M4maths Previous Puzzles

10 Apr 2014 Quadratic equation

Two circle touch externally. The sum of their areas is 130 Pi sq.cm. and the distance between their centres is 14 cm. Find the radii of two circles.

OPtion

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

Solution

Let the radius of one circle be r
then radius of other circle be 14-r
According to the given condition
Pi*r^2 + Pi*(14-r)^2 = 130
Solve
We get
x = 11 or 3

Option 3)

Correct Option: 7 Best Solution (49)

Ann Theressa   11 years AGO

Let the radius of circles be x and y

Area of 2 circles = 130*pi
pi*x^2 + pi*y^2 = 130*pi
x^2 + y^2 = 130 ----(1)

Sum of their radius = 14
x + y = 14 ----(2)

(x+y)^2 = x^2 + y^2 + 2xy
14^2 = 130 + 2xy
2xy = 66
xy = 33 ----(3)

From (2) and (3)

x*(14-x) = 33
x^2 -14x + 33 = 0
(x-11)(x-3) = 0
x = 11,3

Radii of 2 circles are 11cm and 3cm

Like? Yes (70) | No (26)  

AMIT KUMAR   11 years AGO

let 1st radius = r cm
2nd will be = 14-r cm
a/c question,
(pi*r^2) + (pi*(14-r)^2) = 130 pi
=> r^2 + (14-r)^2 = 130
=> 2*r^2 - (28*r) + 66 = 0
=> r^2 - 14*r + 33 = 0
=> r^2 - 3*r - 11*r + 33 = 0
=> r(r - 3) -11(r-3) = 0
r = 3 or r = 11
if r = 3 then 2nd will be = 11
r = 11 then 2nd will be = 3 ans

Like? Yes (34) | No (7)  

Gitanjali K   11 years AGO

Let r, represent the measure of the radius of one of the circles,

then 14-r, must represent the radius of the other circle.

Therefore, the area of the first circle is πr^2,
and the area of the second circle is π(14-r)^2, and given that the sum of these areas is 130π cm^2
Therefore,
πr^2+ π(14-r)^2=130π
r^2+ (14-r)^2=130
2r^2-28r+66=0
r^2-14r+33=0
therefore r=3 or r=11

Like? Yes (16) | No (4)  

Clair Christ D   11 years AGO

pi(a^2+b^2)=130pi
a^2+b^2=130
a+b=14
(a+b)^2=a^2+b^2+2*a*b
196=130+2ab
ab=33
a+b=33
11,3

Like? Yes (10) | No (2)  

Dr.Dipin Singh   11 years AGO

(x^2+y^2)*pi=130 *pi
(x^2+y^2)=130
x+y=14
solving
x=11 cms,y=3 cms

Like? Yes (15) | No (10)  

Surya Narayana K   11 years AGO

Answer : 11 cm and 3 cm (Option 7)
Let r1 and r2 be the radius of two circles and given that they are touch externally.

So, distance between their centers is r1+r2=14cm --(1)
and sum of their areas is Pi(r1^2+r2^2)=130Pi sq.cm --(2)

Substitute r2=14-r1 (from (1)) in (2),then
r1^2+(14-r1)^2=130
=>r1^2+r1^2+196-28r1=130
=>2r1^2-28r1+66=0
=>r1^2-14r1+33=0
=> r1= 11 or 3
then r2= 3 or 11.

The radii of two circles : 11 cm and 3 cm (Option 7).

Like? Yes (15) | No (10)  

vishesh saxena   11 years AGO

let areas of two circle be a1 & a2,
given: a1+a2= 130 Pi,
radius be ri & r2;
given: r1+r2=14;
area of circle= pi(r)^2;
=> a1+a2= pi(r1^2+r2^2);
calculating r1r2 from here. solving further gives
r1= 11,3; r2= 3,11;

Like? Yes (13) | No (8)  

vishal gupta   11 years AGO

sum of areas=130pi
pi(r1)^2+pi(r2)^2=130pi
(r1)^2+(r2)^2=130 (first equation)
distance between centres (touch externally)=14
r1+r2=14 (second equation)
solving equqtion 1 & 2
r1=11.
r2=3.

Like? Yes (3) | No (1)  

shree   11 years AGO

(3.14)*121+(3.14)*9=130(3.14)

Like? Yes (1) | No   

arnold   11 years AGO

simply add two values
we have now two option
option 1 and option 7
take sum of the square of both number
option 7 gets us the answer

Like? Yes (1) | No   

Satish Zod   11 years AGO

only sr no. 7 fits the situation.

Like? Yes (1) | No   

subramani   11 years AGO

In given Answers only 1 and 7 is equal to 14 cms. Which is Distance b/w. their centres.
If apply both in Area of Circle ,7th Answer is equal to 130 Pi Sq.cm.

Like? Yes (1) | No   

Govind   11 years AGO

x^2+y^2=130
x+y= 14
xy = (14^2-130)/2= 33
x=11, y=3

Like? Yes (2) | No (1)  

karthik hariharan   11 years AGO

3.14x^2+3.14y^2=130*3.14
so x^2+y^2=130;
radius distance 14
so x+y=14
x=y-14
substitute the eqn and get radius 11,3
that is 7th option

Like? Yes (1) | No   

C.Aarthi   11 years AGO

let x, y be the two radii..
And since it touch externally
x+y =14
Since sum of the area=130
x^2 pi + Y^2 pi =130 pi
x^2 + Y^2=130
x^2 +(14-X)^2 =130
x^2 +196+X^2 -28x=130
2X^2 -28x +66 =0
x^2 - 14 x + 33=0
x=11, 3
y=11,3
thus radii is 11,3

Like? Yes (3) | No (2)  

Kaustubh karekar   11 years AGO

by solving the equation we get the answer as 11 and 3

Like? Yes (2) | No (1)  

Roopali   11 years AGO

let r1=x,then r2=14-x
pi(r1)square+pi(r2)square=13pi
on solving we get 11,3 Ans.

Like? Yes (2) | No (1)  

Anupam Banerjee   11 years AGO

Let the radii of the two circles be r1 & r2;
so, pi r1^2+ pi r2^2 = 130 pi sq.cm.
or, r1^2 + r2^2=130;

and we have, r1+r2=14cm.
or, r1^2 + 2.r1.r2 + r2^2 = 196 sq. cm;
or, 130 + 2.r1.r2 = 196;
or, 2.r1.r2 = 66;

r1^2 + r2^2 = 130;
or, (r1-r2)^2+2.r1.r2 = 130;
or, (r1-r2)^2 = 130-66;
or, r1-r2 = sqrt 64;
or, r1-r2 = 8;

and r1+r2=14;

so,2.r1 = 22;
or, r1 = 11cm;

and 2.r2 = 6;
or, r2 = 3cm.

Like? Yes (2) | No (1)  

Md Shahid Akhter   11 years AGO

r1+r2=14
pi(r1^2+r2^2)=130*pi
r1^2+r2^2=130
we know
r1^2+r2^2=(r1+r2)^2-2*r1*r2
putting the values
130=14^2-2*r1*r2
r1*r2=33
r1=11,r2=3 or r1=3,r2=11
Option 7 is correct one

Like? Yes (2) | No (1)  

Santosh Vishwakarma   11 years AGO

suppose radius are x and (14-x)
sum of areas is=>Pi*x*x+Pi*(14-x)*(14-x)=130Pi
=>x*x+(14-x)*(14-x)=130
=>x=11,3

Like? Yes (2) | No (1)  

Arindrajit Chakraborty   11 years AGO

Easy as pie..can be done by option testing (11)^2 + (3)^2=130

Like? Yes (3) | No (2)  

Patel Yogesh Rameshbhai   11 years AGO

distance between two centers of circles 14 cm so there are two possibilities from given option
the radius of two cicle are either 9,5 or 11,3
sum of area is 130pi sq.cm so if we take radius 11,3 then we get exact sum of area
so right answers is 11,3

Like? Yes (6) | No (5)  

Pragnesh patel   11 years AGO

there are two possibilities from given option
the radius of two cicle are either 9,5 or 11,3
sum of area is 130pi sq.cm so if we take radius 11,3 then we get exact sum of area
so right answers is 11,3

Like? Yes (3) | No (2)  

Vinotha   11 years AGO

given sum of their area=130(pi)
r^2+R^2=130;
r+R=14
(r+R)^2=196
solving we get r,R=11,3

Like? Yes (2) | No (1)  

neerav   11 years AGO

π(14-x)^2 + π(x^2) = 130 π
π(196+x^2 -28x) + πx^2 = 130 π
π(196+2πx^2-28x)=130π
196+2πx^2-28x=130
66+2x^2-28x=0
33+x^2-14x=0
33-11x+x^2-3x=0
-11(x-3)+x(x-3)=0
(x-3)(x-11)=0
x=3 & x=11

