M4maths Previous Puzzles

02 Dec 2015 Time and Work

If 9 boy working 15/2 hours a day can finish a work in 20 days, then how many days will be taken by 12 boy working 6 hours a day to finish the work, it being given than 3 boy of latter type work as much as 2 boy of he former type in the same.

OPtion

1) 19/2
2) 21/2
3) 17/2
4) 11
5) 15
6) 25/2
7) 27/2
8) 19
9) 9
10)13

Solution

More boy, less days (Direct)
Less hours a day, more days (Indirect)
More speed, less days (Indirect)
Men 12 : 9

Hr/day 6:15/2

Speed 1/2 : 1/3

:: 20 : x

x = 9* 15/2 * 1/3 * 20 * 1/12 * 1/6 * 2 = 25/2 days

Option 6)

Correct Option: 6 Best Solution (10)

MRITUNJAY   9 years AGO

B1/B2=3/2
so capacity are in ratio 3:2.
m1*d1*h1/c1=m2*d2*h2/c2
9*15*20/2*3=12*6*x/2

solve x=25/2

option 6

Like? Yes (2) | No   

ROHIT KUMAR DUBEY   9 years AGO

from formulae
M1*D1*T1/W2 =M2*D2*T2/W1.....(i)
also given
Efficiency 3boys of latter =2boys of former
i.e B2=3B1/2....(ii)
Put all value in eq(i)
9*7.5*20/1=(12*3B1/2 *6*X)/1......(let X is working day of 2nd type)
solving this X=25/2

Like? Yes (2) | No   

vishal varma   9 years AGO

we know that 3 boys latter can do work as much as 2 of former type therefore w2=(3/2)*w2
substitute in formula (m1 d1 h1)/w1 =(m2 d2 h2)/w2
9* 15/2 *20=3/2 * w2 *12*6
w2=25/2

Like? Yes (2) | No   

ABHINAV PUSHPAM   9 years AGO

capacity are in ratio
3:2.
m1*d1*h1/c1=m2*d2*h2/c2
9*15*20/2*3=12*6*x/2

solve x=25/2

option 6)

Like? Yes (1) | No   

Deepak kumar singh   9 years AGO

2/5, 3/5
9*15/2*2/5 men-hour completes in 20 days
=> 12*6*3/5 ,, ,, ,,
( 20*9*15/2*2/5)/(12*6*3/5)

=> 25/2 answer

Like? Yes (1) | No   

RAHUL KUMAR   9 years AGO

B1/B2=3/2
so, capacity are in ratio 3:2.

m1*d1*h1/c1=m2*d2*h2/c2

9*15*20/2*3=12*6*x/2

solve x=25/2

option (6) is correct ans.

Like? Yes (1) | No   

PRABHAT KUMAR   9 years AGO

A/c=(former type boy B1) /(latter type boy B2) =2/3
3(B1)=2(B2)
Now latter type 12 boy =formal type 8 boy
18()

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

Former 9 boy's total work in hours=9*20*15/2=1350
If firmer boy's work=B1 & latter boy's work=B2, Given 2B1=3B2 or B1=3B2/2, So 12 B2's work=3*12/2=18 B1's work.
So days needed to complete the 1350 hr. of work by 12 latter type =18 former type=1350/(6*18)=25/2
Option 6)

Like? Yes (1) | No   

REYAZ FAHIM ANSARI   9 years AGO

according to chain rule
12*6*3*x=9*15/2*2*20
x=25/2

Like? Yes (1) | No   

harshavi   9 years AGO

we have the formula (m1*h1*d1/w1)=(m2*h2*d2/w2)
so we can say that 15*9*20/(2*b1)=12*6*d/(2/3*b1){since the work done by the latter type of 3 boys wold be equal to 2 boys of first type}
thus d=25/2

Like? Yes (1) | No   

01 Dec 2015 Value of x

If 2^(x) = 4^(y) = 8^(z) and 1/(2x) + 1/(4y) + 1/(4z) = 4
then the value of x is

OPtion

1) 7/24
2) 7/36
3) 7/18
4) 7/48
5) 7/87
6) 7/15
7) 7/32
8) 7/12
9) 7/29
10)7/16

Solution

Let 2^x=4^y=8^z=k
then 2 = k^(1/x)
4 = k^(1/y) and 8 = k^(1/z)
Now,
8*4 = 32 = 2^5
k^(1/z).k^(1/y) = k^(5/x)
1/y + 1/z = 5/x
Solve
7/16
Option 10)

Correct Option: 10 Best Solution (30)

Pragati Rode   9 years AGO

2^(x) = 4^(y) =8^(z)
convert all base in same variable
from this equation, we get 2^(x) = 2^(2x) = 2^(3x)
y =x/2 and z=x/3
put this value in second equation, we get
1/2x + 1/(4*x/2) + 1/(4*x/3)=4

1/2x + 1/2x + 3/4x = 4
by solving this we get, 7 = 16x
x = 7/16
so answer is option 10)

Like? Yes (43) | No (15)  

Yadu Vir Singh   9 years AGO

Given:- 2^x = 4^y = 8^z
=> 2^(x) = 2^(2 y) = 2^(3 z)
So, => x = 2 y = 3 z
Putting these values in 2nd equation, it will be:-
1/(2 x) + 1/(2 x) + 3/(4 x) = 4
16 x = 7
x = 7/16

Like? Yes (40) | No (23)  

Rama lakshmi   9 years AGO

2^(x)=4^(y)=8^(z)
So x=2y=3z;
1/(2x)+1/(4y)+1/(4z)=7/(4x)=4
x=7/16

Like? Yes (17) | No (8)  

THOWSIF   9 years AGO

2^(x)=2^(2y)=2^(3z)
x=2y=3z
1/2x + 1/4(x/2) + 1/4(x/3)
x=7/16

Like? Yes (13) | No (7)  

Suraj Mishra   9 years AGO

2^x =4^y=8^z
taking log
x(log(2)=2y(log(2)=3z(log(2)
and log will cancel out
so x=2y=3z
puting this value in 2nd equation
1/(2x)+ 1/2x+ 3/4x= 4
which gives x=7/16

Like? Yes (8) | No (6)  

SHUBHAM   9 years AGO

using property of log,we get the equation
xlog2=2ylog2=3xlog2
thus X=2y=3z
and now this we get after substituting,X=7/16

Like? Yes (6) | No (5)  

