
MOHIT = OQJKV = 15+17+10+11+22 = 75
UDITA = WFKVC = 23+6+11+22+3 = 65
Similalry
NIRAJ = PKTCL = 16+11+20+3+12 = 62

OR

ARUN = (A + R + U + N) + 2*(number of letters) = 54 + 8 = 62
MOHIT = (M + O + H + I + T) + 2*(number of letters) = 65 + 10 = 75
UDITA = (U + D + I + T + A) + 2*(number of letters)= 55 + 10 = 65
Similary
NIRAJ = (N + I + R + A + J) + 10 = 52 + 10 = 62

option 5)

Correct Option: 5 Best Solution (44)

Akanksha Mishra   10 years AGO

Given;
ARUN = 62
MOHIT = 75
UDITA = 65

Firstly find the position of letters;
A(1),B(2),C(3)...................X(24),Y(25),Z(26)


1] ARUN = A(1) + R(18) + U(21) + N(14)
ARUN = 54
Total no of letters of ARUN = 4
Multiply 2 with the no of letters = 4*2 = 8
ARUN = 54 + 8 = 62


2] MOHIT = M(13) + O(15) + H(8) + I(9) + T(20)
MOHIT = 65
Total no of letters of MOHIT = 5
Multiply 2 with the number of letters = 5*2 =10
SO add 10
MOHIT = 65 + 10 = 75

Similarly SoLve NIRAJ;

4] NIRAJ = N(14) + I(9) + R(18) + A(1) + J(10)
NIRAJ = 52
Total no of letters of NIRAJ = 5
Multiply 2 with the number of letters = 5*2 =10
SO add 10

NIRAJ = 52 + 10 = 62
SO answer will option 5

Like? Yes (53) | No (4)  

Dr.Dipin Singh   10 years AGO

ARUN = 62.....4 letters....1+8+21+14+4*2=62
MOHIT = 75
UDITA = 65
Then
NIRAJ =14+9+18+1+10+5*2=62 as there are 5 letters.

Like? Yes (6) | No (3)  

Snehasis   10 years AGO

ARUN = 1+8+21+14 = 54
Now, ARUN has 4 characters therefore 4*2=8
Then, 54 + 8 = 62
Similarly,
MOHIT = 13+15+8+9+20 = 65
Now, MOHIT has 5 characters therefore 5*2=10
Then, 65 + 10 = 75
Likewise,
UDITA = 21 + 4 + 9 + 20 + 1 = 55
Now, again UDITA has 5 characters therefore 5*2 = 10
Then, 55 + 10 =65

So, for NIRAJ we will have,
14 + 9 + 18 + 1 + 10 = 52
and, also NIRAJ had 5 characters therefore 5*2=10
Therefore the required answer is,
NIRAJ = 52 + 10 = 62

Like? Yes (6) | No (3)  

rajesh   10 years AGO

it is the best solution

Like? Yes (1) | No   

M.Sindhu Priya   10 years AGO

according to allocation of 26 numbers in alphatets.
arun=1+18+21+14=54,here 4 letters so *2 for each leteert will get 8 total 54+8=62;
mohit =13+15+8+9+20=65 and +10 =75;
udita =21+4+9+20+1=55 and add +10=65;
from this niraj=14+9+18+1+10=52 and +10 =62

Like? Yes (1) | No   

khushboo   10 years AGO

if character of name is one then add 2 to sum of character of name
so for 5 character add 10 to total no of character of name
if we take A=1,B=2,C=3......soon.
A+R+U+N=54
M+O+H+I+T=65
then N+I+R+A+J=52
then add 10+52=62

Like? Yes (1) | No   

srinivetha   10 years AGO

arun=1+18+21+4=54+twice the number of letters=54+8=62
mohit=13+158+9+20=65+twice the number of letters=75
udita=21+4+9+20+1=55+twice the number of letters=65
niraj=14+9+18+1+10=52+twice the number of letters=62

Like? Yes (1) | No   

Kaustubh karekar   10 years AGO

the sequence is
n=14, i=9, r=18, a=1,j=10
and no of letters in the name=5
so, 5*2=10
so the sequence is
14+9+18+1+10 =52*10=62

Like? Yes (1) | No   

TRIMURTULU   10 years AGO

Arrange english alphabets with corresponding number system,like
A1,B2,C3,D4,E5,F6,G7,H8,I9,J10,K11,L12,M13,N14,O15,P16,Q17,R18,S19,T20,U21,V22,W23,X24,Y25,Z26
ARUN=(1+18+21+14)+(2*NO.OF LETTERS IN WORD"ARUN")=62
MOHIT=(13+15+08+9+20)+(2*NO.OF LETTERS IN WORD"MOHIT")=75
UDITA=(21+4+9+20+1)+(2*NO.OF LETTERS IN WORD"UDITA")=65
THEN
NIRAJ=(14+9+18+1+10)+(2*NO.OF LETTERS IN WORD"NIRAJ")=62

Like? Yes (1) | No   

sravani   10 years AGO

arun=1+18+21+14=54+(2*4=8)=62 no.of letters in arun=4.
mohit=13+15+8+9+20=65
65+2(5=m,o,h,i,t)=75
uditha=21+4+9+20+1=55
55+2(5)=65so niraj=14+9+18+1+10=52
52+2(5)=62

Like? Yes (2) | No (1)  

K.SASIBENINA   10 years AGO

because option 5 satisfies the condition

Like? Yes (1) | No   

ananthavalli   10 years AGO

N+I+R+A+J=14+9+18+1+10=52+10=62

Like? Yes (1) | No   

Vinay Sipani    10 years AGO

Let A=1 , B=2 , C=3 ... Z=26

ARUN = (A + R + U + N) + 2*(No. Of letters)
= 54 + 8 = 62

MOHIT = (M + O + H + I + T) + 2*(no. Of letters)
= 65 + 10 = 75

UDITA = (U + D + I + T + A) + 2*(No of letters)
= 55 + 10 = 65

=>NIRAJ = (N + I + R + A + J) + 10
= 52 + 10 = 62

Like? Yes (2) | No (2)  

Sajal Garg   10 years AGO

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)
ARUN=(1+18+21+14)+(no.ofalphabet*2)=54+(4*2)=62
MOHIT=(13+15+8+9+20)+(no.ofalphabet*2)=65+(5*2)=75
UDITA=(21+4+9+20+1)+(no.ofalphabet*2)=55+(5*2)=65
NIRAJ=(14+9+18+1+10)+(no.ofalphabet*2)=52+(5*2)=62

Answer= option 5 (62)

Like? Yes (2) | No (2)  

imran   10 years AGO

Ans is 62

In ARUN, A=01; R=18; U=21; N=14;
Total=54. but take 54+(2*4)=62
4 is taken because ARUN has 4 characters. so add (2*4)

Similarly
for MOHIT, 13+15+8+9+20=55
and 55+(2*5)=65
Since MOHIT has 5 characters, add (2*5)

for NIRAJ=14+9+18+1+10=52
then 52+(2*5)=62

Like? Yes (2) | No (2)  

Praveen Kumar Devendran   10 years AGO

The most useful tool is a substitution cipher where each letter of the alphabet is represented by a number which corresponds to that letter's position in the alphabet. In simplest terms, this can be written as A=1, B=2 ... Z=26. Since A is the first letter of the alphabet, it is represented by the number 1. B, the second letter, is represented by 2. Z, the last of the 26 letters in the alphabet, is represented by 26.
A+R+U+N = 1+18+21+14 = 54
total letter in the word is 4 by multipy the letter in 2 and will get 8 and adding the multiplied letter value to the summed value of letter will get 62
and similarly we get all the thing