Like? Yes (2) | No (1)  

Raja   11 years AGO

11+3=14 ; 11^2+3^2=130

Like? Yes (2) | No (1)  

MP    11 years AGO

Two circle touch externally. The sum of their areas is 130 Pi sq.cm. and the distance between their centres is 14 cm.
pi*(R^2+r^2)=130 *pi
(R^2+r^2)=130
R+r=14
solving these eqns, we get
R=11 cms,r=3 cms

Like? Yes (2) | No (2)  

Aman   11 years AGO

pi*(x^2+y^2)=130 *pi
(x^2+y^2)=130...(1)
x+y=14...........(2)
solving (1) & (2)
x=11 cms,y=3 cms

Like? Yes (5) | No (5)  

llamar   11 years AGO

take circle1 radius=r,circle2 radius=R
given pir^2+piR^2=130pi
=>r^2+R^2=130---(1)
given r+R=14---(2)
by solving (1)&(2)
we get r-R=8---(3)
by solving (2)&(3)
we get r=11,R=3

Like? Yes (7) | No (7)  

Rahul Kesharwani   11 years AGO

As circles touch each other externally, it means their area are seprated,
So Area of Circle = pi * r^2

i. e. 11^2 * pi + 3^2 * pi = 130* pi

Like? Yes (4) | No (4)  

ASHISH MITTAL   11 years AGO

By solving we get r1*r2=33
And answer which give this value is 11 and 3

Like? Yes (3) | No (3)  

K.Bhagyasri   11 years AGO

circle area formula= pi*(r^2)
sum of two circle areas :
pi(r1^2+r2^2)=130*pi sq.cm--------(1)


form (1)
r1^2+r2^2=130--------------(2)

r1+r2=14 cm------------(3)
square on both sides

(r1+r2)^2 =14^2
r1^2+r2^2+2r1r2 = 196
130+2r1r2 = 196
r1r2=33
r1=33/r2 substitute in (3)

r1+r2=14
(33/r1)+r1=14

r1^2-14r1^2+33=0

sol : 11,3

Like? Yes (5) | No (5)  

Mukesh Nayal   11 years AGO

area of two circles is 130 pi
so,pi*(r1)*(r1)+pi*(r2)*(r2)=130*pi
r1^2+r2^2=130
adding and subtracting 2*r1*r2 we have,
(r1+r2)^2-2*r1*r2=130
since r1+r2=14(given)
so,14^2-130=2*r1*r2
66=2*r1*r2
r1*r2=33
so only one possibility of r1 and r2 which is 11 and 3 which equates 33.

Like? Yes (4) | No (4)  

Ghanshyam Gulati   11 years AGO

r1^2+r2^2=130
r1+r2=14

on solving r1=11
r2=3

Like? Yes (2) | No (2)  

Saraswathy   11 years AGO

Let "r" be the radius of one circle
Then radius of another circle is (14-r)
area of the circle is πr^2

Given sum of 2 areas is 130π
πr^2 + π(14-r)^2 = 130π
π[r^2 + 196 + r^2 - 28r] = 130π
2r^2 - 28r = -66
r^2 - 14r + 33 = 0

Factorize this...
You ll get r = 11, 3

Radius of 1st circle is 11
Radius of 2nd circle is (14 - 11) = 3

So option (7)

Like? Yes (2) | No (2)  

Naveen Ganesh   11 years AGO

Given r1 + r2 = 14
r1^2 + r2^2 = 130
looking at answer options, only 1 and 7 sum up to 14, among them only 7 satisfies the second equation

Like? Yes (1) | No (1)  

RAKESH   11 years AGO

pi(r1^2+r2^2)= 130pi

=> r1^2 + r2^2 = 130

& r1 + r2 = 14 => r1,r2=11,3

Like? Yes (2) | No (2)  

akash kumar    11 years AGO

for 2nd cndition to be fulfilled we have only option 1&7.put values of option 1&7 .we find option 7 is correct

Like? Yes (2) | No (2)  

karthika kumar   11 years AGO

Let A1 & A2 be area of two circles.
Given A1+A2=130pi sq.cm
=> piR1^2+piR2^2=130pi => R1^2+R2^2=130
Also given distance between their centres is 14 cm and the two circle touch externally ,therefore R1+R2=14
(R1+R2)^2=R1^2+R2^2+2R1R2
=> 14^2=130+2R1R2
=> R1R2=33
thus we have R1+R2=14 and R1R2=33
therefore R1=11 and R2=3

Like? Yes (2) | No (2)  

hanamant k   11 years AGO

A=A1+A2 =>given
130*pi= pi*r^2 + pi*(14-r)^2

2r^2-28r+66=0
so,
r=3 or r=11

Like? Yes (1) | No (1)  

puja   11 years AGO

radius=r,o
r^2+o^2=130
and
r+o=14
so
from, equating both eqn. in below
(r+o)^2=14^2
r^2+o^2+2ro=196,
r=11 and o=3

Like? Yes (1) | No (1)  

Rajitha   11 years AGO

Distance between the centers is 14 cm.
assume r be the radius of one circle.
the radius of another circle will be (14-r).
Area of the circle is pi*r^2.
Sum of the two circles is= (pi*r^2)+(pi*(14-r)^2)=130 pi
pi(r^2+(14-r)^2)=130 pi
r^2+(14-r)^2=130
2 r^2-28 r+66=0
r^2-14 r+33=0
(r-3)(r-11)=0
r=3 or r=11
the radius of one circle is 3 and the radius of another circle is 11.

Like? Yes (1) | No (1)  

TUSHAR    11 years AGO

r1+r2=14,r1^2+r2^2=130pi. solve equation, we get r1=11 r2=3

Like? Yes (1) | No (1)  

vishnuvardhan   11 years AGO

C1C2=R1+R2
14=R1+R2
R1^2+R2^2=130
(R1+R2)^2=196
R1R2=33
R1=11,R2=3

Like? Yes (1) | No (1)  

NIRUPOM GHOSH   11 years AGO

pi(r1^2+r2^2)=130pi and r1+r2=14 solving r1 and r2 are 11,3.

Like? Yes (1) | No (1)  

nalini singh   11 years AGO

pi*r1^2+pi*r2^2=130*pi
r1+r2=14
solve this two equation
r1=11,r2=3

Like? Yes (1) | No (1)  

ravi prakash   11 years AGO

area of circle=pir^2.
sum of radius=distance between theire centre.
11+3=14.
area would be=pi(R^2+r^2)
=pi(11^2+3^2)
=130pi.

Like? Yes (2) | No (3)  

Mahalakshmi   11 years AGO

let radius of one circle be r and another (14-r)
area of circle is =3.14*r^2
for area of two circles touching eachother will be
3.14r^2+3.14(14-r^2)=130*3.14
solving this equation we will get
(r-11)(r-3)=0
r=11,3
so ans is 7)11,3

Like? Yes (1) | No (3)  

Saidinesh   11 years AGO

11+3=14 distance of two radius,
121pi+9Pi=130Pi;
it satisfies option 6

Like? Yes (1) | No (4)  

09 Apr 2014 Quadratic equation

Solve
2x^4+9x^3+8x^2+9x+2 = 0

OPtion

1) x = -2 (+-) sqrt 5
2) x = -2 (+-) sqrt 3
3) x = -4 (+-) sqrt 5
4) x = -3 (+-) sqrt 3
5) x = -5 (+-) sqrt 3
6) x = -2 (+-) sqrt 7
7) x = -2 (+-) sqrt 2
8) x = -3 (+-) sqrt 3
9) x = -4 (+-) sqrt 3
10)None of these

Solution

In this equation
divide the equation by x^2 to get,
2[x^2+(1/x^2)]+9[x+(1/x)]+8 = 0
Let [x+(1/x)] = t
solve
x = -2 (+-) sqrt 3

Option 2)

Correct Option: 2 Best Solution (8)

Ann Theressa   11 years AGO

Divide the equation by x^2

2x^2 + 9x + 8 + 9/x +2/x^2 = 0
2(x^2+1/x^2) + 9(x+1/x) +8 = 0

put (x+1/x) = y
so, (x^2 + 1/x^2) = y^2 - 2

2(y^2 - 2) +9y + 8 = 0
2y^2 + 9y + 4 = 0

Solving y = -4 , -1/2

when y = -4
x + 1/x = -4
x^2 + 4x + 1 = 0

x = -4(+-)sqrt(16-4)/2

x = -2(+-)sqrt 3

Like? Yes (13) | No (6)  

Kaleem Pasha   11 years AGO

Answer 2

Like? Yes (3) | No (2)  