Pooja Shaw   9 years AGO

2^(x)=4^(y)=8^(z)
2^x=2^(2y)=2^(3z)
x=2y=3z
y=x/2 z=x/3
1/2x+1/4x+1/4z=4
1/2x+1/2x+3/4x=4 (putting the value of y,z)
(2+2+3)/4x=4
7=16x
x=7/16
thus,correct option is 10)7/16

Like? Yes (3) | No (2)  

MRITUNJAY   9 years AGO

2^x=2^2y=2^3z

x=2y=3z=k

x=k,y=k/2,z=k/3
1/2k+1/4*k/2+1/4*k/3=4

k=7/16

x=7/16

Like? Yes (3) | No (3)  

ABHINAV PUSHPAM   9 years AGO

2^(x) = 4^(y) = 8^(z)
or, 2^(x) = 2^(2y) = 2^(3z)
x = 2y = 3z
therefore, 1/(2x) + 1/(4y) + 1/(4z) =4
1/(2x)+ 1/(2*2y) + 1/(4/3*3z) =4
1/2x + 1/2x + 1/(4/3x) = 4
so, X= 7/16
option (10) is correct.

Like? Yes (6) | No (6)  

anuradha   9 years AGO

since equating powers from first condition we get, x=2y=3z
substituting the value of x in second condition, we get 7/16

Like? Yes (2) | No (2)  

Jamil Ahmed   9 years AGO

2^(x)=4^(y)=8^(z)
Therefore, x=2y=3z.

By putting the values of y & z in 1/2x +1/4y + 1/4z = 4,
we get 1/2x + 1/2x + 3/4x = 4
or 7/4x = 4
Therefore, x = 7/16.

Like? Yes (2) | No (2)  

Deepak kumar singh   9 years AGO

2^x=2^2y=2^3z => x=2y=3z
=>1/2x+2/4 x+3/4x=4
=>x=7/16

Like? Yes (1) | No (1)  

Ritesh kumar   9 years AGO

ans=7/16
2^x=4^y, 2^x=8^z
log2^x=log4^y , log2^x=log8^z
xlog2=ylog4 , xlog2=zlog8
on solving this
x=2y, x=3z
put in the eq..
and then found x= 7/16

Like? Yes (4) | No (4)  

Venkateswarlu   9 years AGO

from the first equation==>x=2y=3z;
y=x/2;z=x/3;
(1/2x)+(2/4x)+(3/4x)=4 ==> (1/x)+(3/4x)=4
(4+3/4x)=4 ==> 7=16x ==> x=7/16.

Like? Yes (5) | No (5)  

CHANDAN KUMAR   9 years AGO

given
2^x=2^2y =2^3z

so,
x=2y=3z..... (I)

now,
1/2x + 1/4y +1/4z =4
solving we get,
z= 7/48

but
x =3z =3×7/48 = 7/16
correct options are 10) 7/16

Like? Yes (2) | No (2)  

RAHUL KUMAR   9 years AGO

2^(x) = 4^(y) = 8^(z)
or, 2^(x) = 2^(2y) = 2^(3z)
x = 2y = 3z
therefore, 1/(2x) + 1/(4y) + 1/(4z) =4
1/(2x)+ 1/(2*2y) + 1/(4/3*3z) =4
1/2x + 1/2x + 1/(4/3x) = 4
so, X= 7/16
option (10) is correct.

Like? Yes (5) | No (6)  

T Purushotham   9 years AGO

by x=2y,x=3z replacing in the euation we get x=7/16
1/2x+1/4*(x/2)+1/4*(x/3)=4;
x=7/16

Like? Yes (3) | No (4)  

Devendra Marghade   9 years AGO

2^(x)=4^(y)=2^(2*y), x=2y or y=x/2 and
2^(x)=8^(z)=2^(3*z), x=3z or z=x/3
Given 1/(2x)+1/(4y)+1/(4z)=4
So substituting values of y, z in terms of x we get
1/(2x) + 2/(4x) + 3/(4x) =4
7/4x=4, x=7/16
Option 10)

Like? Yes (2) | No (3)  

Nandkishor jaiswal   9 years AGO

given that
2^x=4^y=8^z
so x=2*y=3*z
from equation 1/2x+1/4y+1/4z=4
x=7/16

Like? Yes (1) | No (2)  

teja srinivas   9 years AGO

2^(x)=4^(y)=8^(z)
this will become x=2y=3z
by simplification
x=7/16

Like? Yes (1) | No (2)  

vamsi krishna   9 years AGO

we got x=2y=3z
y=x/2, z=x/3
1/2x + 1/2x + 3/4x = 4
7/4x = 4
x = 7/16

Like? Yes (2) | No (3)  

REYAZ FAHIM ANSARI   9 years AGO

2^x =4^y=8^z by taking log we get x=2y, 2y=3z, x=3z
after putting the values of y & z in terms of x we get,
x= 7/16

Like? Yes (1) | No (2)  

SHIVAM KUMAR KAUSHIK   9 years AGO

2^x=2^2y=2^3z
x=2y=3z
put the vaule in given equation,
1/2x +2/4x +3/4x =4
x=7/16

Like? Yes (1) | No (2)  

TARUN CHAUHAN   9 years AGO

2^x=8^z
taking log z=x/3
similarly y=x/2

put it in eq. we get
1/2x+1/2x+3/4x=4

x=7/16

Like? Yes (1) | No (2)  

vishal varma   9 years AGO

here from given equation we get x=2y=3z
from that y=x/2 ; z=x/3
substitute these values in equation 2 and by soving we get
x=7/16

Like? Yes (2) | No (4)  

chetan   9 years AGO

From 1st condition we get x=2y=3z.
From 2nd condition we get 2yz+xz+xy=16xyz.
2yz+2yz+3yz=16xyz
=> x=7/16.
So option 10 is correct.

Like? Yes (2) | No (4)  

armaan   9 years AGO

x=2y=3z
then palce all the value of x,y and z in the given quation..where we find 7/16

Like? Yes (1) | No (3)  

PRABHAT KUMAR   9 years AGO

2^x=2^2y=2^3z
x=2y=3z=k
x:y:z=6k=3k=2k
Putting value in given equation we can get value of k
1/12k+1/12k+1/8k=4
k=7/96
Therefore x=6k=6*(7/96)=7/16
Option 10

Like? Yes (1) | No (4)  

harshavi   9 years AGO

2^(x)=2^(2y)=2^(3y)
therefore x=2y=3z
thus by substituting we get
(1/2x)+(1/2x)+(3/4x)=4
(7/4x)=4
x=7/16

Like? Yes (1) | No (4)  

Nimish   9 years AGO

when bases are same powers are equal Therefore 2^x=2^2y=2^3z
evaluate x in terms of y and x in terms of z put in 2nd equation and solve it

Like? Yes (1) | No (5)  

30 Nov 2015 Sequence

What is the next number of the following sequence
26, 51, 100, 197, 389,......