Like? Yes (1) | No (1)  

Parthi Smith   10 years AGO

A+R+U+N=1+18+21+14=54+(NO.of letters*2)=62
similarly
N+I+R+A+J=62

Like? Yes (1) | No (1)  

jayabharathy   10 years AGO

a=1,b=2,c=3,....z=26
(A+R+U+N)+num of letters *2
(1+21+18+14)=54+(4*2)
=62
so that NIRAJ
(14+9+18+1+10)+(5*2)=62

Like? Yes (1) | No (1)  

karthik hariharan   10 years AGO

n=14
i=9
r=18
a=1
j=10

14+9+18+1+10=52....
niraj is a5 letter word 5*2=10
52+10=62
so 5th option

Like? Yes (2) | No (2)  

arun prakash singh   10 years AGO

niraj= 62

Like? Yes (1) | No (1)  

Devendra Marghade   10 years AGO

If letters A,B,C,....,Z are numbered as 1,2,3,..,26 respectively
Coding=Sum of the letters+2(Number of letters in the word)
ARUN=(1+18+21+14)+2(4)=54+8=62
MOHIT=(13+15+8+9+20)+2(5)=65+10=75
UDITA=(21+4+9+20+1)+2(5)=55+10=65
NIRAJ=(14+9+18+1+10)+2(5)=52+10=62

Like? Yes (2) | No (3)  

Rajasekhar Reddy   10 years AGO

A-1,B-2,...............Z-26
NIRAJ = N+I+R+A+J+(2*NUMBER OF LETTERS) = 62

Like? Yes (1) | No (2)  

nilu kumari   10 years AGO

WE ARE TAKING THE POSITIONS OF THE LETTERS IN THE ALPHABET.
ARUN = 1+18+21+14 + 2*no. of letters in ARUN
= 54+2*4
= 54+8
= 62
MOHIT =13+15+8+9+20 + 2*no. of letters in MOHIT
=65+2*5
=65+10
=75
UDITA =21+4+9+20+1 +2*no. of letters in UDITA
=55+2*5
=65
NOW FOR NIRAJ
NIRAJ =14+9+18+1+10 + 2* no. of letters in NIRAJ
=52 + 2*5
=62

Like? Yes (1) | No (2)  

Priyanka   10 years AGO

ADD POSITION OF ALL THE ALPHABETS+10 GIVES THE ANSWER.SO THE ANSWER FOR NIRAJ IS 52+10=62.

Like? Yes (1) | No (2)  

Lavankumar Reddy   10 years AGO

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.

ARUN = 1+18+21+14 = 54+2*4 (count of Letters) =54+8=62
MOHIT = 13+15+8+9+20 = 65 + 2*5 = 65+10 = 75
UDITA = 21+4+9+20+1 = 55 + 2*5 = 55+10 = 65

NIRAJ = 14+9+18+1+10 = 52 + 2*5 = 52+10 = 62

Like? Yes (2) | No (3)  

Varunraj Reddy   10 years AGO

ARUN=1+18+21+14=54
54+(4*2)=62

MOHIT=13+15+8+9+20=65
65+(5*2)=75
..........

Similarly,
NIRAJ=14+9+18+1+10=52
52+(2*5)=62 option 5)

Like? Yes (1) | No (2)  

K Venkata Subba Rao   10 years AGO

ARUN=54+(4*2) /*4 CHAR'S*/==> 62
MOHIT=65+5*2=75
UDITA=55+5*2=65
NIRAJ=52+5*2=62

Like? Yes (1) | No (2)  

SONU GUPTA   10 years AGO

ARUN=(1+18+21+14)=54+4*2=62;
MOHIT=(13+15+8+9+20)=65+5*2=75
then
NIRAJ=(14+9+18+1+10)=52+5*2=62

Like? Yes (1) | No (2)  

khotha sireesha   10 years AGO

For ARUN = 1+18+21+14 = 54+(letters are four so 4*2=62).
MOHIT = 13+15+8+9+20 = 65+(5letters*2=10)=75
UDITA= 21+4+9+20 = 55+(5letter*2=10)=65
so ANSWER IS
NIRAJ = 14+9+18+1+10=52+(5letters*2=10)=62

Like? Yes (1) | No (2)  

achyuth   10 years AGO

ARUN=1+18+21+14=54+8=62.
.....
....
NIRAJ=14+9+18+1+10=52+10(LETTERS*2)=62
ANS=62
OPT=5

Like? Yes (1) | No (2)  

SID PRASANTH KUMAR   10 years AGO

mohit=13+15+8+9+20=65+10=75
14+9+18+1+10=52+10=62

Like? Yes (1) | No (2)  

Murugan D   10 years AGO

answer is 4
Arun=54+8=62
mohit=65+10=75
udita=55+10=65
niraj=52+10=62
so the answer is 62(option :5)

Like? Yes (1) | No (2)  

Sudipta Mukherjee   10 years AGO

position wise addition + (5*2)=52+10=62

Like? Yes (1) | No (2)  

neeraj pandey   10 years AGO

sum of al letter and +10

Like? Yes (1) | No (2)  

soundar   10 years AGO

sum all alphabet position's value... then add 8 to 4 digit name.... if the name is 5 digit add 10...

Like? Yes (1) | No (2)  

Chithra   10 years AGO

NIRAJ=14+9+18+1+10+10=62

Like? Yes (1) | No (2)  

Sapan Suman   10 years AGO

sum of all alphabet position value +10

Like? Yes (2) | No (3)  

ANUP KUMAR   10 years AGO

ARUN=1+18+21+14=54+2*(NO. OF LETTERS)=54+2*4=62
MOHIT=13+15+8+9+20=65+2*5=75
UDITA=21+4+9+20+1=55+2*5=65
NIRAJ=14+9+18+1+10=52+2*5=62

Like? Yes (2) | No (3)  

Komala.G   10 years AGO

Option 5
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)
For every letter in a word add 2
A(1)+R(18)+U(21)+N(14) = 54+8 = 62 ARUN
M(13)+O(15)+H(8)+I(9)+T(20) = 65+10 = 75 MOHIT
U(21)+D(4)+I(9)+T(20)+A(1) = 55+10 = 65 UDITA
Similarly
N(14)+I(9)+R(18)+A(1)+J(10) = 52+10 = 62 NIRAJ

Like? Yes (1) | No (2)  

priyanka   10 years AGO

13+9+17+1+10=62

Like? Yes (1) | No (2)  

MD SOHAIL AKHTER   10 years AGO

n=14,i=9,r=18,a=1,j=10
niraj=14+9+18+1+10+2*no.of letters (i.e 5)= 62

Like? Yes (1) | No (2)  

Inderjeet Singh   10 years AGO

i used a as x and b as x+1 and so on....
and using arun =65
i found out x=3
then i put x=3
x + x+17 +x+20 +x+13 = NIRAJ which comes out 62

Like? Yes (1) | No (3)  

sivaji ragala   10 years AGO

ARUN=1+18+21+14=54+8(IT IS FOUR LETTER WORD SO ADD 8 i.e(4*2))
MOHIT=13+15+8+9+20=65+10(IT IS five LETTER WORD SO ADD 10 i.e(5*2))
UDITA=21+4+9+20+1=55+10(IT IS five LETTER WORD SO ADD 10 i.e(5*2))
NIRAJ=14+9+18+1+10=52+10=62(IT IS five LETTER WORD SO ADD 10 i.e(5*2))

Like? Yes (1) | No (3)  

sowmya   10 years AGO

ans is 62

Like? Yes (1) | No (3)  

20 Feb 2014 Solve

Seven boys are arranged in a row in such a way that
- Ravi is third from Kapil.
- Lakhan and Shyam are together.
- The modulus of difference between the position of Sumit and Ram is 5.
- Geometrical mean of position of Lakhan and Ram is 2.
- Mohan is nearer to Ravi than Kapil.
Arrange all the boys in a correct sequence such that they follow the above written conditions.

