The sum of ages of Mac and Jabi is 26 times the difference of their ages. Maximum possible differenc

Supriya Mondal   9 years AGO

There are 5 possibilities that satisfy above conditions
They are-
(135,125)
(108,100)
(81,75)
(54,50)
(27,25)

Like? Yes (11) | No (1)  

Gurram Mahendra ranganath   9 years AGO

M+J=26(M-J)
by solving we get,M=27*J / 25
hence,possible values of J are:25,50,75,100,125
hence,ans is 5.

Like? Yes (8) | No (1)  

Yathiraj   9 years AGO

From the puzzle statement, (M + J) = 26(M - J)
J = (25/27)M
For J to be integer, M should be multiples of 27.
Mac: 27, Jabi: 25
Mac: 54, Jabi: 50
Mac: 81, Jabi: 75
Mac: 108, Jabi: 100
Mac: 135, Jabi: 125
No. of possibilities: 5

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Subhadip Samanta   9 years AGO

suppose the age of Mac=x and age of Jabi=y
as per the question (x+y)=26(x-y)
and max(x-y)=10
also the values of x and y should be integer.
to satisfy these conditions (x-y) should be a even no.
(x-y) should not be 0 because in that case both x and y should be 0
the possible values of (x-y) should be 2,4,6,8,10
for these values there should be 5 combinations of x and y
so that bpth x and y are integers.
So the ans is option 5)

Like? Yes (4) | No (1)  

Dharmraj Sharma   9 years AGO

x=(27/25)*y
so for x to be integer y must be multiple of 25.
so possible pairs
(x,y)
(27,25)
(54,50)
(81,75)
(108,100)
(135,125)

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surya   9 years AGO

m+j=26(m-j)
said max possible difference is 10
so for only even values of (m-j) the ages may be possible . so no of possiblities is 5.

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Abhishek Sumit   9 years AGO

possible ages of two person={27,25},{54,50},{81,75},{108,100},{135,125}

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Surit Datta   9 years AGO

five pairs satisfies the above condition
which are
135,125
108,100
81,75
54,50
27,25
so option 5
that's it

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b.subbarao   9 years AGO

let mac and jabi ages are be =x and y' as per the data
x+y=26(x-y) so 25x-27y=0 27+25=26*2=52,------(1)
54+50=26(4)=104--------------(2)
81+75=26(6)=156--------------(3)
108+100=26(8)=208-----------(4)
135+125=26*10=260----------(5)

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A.SOMA SUNDARI   9 years AGO

If the diff of two ages are in odd no then there is no possibilities for the sum of two ages are equal to 26 times of the diff of the ages.
So the even num upto 10 is 2,4,6,8,10
therfore the no of possibilities is 5

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SAKTHEESVARI S   9 years AGO

difference of their age must be an even num,then only the condition is satisfied, the max possibility of their age difference is 10, so all the resultant possibilities are 2,4,6,8,10,totally 5 possibilities.

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Nidadavolu.Suneetha   9 years AGO

The maximum difference is 10.so the diff. in b/w 0 to 10.
for example take diff. is 1.then (age of mac+age of jabi)=26(1)
so the diff. b/w two of them is 1 but it not possible.If diff. is 2 sum of ages is 2(26)=52 bcoz take one person age is 25&another is 27.It satisfies the diff.
The diff. 2,4,6,8,10 satisfies the condition.So max. no of possibilities of their ages are 5

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ABHINAV PUSHPAM   9 years AGO

5 is asnwer.
=> x+y=26(x-y)
=> 25x=27y
=> x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125
So, answer ie 5.
option 5)

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RAKESH   9 years AGO

M+J = 26*(M-J)

M-J

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Saurabh Kumar Gupta   9 years AGO

Let M > J
(M+J) = 26*(M-J)
Now (M-J) can take values between 1 to 10
Out of these only 2, 4, 6, 8, 10 satisfy the condition, i.e. for eg. if their age difference is 2 years then M+J = 52 and so M & J can be 27 and 25 respectively.
But if the difference is 3 then M+J = 78, now we cant get their ages with difference of 3 years.
So total possible value pairs = 5

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praveen kumar   9 years AGO

5 such ages can be

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sompon temtheerakij   9 years AGO

If diff. 1, ages of the two persons = 24.5+25.5
If diff. 2, ages of the two persons = 25 + 27
If diff. 3, ages of the two persons = 37.5+40.5
If diff. 4, ages of the two persons = 50+54
If diff. 5, ages of the two persons = 62.5+67.5
If diff. 6, ages of the two persons = 75+81
If diff. 7, ages of the two persons = 87.5+94.5
If diff. 8, ages of the two persons = 100+108
If diff. 9, ages of the two persons = 112.5+121.5
If diff. 10, ages of the two persons = 125+135
Ages are taken in integers , possibilities for ages of the two persons are 5.