OPtion

1) 772
2) 767
3) 786
4) 774
5) 768
6) 779
7) 789
8) 728
9) 771
10)759

Solution

26
26*2 - 1 = 52-1 = 51
51*2 - 2 = 102-2 = 100
100*2 - 3 = 200-3 = 197
197*2 - 5 = 394-5 = 389
Similarly
389*2 - 7 = 778-7 = 771
Option 9)

Correct Option: 9 Best Solution (10)

rakhi    9 years AGO

26x2=52-1=51
51x2=102-2=100
100x2=200-3=197
197x2=394-5=389
389x2=778-7=771

Like? Yes (6) | No (2)  

Meenakshi   9 years AGO

(double of given number) - (Consecutive Prime number Staring with 1) = Next number
(26+26) -1 = 51
(51+51) -2 = 100
(100+100) -3=197
(197+197) -5=389
(389+389) - 7=771

Like? Yes (3) | No (1)  

harshavi   9 years AGO

the series is as
26*2-1,
51*2-2
100*2-3
197*2-5
389*2-7=771

Like? Yes (1) | No   

sri   9 years AGO

26*2=52-1;
51*2=102-2;
100*2=200-3;
197*2=394-5;
389*2=778-7=771;

Like? Yes (2) | No (2)  

chaitanya krishna   9 years AGO

difference is 25 multiply with 2 and subtract one for the following two consecutive numbers after that subtract two

Like? Yes (3) | No (3)  

Ritesh kumar   9 years AGO

26 51 100 197 389 ... next is 771
double the number and subtract the prime number
389*2=778
778-7=771

Like? Yes (2) | No (2)  

Pragati Rode   9 years AGO

second number = (first number * 2 ) - 1
third number = (second number * 2 ) - 2
fourth number = (third number * 2 ) - 3
fifth number = (fourth number * 2 ) - 5
so
sixth number = (fifth number * 2 ) - 7
i.e. (389 * 2 ) - 7 = 771
option 9)

Like? Yes (1) | No (1)  

sahil arora   9 years AGO

26*2-1=51
51*2-2=100
100*2-3=197
197*2-5=389
389*2-7=771
prime number are substracted

Like? Yes (2) | No (2)  

jalaj   9 years AGO

26 *2=52-1=51
51*2=102-2=100
100*2=200-3=197
197*2=394-5=389
389*2=778-7=771

Like? Yes (1) | No (2)  

RAHUL KUMAR   9 years AGO

26*2-1=51
51*2-2=100
100*2-3=197
197*2-5=389
389*2-7=771
option (9) is correct.

Like? Yes (1) | No (3)  

27 Nov 2015 Probability

In a chess game a man plays 16 moves, one move to every piece in the following manner:-
- King can move 1 step in any direction
- Queen can move in any direction
- Rook can move straight in any direction
- Bishop can move in slant squares
- Knight can move 2 forward then 1 turned step
- Every pawn can move 1 or 2 step forward
What is the probability that after that move all the piece are arranged in third and fourth row if initially they are arranged in first and second row as the rule of game.

OPtion

1) 3/5
2) 1/2
3) 1/3
4) 1/4
5) 4/5
6) 5/6
7) 2/3
8) 3/4
9) 2/5
10)0

Solution

Since King can move only one place in any direction hence it is impossible to take it in third or fourth row.

Option 10)

Correct Option: 10 Best Solution (3)

ABHINAV PUSHPAM   9 years AGO

Since King can move only one place in any
direction hence it is impossible to take it in third or fourth row.

Like? Yes (3) | No   

RAHUL KUMAR   9 years AGO

Since King can move only one place in any direction hence it is impossible to take it in third or fourth row.

Like? Yes (2) | No (1)  

MRITUNJAY   9 years AGO

Since King can move only one place in any
direction hence it is impossible to take it in third or fourth row.

Like? Yes (2) | No (1)  

26 Nov 2015 Volume

Spherical bullets can be made out of a solid lead cube whose edge measures 33 cm. Each bullet has a radius of 1.5 cm and is required to be covered by another material of thickness 2mm. Find the volume of material required.

OPtion

1) 16776.72 cm^3
2) 15366.42 cm^3
3) 18576.12 cm^3
4) 16376.62 cm^3
5) 12346.82 cm^3
6) 14372.44 cm^3
7) 12444.66 cm^3
8) 16678.56 cm^3
9) 16736.26 cm^3
10)None of these

Solution

We have
cube edge length a = 33 cm
r = 1.5 cm
Volume of one bullet = 4/3 Pi r^3
= 4/3 *22/7 * (3/2)^3 = 11/7 * 9
Volume of cube = a^3 = 33^3
Let the number of bullets that can be made be n
n = Volume of cube / volume of one bullet
= 33*33*33/ (99/7) = 11*33*7 = 2541
So 2541 bullet can be made
Volume of material required = 2541*(4*Pi/3 (1.5+0.2)^3 - 4*Pi/3(1.5)^3)
= 16376.62 cm^3

Correct Option: 4 Best Solution (6)

Pooja Shaw   9 years AGO

edge of the cube=33 cm
vol of the cube=33^3 cm^3=35937 cm^3
radius of bullet=1.5 cm
vol of bullet=4/3*pi*(1.5)^3 cm^3=297/21 cm^3
radius of bullet with outer layer=1.7cm
vol of bullet with outer layer=4/3*pi*1.7^3 cm^3
=432.344/21
therefore,volume of the material=432.344/21-297/21
lets no of bullets is n
therefore,n*297/21=35937
solving this we get n=2541
therefore,volume of material in 2541 bullets=2541(432.344/21-297/21)=16376.62 cm^3
Thus correct answer is 4)

Like? Yes (4) | No   

Vinay Sipani    9 years AGO

volume of cube = 33^3 = 35937 cm^3
Volume of 1 bullet = (4/3)*pi*1.5^3 = 9*pi/2 cm^3
=> Number of bullets = 35937/(9*pi/2) = 2542