OPtion

1) Ram, Mohan, Shyam, Lakhan, Ravi, Sumit, Kapil
2) Ram, Mohan, Shyam, Lakhan, Kapil, Sumit, Ravi
3) Ram, Mohan, Lakhan, Shyam, Ravi, Sumit, Kapil
4) Mohan, Ram, Shyam, Lakhan, Ravi, Sumit, Kapil
5) Lakhan, Mohan, Shyam, Ram, Ravi, Sumit, Kapil
6) Lakhan, Mohan, Shyam, Ram, Kapil, Sumit, Ravi
7) Shyam, Lakhan, Ravi, Sumit, Ram, Kapil, Mohan
8) Ram, Ravi, Sumit, Kapil, Mohan, Shyam, Lakhan
9) Sumit, Mohan, Shyam, Lakhan, Ravi, Ram, Kapil
10)None of these

Solution

Geometrical mean of position of Lakhan and Ram is 2
=> Their positions are necessarily 1st and 4th
Since the modulus of difference between the position of Sumit and Ram is 5
=> Ram is necessarily 1 (if Ram would be at 4th position then it would be difficult to place Sumit)
It is very clear now to say that Sumit is 6th and Lakhan is 4th
Lakhan and Shyam are together
=> Shyam is either 3rd or 5th

If Shyam would be at 5th position then the statement "Ravi is third from Kapil." would be impossible. Therefore it should be 3rd
Ravi and Kapil will take 5th and 7th place
Mohan is at 2nd finally
Since Mohan is nearer to Ravi than Kapil
Therefore Ravi will be 5th.

Correct Option: 1 Best Solution (12)

Snehasis   10 years AGO

Answer is 1st option.

Its better to go through options:

As Kapil is last in the sequence and if we count in left side from Kapil then Ravi is 3rd from Kapil.

Lakhan is in 4th position and Shayam in 3rd so they are together.

Ram\\\'s position is 1st and Sumit\\\'s position is 6th, therefore difference is -5(i.e. 1-6) and its modular difference is 5(|-5| = 5)

Geometric mean can be find as, GM = root of Lakhan and Ram position.
Therefore, GM = root of 4*1 = 2.

Mohan is nearer to Ravi than kapil as Mohan is at 2nd position and Ravi is in 5th and Kapil is in 7th position therefore 2 is nearer to 5 than 7.

therefore all the condition is proved and the correct choice is Option 1.

Like? Yes (11) | No   

KRISHNAN   10 years AGO

Solving by the options given

Like? Yes (2) | No   

R.RAVIKUMAR   10 years AGO

The option 1 follows the given instructions...
- Ravi is third from Kapil.
- Lakhan and Shyam are together.
- The modulus of difference between the position of Sumit and Ram is 5.
- Geometrical mean of position of Lakhan and Ram is 2.
- Mohan is nearer to Ravi than Kapil.

Like? Yes (1) | No   

sravani   10 years AGO

only option 1 satisfys all the conditions given in the problem.

Like? Yes (1) | No   

nilu kumari   10 years AGO

because option 1 satisfies all the conditions

Like? Yes (1) | No   

shilpi   10 years AGO

2 is not satisfying 5th condition
3 is not satisfying 4th condition
4 is not satisfying 3rd condition
5 is not satisfying 2nd condition
6 is not satisfying 2nd condition
7 is not satisfying 5th condition
8 is not satisfying 5th condition
9 is not satisfying 4th condition


only option 1 is satisfying all the conditions so the answer is 1

Like? Yes (1) | No   

Mayuresh Roy   10 years AGO

Because it satisfies all the conditions.

Like? Yes (1) | No   

imran   10 years AGO

Here Ravi is third from kapil
Shyam and Lakhan are together
Position of Sumit=6, Ram=1... So Difference=5
GM of position of Lakhan and Ram::
Position of lakhan=4
Position of Ram=1
GM= (4*1)^(1/2)=2
Mohan is nearer to ravi than kapil

Like? Yes (1) | No   

santhosh   10 years AGO

all conditions in 1

Like? Yes (1) | No   

vinod   10 years AGO

First of all given that gm of Lakshman and ravi are 2 so from that ravi is at pos-1 and Laksman at pos-2(sqrt(4*1)=2)) and from 4th point the sumit must be in 6 th place.

Like? Yes (1) | No   

shruti srivastava   10 years AGO

solve through options...

Like? Yes (1) | No   

Leelanand Kumar Yalla   10 years AGO

According to the options it follows 1st option....

Like? Yes (1) | No   

19 Feb 2014 Solve

In a room of dimensions l*b*h = 16*14*12, bricks of dimensions 6*4*2 are placed in such a way that the room is occupied completely. Total number of bricks in a layer on the floor is :-

OPtion

1) 12
2) 56
3) 24
4) 8
5) 48
6) 82
7) 14
8) 28
9) 128
10)None of these

Solution

On the basis of dimensions of brick it is necessary to place the brick in the room in such a way that
l = 4
b = 2
h = 6
( If we place the brick in any other manner then it is not possible to fill the room completely)
Now area of floor = 16*14 = 224
And area of brick = 4*2 = 8
Hence Number of bricks = 224/8 = 28

Option 8)

Correct Option: 8 Best Solution (4)

Devendra Marghade   10 years AGO

Volume of room=16*14*12 unit^3
Total bricks occupied=(16*14*12)/(6*4*2)=56
Height of 12 units can be covered with either brick height=2,4 or 6 units.
For the height of 12 unit,if brick is placed with height of 2 unit,then 6 layers will be there.But division 56/6 is not possible.
If brick is placed with height of 4 units, then 3 layers will be there,but division 56/3 is not possible.
If brick is placed with height of 6 unit,it can form 2 layers, so number of bricks on the floor=56/2=28

Like? Yes (9) | No   

KRISHNAN   10 years AGO

Volume of the room is l*b*h=16*14*12 = 2688
Volume of the brick is 6*4*2=48
No of bricks required is 2688/48 = 56
Height of the brick is 2 therefore the number of brick in a layer is 56/2 or 28 so option 8 is correct

Like? Yes (8) | No (3)  

shruti srivastava   10 years AGO

number of bricks=(16*14)/(4*2)=28
Bricks should be arranged on the floor in such a way that length of brick becomes 4 and width of brick becomes 2.