AVIJIT   11 years AGO

if we calculte thish equation we will get 2 real roots and 2 complex roots and these are
x1 = -0.25+0.9682458i
x2 = -0.25-0.9682458i
x3 = -0.2679492
x4 = -3.7320508
so if we go for 2nd option then -2(+-)sqrt(3)
we got -3.7320 nad -0.267 that same with x3 nad x4

Like? Yes (2) | No (1)  

Md Shahid Akhter   11 years AGO

on factorising we get (2x^2+x+2)(x^2+4x+1)=0
for second factor i.e
x^2+4x+1=0
x=-2(+-)sqrt(3)
So,
Ans is 2nd option

Like? Yes (3) | No (2)  

ABY   11 years AGO

if we calculte thish equation we will get 2 real roots and 2 complex roots and these are
x1 = -0.25+0.9682458i
x2 = -0.25-0.9682458i
x3 = -0.2679492
x4 = -3.7320508
so if we go for 2nd option then -2(+-)sqrt(3)
we got -3.7320 nad -0.267 that same with x3 nad x4

Like? Yes (1) | No (1)  

ABYSUP   11 years AGO

if we calculte thish equation we will get 2 real roots and 2 complex roots and these are
x1 = -0.25+0.9682458i
x2 = -0.25-0.9682458i
x3 = -0.2679492
x4 = -3.7320508
so if we go for 2nd option then -2(+-)sqrt(3)
we got -3.7320 nad -0.267 that same with x3 nad x4

Like? Yes (1) | No (2)  

avi   11 years AGO

if we calculte thish equation we will get 2 real roots and 2 complex roots and these are
x1 = -0.25+0.9682458i
x2 = -0.25-0.9682458i
x3 = -0.2679492
x4 = -3.7320508
so if we go for 2nd option then -2(+-)sqrt(3)
we got -3.7320 nad -0.267 that same with x3 nad x4

Like? Yes (1) | No (2)  

MD SOHAIL AKHTER   11 years AGO

by substituting the values given in option we get
x=-2(+-)sqrt 3
so option (2) is correct

Like? Yes (2) | No (4)  

08 Apr 2014 Solve

Upto how many integers we should take the product 1*2*3*4*5.......... to make 31 zeros at last of the number.

OPtion

1) 140
2) 130
3) 125
4) 110
5) 100
6) 105
7) 135
8) 115
9) 145
10)None of these

Solution

For obtaining 31 zeros at last of the number, 31 powers of 5 is necessary. Product 1*2*3*4*5�����upto 125
contain 5, 10 ,15���. 125. These 25 numbers are multiple of 5, besides it 25, 50, 75, 100, 125 contain a second 5 and 125 contains a third 5 hence total number of 5 = 25+5+1 = 31

Option 3)

Correct Option: 3 Best Solution (24)

Dr.Dipin Singh   11 years AGO

No of zeros at last of product 1*2*3*4*5.......n = n/5 +n/25+n/125 =31
n(25+5+1)=31*125
31*n=31*125
n=125 .... option 3

Like? Yes (10) | No (3)  

Aman   11 years AGO

As
125/5 +125/25 +125/125= 25+5+1=31
hence we should take 125 integers for the product 1*2*3*4*5.......... to make 31 zeros at last of the number.

Like? Yes (9) | No (3)  

Saranjit Singh   11 years AGO

zeros at last of number in n!= n/5 +n/25+n/125+n/625
when n=125,
no. of zeros= 125/5+125/25+125/125=31

Like? Yes (7) | No (2)  

Pargat Pal Singh   11 years AGO

1*2*3*4*5.....n = n!
no of zeros at end of n! is given by
n/5 +n/25+n/125.
so n/5 +n/25+n/125 = 31
n=125

Like? Yes (7) | No (3)  

llamar   11 years AGO

1-10 you get 2 zero
10-20 u get 2 as well
...............upto 100 u get 21 zeros then after
100-110=>4 zero
110-120=>4 zeros
125 u get 2 zeros

Like? Yes (1) | No   

vishnuvardhan   11 years AGO

125!=
[125/5]+[125/25]+[125/125]=25+5+1=31

Like? Yes (1) | No   

Vaishali Dhiman   11 years AGO

ans is 125

Like? Yes (1) | No   

AMAN   11 years AGO

125/5=25,125/25=5,125/125=1,add them 31

Like? Yes (1) | No (2)  

ashish kumar   11 years AGO

for this the rule is
(N/5)+(N/5^2)+(N/5^3)+(N/5^4)+.......
take N =125 from option u will dirtectly able to tel l that this is answer.
(125/5)+(125/5^2)+(125/5^3)+.....
25+5+1=31
take only integers

Like? Yes (3) | No (4)  

MD SOHAIL AKHTER   11 years AGO

To calculate no. of zeroes let the no. be 125 then
125/5=25
25/5=5
5/5=1
now, adding all the quotient-25+5+1=31
total no. of zeroes =31
ans- 125

Like? Yes (3) | No (6)  

Md Shahid Akhter   11 years AGO

To calculate no. of zeroes let the no. be 125 then
125/5=25
25/5=5
5/5=1
now, adding all the quotient-25+5+1=31
total no. no. of zeroes =31
the no. will be 125

Like? Yes (2) | No (5)  

ASHISH MITTAL   11 years AGO

125/5=25
25/5=5
5/5=1
25+5+1=31

Like? Yes (1) | No (4)  

Ann Theressa   11 years AGO

For finding the number of trailing zeros at the end of a factorial(let it be n!), the equation is,

number of trailing zeros
= n/5 + n/5^2 + n/5^3+....+n/5^k
where 5^(k+1) > n

Here number of trailing zeros = 31

n/5 + n/25 + n/125+ ... = 31

Solving n = 125

Like? Yes (1) | No (4)  

AVIJIT   11 years AGO

125/5=25
125/25=5
125/125=1
25+5+1=31

Like? Yes (1) | No (4)  

ABY   11 years AGO

125/5=25
125/25=5
125/125=1
25+5+1=31

Like? Yes (1) | No (4)  

ABYSUP   11 years AGO

125/5=25
125/25=5
125/125=1
25+5+1=31

Like? Yes (1) | No (4)  

mohit gupta   11 years AGO

125!= to calculate zero in product we formula divide 125/5=25
125/25=5
125/125=1
total 25+5+1=31
hence c option

Like? Yes (1) | No (4)  

Mukesh Nayal   11 years AGO

the method is like this
125/5+125/5*5+125/5*5*5=31
no of zeroes are found by calculating the number of 5 for the factorial of a number.

Like? Yes (3) | No (6)  

VIKASH RAY   11 years AGO

We know to find out the zeros last in the number we should divide the number by 5
here the number of zeroes is 31 ....
so the last in the required sequence is---
31*5=125
so option 3 is correct.

Like? Yes (1) | No (4)  

Surya Narayana K   11 years AGO

Ans is 125 (option 3).
The formula for number of zeros in n! is n/5+(n/5^2)+(n/5^3)+....

Consider 140 = 140/5+140/25+140/125=28+5+1=34
130=130/5+130/25+130/125=26+5+1=32
125=125/5+125/25+125/125=25+5+1=31.

So answer is 125.

Like? Yes (1) | No (4)  

Sivanjali   11 years AGO

The formula for number of zeros in n! is n/5+(n/5^2)+(n/5^3)+....

Consider
125=125/5+125/25+125/125=25+5+1=31.

So answer is 125.

Like? Yes (1) | No (4)  

AJAY KUMAR   11 years AGO

ans=125
125/5-quotient 25
25/5-quotient 5
5/5-quotient 1
total no of zeros=25+5+1=31

Like? Yes (1) | No (4)  

shree   11 years AGO

because 4 multi give the 0 up to 10

Like? Yes (1) | No (6)  

Rajat Goyal   11 years AGO

Best Option

Like? Yes (1) | No (6)  

07 Apr 2014 Sequence

What is the next number of the following sequence
4,7,5,7,6,9,7,13,8,10,9,....

OPtion

1) 19
2) 18
3) 15
4) 14
5) 13
6) 11
7) 10
8) 9
9) 7
10)None of these

Solution

Answer is 9
[4] = 4^2 = 16 = 1+6 =
[7]
[5] = 5^2 = 25 = 2+5 =
[7]
[6] = 6^2 = 36 = 3+6 =
[9]
[7] = 7^2 = 49 = 4+9 =
[13]
[8] = 8^2 = 64 = 6+4 =
[10]
[9] = 9^2 = 81 = 8+1 =
[9]

[4,7] [5,7] [6, 9] [7, 13] [8, 10] [9, 9]
4,7,5,7,6,9,7,13,8,10,9,[9]

Option 8) 9

Correct Option: 8 Best Solution (2)

karthika kumar   11 years AGO

4,7,5,7,6,9,7,13,8,10,9,... ANS:9
alternate no in the series are obtained by adding 1 (4,..,5,..,6,..,7,..,8,..,9,..)
..,7,..,7,..,9,..,13,..,10,..,__ in this series
7+0=7
7+2=9
9+4=13
13+6=19=1+9=10 (if the ans exceed 13)
10+8=18=1+8=9

Like? Yes (3) | No (7)  

KEERTHIPALLE SUDHEERKUMAR   11 years AGO

4,5,6,7,8,9
7,7,9,13,10,9
7-7=0,9-7=2,13-9=4,13-10=3,10-x=1 in the place we can take 9...