Volume of 1 material over bullet = (4/3)*pi*(1.7^3 - 1.5^3) = 769*pi/375 = x

=> Volume of material = 2542 * x
= 16376.42
=> option 4

Like? Yes (1) | No   

harshavi   9 years AGO

the volume of cube=33^3
the volume of spheres made =n*4*3.14*(1.5)^3
both are equal
thus we get n=2541(no of bullets that can be made)
thus for each bullet the volume of materail required is
4*3.14*((1.7)^3-(1.5)^3)/3=6.445cu.cm
thus for 2541 bullets we require 2541*6.445=16376.62 cm^3

Like? Yes (1) | No   

anand modi   9 years AGO

BY VOLUME CONSERVATION,
SUPPOSE N = NUMBER OF BULLET FORMED,
SO,
33^3 =N*(4/3)*3.1415*(1.5)^3,
AFTER SOLVING
N=2524,
THEREFORE VOLUME REQUIRED=N*(4/3)*3.1415*(1.7^3-1.5^3),
16376.62cm^3
option 4)

Like? Yes (1) | No   

bhaskar   9 years AGO

(33*33*33/(4*pi*1.5*1.5*1.5/3))*(4*pi(1.7*1.7*1.7-1.5*1.5*1.5)/3)=16376.62

Like? Yes (1) | No   

anjali gupta   9 years AGO

vol of cube=33*33*33=35937 cm^3
vol of sphere= 4*22*(1.5)^3/(3*7) cm^3
no. of sphere= vol of cube/vol of sphere=2541
vol of material around sphere= 2541*(4/3)(22/7)(1.7^3-1.5^3) cm^3
=16376.62 cm^3

Like? Yes (1) | No   

25 Nov 2015 Area

A rectangular plot is half as long again as it is broad. The area of the plot is 2/3 hectares. The length of the plot is

OPtion

1) 100 m
2) 66.66 m
3) 33.33 m
4) 100/Sqrt(3) m
5) 73.33 m
6) 90 m
7) 50/Sqrt(3) m
8) 120 m
9) 125.5 m
10)None of these

Solution

Let breadth = x
Then length = 1.5 x in metres
Then x*3/2 x = 2/3*1000
So, x = 2/3*100 = 200/3
Therefore Length of plot = 3/2 * 200/3 = 100 m
Option 1)

Correct Option: 1 Best Solution (12)

ABHINAV PUSHPAM   9 years AGO

Let breadth = b meters.
then, length = 3b/2 meters
∴ b x 3b/2 = 2/3 X 10000
⇒ b = (4 x 10000)/9
⇒ b = ( 2 X 100)/3 m
∴ Length = (3/2) x (2/3) x
100 m
= 100 m
option (1) is correct

Like? Yes (14) | No (2)  

Pooja Shaw   9 years AGO

The length is one and a half time the width
If width is x then length is 1.5x
Area=length*width
area=1.5x*x=1.5x^2=2/3(10000)
or,3/2x^2=2/3(10000)
or,x^2=40000/9
or,x=sqrt(40000/9)=66.667=width
length is 1.5 times the width=66.667*1.5=100.0005=100
so the correct answer is 100
thus,the correct option is 1)100m

Like? Yes (9) | No (6)  

Jaikumar Bhatia   9 years AGO

Let Breadth=x Meter
Length=y=x+1/x=3/2x meter
Therefore
(x+3/2x)=(2/3)*10000
x^2=(4/9)*10000
x=[(2/3)*100] m
Length=[(3/2)*(2/3)*100 m
=100 m
Ans is 100 meters
So Option is (1)

Like? Yes (3) | No   

Jamil Ahmed   9 years AGO

Let b metres be the breadth of the plot.
Then 3b^2/2 = 2*10000/3
Therefore, b = 200/3 mtrs.
Hence, length of the plot = 200/3 * 3/2 mtrs
= 100 mtrs

Like? Yes (2) | No   

MRITUNJAY   9 years AGO

Let breadth = b meters.
then, length = 3b/2 meters
? b x 3b/2 = 2/3 X 10000
? b = (4 x 10000)/9
? b = ( 2 X 100)/3 m
? Length = (3/2) x (2/3) x
100 m
= 100 m
option (1) is correct

Like? Yes (3) | No (1)  

ramji   9 years AGO

length=b+b/2
breadth=b
3b^2/2=2/3 hectares
so l is 100m

Like? Yes (2) | No (1)  

Chetharajupalli.Veerendar   9 years AGO

let a=breadth

then length =(3/2)a
then ,area=(length)*(breadth)=2*10000
20000=(3/2)a*(a)
a^2=(2/9)*20000
a=sqrt(40000/9)
a=66.667 m
then length =(3/2)*a
=(3/2)*66.667
=100 m

Like? Yes (3) | No (2)  

Mani kumar   9 years AGO

length = 1.5 * Area
area = l * b
let width = x
area = 1.5 * (x * x)
x2= 40000/9
x = 66.667
length = 1.5(x)
length = 100

Like? Yes (3) | No (2)  

N.Vennela Rani   9 years AGO

let b=x meters then l=1.5x meters
area=l*b=1.5*x^2=(2/3)*10000
=>x=100 meters

Like? Yes (2) | No (1)  

RAHUL KUMAR   9 years AGO

Let breadth = b meters.
then, length = 3b/2 meters
∴ b x 3b/2 = 2/3 X 10000
⇒ b = (4 x 10000)/9
⇒ b = ( 2 X 100)/3 m
∴ Length = (3/2) x (2/3) x
100 m
= 100 m
option (1) is correct

Like? Yes (3) | No (3)  

Surabhi Yadav   9 years AGO

let breadth=x metres
length=3/2*x*metres
x*3/2x=2/3*10000
x^2=4/9*10000
x=2/3*100 m
length=3/2*2/3*100=100m

Like? Yes (1) | No (1)  

chetan   9 years AGO

Let Breadth=x.
length=3*x/2.
(3x/2)*x=2*10000/3.
x=2*100/3.
then length=100 m.
so option 1 is correct.

Like? Yes (1) | No (1)  

24 Nov 2015 Area

A copper wire when bent in the form of a hexagon encloses an area of 726*Sqrt(3) cm^2. If the wire is bent to form a circle, find the area enclosed by it.