Like? Yes (1) | No   

viveka   10 years AGO

in 16-14-12 place bricks verticalli as 4-2-6
it forms two layers ie is 12/6
it require 4 bricks along length 16 on 16/4
it require 7 bricks along bredth 14/2

so 7*4=28 bricks in layer 1 ie bottom layer

Like? Yes (1) | No   

18 Feb 2014 Quadratic equation

Solve {x/(x+1)}^2 - 5{x/(x+1)} + 6 = 0

OPtion

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

Solution

Solve
x = -2, x = -3/2

Correct Option: 9 Best Solution (22)

Dr.Dipin Singh   10 years AGO

{x/(x+1)}^2 - 5{x/(x+1)} + 6 = 0
let
x/(x+1)=y;
then
y^2-5y+6=0
y=2,3
Then
x/(x+1)=y =2; x= -2
x/(x+1)=y=3; x=-3/2

Like? Yes (37) | No (5)  

aviral bhargava   10 years AGO

let (x/x+1)=a
we get the equation as a^2-5a+6=0
=> a^2-3a-2a+6=0
=> (a-3)(a-2)=0
On solving the above given equation we get the values of a as
a=3 or 2
Since a= (x/x+1)
Now finding the values of x
Firstly,

x/x+1=2
From above
We get x=-2

Keeping (x/x+1)=3
We get x=-3/2
Hence the answer is -2,-3/2

Like? Yes (12) | No (5)  

kiruthika   10 years AGO

The Quadratic equation is
{x/(x+1)}^2 - 5{x/(x+1)} + 6 = 0

{x/(x+1)}^2 - 2{x/(x+1)} - 3{x/(x+1)} + 6 = 0
Taking {x/(x+1)} outside we get,

{x/(x+1)}[{x/(x+1)}- 2]- 3[{x/(x+1)}-2] = 0

[{x/(x+1)}-3][{x/(x+1)}-2] = 0,then we get

==> [{x/(x+1)}-3] = 0
{x/(x+1)} = 3
x = 3x + 3
-2x = 3
x = -3 /2, similarly the other one

==> [{x/(x+1)}-2] = 0
{x/(x+1)} = 2
x = 2x + 2
-x = 2
x = -2
So the answer for the quadratic equation is
option 9 ==> x =-3/2 and x=-2

Like? Yes (9) | No (2)  

Varunraj Reddy   10 years AGO

Given equation
{x/(x+1)}^2-5{x/(x+1)}+6=0
by simplifying we get
2x^2+7x+6=0
by solving we get values of x=-2,-3/2 option 9)

Like? Yes (8) | No (4)  

Saidinesh   10 years AGO

X^2/(X+1)^2-5(x)/(X+1)+6=0;
x^2-5X(x+1)+6(X+1)^2=0;
x^2-5x^2-5x+6x^2+6+12x=0;
2x^2+7x+6=0;
SOLVING THIS
2X(X+2)+3(X+2)=0;
THEN
X=-2;
X=-3/2;

Like? Yes (9) | No (5)  

Lavankumar Reddy   10 years AGO

{x/(x+1)}^2 - 5{x/(x+1)} + 6 = 0

Let a=x/x+1

a^2-5a+6=0

a^2-3a-2a+6=0

(a-2)(a-3)=0

then subsitute a value

{(x/(x+1))-2}{(x/(x+1)-3}=0

{x-3x-3/(x+1)}{-x-2x-2/x+1}=0

{-2x-3/(x+1)}{-x-2/(x+1)}=0

(-2x-3)(-x-2)/(x+1)^2=0

(-2x-3)(-x-2)=0

-2x-3=0 and -x-2=0

-2x=3 and -x=2

x=-3/2 and x=-2

Like? Yes (6) | No (3)  

Murali Konduru   10 years AGO

take ((x/x+1)) as y i the above equation so the equation becomes y^2-5y+6=0 so by solving the equation we get y=3,2 and equating y=(x/(x+1))
3=x/x+1,2=x/x+1
we get x=-2,x=-3/2 so option 9 is the correct answer

Like? Yes (3) | No (2)  

Grishma Shah   10 years AGO

here take x/x+1=y
thus putting value in equation here y^2-5y+6=0
here solving,we get y=3,y=2
thus x/x+1=3 , x/x+1=2
after solving both equations,we get x=-3/2 and x=-2
so option 9 is correct.

Like? Yes (6) | No (5)  

Devendra Marghade   10 years AGO

{x/(x+1)}^2 - 5{x/(x+1)} + 6 = 0
2x^2+7x+6=0
(x+2)(2x+3)=0
x=-2, -3/2

Like? Yes (4) | No (3)  

ASHISH KUMAR MISHRA   10 years AGO

let Y= x/(x+1); now y^2-5y+6=0;
Now y=2,3
x/(x+1)=2 => x=-2 and x/(x+1)=3 => x=-3/2;
so ans is -2,-3/2

Like? Yes (3) | No (2)  

Anantharaman   10 years AGO

lets assume y= (x/x+1)

so eqn => y^2-5y+6=0

Solving we will get y=3,2

Now substituting (x/x+1)=y

1) (x/x+1)=3
x=3x+3
2x=-3
x=-3/2

2) x/x+1=2
x=2x+2
x=-2

So solution of x : -2, -3/2
Option 9

Like? Yes (4) | No (3)  

sravani   10 years AGO

x/x+1=y
y^2-5y-6=0
y=2,3
and substitute these values and we get -2,-3/2

Like? Yes (3) | No (2)  

Mahalakshmi   10 years AGO

{(x/(x+1))-3}{(x/(x+1))-2}=0 equation we will get
solving this get x=-2,-3/2

Like? Yes (1) | No (1)  

Vaishali Dhiman   10 years AGO

ans is option 9
{x/(x+1)}^2 - 5{x/(x+1)} + 6 = 0

taking LCM
{x^2-5(x+1)+6(x+1)^2}/(x+1)^2=0
{x^2-5(x+1)+6(x+1)^2}=0
2x^2+7x+6=0
2x^2+3x+4x+6+0
x(2x+3)+2(2x+3)=0
(x+2)(2x+3)=0
x=-2, -3/2

Like? Yes (1) | No (1)  

Srinath   10 years AGO

SOLVE X^2-5X(X+1)+6(X^2+2X+1)=0
2X^2+7X+6=0
X=-7+(49-48)/4
THEN WE GET X=-1.5, -2

Like? Yes (3) | No (3)  

dhyanendra mani tripathi   10 years AGO

let (x/x+1)=y
we get y^2-5y+6=0
solving we get y= 2,3
=> (x/x+1)=y= 2 and 3
hence we get
x/x+1=2 => x= -2
x/x+1=3 => x= -3/2

Like? Yes (2) | No (2)  

ravinder reddy   10 years AGO

take x/(x+1)=y
then it becomes
y^2-5y+6=0
by factorization
y=2,3
now take y=2
substitute y value
x/(x+1)=2
x=-1 , similarly with y=3 then x=-3/2

Like? Yes (2) | No (2)  

DHAKAD NEETENDRA SINGH   10 years AGO

let x/(x+1)=t

t^2-5t+6=0
(t-3)(t-2)=0
t=3,2

x/(x+1)=3,x/(x+1)=2
x=-2,-3/2

Like? Yes (2) | No (2)  

Suman   10 years AGO

2x^2+7x+6=0
so x=-2,-3/2

Like? Yes (2) | No (3)  

viveka   10 years AGO

let y be x/x+1
y2-5y+6=0

y=3
x/x+1=3
x=-3/2

y=2
x/x+1=2
x=-2

Like? Yes (1) | No (2)  

sivaraman   10 years AGO

x/x+1=3
or
x/x+1=2 x=2x+2
x=3x+3 x=-2
x=-3/2

Like? Yes (1) | No (2)  

Sajal Garg   10 years AGO

9) -2, -3\2
a) (-2/-1)^2-5(-2/-1)+6=4-10+6=0
b) [(-3/2)/{(-3/2)+1)}]^2 -5[(-3/2)/{(-3/2)+1)}]+6 = {(-3/2)/(-1/2))}^2-5{(-3/2)/(-1/2))}+6=3^2-5(3)+6=9-15+6=0

Like? Yes (1) | No (3)  

17 Feb 2014 Sequence

What is the next number of the following sequence
31,28,31,30,31,30,31,31,.....