Like? Yes (3) | No (8)  

04 Apr 2014 Coded language

If in coded language
DARPAN = 306
DEVENDRA = 204
BHUWNESH = 721
Then
DHERAJ = ?

OPtion

1) 301
2) 327
3) 132
4) 299
5) 286
6) 305
7) 367
8) 787
9) 309
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)
DARPAN = (D*A)+(R*P)+(A*N) = (4*1)+(18*16)+(1*14)
= 4 + 288 + 14 = 306
DEVENDRA = (D*E)+(V*E)+(N*D)+(R*A)
= (4*5)+(22*5)+(14*4)+(18*1) = 20 + 110 + 56 + 18 = 204
BHUWNESH = (B*H)+(U*W)+(N*E)+(S*H)
= (2*8)+(21*23)+(14*5)+(19*8) = 16 + 483 + 70 + 152 = 721
Similarly
DHERAJ = (D*H)+(E*R)+(A*J) = (4*8)+(5*18)+(1*10)
= 32 + 90 + 10 = 132

Option 3)

Correct Option: 3 Best Solution (9)

ANUP KUMAR   11 years AGO

D(4),A(1),R(18),P(16),A(1),N(14)
(4*1)+(18*16)+(1*14)=4+288+14=306
SIMILARLY,
D(4),H(8),E(5),R(18),A(1),J(10)
(4*8)+(5*18)+(1*10)=32+90+10=132

Like? Yes (16) | No (3)  

Mukesh Nayal   11 years AGO

take numeric values of alphabet like a=1..and so on
then you can say:
DxA+RxP+AxN=306
likewise
DXH+ExR+AxJ=8X4+5X18+1x10=132

Like? Yes (5) | No (2)  

llamar   11 years AGO

change alphabets into numerals in the following way i.e a=1.multiply 1st,2nd&3rd,4th&5th,6th......finally add the results .we will get ans 132

Like? Yes (1) | No   

S L PRASANNA P   11 years AGO

Logic in that is first he wrote alphabetical numbers for all letters in that name and multiplied each 2 numbers which are successive and
add all those. Like
D H E R A J
4 8 5 18 1 10
result:(4*8)+(5*18)+(1*10)=132

Like? Yes (1) | No   

Devendra Marghade   11 years AGO

Numbering letters A=1,B=2,C=3,...., z=26
Add the pair of numbers corresponding to letters

DARPAN =(4)(1)(18)(16)(1)(14)
=[4*1]+[18*16]+[1*14]= 306

DEVENDRA =(4)(5)(22)(5)(14)(4)(18)(1)
=[4*5]+[22*5]+[14*4]+[18*1]= 204

BHUWNESH =(2)(8)(21)(23)(14)(5)(19)(8)
=[2*8]+[21*23]+[14*5]+[19*8]= 726

DHERAJ =(4)(8)(5)(18)(1)(10)
=[4*8]+[5*18]+[1*10]=132

Like? Yes (3) | No (3)  

Soham Dalwadi   11 years AGO

According to A(1),B(2),C(3)....Z(26)

Take, D A R P A N equals
4 1 18 16 1 14

here SUM OF PROCUCTS is consider
so (4x1)+(18x16)+(1x14)=306

similarly
DHERAJ = (4x8)+(5x18)+(1x10)=132

Like? Yes (2) | No (2)  

champavat jaydipsinh padamsinh   11 years AGO

According to A(1),B(2),C(3)....Z(26)

Take, D A R P A N equals
4 1 18 16 1 14

here SUM OF PROCUCTS is consider
so (4x1)+(18x16)+(1x14)=306

similarly
DHERAJ = (4x8)+(5x18)+(1x10)=132

Like? Yes (3) | No (3)  

srilakshmi   11 years AGO

addition of multiplication of adjacent numbers

Like? Yes (2) | No (3)  

Uma shankar   11 years AGO

alphabets A to Z coded as 1 to 26,
for DARPAN= (D*A)+(R*P)+(A*N)
As in coded language (4*1)+(18*16)+(1*14)=306
similarly for DHERAJ= 132

Like? Yes (1) | No (3)  

03 Apr 2014 Find the distance

A man moves 4 m. towards east then 3 m. towards left after then 8 m. towards right then 6 m. towards right then 16 m. towards right. Finally he moves 4m towards north followed by 4 m. towards right. What is his distance from the starting point.

OPtion

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

Solution

a graph to obtain the result.
Ans is 1 m

Option 9)

Correct Option: 9 Best Solution (17)

MADHU   11 years AGO

4E 3N 8E 6S 16W 4N 4E

16E 7N 16W 6S

add distance of same direction and cancel them with opposite directions then we get
1m.
option is 9

Like? Yes (14) | No (1)  

Devendra Marghade   11 years AGO

If Move along 'X' axis is considered as horizontal and along 'Y' axis as Vertical movement.

Horizontal move=(+4),(+8),(-16),(+4)=0
So man will be at same horizontal position

Vertical move=(+3),(-6),(+4)=+1
Man will be at 1m. towards north from starting point

Like? Yes (2) | No   

srija   11 years AGO

4m towards east.
then 3m left mean towards north from now..
then 8m rgt means towards east..
then 6m rght means towards south..
then 16m rght means towards west..
after then 4m towards north..
finallyy 4m rht means towards east..
distance from starting point is 1m

Like? Yes (2) | No   

shree   11 years AGO

because this only 1 km remaining............ from starting in north

Like? Yes (2) | No   

Rashmi bharti   11 years AGO

ans=1m

Like? Yes (2) | No   

ASHISH MITTAL   11 years AGO

after solving we get 1 m

Like? Yes (2) | No   

Aakash Jain   11 years AGO

If we draw figure that he moves then finally he is 1 m. away from starting point.

Like? Yes (1) | No   

mani chaitanya   11 years AGO

if u represent given data in pictorial representation, then u can easily get it

Like? Yes (1) | No   

Dinesh thatti   11 years AGO

by solving above directions we get 1m distance from
starting point towards north direction

so answer is 1m

Like? Yes (1) | No   

megha   11 years AGO

Starting point is 1m south from ending point

Like? Yes (1) | No   

muni   11 years AGO

ans is 1 m up

Like? Yes (1) | No   

nikhil   11 years AGO

HE WILL BE ONLY 1M FROM THE INITIAL POINT AS THE WHOLE JOURNEY IS ROUND FIGURE

Like? Yes (1) | No   

karthika kumar   11 years AGO

----8--.
....4.. 3| |
4 .--4-| |6
| |
|__________16______|

Like? Yes (1) | No   

ashish kumar   11 years AGO

bcz by drawing all the distances,and by taking conisederation of all the four directions we got that the man is 1m away from the starting point

Like? Yes (1) | No   

Satish Zod   11 years AGO

From starting point,
Total distance travelled towards north=
4 + 8 - 16 + 4 = 0

Total distance travelled towards east=
3 - 6 + 4 = 1

Like? Yes (1) | No   

hanamant k   11 years AGO

there is only 1m dist from starting to end point

Like? Yes (1) | No   

ravi prakash   11 years AGO

finally he is 1 m apart from starting position towarts north.