OPtion

1) 1510 cm^2
2) 1426 cm^2
3) 1396 cm^2
4) 1236 cm^2
5) 1386 cm^2
6) 1358 cm^2
7) 1378 cm^2
8) 1478 cm^2
9) 1486 cm^2
10)None of these

Solution

Using area we can compute length of the side of the hexagon. Then by equating perimeter of hexagon to circumference we can get radius of circle to compute area.
We have
Area of hexagon = 726*Sqrt(3) cm^2
Let side length be a.
Area = 3*Sqrt(3)/2 * a^2 = 726*Sqrt(3)
a^2 = 484
a = 22 cm
Perimeter of hexagon = circumference (=length of wire)
6a = 2 Pi r
r = 21 cm
Area = Pi r^2
= 1386 cm^2
Option 5)

Correct Option: 5 Best Solution (24)

bhaskar   9 years AGO

1.5*sqrt(3)*a*a=726*sqrt(3)
a=22
6a=132
132=2*pi*r
r=21
area=pi*r*r=1386

Like? Yes (1) | No   

Deepak kumar singh   9 years AGO

6* sqrt(3)*a^2/4=726*sqrt(3)
=> a=22
perimeter of hexagon=perimeter of circle
=> 6*22=2*pi*r
=>r=21
Area of circle= Pi*(21)^2
= 1386

Like? Yes (1) | No   

ABHINAV PUSHPAM   9 years AGO

area of hexagon = 3sqrt(3)*a^2/2
726sqrt(3)= 3sqrt(3)*a^2/2
so, A= 22cm
perimeter of hexagon = 6*22= 132
perimeter of hexagon = circumference of circle
2pi*r= 132
r=21
now, area = pi*r^2= 1386 cm^2
option 5)

Like? Yes (1) | No   

poojasaisri   9 years AGO

area of hexagon=(3*sqrt(3)/2)*a^2
given area of hexagon=726*sqrt(3)
by solving we get a=22
circumference of circle= perimeter of hexagon
2*pi*r=6*a
solving r=21
area of circle=pi*r^2
=(22/7)*21*21
=1386 cm^2

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

area of hexagon = 3sqrt(3)*a^2/2
726sqrt(3)= 3sqrt(3)*a^2/2
so, A= 22cm
perimeter of hexagon = 6*22= 132
perimeter of hexagon = circumference of circle
2pi*r= 132
r=21
now, area = pi*r^2= 1386 cm^2
option 5)

Like? Yes (1) | No   

RAHUL KUMAR   9 years AGO

area of hexagon = 3sqrt(3)*a^2/2
726sqrt(3)= 3sqrt(3)*a^2/2
so, A= 22cm
perimeter of hexagon = 6*22= 132
perimeter of hexagon = circumference of circle
2pi*r= 132
r=21
now, area = pi*r^2= 1386 cm^2
option 5)

Like? Yes (1) | No   

Pooja Shaw   9 years AGO

area of hexagon=(3sqrt3/2)a^2=726*sqrt3
by solving this we will get a=22
thus circumference of the circle=22*6=132
132=2piR
solving this we will get R=21.008
then calculating the area piR^2=pi(21.008)^2=1386 cm^2
thus 5)1386 cm^2 is the correct answer

Like? Yes (1) | No   

NAVEEN KUMAR   9 years AGO

area of hexagonal=(3*sqrt(3)*S)/2=726*sqrt(3)
so, S=22 c.m
perimeter of hexagonal=perimeter of circle=2*pi*r
=>22*6=2*(22/7)*r
=>r=21cm
Hence,Area of circle=(22/7)*21*21=1386 cm^2

Like? Yes (1) | No   

Isha Saxena   9 years AGO

given area of hexagon=726*sqrt(3) cm2
and area of hexagon=3*sqrt(3)*(1/2)*(a^2) cm2
solving the above equation we get a=22 cm
where a is the length of a side of hexagon
thus the total length of the copper wire=22*6=132 cm
now for a circle
2*(22/7)*radius=132 cm
thus radius=11*6*(7/22)
=21 cm
thus area =(22/7)*(radius^2)
=(22/7)*21*21
1386 cm^2=answer

Like? Yes (1) | No   

Jaikumar Bhatia   9 years AGO

Area of Hexagon=[3*Sqrt(3)/2]*a^2......(a is side of hexagon)
726*Sqrt(3)=[3*Ssrt(3)/2]*a^2...............(i)
Simplify (i)
a=Sqrt(484)
Perimeter of hexagon=6*a=6*Sqrt(484)...................(ii)
Perimeter of Circle=Pi*d=Perimeter of Hexagon.....(iii)
Pi*d=6*Sqrt(484)....................................from (ii)&(iii)
d=(6/Pi)*Sqrt(484).....................................(iv)
Area of Circle=(Pi/4)*d^2
=1386..............................(by putting the value of 'd' from (iv)
So ans is 1386Cm^2
Option is (5)

Like? Yes (1) | No   

Nandkishor jaiswal   9 years AGO

area of hexagon is = 3/2*sqrt(3)*a*a
or, 726*sqrt(3)=3/2*sqrt(3)*a*a
a=22 cm
so side of hexagon= 22 cm
now perimeter of hexagon=perimeter of circle
or, 6*a=2*pi*r
or, 6*22=2*pi*r
so, r=21
now area of circle will be
pi*21*21=1386

Like? Yes (1) | No   

ashok   9 years AGO

side length of hexagon=22(since hexagon forms six equilateral triangles)
radius of circle=66*7/22(length of the wire equal to perimeter of circle)
hence area=1386

Like? Yes (1) | No   

ramakrishnateja   9 years AGO

are aof hexagon is 726 sqrt(3).
but we know that area is 3 sqrt(3)l^2/2. so, equating two we get l^2=484. l=22. perimetr=6l=6*22.
for cirlce it is 2*pi*r=6*22. hence r=6*22/2*pi.
are of circle ispi*r^2=1386cm^2.

Like? Yes (1) | No   

Shubham Raj   9 years AGO

area of hexagon=(3*sqrt(3))* a^2/2
so,side of hexagon=22 cm
perimeter=22*6=132 cm =perimeter of circle=2r*3.14
so,,radius of circle=66/3.14
area of circle=1387.26

Like? Yes (1) | No   

Rhythm   9 years AGO

6x area of equilateral triangle=726sqrt(3).
side= 22cm
length of wire=132cm
area of circle= 1386 cm^2

Like? Yes (1) | No   

sumit kumar   9 years AGO

a/c to question
6*area of equilateral triangle=726*3^(1/2)
side=22cm
now have to find length of copper wire which is 22*6=132 cm
radius of circle which will form with same length
so, 2*22/7*r=132
r=21 cm
therefore, area enclosed by circle is 22/7*r^2=1386 cm^2.