OPtion

1) 28
2) 30
3) 31
4) 38
5) 42
6) 48
7) 50
8) 51
9) 52
10)None of these

Solution

This is series for no of day in a month starting from January itself.
Jan - 31
Feb - 28
Mar - 31
Apr - 30
May - 31
Jun - 30
Jul - 31
Aug - 31
then
Sep - 30
option 2)

Correct Option: 2 Best Solution (44)

Akanksha Mishra   10 years AGO

Given sequence
31,28,31,30,31,30,31,31,.....
It's a sequence of number of days in a month of a year
January - 31 days
February - 28 days
March - 31 days
April - 30 days
May - 31 days
June - 31 days
July - 30 days
August - 31 days
similarly ;
september- 30 days

so answer will be 30 (option 2)

Like? Yes (35) | No (5)  

Dr.Dipin Singh   10 years AGO

31,28,31,30,31,30,31,31,.....
No of days in January --31
no of days in Feb ( Non Leap year)= 28
no of days in March = 31
..
..
no of days in September =30 .. option 2

Like? Yes (5) | No (3)  

dharmaveer   10 years AGO

number of days in a month... starting from january, feb, march, ...

Like? Yes (3) | No (1)  

Venkatesh KK   10 years AGO

The sequence follows number of days in month.

i.e, 31(Jan)
28(Feb)
31(Mar)
30(Apr)
31(May)
30(Jun)
31(Jul)
31(Aug)
30(Sep)

Hence answer is Option 2.

Like? Yes (4) | No (3)  

Kaustubh karekar   10 years AGO

lets take first 3 no.
31+28+31=90
then, further 3 no
30+31+30=91
so,
31+31+?=92
so, required no is 30

Like? Yes (1) | No   

Saidinesh   10 years AGO

It show's No of days in the month Sequeance
Jan -31;
Feb -28;
Mar -31;
Apr -30;
May -31;
Jun -30;
July-31;
Aug -31;
Sep -30(that's the answer)
* Consider Non-Leap Year

Like? Yes (2) | No (1)  

Rituparna Chakraborty   10 years AGO

First term:31,January has 31 days.2nd term:28,February has 28 days if the year is not leap year.3rd term:31,March has 31 days.4th term:30,April has 30 days.5th term:31,May has 31 days.6th term:30,June has 30 days.7th term:31,July has 31 days.8th term:31,August has 31 days.So 9th term:September,it has 30days.So ans is 30.

Like? Yes (1) | No   

SADINENI RAGHAVENDRA BABU   10 years AGO

No.of days in mounths in a year.

Like? Yes (1) | No   

Dinesh thatti   10 years AGO

they are as follows
JANUARY-31 days
February-28 days
MARCH- 31 days
APRIL- 30 days
MAy- 31 days
JUNE- 30 days
JULY- 31 days
AUGUST- 31 days
SEPTEMBER 30 days
so answer is september 30

Like? Yes (2) | No (1)  

Sajal Garg   10 years AGO

Jan =31 days
feb = 28 days ....
sep = 30 days

Like? Yes (1) | No   

ch rajesh   10 years AGO

january =31
feb=28
----------
sep 30

Like? Yes (2) | No (1)  

rajesh   10 years AGO

january=31



sep=30

Like? Yes (1) | No   

rajya lakshmi   10 years AGO

descending then next ascending order

Like? Yes (1) | No   

Harshit Daga   10 years AGO

First three numbers of the series(31+28+31)=90
Next three numbers (30+31+30)=91
Then sum of next three numbers will be 92
And we have 2 numbers available 31, 31
So, the next number will be 30..

Which means option 2 is correct..

Like? Yes (2) | No (1)  

Aakash Jain   10 years AGO

31 = Days in January
28 = Days in February
31 = Days in March
30 = Days in April
31 = Days in May
30 = Days in June
31 = Days in July
31 = Days in August
30 = Days in September
So, 30 is Answer....

Like? Yes (1) | No   

Jigyasa Kumar   10 years AGO

Answer = 30
The days in a month is the sequence followed.
So starting from jan, feb, mar, april, may, jun,jul, aug, next comes sept which has 30 days

Like? Yes (2) | No (2)  

abhilasha   10 years AGO

Answer is 30
They are the days in a month so from jan, feb, mar, april, may, jun, jul, aug, next comes sept which has 30 days..

Like? Yes (1) | No (1)  

Aarush Singh   10 years AGO

these are the days in a month so from jan,feb, mar, april, may, jun,jul, aug, next comes sept which has 30 days

Like? Yes (1) | No (1)  

Suyog   10 years AGO

Jan- 31
Feb- 28
Mar- 31
Apr- 30
May- 31
June- 30
July- 31
Aug- 31
Sep- 30
So the next no in sequence is 30

Like? Yes (2) | No (2)  

shivam tiwari   10 years AGO

jan 31 days
feb 28 days
mrh 31 days
apr 30 days
may 31 days
jun 30 days
july 31 days
augest 31 days
so next number sept 30 days

Like? Yes (1) | No (1)  

Srinath   10 years AGO

January has 31 days.
February has 28 days.
March has 31 days.
April has 30 days.
like this, for september 30 days.

Like? Yes (1) | No (1)  

Devendra Marghade   10 years AGO

The sequence is of days of months of January,February,March,April,May,June,July,Aug
So next month is September=30

Like? Yes (1) | No (1)  

Rajasekhar Reddy   10 years AGO

the sequence is no.of days of each month from january on wards

Like? Yes (1) | No (1)  

Murali Konduru   10 years AGO

The above number sequence are the sequence of number of days in each month and the option corresponds to September month and has 30 days.
So answer is option 2

Like? Yes (2) | No (2)  

Gokulnath   10 years AGO

Jan-30,Feb28,March31,april30,may31,june30,july31,aug31,......Sep=30

Like? Yes (1) | No (1)  

KRISHNAN   10 years AGO

Days of the month from Jan to Dec
jan 31
feb 28
Mar 31
Apr 30
May 31
June 30
July 31
Aug 31
Sept 30 so option 2 is correct

Like? Yes (1) | No (1)  

Mrinal Kishor   10 years AGO

sequence of months
like, jan=31 days
feb=28 days
march=31 days
april=30 days
may=31 days
june=30 days
july=31 days
aujust=31 days
so, september=30 days

Like? Yes (1) | No (1)  

rajani   10 years AGO

sum of 3 numbers in sequence 90,91,.. so the last no is 30

Like? Yes (1) | No (1)  

ASHISH KUMAR MISHRA   10 years AGO

jan=31 days; feb=28days ; march=31days ; april= 30days ; may =31 days ; jun=30days ; july=31days ;aug =31 days; sept=30days; oct=31days;
Implies series of 31,28,31,30,31,30,31,31,30,31...
so Answer is 30

Like? Yes (1) | No (1)  

Rajat Goyal   10 years AGO

These are the no. of days in the month.

Like? Yes (1) | No (1)  

jyoti   10 years AGO

september

Like? Yes (1) | No (1)  

yonesh   10 years AGO

it is month days followed by..
jan feb mar apr may jun jul aug

so prob next sep = 30 days

ans : 30

Like? Yes (1) | No (1)  

OM PRAKASH JHA   10 years AGO

jan 31 days
feb 28 days
mrh 31 days
apr 30 days
may 31 days
jun 30 days
july 31 days
augest 31 days
so next number is sept 30 days

Like? Yes (1) | No (1)  

OM PRAKASH JHA   10 years AGO

jan 31 days
feb 28 days
mrh 31 days
apr 30 days
may 31 days
jun 30 days
july 31 days
augest 31 days
so next number is sept 30 days

Like? Yes (1) | No (1)  

ATUL   10 years AGO

jan 31 days
feb 28 days
mrh 31 days
apr 30 days
may 31 days
jun 30 days
july 31 days
augest 31 days
so next number sept 30 days

Like? Yes (1) | No (1)  

IFFAT IRAM   10 years AGO

no. of days in a month
next is 30 for the month of september

Like? Yes (1) | No (1)  

haritha   10 years AGO

by counting no of days of a month in non leap year

Like? Yes (1) | No (1)  

Srinivasa Vamsi Krishna Dodda   10 years AGO

No of days in a Month :
Jan = 31
feb = 28
mar = 31
apr = 30
may = 31
jun = 30
jul = 31
aug = 31
sep = 30 (OUR ANSWER)

Like? Yes (1) | No (1)  

Leelanand Kumar Yalla   10 years AGO

it follows the days of month. So september has 30 days..