Like? Yes (1) | No   

02 Apr 2014 Solve

A carom board is of size 2.5 m � 2.5 m. Striker is placed on a side at a position 0.5 m from the left corner. For overcoming the condition of foul it is necessary to touch the queen placed on right side at a position 0.5 m from the side where striker is placed. If player wants to make a two collision rebound then he should make an impact on the first side at a distance of

OPtion

1) 0.8 m from corner
2) 0.75 m from corner
3) 0.5 m from corner
4) 1 m from corner
5) 0.6 m from corner
6) 1.25 m from corner
7) 1.4 m from corner
8) 1.3 m from corner
9) 1.5 m from corner
10)None of these

Solution

Let striker makes first impact at a distance x from the corner and second at a distance y from other corner.
First draw a plot for this condition. Then according to plot
x/0.5 = (2.5-x)/y => xy = 0.5(2.5-x) ..... Equation 1
and y/(2.5-x) = (2.5-y)/2 => 2y =(2.5-x) (2.5-y) .....Equation 2
On solving these equations we get x = 0.75 m

Option 2)

Correct Option: 2 Best Solution (2)

RAKESH   11 years AGO

0.5 + 0.5/2 = 0.75 m

Like? Yes (10) | No (3)  

Ann Theressa   11 years AGO

Distance from left corner of 1st side = 0.5 + 0.5/2
=0.5 + 0.25 = 0.75m from corner

Like? Yes (4) | No (1)  

01 Apr 2014 Quadratic Equation

Solve
4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28

OPtion

1) x = -159/165, 1/3
2) x = -164/165, 1/4
3) x = -169/165, 1/4
4) x = -169/165, 1/3
5) x = -144/165, 1/3
6) x = 174/165, 1/3
7) x = -174/165, 1/3
8) x = 154/169, 1/3
9) x = -165/169, 1/3
10)None of these

Solution

Put Sqrt[x/(x+1)] = t then solve
we get
x = -169/165, 1/3

Option 4)

Correct Option: 4 Best Solution (37)

Dr.Dipin Singh   11 years AGO

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28
let
Sqrt[x/(x+1)] =y
then eqn becomes
4y+13/y=28
solving, we get
y= -13/2 , 1/2 = Sqrt[x/(x+1)]
hence x= -169/165, 1/3

Like? Yes (11) | No (4)  

Surya Narayana K   11 years AGO

16*(x/x+1) + 169*(x+1/x) + 104 = 784
After solving,
495x^2 + 342x - 169 = 0
x = -169/165 and 1/3

Like? Yes (2) | No (1)  

TUSHAR    11 years AGO

put Sqrt[x/x+1]=p,
equation become, 4p+13/p=28
the root of equation is p1=13/2,p2=1/2
i.e. x1=-169/165, x2=1/3

Like? Yes (1) | No   

vishnuvardhan   11 years AGO

let x+1/x=t
4/sqrt(t)+13*sqrt(t)=28
t=4,4/169
x=1/3,-169/165

Like? Yes (1) | No   

nikhil   11 years AGO

on solving above given equation we get x=1/3 and on hit and trial method we get -169/165

Like? Yes (1) | No   

Aakash Jain   11 years AGO

when we solve this equation, this equation is looks alike 495*X^2+342*X-169=0. when we solve this equation,the result is X=-169/165,1/3

Like? Yes (1) | No   

suryam surada   11 years AGO

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28

>>square operation both sides..then we will get
495*x^2+342*x-169=0;

>> if you substitute all the options..
4 th option satisfy the equation

Like? Yes (1) | No   

arnold   11 years AGO

square on both side
we get
16(x/x+1)+169(x+1/x)+104=784
16(x/x+1)+169(x+1/x)-680=0
take LCM
we get
495(x^2)+395(x^2)-169=0
calculate quadratic root of above equation
x= -169/16, 1/3

Like? Yes (1) | No   

VIKASH RAY   11 years AGO

The given equation is,
4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28
Now,Let, [x/(x+1)]=y-----------------(1)
Then the given equation becomes,
4*sqrt(y)+ 13* sqrt(1/y) = 28
Now solving this equation we get,
y=1/2, 13/2
Now put the value of y in the equation number 1 we get,
x=1/3, -169/165
Hence option number 4 is correct.

Like? Yes (1) | No   

Ann Theressa   11 years AGO

put sqrt(x/(x+1)) = y

4y + 13/y = 28
4y^2 -28y + 13 = 0

y = 13/2 , 1/2

x/(x+1) = y^2
x/(x+1) = (13/2)^2
x/(x+1) = 169/4
x = -169/165

x/(x+1) = y^2
x/(x+1) = 1/2^2
x/(x+1) = 1/4
x = 1/3

Like? Yes (1) | No   

ARJYA HALDER   11 years AGO

Let Sqrt*[x/(x+1)]=a
So, the equation becomes 4a+(13/a)=28
which gives a=(1/2),(13/2)
which gives x=(1/3),(-169/165)

Like? Yes (1) | No   

The Trojan   11 years AGO

complex number

Like? Yes (1) | No   

Naveen Ganesh   11 years AGO

Let x/x+1 = a^2 then equation becomes 4a + 13/a = 28
=> 4a^2 - 28a + 13=0
=> a= 13/2 , a=1/2
Substituting both the values of a we get, x=-169/165 and x=1/3

Like? Yes (1) | No   

Rajasekhar Reddy   11 years AGO

let sqrt[x/x+1] = P

then it becomes

4P^2-28P+13 = 0

SOLVING WE GET

P = 13/2 AND 1/2

EQUATING SQRT[X/X+1] TO 13/2 AND 1/2 WE GET

X = -169/165 AND 1/3

Like? Yes (1) | No   

srija   11 years AGO

by simplifyng given quatr eqn we will get the followng eqn...
495x*2+342x-169=0
after solving the eqn we can get the roots -169/165 , 1/3

Like? Yes (1) | No   

gaurav bhatt   11 years AGO

Squaring on both sides:
(4 *sqrt(x/x+1)+(13*sqrt(x+1/x))2=(28)2
16(x/x+1)+169(x+1/x)+104=784
16(x/x+1)+169(x+1/x)=680
16x2+169(x+1)2=680(x(x+1))
495x2+342x-169=0;

Finding factors:
X=-169/165 and x=1/3

Like? Yes (1) | No   

Ragaven   11 years AGO

solving the quadratic equation we get the answer (-169/165,1/3)

Like? Yes (2) | No (1)  

Mukesh Nayal   11 years AGO

Squaring on both sides,
(4 *sqrt(x/x+1)+13*sqrt(x+1/x))2=(28)2
16(x/x+1)+169(x+1/x)+104=784
16(x/x+1)+169(x+1/x)=680
Taking lcm on left hand sides,
16x2+169(x+1)2=680(x(x+1))
On solving ,
495x2+342x-169=0;
Factorizing using shridharacharya method,
X=-b+sqrt(b2-4ac)/2a
We got,
X=-169/165 and x=1/3

Like? Yes (1) | No   

sarvesh jhawar   11 years AGO

square both the side and then some manipulation
495*x*x+342*x-169=0
which give solution 4.

Like? Yes (1) | No   

Udhaya kumar   11 years AGO

let u=(x)/(x+1)
16u^2-680u+169=0
u=0.25,42.25
so x=-(169/165),(1/3)....

Like? Yes (2) | No (1)  

Vivek Jain   11 years AGO

4t+13/t=28
where t=Sqrt[x/(x+1)]
t=13/2 and 1/2
so x=-169/165 and 1/3

Like? Yes (1) | No   

Saikat Guin   11 years AGO

4* sqrt[x/x+1]+13sqrt[x+1/x]= 28
=> 16[x/x+1]+ 169[x+1/x] +104= 28
=> 16x^2+169x^2+169+338x= 680x^2+680x
=> 495x^2+342x- 169=0
=> x={-342±√(342^2+ 4*495*169)}/2*495 \\\\ as x={-b±√(b^2-4*a*c)}/ 2a
=> x= (-342±672)/990
=> x= -169/165 or 1/3

Like? Yes (1) | No   

Rajat Goyal   11 years AGO

Best Option

Like? Yes (1) | No   

llamar   11 years AGO

take squrt[x/(x+1)]=p& do simplifications as follows
4p^2-28p+13=0 =>4p^2-2p-26p+13=0
at the end u will get x=-169/165,1/3

Like? Yes (1) | No   

RAKESH   11 years AGO

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28

Sqrt[x/(x+1)] = y then

4y + 13/y = 28 => y=1/2 , 13/2

or, Sqrt[x/(x+1)] = 1/2 , 13/2 => x= -169/165, 1/3

Like? Yes (3) | No (3)  

shilpi   11 years AGO

put x/(x+1) = y
and solve the equation then

Like? Yes (1) | No (1)  

Sivanjali   11 years AGO

After squaring on both sides the equation will be

16*(x/x+1) + 169*(x+1/x) + 104 = 784
16x^2 + 169(x+1)^2 + 104(x*(x+1))=784(x*(x+1))
495x^2 + 342x - 169 = 0
On solving we get
x = -169/165 and 1/3

Like? Yes (1) | No (1)  

J.DHANDAPANI   11 years AGO

Let Sqrt[x/(x+1)=t.
On sub in the given equation we get,
4t^2-28t+13=0.
on solving which gives t=6.5 , 0.5.
===>Sqrt[x/(x+1)]=6.5.
x/(x+1)=(6.5)^2
41.25 x=-42.25
x=-1.0242
=-169/165 and
===>Sqrt[x/(x+1)]=0.5.
x/(x+1)=(0.5)^2.
3x=1.
x=1/3.