Like? Yes (1) | No   

bhargavi b s   9 years AGO

by using the given data we can find side of the hexagon i.e.,a=22.by using this side we can find perimeter of the hexagon p=6(22)=132 that itself the perimeter of circle using that we can find radius r=21 finally the area of cicle a=1386 cm^2

Like? Yes (1) | No   

Saurabh Agrawal   9 years AGO

area of hexagon = (3 *sqrt(3)*a^2)/2=726*sqrt(3)cm^2
rom eq. =22cm
perimeter of hexago = 6*a=6*22=132cm=perimeter of circle
132=2*22*r/7
r=21cm
area of circle =(22*r^2)/7
area= (22*21^2)/7=1386cm^2

Like? Yes (1) | No   

shubham gupta   9 years AGO

726sqrt3÷(sqrt3/4)=a^2

A=22
So perimeter or hexa is 6*22=132

132=2'π४
४=21
Area =π४^2
=1386

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

If side of hexagon=a, then Area=(3/2)*Sqrt(3)*a^2=726*Sqrt(3)
a=22 and total length of side=6*a=6*22=132=Perimeter of circle.
So 2*Pi*R=132, R=132/(2*Pi)=66/Pi
Area enclosed by circle=Pi*R^2=Pi*[66/Pi]^2=66^2/Pi=1386
Option 5)

Like? Yes (1) | No   

Mukul   9 years AGO

Area of Hexagon = 3*sqrt3*a/2
a = sqrt(484)
perimeter of hexagon = 6*a
perimeter of circle = 2*pie*r
since we are using same copper wire,
6*a=2*pie*r
hence, r= 3*a/pie where a = sqrt(484)

Now, area of circle = pie*r*r
putting the value of r and a we get the area of circle to be,
1386.55 cm^2

Like? Yes (1) | No   

N.Vennela Rani   9 years AGO

area of hexagon=(3*3^(1/2)*a^2)/2=726*3^1/2
=>a=22,perimeter of the hexagon=6a=132
perimeter of the circle=2*pi*r=132
=>r=21.02
=>pi*r^2=1386 cm^2

Like? Yes (1) | No   

Surabhi Yadav   9 years AGO

Area of hexagon=(3*sqrt3)/2*side square
726*sqrt3=(3*sqrt3)/2*side square
side(a)=22cm
Perimeter of hexagon=6a
=6*22=132cm
Area of circle=Perimeter of hexagon
2*pi*radius=132
radius(r)=21cm
Area of circle=pi*r*r
=22/7*21*21
=1386cm^2

Like? Yes (1) | No   

Jamil Ahmed   9 years AGO

Each side of the hexagon = sqrt {726*sqrt(3)*4/6*sqrt(3)}
=22 cm

Length of the wire = Perimeter of the hexagon = 22*6 = 132 cm

Hence, radius of the circle = 132*7/2*22 = 21 cm

Therefore, area of the circle = 22*21^2/7 = 1386 cm^2

Like? Yes (1) | No   

23 Nov 2015 Inerest

What should be the least number of years in which the simple interest on Rs.2600 at 20/3% will be exact number of rupees ?

OPtion

1) 1 years
2) 2 years
3) 3 years
4) 4 years
5) 5 years
6) 6 years
7) 7 years
8) None of these

Solution

SI = Rs.(2600 * 20/3 * 1/100 * T)
which is an exact number of rupees when the least value of T is 3 years

Option 3)

Correct Option: 3 Best Solution (26)

Akshaya   9 years AGO

SI = Pnr/100, P=Interest, n= Number of years, r = rate of interest
Here P = 2600, r = 20/3
For n=1 and n=2, the interest amount will be 173.33 and 346.66 respectively.
For n = 3, Interest = 520
The least number of years at which the interest amount is whole number is Rs.520.
Hence, the answer is option (3).

Like? Yes (13) | No (6)  

Devendra Marghade   9 years AGO

Simple interest on Rs 2600 at 20/3% for 'x' years =2600*20*x/(3*100)=520x/3
For exact number of rupee, x=3,6,9,...
So least number of years=3
Option 3)

Like? Yes (6) | No (1)  

kriti kankariya   9 years AGO

to get exact value denominator should completely divide the numerator
here si=2600*20*t / (3*100)
as 2600 is divisible by 100
but neither of them is divisible by 3
so taking time as multiple of 3 will give exact value
as we need least time so it will be 3*1 = 3yrs

Like? Yes (5) | No   

sumit kumar   9 years AGO

exact no of rupees means not decimal amount.this will happen at 3 and 6 but according to question least no will 3. so option is 3.

Like? Yes (5) | No   

MRITUNJAY   9 years AGO

s.I =2600*20/3*t/100

so s.I be exact number t will be 3 years.

Like? Yes (2) | No   

lokesh kumar   9 years AGO

Because of when applying formula of simple interest the 3 will be divided get the SI in number of rupees.

Like? Yes (2) | No   

Venkateswarlu   9 years AGO

interest for 1 year=920/300)*2600=173.3333
after three years interest will be=173.3333*3=520.(exact no. of rupees)

Like? Yes (1) | No   

Deepanshu Chaturvedi   9 years AGO

SI= Rs. 2600*20/3*t/100= Rs 520/3
so t=3 years

Like? Yes (2) | No (1)  

Chandrajeet Yadav   9 years AGO

(2600*20/3*x)/100 = (520*x)/3
if we we put x =3 then it will be exact number of rupees.

(520*3)/3 = 520

Like? Yes (1) | No   

Jaikumar Bhatia   9 years AGO

I=Prt=2600*[(20/3*100)]*3=520
ans is 3 years
option is (3)

Like? Yes (1) | No   

ramakrishnateja   9 years AGO

si=p*t*r/100. si=2600*t*20/(3*100)=26*20*t/3.
for si to be equal to rupees then the nr should be multiple of 3. so least possible value is 3.

Like? Yes (1) | No   

Anil kumar   9 years AGO

simple interest= PRT/100
I=2600*20*3/(100*3)=520

Like? Yes (1) | No   

JENSI KARUNYA.S   9 years AGO

S.I=(2600*20/3*1/100)*T
=(520/3*T)
only if T=3 years we will get exact rupees.
so T=3 years

Like? Yes (2) | No (1)  

Rahul   9 years AGO

Put values P = 2600, R = 20/3 in I = PRN / 100. Value of N which makes the denominator 1 is N = 3. Hence, 3 years.

Like? Yes (1) | No   

chetan   9 years AGO

Simple interest=2600*20/3*1/100*t.
S.I=520/3*t.
t=3.
so option 3 is correct.

Like? Yes (1) | No   

Neha Giri   9 years AGO

As SI=P*R*T/100;
when t=1
Si=(2600*20*1)/(3*100)
which is not equal to an exact rupees.
Same when t=T=2 yrs.
When T=3 yrs,
SI=(2600*20*3)/(3*100)
=520, an exact number of rupees.