Like? Yes (1) | No (1)  

karthik hariharan   10 years AGO

jan-31
feb-28
mar-31
apr-30
may-31
jun-30
jul-31
ang-31

Like? Yes (1) | No (1)  

VIJAYA KUMAR   10 years AGO

Answer is 30..
Its simple because No of Days in the months jan=31
feb=28
march=31
apr=30
may=31
june=30
july=31
aug=31
sep=30

Like? Yes (1) | No (1)  

saranya sankar   10 years AGO

January has 31 days.
February has 28 days.
March has 31 days.
April has 30 days.
May has 31 days.
June has 30 days.
July has 31 days.
August has 31 days.

September has 30 days.

Like? Yes (1) | No (1)  

S.SHAFI   10 years AGO

Given series is number of days in a month..
January - 31
February - 28
March - 31
April - 30
May - 31
June - 30
July - 31
August - 31
September - 30 --->>>> ans) option 2
October - 31
November - 30
December - 31

Like? Yes (1) | No (2)  

aviral bhargava   10 years AGO

these are the number of days in a year which is not a leap year .

Like? Yes (1) | No (2)  

14 Feb 2014 Coded language

If in coded language
AMAN = 33
RAJU = 54
NITU = 68
Then
RENU = ?

OPtion

1) 36
2) 42
3) 48
4) 52
5) 55
6) 58
7) 62
8) 66
9) 70
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)
AMAN = (1+13+1+14)+4 = 33
RAJU = (18+1+10+21)+4 = 54
NITU = (14+9+20+21)+4 = 68
Then
RENU = (18+5+14+21)+4 = 62

Correct Option: 7 Best Solution (76)

Akanksha Mishra   10 years AGO

If in coded language
AMAN = 33
RAJU = 54
NITU = 68

1] AMAN = 33
A(1)+ M(13)+ A(1)+ N(14) = 29
ADD 4 in the total = 29 + 4 = 33

2] RAJU = 54
R(18)+A(1)+J(10)+ U(21) = 50
ADD 4 in the total = 50 + 4
= 54

3] NITU = 68
N(14)+ I(9)+ T(20)+U(21)= 64
ADD 4 in the total= 64 +4
= 68
Similarly
solve RENU = ?
R(18)+ E(5)+ N(14)+ U(21)= 58
ADD 4 in the total= 58 + 4
= 62
So answer will option 7(62)

Like? Yes (44) | No (7)  

Mallikharjuna   10 years AGO

renu = 18+5+14+21=58+4=62

Like? Yes (1) | No   

Vaishali Dhiman   10 years AGO

answer=62

AMAN= 1+13+1+14=29+4=33
RAJU=18+1+10+21=50+4=54
NITU=14+9+20+21=64+4=68
&
RENU=18+5+14+21=58+4=62

Like? Yes (1) | No   

neeraj   10 years AGO

AMAN=1+13+1+14=29+4
RAJU=18+1+10+21=50+4
NITU=14+9+20+21=64+4
RENU=18+5+14+21=58+4=62

Like? Yes (1) | No   

SADINENI RAGHAVENDRA BABU   10 years AGO

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
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

AMAN = 1+13+1+14=29=> 29+4=33
RAJU= 18+1+10+21=50=> 50+4=54
NITU=14+9+20+21=64=> 64+4=68
so
RENU=18+5+14+21=64=> 58+4=62

ANSWER is option 7 (62).

Like? Yes (1) | No   

Neha Jain   10 years AGO

AMAN=1+13+1+14= 29+4=33
RAJU=18+1+10+21= 50+4=54
NITU=14+9+20+21 = 64+4=68
Then
RENU=18+5+14+21 =58+4=62

Like? Yes (1) | No   

jyoti   10 years AGO

as u knoq a to z numbered as 1to26.so a(1)+m(13)+a(1)+n(14)+4=33 ,r(18)+a(1)+j(10)+u(21)+4=54 like wise r(18)+e(5)+n(14)+u(21)+4=62

Like? Yes (1) | No   

lakhan balani   10 years AGO

as a=1,b=2,c=3.......to z=26
now..
a+m+a+n+4=33
r+a+j+u+4=54
n+i+t+u+4=62
so
r+e+n+u+4=62

Like? Yes (1) | No   

nalini singh   10 years AGO

a=1,b=2,c=3,d=4,e=5,f=7........z=26
so
a+m+a+n=29+4=33
r+a+j+u=50+4=54
n+i+t+u=64+4=68
r+e+n+u=58+4=62

Like? Yes (1) | No   

Sruthi Pasumarthy   10 years AGO

In Alphabetical order, R=18,E=5,N=14,U=21. Summation is 58. 58+4=62. Since all the above words are also coded in the same way. Summation+4.
So. Ans=62.

Like? Yes (1) | No   

Gusto Santhosh Myth   10 years AGO

alphabets(a-z) replaced by number (1-26)
A-1, M-13,A-1,N-14=>(1+13+1+14)=29 THEN ADD 4 = (29+4)=33
R-18, A-1, J-10,U-21=>(18+1+10+21)=50 THEN ADD 4=>(50+4)=54
N-14, I-9,T-20,U-21=>(14+9+20+21)=64 THEN ADD 4= 68 SO
R-18, E-5, N-14,U-21=> (18+5+14+21)= 58 THEN ADD 4= 62 ANS: (7)62

Like? Yes (1) | No   

NAGENDRA P   10 years AGO

AMAN=1+13+1+14=29+4=33..
RAJU=18+1+10+21=50+4=54..
NITU=14+9+20+21=64+4=68..
RENU=18+5+14+21=58+4=62..
SO OPTION (7)..

Like? Yes (1) | No   

Rituparna Chakraborty   10 years AGO

Ans is 7)62.
because,AMAN=1+13+1+14=29.[the position of alphabets],Here AMAN=29+4=33
RAJU=18+1+10+21=50,here RAJU=50+4=54
NITU=14+9+20+21=64,here NITU=64+4=68
Then
RENU=18+5+14+21=58,so here RENU=58+4=62.

Like? Yes (1) | No   

G.Madhumitha   10 years AGO

AMAN = 1+13+1+14=29 (29+4=33)
RAJU = 18+1+10+21=50 (50+4=54)
NITYU = 14+9+20+21=64 (64+4=68)
then,
RENU = 18+5+14+21=58 (58+4=62)

So ans is 62

Like? Yes (1) | No   

Aditya Verma   10 years AGO

AMAN=>1+13+1+14=29 add 4 => 29+4 = 33
RAJU=>18+1+10+21=50 add 4 => 50+4 = 54
NITU=>14+9+20+21=64 add 4 => 64+4 = 68
then
RENU=>18+5+14+21=58 add 4 => 58+4 = 62
so answer is option 7

Like? Yes (4) | No (3)  

Varunraj Reddy   10 years AGO

Given that
AMAN=33
ie.,1+13+1+14=29
So 29+4=33

RAJU=54
18+1+10+21=50
So 50+4=54

Similarly,
NITU=64
64+4=68

Then RENU=18+5+14+21=58
So 58+4=62
So Ans is option 7)

Like? Yes (3) | No (3)  