Like? Yes (1) | No (1)  

ADVAITA BASAK   11 years AGO

Solving the equation in general manner the answer become x=-169/165, 1/3

Like? Yes (1) | No (1)  

Devendra Marghade   11 years AGO

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28
Substituting Sqrt[x/(x+1)]=t
4t+(13/t)=28
4t^2-28t+13=0, roots=13/2, 1/2
So Sqrt[x/(x+1)]= 13/2, x=-169/165 &
Sqrt[x/(x+1)]= 1/2, x=1/3

Like? Yes (1) | No (1)  

PARVY GOVIL   11 years AGO

x=-169/165,1/3 i..e 4th option
reason :
let sqrt[x/x+1] = p
we have
4p + 13/p =28
4(p*p) -28p+13 =0
thus solving the above quadratic we get :
p=1/2;13/2
case 1:
p=1/2
1/2 = sqrt(x/x+1)
1/4=x/x+1
x=1/3
case 2:
sqrt(x/x+1)=13/2
x/x+1=169/4
x= -169/165
from case 1 & 2 we get
x= -169/165,1/3

Like? Yes (1) | No (1)  

Kowshik Sarker   11 years AGO

let [x/(x+1)]=k^2
so [(x+1)/x]=1/k^2
putting this the equation becomes
(4*k)+(13/k)=28
which becomes 4*k^2-28k+13=0
solving this quadratic equation we get
k=13/2,1/2
so we get 2 equations to solve
1.x/(x+1)=(13/2)^2 and
2.x/(x+1)=(1/2)
solving this two equation we get the value of x which is x=-169/165 and x=1/3. So option 4 is correct.

Like? Yes (1) | No (1)  

Soham Dalwadi   11 years AGO

suppose Sqrt[x/(x+1)]=a
then eq= 4a^2++13=28a
finding roots we get a=1/2,a=13/2
substituting a
4) x = -169/165, 1/3

Like? Yes (2) | No (2)  

SUPROTIM PAUL   11 years AGO

SOLVING THE EQUATION WE GET THE ANSWER.ONE MAY LET THE SQRT PART AS 'A' FOR EASY TO CALCULATE

Like? Yes (1) | No (1)  

megha   11 years AGO

Answer is 4) x=-169/165, 1/3

4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)] = 28
By Squaring both sides, we get
(4*Sqrt[x/(x+1)] + 13*Sqrt[(x+1)/x)])^2 = 28^2
16*(x/x+1) + (169*(x+1)/x) + 2*4*13*1 = 784
((16x)/(x+1)) + 169 + (169/x) + 104 = 784
((16x)/(x+1)) + (169/x) = 511
16x^2 + 169x + 169 = 511(x^2 + x)
16x^2 + 169x + 169 = 511x^2 + 511x
495x^2 + 342x - 169 = 0
495x^2 + 507x - 165x - 169 = 0
3x(165x + 169) - 1(165x + 169) = 0
(165x + 169) (3x-1) = 0
(165x + 169)= 0
165x=-169, x = -169/165
3x-1=0, x = 1/3

Like? Yes (1) | No (1)  

K.Bhagyasri   11 years AGO

by applying square on both sides
495X^2+342X-169=0

Like? Yes (1) | No (1)  

Nitin Jaiswal   11 years AGO

expand the equation as 4x+13(x+1)=28*x^1/2*(x+1)^1/2
square both sides and find the root of the quadratic equation which is derived i.e,
495x62+342x-169=0
the roots are 1/3 and -169/165.

Like? Yes (1) | No (2)  

31 Mar 2014 Sequence

What is the next number of the following sequence
16, 71, 87, 158, 141, 61, 76, 137, ........

OPtion

1) 132
2) 122
3) 167
4) 139
5) 147
6) 127
7) 114
8) 111
9) 135
10)None of these

Solution

16 = (1+6)(1) = 7(1) = 71 (Sum of digits)(use first digit of number)
71 = (7+1)(7) = 8(7) = 87
87 = (8+7)(8) = 15(8) = 158
158 = (1+5+8)(1) = 14(1) = 141
141 = (1+4+1)(1) = 6(1) = 61
61 = (6+1)(1) = 7(6) = 76
76 = (7+6)(7) = 13(7) = 137
Similarly
137 = (1+3+7)(1) = 111
Option 8)

Correct Option: 8 Best Solution (63)

ANUP KUMAR   11 years AGO

1st digit of previous no. and sum of all digit
= next no.
so,previous no. is 137
ist digit of no.=1 and sum of all digit=1+3+7=11
so,next no. is 111

Like? Yes (4) | No (2)  

sowjanya   11 years AGO

16,71,87,158,141,61,76,137

1+6= 7 it will be next num starting and second digit will be 1st num (16) ten's place number i.e 1 => 71 similarly
7+1=8 => 87
8+7=15=>158
1+5+8=14=> 141
1+4+1=6=>61
6+1=7=>76
7+6=13=>137
1+3+7=11=>111

ans is 111

Like? Yes (2) | No (1)  

Dr.Dipin Singh   11 years AGO

16..1+6=7 and place first digit of previous number next to 7 to get 71 as next number.

similarly next number after 137 (1+3+7=11) will be 111.

Like? Yes (2) | No (1)  

vaibhav rajhns   11 years AGO

16 ,71,87,158,141,61,76,137,?
71 = > (1+6)=7 &leftmost digit 1 i.e 71
87 => (7+1)=8 &leftmost digit 7 i.e 87
so for 137
(1+3+7)=11 &leftmost digit 1 i.e 111

Like? Yes (1) | No   

Pargat Pal Singh   11 years AGO

Add most significant digit as least significant digit by placing it after sum of digits of previous number to get next number.
as 16..1+6=7.. put 1 after 7 to get 71.
71...1+7=8.. put 7 after 8 to get 87.
..
similarly
137..1+3+7=11.. put 1 after 11 to get 111.

Like? Yes (2) | No (1)  

Kaustubh karekar   11 years AGO

The numbers 16, 71, 87, 158, 141, 61, 76, 137 are in sequence as follows:

16(6+1)=7 and first digit 1
the next is 71
and so on
the last number is 137
sum=1+3+7=11 and first digit is 1
so next number in the series is 111

Like? Yes (1) | No   

megha   11 years AGO

Answer is 8: 111
Each number in the series is equal to addition of all the digits of previous number followed by the first digit of the previous number.
After 16,
2nd no. is 71, 1+6=7(all digits of previous no.) followed by 1(1st digit of previous no.)
Similarly 3rd is 87, (7+1=8 & 7)
4th 158, (8+7=15 & 8)
5th 141, (1+5+8=14 & 1)
6th 61, (1+4+1=6 & 1)
7th 76, (6+1=7 & 6)
8th 137, (7+6=13 & 7)
9th 111, (1+3+7=11 & 1)

Like? Yes (2) | No (1)  

Udhaya kumar   11 years AGO

addition of individual digits & 1st digit put last digit........

Like? Yes (2) | No (1)  

Pradeep   11 years AGO

1+6=7 and 1st digit of privious number 71,, similarly 1+3+7=11 and ist digit of 137 === 111

Like? Yes (1) | No   

SAM   11 years AGO

Left most digit followed by is sum of all digits of previous no.
so, series is...
1+6 = 71
7+1 = 87
8+7 = 158
. .
. .
. .
so last no is...1+3+7 = 111

Like? Yes (2) | No (1)  

nalini singh   11 years AGO

1+6=71
7+1=87
8+7=158
1+5+8=141
1+4+1=61
6+1=76
1+3+7=111

Like? Yes (1) | No   

Ragaven   11 years AGO

add numbers between them (1+6,7+1=78)interchange that to get the answer (ie : 78 = 87 ) same step follows for every one and if answer came in 2 digit for 3 digit number them take 1 st number only

Like? Yes (2) | No (1)  

Prabhat yadav    11 years AGO

add both digit of the no.
and put the first digit of the no. on the second position of new no.

Like? Yes (1) | No   

Aayush shah   11 years AGO

1+6+1=7
startinf of first number=1
so,71

Like? Yes (1) | No   

Pritish jha   11 years AGO

Add the sum of digits of the given number and put the first digit of the number in units place.
16 -> 1+6=7 -> 71
71 -> 7+1=8 -> 87
87 -> 8+7=15->158
.................
.................
137-> 1+3+7=11-> 111

Like? Yes (1) | No   

bidisha mandal   11 years AGO

1+6=7,and then the 1st digit 1=71.
7+1=8,and then the 1st digit 7=87.
8+7=15,then the 1st digit 8 =158.
1+5+8=14,then the 1st digit 1=141.
1+4+1=6,then the 1st digit 1=61.
6+1=7,then the 1st digit 6=76.
7+6=13,then the 1st digit 7=137.
so,1+3+7=11,then the 1st digit 1=111.