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Santosh Kumar   9 years AGO

(x - y)26 = (x + y)
25x - 26y =x + y
25x - 27y = 0 ................eqn 1
x - y = 1 to 10 (given) .........eqn 2
if x-y=1
we don't get integer age
we get integer only if
x - y = (2,4,6,8,10)
so, obtion 5

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Himanshu   9 years AGO

Let age of mac= x and age of jabi = y
According to question, sum of their ages=26(difference between their ages)
x+y=26(x-y)
now let diff =1
x-y=1 and x+y =26
solving x =13.5 and y = 12.5
but ages are taken as integers, so this can't be the right one,
now diff = 2
x+y = 52 and x-y=2
on solving x=26 y =25
similarly the maximum possible diff can be 10
taking different values of difference of their ages upto 10, ages of x and y which are integers are considered and no. comes out to be 5.
So 5 is the right option

Like? Yes (1) | No   

Divya jain   9 years AGO

mac+jabi=26(mac-jabi)
mac-jabi 125,135
for 8-> 108,100
6-> 81,75
4-> 50,54
2-> 25,27

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Ajinkya Kulkarni   9 years AGO

5

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ramakrishnateja   9 years AGO

given that the m+j=26(m-j). also given that the diff is max of 10 yrs. so, it should be 1,2,.....upto 10. where ages shouldnot be integers frm these diff.
so let us take diff=1. here 27j=25m.so m:j=27:25 hence m=27x and j=25x. so diff is 2x=1 hence x=1/2 where ages will be 27/2 and 25/2. which is wrong. so 1,3,5,7,9 are omitted. take diff as 2,4,6,8,10. then for diff =2 then 2x=2 x=1. so ages will be m=27,25. so sum =52 which is 26(2)

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purba mondal   9 years AGO

x+y=26(x-y)
if x-y=1,x+y=26,and x=13 y=13 x-y=0; x=14,y=12,x-y=2 so x-y!=1 not possible
if x-y=2,x+y=52 and x=27,y=25,x-y=2 possible
then same as x-y=3 not possible
x-y=4 is possible as x=54,y=50
....
then for x-y=even then possible,as max(x-y)=10 then
x-y=2,4,6,8,10 possible
5 possibilities

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Krishna Gandhi   9 years AGO

M+J=26(M-J)
M-J=1,2,3,….,10
So Case-1: M-J=1
M+J=26*1=26
M-J=1
M+j=26
-----------
2M=27;
M=13.5 ; which isn’t a integer
So, M-J=1 isn’t possible

Case-2: M-J=2
M+J=26*2=52
M-J=2
M+J=52
-----------
2M=54; M=27; J=25
So, ages are in integers hence M-J=2 is possible

Similarly, solve for M-J=3,4,…,10
We get the possible ages in integer for M-J=2,4,6,8,10
Ans: So there are 5 possibilities (i.e. M-J= 2,4,6,8,10)

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ADARSH GAURAV   9 years AGO

5 is the answer.

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Alka Sinha   9 years AGO

Max Possible diff of ages = 10
So, Possibilities of differences is from 1 to 10

Let Mac age = X
Jabi age = Y
Equation : X + Y = 26(X-Y)

Case 1: Let diff of ages = 10
X + 10 + X = 26(X+10-X)
X = 125
Y = 135

When we check other possibilities then it will satisfy the equation only for age diff 10, 8, 6, 4, 2.

So, Answer is 5.

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ANKIT RAMAN   9 years AGO

possible combinations are (25,27),(50,54),(75,81),(100,108),(125,135)

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Nupur   9 years AGO

All the even numbered difference can be mapped with ages.

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ganesh babu   9 years AGO

Only possible ages
(27,25)
(54,50)
(81,75)
(108,100)
(135,125)

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Sabitha.R   9 years AGO

x+y=26(x-y)
25x=27y
x/y= 27/25

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mohini gupta   9 years AGO

only cases where x-y is even

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karthik   9 years AGO

fjhjv

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Devendra Marghade   9 years AGO

Possible ages are total 5
(25,27) , (50,54) , (75,81), (100,108), (125,135)
Ans: 5

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lakshmi aparna   9 years AGO

Let the ages of mac and jabi be M, J.
M+J = 26 (M-J); M-J

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A M ANKIT KALLURAYA   9 years AGO

x+y=26(x-y)
25x=27y
x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125

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Bhavana.K   9 years AGO

guess

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vishal kumar   9 years AGO

=> x+y=26(x-y)
=> 25x=27y
=> x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125
So, answer ie 5

Like? Yes (1) | No   

Sneha   9 years AGO

=> x+y=26(x-y)
=> 25x=27y
=> x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125
So, answer ie 5

Like? Yes (1) | No   

Namrata Gupta   9 years AGO

ans is 5
x+y=26(x-y)
25x=27y
x/y=27/25
so possible sol is 5.

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Jitendra Prasad   9 years AGO

5 is the answer.
=> x+y=26(x-y)
=> 25x=27y
=> x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125
So, answer ie 5

Like? Yes (1) | No   

POOJA SINGH   9 years AGO

x+y=26(x-y)
25x=27y
x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125So, answer is 5

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neethu   9 years AGO

5

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Sourabh Prasad   9 years AGO

x+y=26(x-y)
25x=27y
x/y= 27/25
difference is 2.
So all possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125

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Paras Jain   9 years AGO

its 5 Ans.

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MRITUNJAY   9 years AGO

This is answer

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tasho k   9 years AGO

Trial and error...
25, 27
50,54
81, 75
100, 108
135, 125

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Dolly Singh   9 years AGO

Since their age difference can be max up to 10

Let say age of Max is x

age of Jabi is y

then

A To P, x + y = 26(x-y)

and x - y

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arush gupta   9 years AGO

5 is the answerx+y=26(x-y)=> 25x=27y=> x/y= 27/25difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125So, answer is 5

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Jaspalsingh Parmar   9 years AGO

Option 6 is correct

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Anshul Garg   9 years AGO

M + J = 26( M - J )
And M-J

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RAHUL KUMAR   9 years AGO

ans is 5

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