MP    10 years AGO

AMAN = 33 >> 1+13+1+14=29>> 29 +4=33
similarly
RAJU = 54
NITU = 68
Then
RENU = 62>> 18+5+14+21=58>> 58+4=62

Like? Yes (4) | No (4)  

Dinesh thatti   10 years AGO

A M A N = 1+13+1+14 = 29+ add 4 to obtained result as A M A N has 4 alphabets(4)=33
R A J U = 18+1+10+21 = 50+4=54
N I T U = 14+9+20+21 = 64+4=68
R E N U = 1+5+14+21 = 58 + 4 =62

so option is 7) 62

Like? Yes (2) | No (2)  

Komala.G   10 years AGO

7)62 Ans
Consider A = 1, B = 2, C = 3...............Z = 26.
AMAN 1+13+1+14 = 29+4 =33
RAJU 18+1+10+21 = 50+4 = 54
NITU 14+9+20+21 = 64+4 = 68
Similarly,
RENU 18+5+14+21 = 58+4 = 62

Like? Yes (2) | No (2)  

Srinivasa Vamsi Krishna Dodda   10 years AGO

AMAN (Conver to Numbers and ADD)=29(+4) =33
RAJU (Conver to Numbers and ADD)=50(+4) =54
NITU (Conver to Numbers and ADD)=64(+4) =68
RENU (Conver to Numbers and ADD)=58(+4) =62

Like? Yes (2) | No (2)  

Sukanya   10 years AGO

A + M+ A+ N=1+13+1+14=29
29+4=33

R+A+J+U= 18+1+10+21=50
50+4=54

SIMILARLY
R+E+N+U=18+5+14+21=58
58+4=62.

Like? Yes (2) | No (3)  

Avishek Pattanaik   10 years AGO

A+M+A+N=1+13+1+14=29>>>>>29+4=33
R+A+J+U=18+1+10+21=50>>>>50+4=54
N+I+T+U=14+9+20+21=64>>>>64+4=68
R+E+N+U=18+5+14+21=58>>>>58+4=62

So, option 7 is correct

Like? Yes (2) | No (3)  

Venkatesh KK   10 years AGO

Addition of all the numbers corresponding to the letters and 4...
Ie.,

AMAN = 1+13+1+14 = 29 => 29+4 = 33
RAJU = 18+1+10+21 = 50 => 50+4 = 54

Similarly,

RENU = 18+5+14+21 = 28 => 58+4 = 62 (Option 7)

Like? Yes (2) | No (3)  

kiruthika   10 years AGO

from the alphabets position
AMAN= A+M+A+N = 1+13+1+14 = 29+4 = 33
RAJU= R+A+J+U = 18+1+10+21 = 50+4 = 54
NITU= N+I+T+U = 14+9+20+21 = 64+4 = 68

RENU= R+E+N+U = 18+5+14+21 = 58+4 = 62

Like? Yes (1) | No (2)  

santhosh   10 years AGO

18+5+14+21=58+4=62

Like? Yes (1) | No (2)  

madanpal vishwakarma    10 years AGO

the pattern
AMAN
(1+13+1+14)+4=33

RAJU
(18+1+10+21)+4=54
NITU
(14+9+20+21)+4=68
so
NITU
(18+5+14+21)+4=62
so option 7 is correct

Like? Yes (5) | No (6)  

Sourav Das   10 years AGO

aman = (1+13+1+14)+4=33
raju=(18+1+10+21)+4=54
nitu=(14+9+20+21)+4=68
so,
renu=(18+5+14+21)+4=62 (option 7)

Like? Yes (1) | No (2)  

Manikumar R   10 years AGO

renu=18+5+14+21+4=70

Like? Yes (1) | No (2)  

Paras   10 years AGO

Aman=1+13+1+14+4=33
same added crypt each in alphabet and add 4
18+5+14+21+4=62

Like? Yes (1) | No (2)  

Suyog   10 years AGO

AMAN= 1+13+1+14=29+4=33
RAJU= 18+1+10+21=50+4=54
NITU= 14+9+20+21=64+4=68
RENU= 18+5+14+21=58+4=62
So the ans is option 7

Like? Yes (1) | No (2)  

julee verma   10 years AGO

A+M+A+N=29+4=33
R+A+J+U=50+4=54
N+I+T+U=64+4=68
R+E+N+U=58+4=62

Like? Yes (1) | No (2)  

Younus   10 years AGO

R=18+E=5+N=14+U=21 + 4 = 62

Like? Yes (1) | No (2)  

Grishma Shah   10 years AGO

here,
for AMAN (a=1 ,m=13 ,n=14)
so 1+13+1+14=29
for RAJU(r=18,a=1,j=10)
so 18+1+10+21=50
here after adding 4 we get answer.

thus for RENU (R=18,E=5,N=14,U=21)
so 18+5+14+21=58

thus after adding 4 we get 62.
so option 7 is correct.

Like? Yes (1) | No (2)  

Hanumanth Naik   10 years AGO

AMAN
A=1
M=13
A=1
N=14
SUM OF THESE FOUR NUMBER 1+13+1+14=29
AND THAN ADD EXTRA 4 29+4=33

RAJU
R= 18,A=1,J=10,U=21
SUM=18+1+10+21=50,50+4=54

NITU

N=14,I=9,T=20,U=21
SUM 14+9+20+21=64;64+4=68
THEN

RENU R=18,E=5,N=14,U=21
SUM 18+5+14+21=58;58+4=62

ANS=62

Like? Yes (1) | No (2)  

sivalakshmi   10 years AGO

here we have 2 consider the value of before alphabet like for "A" we have 2 consider value of "B" so aman=2+14+2+15=33
like wise
renu=19+6+15+22=62

Like? Yes (1) | No (2)  

j.b.manasa   10 years AGO

By giving numberings from 1-26 to a-z respectively then
AMAN=1+13+1+14=29+4=33
RAJU=18+1+10+21=50+4=54
NITU=14+9+20+21=64+4=68
similarly for
RENU=18+5+14+21=58+4=62

Like? Yes (1) | No (2)  

Lavankumar Reddy   10 years AGO

A=1, B=2, C=3, D=4, to Z=26
AMAN = A+M+A+N = 1+13+1+14 = 29 + 4 = 33
RAJU = R+A+J+U = 18+1+10+21 = 50 + 4 = 54
NITU = N+I+T+U = 18+5+14+21 = 64 + 4 = 68
Similarly
RENU = R+E+N+U = 18+5+14+21 = 58 + 4 = 62

Like? Yes (1) | No (2)  

aviral bhargava   10 years AGO

in every word we add the positional values and then add 4 afterwards.in AMAN= 1+13+1+14 + 4=33

NITU= 14+9+20+21 + 4=68

Similarly,
RENU= 18+5+14+21 + 4=62
Answer=62

Like? Yes (1) | No (2)  

suresh   10 years AGO

AMAN=29
RAJU=50
NITU=64
RENU=58
ADD 4 WITH D RESULTS OF EACH...U WILL GET IT....33,54,68,62

Like? Yes (1) | No (2)  

Kaleem Pasha   10 years AGO

Assign numbers 1-26 to alphabets A-Z,
Use those values below & add extra 4 to it.
AMAN=34--->(1+13+1+14)+4 =33
RAJU=54--->(18+1+10+21)+4 =54
NITU=68--->(14+9+20+21)+4 =68
RENU=62--->(18+5+1+21)+4 =62

Like? Yes (1) | No (2)  

Krishna Chaitanya   10 years AGO

Given AMAN = 33
RAJU = 54
NITU = 68
Then
RENU = ?