Like? Yes (1) | No   

smily   11 years AGO

1+6=7 which is the first digit in second number and 1 in first number becomes last in second number and this continues so we get answer as
1+3+7=11 and 1 in 135 should come in last of 11 i.e answer is 111

Like? Yes (1) | No   

DHAKAD NEETENDRA SINGH   11 years AGO

add the digits and put the ten's or hundred's digit to the right side of the sum to get the next number
16=6+1=71
71=7+1=87
87=8+7=158
.
.
.
137=1+3+7=11 and 1 = 111 ans...

Like? Yes (1) | No   

Navnath Gaikwad   11 years AGO

addition of digit and append first digit to next no
16=(1+6)append first digit ans 71

Like? Yes (2) | No (1)  

lokesh kumar   11 years AGO

1+6=7 so 2nd term is 71 i.e sum of digits of previous no with the digit at left most place is the once place in the new no.7+1=8:-87
8+7=15:-158 now sequence changes from 141 i.e
1+4+1=6:-61
6+1=7:-76
7+6=13:-137
1+3+7=11:-111

Like? Yes (1) | No   

roshani munde   11 years AGO

the no. aftr 137 will be having 1 at its unit place & the no. before will b the sum of 1+3+7=11

Like? Yes (1) | No   

gaurav bhatt   11 years AGO

16 = 1+6=7 and first digit of previous number =71

71= 7+1= 8 and first digit of previous number = 87

87= 8+7= 15 and first digit of previous number = 158

158= 1+5+8= 14 and first digit of previous number = 141

141= 1+4+1= 6 and first digit of previous number =61

61= 6+1=7 and first digit of previous number = 76

76= 7+6= 13 and first digit of previous number =137

137= 1+3+7= 11 and first digit of previous number = 111

answer= 111

Like? Yes (1) | No   

rahul   11 years AGO

1+7=8
8+3=11
111

Like? Yes (1) | No   

AJAY KUMAR   11 years AGO

1+6=71
7+1=87
8+7=158
......
1+3+7=111

Like? Yes (1) | No   

NIRUPOM GHOSH   11 years AGO

16=(1+6)1=71,87=(8+7)8=158.....137=(1+3+7)1=111

Like? Yes (1) | No   

Pallab Das   11 years AGO

16 (6+1)=71
71 (7+1)=87
87 (8+7)=158
.
.
.
137 (7+3+1)=111

Like? Yes (1) | No   

mansi garg   11 years AGO

16 1+6=7 and the first term of 16=71
7+1=8 and the first term of 71=87
and so on
1+3+7=11 and the first term of 137=111

Like? Yes (1) | No   

ADVAITA BASAK   11 years AGO

137
1+3+7=11
and the 1st digit of previous no i.e. 137 is 1
so the next no will be 111

Like? Yes (2) | No (2)  

Nitin Jaiswal   11 years AGO

the next number is found by summing all the digits of the previous number followed by the first digit of the previous number.
for eg:71=(1+6)*10+1
141=(1+5+8)*10+1
similarly,the next number will be (1+3+7)*10+1

Like? Yes (1) | No (1)  

Subhash chandra pal   11 years AGO

next number of the series = sum of all the digits of previous number and the first digit of previous number

example
previous = 16
next will be= (1+6)1=71

previous=71
next=(7+1)7
=87

so on....

Like? Yes (1) | No (1)  

ashish kumar   11 years AGO

the digit on the left most of the number in sequence is the unit digit for the next number.
16 so 1 is at left most so in 71 ,1 is at the unit digit.
besides the unit digit of the next no in the sequence the rest digit is the sum of previous no.
16=1+6=7 at tens place and 1 at unit's place=71
71=7+1=8 at tens place and 7 at unit place=87
87=8+7=15 at left most place and 8 at units place=158
and so on..........
137=1+3+7=11 at left most place and 1 at unit place=111

Like? Yes (2) | No (2)  

Mahalakshmi   11 years AGO

16 =>(1+6) && 1(1st digit of 16) =71
71 =>(7+1) && 7(1st digit of 71) =87
87 =>(8+7) && 8(1st digit of 87) =158
158=>(1+5+8) && 1 =141
141=>(1+4+1) && 1 =61
61 =>(6+1) && 6 =76
76 =>(7+6) && 7 =137
137=>(1+3+7)&&1 =111
so ans is 8)111

Like? Yes (1) | No (1)  

AVIJIT   11 years AGO

1st =16 ,2nd = (1+6) 1st digit=71
.
4th number 158.5th is (1+5+8)1st digit=141
so trick is sum of all digit and 1st digit
so 137, next will be [(1+3+7 )1st digit of previous num]=111

Like? Yes (1) | No (1)  

lakshman   11 years AGO

16=1+6=7&1(1st digit)=71
.
.

137=1+3+7=11(1st digit)=111

Like? Yes (1) | No (1)  

P.Selva Dhasni   11 years AGO

the ans is 8) 111.

from the following sequence,

the addition of each digit of a number is the first digit of the next number and another digit is obtained by the first digit of the previous number.

here, 1+6=7 and 1(first digit of 16)

7+1=8 and 7(first digit of 71)

8+7=15 and 8(first digit of 87)

1+5+8=14 and 1(first digit of 158)

1+4+1=6 and 1(first digit of 141)

6+1=7 and 6(first digit of 61)

7+6=13 and 7(first digit of 76)

1+3+7=11 and 1(first digit of 137)

thus the ans is 111......:)

Like? Yes (1) | No (1)  

avi   11 years AGO

1st =16 ,2nd = (1+6) 1st digit=71
.
4th number 158.5th is (1+5+8)1st digit=141
so trick is sum of all digit and 1st digit
so 137, next will be [(1+3+7 )1st digit of previous num]=111

Like? Yes (1) | No (1)  

ABY   11 years AGO

1st =16 ,2nd = (1+6) 1st digit=71
.
4th number 158.5th is (1+5+8)1st digit=141
so trick is sum of all digit and 1st digit
so 137, next will be [(1+3+7 )1st digit of previous num]=111

Like? Yes (1) | No (1)  

ABYSUP   11 years AGO

1st =16 ,2nd = (1+6) 1st digit=71
.
4th number 158.5th is (1+5+8)1st digit=141
so trick is sum of all digit and 1st digit
so 137, next will be [(1+3+7 )1st digit of previous num]=111

Like? Yes (1) | No (1)  

Sivanjali   11 years AGO

Given sequence is
16 71 87 158 141 61 76 137 ----
On observing the sequence
-----------------------------------------
case-I

we can get the unit digit from the leftmost digit of the previous number
from 16 we can get unit digit of 71 as 1
from 71 we can get unit digit of 87 as 7
and so on from 137 we can get the unit digit of answer as 1
---------------------------------------------
case II

The left most digits can be obtained by adding the digits of the previous number
from 16 we can get (1+6=7) as the leftmost digit of the 71 as 7
from 71 we can get (1+7=8) as the leftmost digit of the 87 as 8
and so on..........
from 137 we can get (1+3+7=11) as the leftmost digit of the answer as 11
------------------------------------------------
so from caseI and caseII the answer is

111

Like? Yes (2) | No (2)  

ASHISH MITTAL   11 years AGO

addition of digit and then 1st digit
137=11 1=111
option(8)

Like? Yes (1) | No (1)  

Surya Narayana K   11 years AGO

The unit digit is the leftmost digit from the previous number

The leftmost digits are sum of the digits from the previous number

so the answer is 111

Like? Yes (2) | No (2)  

Kowshik Sarker   11 years AGO

16
71=1st digit is (6+1) and last digit is the 1st digit of 16 means 1.
so after 137 it will be 111 because 7+3+1=11 is the first digit and the last digit will be 1 of 137.So the correct option is no.8

Like? Yes (1) | No (1)  

BHAVNISH BHUTANI   11 years AGO

if the number is AB then next number will be (A+B)A
e.g. 16 next number (1+6)1 =>71
71=>(7+1)7=>87
and so on..
137=>(1+3+7)1=>111

Like? Yes (1) | No (1)  

S.Rosyponmary   11 years AGO

1+6=7 previous first digit 1 so=>71
7+1=8 p.f.d 7 =>87
similarly
1+3+7=11 p.f.d 1=>111

Like? Yes (1) | No (1)  

Nilkantha Sen   11 years AGO

1+6=7
So next number after 16 = 71
7+1=8
So next number after 71=87
Hence sum of digits of 137=1+3+7=11
So next number after 137= 111

Like? Yes (1) | No (1)  

sruthi   11 years AGO

here take 16=1+6=7 then take first digit from 16 ie 1 so 71
next 71=7+1=8 then first digit of 71 is 7 so 87
next 87=8+7=15 then first digit is 8 so 158
next 158=1+5+8=14 then first digit is 1 so 141
after 137=1+3+7=11 then first digit is 1 so ans is 111

Like? Yes (1) | No (1)  

Devendra Marghade   11 years AGO

next number=Sum of the digits of previous number and last digit is the first digit of previous number
16: 6+1=7,1
71: 7+1=8,7
87: 8+7=15,8
158: 1+5+8=14,1
141: 1+4+1=6,1
61: 6+1=7,6
76: 7+6=13,7
137:1+3+7=11,1

Like? Yes (1) | No (1)  

Satish Zod   11 years AGO

Add individual numbers to get first number, put first number as second.