Like? Yes (1) | No   

Saurabh Agrawal   9 years AGO

intrest would be 26*20/3 =173.333...
that will be exact value after 3 years i.e 26*20*3/3=26*20=520

Like? Yes (1) | No   

Surabhi Yadav   9 years AGO

Simple interest = Rs. [2600 × 20 / 3 × 1 / 100 × T] = Rs.[  520 / 3 × T]

       Which is an exact number of rupees when T = 3

Like? Yes (1) | No   

Nandkishor jaiswal   9 years AGO

si=p*r*t/100
2600*20/3*t/100
=520/3*t
now t=3 will satisfy the simple interest as a positive integer
option 3

Like? Yes (1) | No   

Rajni   9 years AGO

3 is directly divisible so is 6.. but we are also getting the exact number of rupees.

Like? Yes (1) | No   

Pankaj   9 years AGO

(2600 x 20 x T)/(3 x 100) = SI
(26 x 20 x T) / 3 = SI
SI is interral only if T=3
=> T = 3 ANS.

Like? Yes (1) | No   

Isha Saxena   9 years AGO

interest=(principal*rate*time)/100
=[(2600)*(20/3)*time]/100
=(520*time)/3
thus for the answer to be in exact rupees time should be a multiple of 3.therefore 3 is the minimum no. of years.

Like? Yes (1) | No   

RAHUL KUMAR   9 years AGO

S.I. = Rs. [ 2600 × 20 / 3 × 1 / 100 × T ] = Rs.[ 520 / 3 × T ]
Which is an exact number of rupees when T = 3
option (3) is correct.

Like? Yes (1) | No (1)  

vamsi krishna   9 years AGO

S.I = PTR/100
2600x(20/3)xT = 520/3
For least exact number T=3

Like? Yes (1) | No (1)  

sravani   9 years AGO

it should be a multiple of 3...
3 is the least possible multiple..

Like? Yes (1) | No (1)  

n.sita   9 years AGO

the answer should a multiple of 3 and 3 is the least multiple .

Like? Yes (1) | No (1)  

20 Nov 2015 Number of bricks

In a rectangular park, there is an elliptical flower bed in the middle and the border. The side lengths of the park are 30m and 20m. Major axis & minor axis are of length 14m and 10m respectively. Around the flower bed in the middle there is a walking path 1.4m wide and in the middle of each side there is entrance of 2m width to the park upto the walking path. At the border the width of flower bed is 1m. Rest of the park is grass area. In the entire walking path is covered by bricks of dimension 10*20 cm^2, find the approximate number of bricks required.

OPtion

1) 3978
2) 4858
3) 5928
4) 4688
5) 3656
6) 4988
7) 3898
8) 3624
9) 4868
10)4818

Solution

By using figure solve it
The length of walking path from entrance of smaller side to the walking path around the flower bed is
= 30-14-1.4-1.4 / 2 = 6.6 m
and the length of other straight walking path is
= 20-10-1.4-1.4 / 2 = 3.6 m
Area of straight waling path (approx)
= 2(6.6*2) + 2(3.6*2) = 40.8 m^2
Area of elliptical (waling path)
= 22/7 (7+1.4)(5+1.4)-(22/7) *5*7
= 168.96 - 110 = 58.96 m^2
Total walking area = 58.96 + 40.8 = 99.76 m^2
Area covered by each brick = 200*10^-4 m^2 = 0.02 m^2
Number of bricks required = 99.76/0.02 = 4988 (approx)
Option 6)

Correct Option: 6 Best Solution (2)

Jaikumar Bhatia   9 years AGO

Inner elliptical of pathway
a1=14/2=7m
b1=10/2=5m
A1=Pi*a1*b1=Pi*7*5=110m^2.............................(i)
Outer elliptical of pathway
a2=[(14+1.4+1.4)/2]=8.4m
a2=[(10+1.4+1.4)/2]=6.4m
A2=Pi*a2*b2=Pi*8.4*6.4=168.96m^2........................(ii)
Therefore
Area of elliptical Pathway=A(p1)
=A2-A1
=168.96-110.....................................from(i)&(ii)
=58.96m^2=A(p1)................................................(iii)
Sides of Rectangular park
x=30m
y=20m
Lenght of horizontal entrance
L(h)=x-(14+1.4+1.4)=13.2m........................(iv)
Length of vertical entrance
L(v)=y-(10+1.4+1.4)=7.2m...........................(v)
w=width of entrance=2m...........................(vi)
Therefore
Area of entrance pathway=A(p2)
=[L(h)*w]+[L(v)*w]
=[13.2*2]+[7.2*2].............from(iv),(v),(vi)
=26.4+14.4
=40.8=A(p2)...........................................(vii)
Therefore
Total pathway area is
A(p)=A(p1)+A(p2)
=58.96+40.8...................from(iii)&(vii)
=99.76m^2
Area of one brick=A(br)
=10*20=200cm^2=0.02m^2
Therefore
No of bricks=A(p)/A(br)
=99.76/0.02
=4988 nos
Ans is 4988 nos

so option is (6)

Like? Yes (2) | No   

bhaskar   9 years AGO

(pi((9.4*7.4)-(8*6))+(30-18.8+20-14.8)*2)/.02=4988

Like? Yes (1) | No   

19 Nov 2015 Time and Work

Mukesh and Chandan can complete a task in 30 days when working together. After Mukesh and Chandan have been working together for 11 days, Chandan is called away and Mukesh, all by himself completes the task in the next 28 days. Had Mukesh been working alone, the number of days taken by him to complete the task would have been

OPtion

1) 379/9
2) 874/11
3) 520/11
4) 590/11
5) 859/9
6) 595/9
7) 444/9
8) 840/19
9) 533/9
10)None of these

Solution

(M+C)'s 11 days task = 11/30
Remaining work = 1 - 11/30 = 19/30
19/30 work is done by M in 28 days.
Whole work will be done by M in (28*30/19) = 840/19
Option 8)

Correct Option: 8 Best Solution (21)

BONAM SRINIVAS   9 years AGO

let mukesh=x, chandan=y
now x+y can complete work in 30 days. x+y one day work=1/30
given x+y worked for 11 days so,x+y 11 days work=11(1/30).
now remainini work=1-11/30 (since total work is one) i.e.,1-11/30=19/30.
this 19/30 work is done by mukesh in 28 days(given remaining work is done by mukesh)
19/30 work-------->28 days
1 work -------------->???
(1*28)/(19/30)=(28*30)/19=840/19=ans.