AMAN=A+M+A+N=29 but given AMAN=33 therefore 4 should be added to result to get in that coded language.
Similarly RAJU=50+4=54,NITU=64+4=68
Therefore RENU=R+E+N+U=18+5+14+21=58+4=62

Like? Yes (1) | No (2)  

KRISHNAN   10 years AGO

THE CODE IS SUM OF THE ALPHABETS PLUS 4
Like AMAN= 1+13+1+14=29+4=33
RAJU= 18+1+10+21=50+4=54
NITU= 14+9+21+21=64+4=68
SO RENU= 18+5+14+21=58+4=62 So option 7 is the correct answer

Like? Yes (1) | No (2)  

aditi bose   10 years AGO

add the position of the letters in the word ans then add 4
R = 18
E = 5
N = 14
U = 21
18+5+14+21= 58
58+4 = 62

Like? Yes (1) | No (2)  

rajani   10 years AGO

aman---1+13+1+14
raju---18+1+10+21=29+4
nitu---14+9+20+21=50+4
renu---18+5+14+21=58+4

for every solution one more 4 is added so 58+4 =62 for renu

Like? Yes (1) | No (2)  

T G   10 years AGO

aman=1+13+14+1=29 29+4=33
raju=18+1+10+21=50 50+4=54
nitu=14+9+20+21=64 64+4=68
renu=18+5+14+21=58 58+4=62
so ans is 62

Like? Yes (2) | No (3)  

saikrishna   10 years AGO

R+E+N+U=14+9+20+21=58+4=62

Like? Yes (1) | No (2)  

S Prakash   10 years AGO

A=1,B=2............Z=26

AMAN=1+13+1+14=29
=29+4
=33
RAJU=18+1+10+21=50
=50+4
=54
NITU=14+9+20+21=64
=64+4
=68
RENU=18+5+14+21=58
=58+4
=62

Like? Yes (3) | No (4)  

mahesh   10 years AGO

sum of the alphabet positions and addtn of 4

Like? Yes (1) | No (2)  

ravindra   10 years AGO

RENU=18+5+14+21=58
ANSWER+4=62

Like? Yes (1) | No (2)  

Sudipta Mukherjee   10 years AGO

according to the Positon A+M+A+N+4=33
R+A+J+U+4=54
N+I+T+U+4=68
SO, R+E+N+U+4=62

Like? Yes (1) | No (2)  

Suman   10 years AGO

sum total of letter position + 4

Like? Yes (1) | No (2)  

Himanshu Bisht   10 years AGO

following are d reason...
lets A =1 b=2 c=3......z=26;
AMAN = 1+13+1+14=29+4 = 33;
RAJU= 18+1+10+21= 50+4 = 54;
.
.
.
So RENU = 18+5+14+21 = 58+4 = 62;
so option 7 is correct

Like? Yes (1) | No (2)  

Arpit   10 years AGO

+4 ...sum up all alphabets values

Like? Yes (1) | No (2)  

SAGAR ANAT   10 years AGO

A+M+A+N= 1+13+1+14=29+4=33
R+A+J+U=18+1+10+21=50+4=54
N+I+T+U=14+9+20+21=64+4=68
THEN
R+E+N+U=18+5+14+21=58+4=62

Like? Yes (1) | No (2)  

dishant   10 years AGO

if we add numerically..
A=1 B=2....
AMAN = 29 (+4) =33
RAJU = 50 (+4) =54
NITU = 64 (+4) =68
THEN WE FIND NUMBER EQUIVALENT IS SUM OF NUMERICAL VALUE PLUS 4.
SO
RENU=58 (+4) = 62

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Ashish Kumar Singh   10 years AGO

Pattern is A+M+A+N (1+13+1+14)+4=33
N+I+T+U (14+9+20+21)+4=68...........

SO for R+E+N+U (18+5+14+21)+4= 58+4=62

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K Venkata Subba Rao   10 years AGO

AMAN=>1+13+1+14 =29+4=33
RAJU=>18+1+10+21 =50+4=54
//ly
RENU=>18+5+14+21 =58+4=62

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hemavathi   10 years AGO

RAJU= 1+13+1+14+4(WORDS)=33
same like, RENU= 18+5+14+21+4=62

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vinod   10 years AGO

adding the alphabets corresponding no and +4 for it.....

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jayabharathy   10 years AGO

a=1, b=2,.....,z=26
sum of values of all alphabets + num of alphabets so that
R=18,E=5,N=14,U=21
(18+5+14+21)+4=62

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Nidhi Raturi   10 years AGO

A=1, M=13, N=14
then sum of AMAN= 29. In the sum of AMAN when we add 4 it becomes 33
Same pattern is applied for rest.
Therefor, for RENU : R=18, E=5, N=14, U=21 sum of RENU=58 then add 4 it becomes 62 which is the answer

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DHIRENDRA    10 years AGO

DECODE THE LANGUAGE
A=1,B=2,C=3.............Z=26
AMAN = 33 | 1+13+1+14=29+4=33
RAJU = 54 | 18+1+10+21=50+4=54
NITU = 68 | 14+9+20+21= 64+4=68
SAME AS
RENU = 18+5+14+21= 58+4= 62

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R.RAVIKUMAR   10 years AGO

a-1,b-2,....z-26
AMAN+4=33
RAJU+4=54
NITU+4=68
RENU+4=62

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Rajasekhar Reddy   10 years AGO

A-1,B-2,.........Z-26
RENU = R+E+N+U+4 = 62

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Santan kumar   10 years AGO

Aman= 1+13+1+14=29 +4=33
RAJU= 18+1+10+21=50 +4=54
NITU= 14+9+20+21=64 +4=68
then
RENU=18+5+14+21=58 +4=62

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VENKATA SIVA AVINASH A   10 years AGO

ans:-RENU=62

AMAN=1+13+1+14=29+4=33
RAJU=18+1+10+21=50+4=54
NITU=14+9+20+21=68+4=68
RENU=18+5+14+21=58+4=62

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Saidinesh   10 years AGO

A=1;B=2;c=3;.........z=26
A+M+A+N=1+13+1+14=29+4(add common to all)=33;
R+A+J+U=18+1+10+21=50+4(common to all)=54;
N+I+T+U=14+9+20+21=64+4=68
then
R+E+N+U=18+5+14+21=58+4=62
(add 4 to all the words)

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K PAIDAM NAIDU   10 years AGO

add all alphabet numbers +4
...(r-18,E-5,N-14,U-21)+4=62

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gopinath   10 years AGO

reason is that each letter of the word having the numerical value...from the question RENU-58,here with the final value 4 is just added,so 62 is the answer

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vamsi goli   10 years AGO

aman=01+13+01+14=29+4=33
raju=same 50+4
nitu=64+4
then now renu it is 58+4=62

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Dinesh Tripathi   10 years AGO

AMAN(1+13+1+14=29+4)=33
RAJU(18+1+10+21=50+4)=54
NITU(14+9+20+21=64+4)=68)
HENCE
RENU=(18+5+14+21=58+4)=62

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arunkumar   10 years AGO

renu=18+5+14+21
58+4=62

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sravani   10 years AGO

R+E+N+U+4=62

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Priya Dharshini   10 years AGO

A+M+A+N=1+13+1+14=29+4=33;
SIMILARLY ,....
N+I+T+U=14+9+20+21=64+4=68;
SO...., R+E+N+U=21+14+18+21=58+4=62;
ANS IS 62

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vky   10 years AGO

R=18,E=5,N=14,U=21=> 18+5+14+21+4=62

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K.Shyam Prasad Rao   10 years AGO

Aman=1+13+1+14=29+4=33
raju=18+1+10+21=50+4=54
nitu=14+9+20+21=64+4=68
renu=18+5+14+21=58+4=62

Like? Yes (1) | No (3)