Like? Yes (1) | No (1)  

SUMIT KUMAR   11 years AGO

like16=1+6=7 n fist digt1 so become 71

Like? Yes (1) | No (1)  

KALAIYARASI PADHMANABAN   11 years AGO

16= 6+1= 7 and put the left most digit 1=71
71= 7+1 =87
similarly
137= 1+3+7=11 and leftmost digit is 1 so,=111

Like? Yes (1) | No (1)  

shashi   11 years AGO

sum of total digits of previous and including previous first digit

Like? Yes (1) | No (1)  

kushagra gupta   11 years AGO

because here we are adding the digit of individual number and then writing the result along with the left most no

Like? Yes (1) | No (1)  

Soham Dalwadi   11 years AGO

In Sequence,
next no. = (sum of digit)first digit of no.
ex.
16
71=(1+6)1
87=(7+1)7
so on...
137 will split (1+3+7)1 =111

answer is 111

Like? Yes (1) | No (1)  

champavat jaydipsinh padamsinh   11 years AGO

16
71=(1+6)1
87=(7+1)7
so on...
137 will split (1+3+7)1 =111

answer is 111

Like? Yes (1) | No (1)  

Hemalatha   11 years AGO

1+3+7=11 den first num comes..so 111..

Like? Yes (1) | No (1)  

Vivek Jain   11 years AGO

the sum of digits becoome the first part of the number and the other part is the first digit.
16 = > (1+6)1=71
71 = > (7+1)7=87
87 = > (8+7)8=158
similarly,
137 = > (1+3+7)1=111

Like? Yes (1) | No (1)  

Narra Sirisha   11 years AGO

first 16
then sum of the digits 1+6 and first digit 71
then 7+1 and first digit 87
and so on
137 sum of the digits 1+3+7=11 and first digit 1
ans is 111

Like? Yes (1) | No (1)  

Dinesh thatti   11 years AGO

add the digits of previous number that will be the starting digit of next number

previous number 1st digit comes at last digit of next number

16=1+6= 7 which is next digit starting number
1 in 16 comes at end of next number
so for 16 next number is 71
for 71 it is 87


for 137
next number is 111
so answer is 111 which is 8th option

Like? Yes (1) | No (1)  

susmit   11 years AGO

First digit of the number is the last digit of the next number.
Adding the digits will give remaining digits before the last digit.
For e.g. 1 + 6 = 7, Hence the next number is 71
7 + 1 = 8, Hence the next number is 87
Therefore 1 +3 + 7 = 11, Hence the next number is 111

Like? Yes (1) | No (1)  

anubhav verma   11 years AGO

1+6=7and 7-6=1=====>71
7+1=8&8-1=7======>87
8+7=15&15-7=8=====>158
1+5+8=14&14-(5+8)=1========>141
1+4+1=6&6-(4+1)=1====>61
6+1=7&7-1=6======>76
7+6=13&13-6=7======>137
1+3+7=11&11-(3+7)=1=======>111

Like? Yes (1) | No (1)  

saurabh   11 years AGO

here 16 , 1+6=7 and then 1st integer 71
as so on
to
1+3+7=11 and 1st integer is 1 of first 111

Like? Yes (1) | No (1)  

sayeda fatma    11 years AGO

ans:111
1+6 is 7
keeping the first digit as it is on unit place we get 71
Similarly
1+3+7 is 11
so we get 111

Like? Yes (1) | No (2)  

Saraswathy   11 years AGO

Each number is obtained by
first digit - sum of digits of previous number
Last digit - first digit of previous number

Like? Yes (1) | No (2)  

28 Mar 2014 Coded language

If in coded language
SHANKAR = 2016
NEERAJ = 2520
RAMESH = 144
Then
RAJAT = ?

OPtion

1) 3640
2) 2640
3) 8096
4) 5760
5) 5040
6) 6760
7) 3600
8) 3024
9) 5760
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)
SHANKAR = 19 8 1 14 11 1 18 = 8*14*18 = 2016 (Multiplication of even)
NEERAJ = 14 5 5 18 1 10 = 2520
RAMESH = 18 1 13 5 19 8 = 144
Similarly
RAJAT = 18 1 10 1 20 = 3600

Option 7)

Correct Option: 7 Best Solution (16)

Dr.Dipin Singh   11 years AGO

Product of positional value of letters having even positional value in English Alphabets.
Like B=2; ;D=4; F=6..
So RAJAT = 18(R)*10(J)*20(T) =3600

Like? Yes (18) | No (6)  

Akanksha Mishra   11 years AGO

Given
SHANKAR = 2016
NEERAJ = 2520
RAMESH = 144

A(1),B(2),C(3)................,Y(25),Z(26)

1].SHANKAR
S(19),H(8),A(1),N(14),K(11),A(1),R(18)
Multiply the even position values
= 8*14*18
= 2016

Similarly
RAJAT
R(18),A(1),J(10),A(1),T(20)
Multiply the even position values
= 18*10*20
= 3600

SO Answer will be 3600(option 7)

Like? Yes (12) | No (6)  

RAKESH   11 years AGO

multiply only even no.s assigned to alphabets

Like? Yes (9) | No (8)  

Santosh Vishwakarma   11 years AGO

A(1),B(2)............Z(26)
1) S H A N K A R
(18)(8)(1)(14)(11)(1)(18)
multiplication of even numbers(18*8*14)=2016
note:take one number only one time in multipy
R(18)A(1)J(10)A(1)T(20)....18*10*20=3600.

Like? Yes (1) | No   

Udhaya kumar   11 years AGO

product of even alphanumerics...
RAJAT = R * J * T = 18*10*20=3600.....

Like? Yes (4) | No (5)  

P.Selva Dhasni   11 years AGO

the ans is 7) 3600

multiply the even numbers

SHANKAR => 19 8 1 14 11 1 18 =>8*14*18 => 2016

NEERAJ => 14 5 5 18 1 10 => 14*8*10 =>2520

RAMESH => 18 1 13 5 19 8 => 18*8 =>144

RAJAT =>18 1 10 1 20 =>18*10*20 =>3600

Like? Yes (3) | No (6)  

Ragaven   11 years AGO

product of even alpha numerics...
RAJAT = R * J * T = 18*10*20=3600.....

Like? Yes (2) | No (5)  

J.DHANDAPANI   11 years AGO

EVEN NOS ARE MULTIPLIED
ANS:18X10X20=3600

Like? Yes (1) | No (4)  

Mithra MR   11 years AGO

EVEN NOS ARE MULTIPLIED
ANS:18X10X20=3600

Like? Yes (1) | No (4)  

Satish Zod   11 years AGO

Consider A=1, B=2,............Z=26.
Multiply even numbers corresponding to alphabets in name.

Like? Yes (1) | No (4)  

Ann Theressa   11 years AGO

write the numbers corresponding to alphabets.ie
A = 01
B = 02
C = 03
.
.
.
Z =26
and multiply the even numbers
SHANKAR
19, 08, 01, 14, 11, 01, 18
8*14*18 = 2016

NEERAJ
14, 05, 05, 18, 01, 10
14*18*10 = 2520

RAMESH
18, 01, 13, 05, 19, 08
8*18 = 144

RAJAT
18, 01, 10, 01, 20
18*10*20 = 3600

Like? Yes (3) | No (7)  

ASHISH MITTAL   11 years AGO

if a=1,b=2,c=3 and so on....
in given word answer is obtained by multiply the digit which is even
so for rajat the even no is 18,10,20 and their multiplication will be 3600...so answer will be option (7)

Like? Yes (1) | No (5)  

azharshahid01@gmail.com   11 years AGO

In coded Language,
A=1,
B=2,
...
...
z=26.
Multiply all even numbers in a word.
SHANKAR = 19,8,1,14,11,1,18 => 8*14*18= 2016
NEERAJ = 14,5,5,18,1,10 => 14*18*10 = 2520
RAMESH = 18,1,13,5,19,8 => 18*8 = 144
Similarly,
RAJAT = 18,1,10,1,20 => 18*10*20 = 3600
Thus answer is option 7) 3600

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ADVAITA BASAK   11 years AGO

R=18, A=1, J=10, A=1, T=20

Multiplying the even numbers.....

18*10*20=3600

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hanamant k   11 years AGO

multiply only even digit of letters sum is th result
exmple
R-18 A-1 J-10 A-1 T-20

so
18*10*20=3600

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muni   11 years AGO

36 first two digits are perfect square and divisible by 4 and 9

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