Like? Yes (5) | No (1)  

NAVEEN KUMAR   9 years AGO

workdone by mukesh&chandan in 11 days=11/30
hence , remaining work=1-11/30=19/30
A/Q..in 28 days mukesh complete the 19/30 part of work
hence,workdone of mukesh in 1 day=19/(28*30)=19/840
So,mukesh can complete the work in 840/19 days

Like? Yes (2) | No   

CHANDAN KUMAR   9 years AGO

m+c=30
m=28
toatal work=840
R(m+c)=840/30=28/day
work in 11 days by both=28x11=308
rem. work=840-308=532
rm=532/28=19/day
m alone=840/19

Like? Yes (2) | No   

MURUGESAN.P   9 years AGO

total number of units=30*11*28=9240 units.
both of them 1 day unit=9240/30=338 unit
11 days they work together,
so they complete=11*308=3388
remaining work=9240-3388= 5852 unit
Mukesh completes the remaining unit in 28 days,
so his one day unit is 5852/28=209 unit
he will complete the all the unit in( 9240/209) days
after simplification it will become 840/19 days
so,option 8:840/19 is correct

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

total work done by both mukesh and Chandan=1/30
work in 11 days=11/30
remaining work=19/30
time taken by mukesh alone to complete work is=28*30/19=840/19

option 8

Like? Yes (1) | No   

Riteswari sahu   9 years AGO

1 day work of both = 1/30
11 day work of both = 11/30
remaining work = (1-11/30)= 19/30
mukesh completed it in 28 day = (19/30)*28= 840/19

Like? Yes (1) | No   

ramakrishna   9 years AGO

mukesh and chandana can do the work =30 days
in11 days they completed =11 th part of work
remaining work=19
19 th part of work completed in 28 days by mukesh
so mukesh can complete the work in 30*28/19=840/19

Like? Yes (1) | No   

SHUBHAM   9 years AGO

(1/X + 19/(30*28))=1/30
1/X=3/280,
(3/280+1/y=1/30),
y=840/19

Like? Yes (1) | No   

chetan   9 years AGO

(M+C)'s 11 days work =(1*11)/30=11/30.
Remaining work=1-(11/30)=19/30.
Whole work will be done in (30*28)/19=840/19.
So option 8 is correct.

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Jagadeesh   9 years AGO

M+C=30 days
1 day work (M+C)=1/30 days
11 days work both together=11/30
remaining work=1-11/30=19/30
19/30 days work done in 28 days
1 day work=28*30/19
11/30 day work=28*11/19
total work=28*11/19 +28
equals to 840/19

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Pooja Shaw   9 years AGO

Mukesh+Chandan can complete the work in 30 days
in 1 day Mukesh+chandan can complete 1/30 of the work
There fore,in 11 days Mukesh +chandan will complete 11/30 of the work
Remaining work=1-11/30=19/30
19/30 of the work is completed in 28 days by Mukesh
therefore the whole work will be completed in (28*(30/19))=840/19 days.
Thus correct option is 8)840/19

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aditya vasudeva   9 years AGO

one day work of mukesh and chandan combined = 1/30
Let one day work of mukesh = 1/m and one day work of chandan = 1/c
one day work of chandan = 1/30 - one day work of mukesh
11*(1/m+1/30-1/m) = 11/30
Remaining work = 1-11/30 = 19/30
This remaining work is done by Mukesh in 28 days ...
28 * 1/m = 19/30
solving above equation we get m = 840/19 Answer...

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Abhisheak Kumar Missu   9 years AGO

m+c in 1 day= 1/30
m+c in11 day= 11/30
remaining work=1-11/30=19/30
mukesh does 19/30 work in 28 days
so 1 work will done in 28*30/19

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sushma g   9 years AGO

mukesh +chandan can do a work in 30days=1/30
mukesh and chandan working together,hence amount of work done by both=11/30
amount of work done=1-11/30=19/30
remaining work is completed by mukesh in 28 days=19/30*28=19/840
days taken by mukesh alone to complete is 840/19

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Akshaya   9 years AGO

Mukesh + Chandan = 30 days

(M + C)'s one day work = 1/30
For 11 days they are working together,
(M+C)'s 11 days work = (1/30)*11 = 11/30
Remaining work = 1 - 11/30 = 19/30
Now, 19/30 work is done by Mukesh in 28 days.
Therefore, the entire work will be done by Mukesh in 28 * (30/19) days.
i.e. Mukesh will take 840/19 days to complete the entire work.

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Nandkishor jaiswal   9 years AGO

from the question
30(1/m+1/c)=1
or, 1/c=1/30-1/m ---------1
and 11(1/m+1/30-1/m)+28/m=1
or, 28/m=1-11/30
m=30*28/19=840/19

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Jaikumar Bhatia   9 years AGO

Let's
Mukesh=A
Chandan=B

(A+B)'s 11 day's work=[(1/30)*11]=11/30
Remaining work=[1-(11/30)]=19/30
now 19/30 work is done by A in 28 days
Then whole work will be done by A in [28*(30/19]=840/19

so ans is 840/19
Option is (8)

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Devendra Marghade   9 years AGO

1 day work of Mukesh & Chandan is 1/M & 1/C respectively, then, 1/M+1/C= 1/30 ---(i)
Also 11[(1/M)+(1/C)]+28/M=1 or 29/M +11/C=1 ---(ii)
Substituting 1/C=(1/30)-(1/M) in (ii), we get 28/M=19/30 or
1/M=19/840
So Mukesh working alone can complete the task in 840/19 days Option 8)

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V CHANDRA SHEKHAR   9 years AGO

(MUKESH+CHANDAN)'S11 DAYS WORK=(1/30)*11=11/30

REMAINING WORK=1-11/30=19/30

NOW 19/30 WORK IS DONE BY MUKESH IN 28 DAYS ,

THEREFORE ,THE WHOLE WORK WILL BE DONE BY
MUKESH IN (28*30)/19 DAYS =840/19 DAYS

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Saurabh Agrawal   9 years AGO

left work of 19 days done by mukesh in = 28 days
i.e. work of 1 day can be done by mukesh is =28/19 days
i.e. work of 30 days can be done by mukesh alone is =(28/19)*30=840/19

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Ankusj   9 years AGO

Let time taken by mukesh be x days and chandan be y days.
So, 1/x+1/y= 1/30
Now, 11*(1/x+1/y) +28/x=1
Solving this..x= 840/19

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