Latest TCS Aptitude Question SOLUTION: (68-a)(68-b)(68-c)(68-d)(68-e) = 725 ? Find a+b+c+d= ??

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(68-a)(68-b)(68-c)(68-d)(68-e) = 725 ? Find a+b+c+d= ??

Asked In TCS Amit Wagh (9 years ago)

Unsolved Read Solution (13)

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Read Solution (13) :

SUBHADEEP BANIK  (9 years ago)

by factorizing 725, we get prime factors 5,5 and 29 i.e.,
725=5*5*29
therefore we can assume that the expressions enclosed within each pair of
parentheses results into two 5's and a single 29. The remaining to expressions are bound to result into two 1's.
like let's say,
(68-a)=5
(68-b)=5
(68-c)=29
(68-d)=1 and
(68-e)=1
for the above to be true, value of
a=63
b=63
c=38
d=67
e=67
therefore
a+b+c+d=38+63+63+67
=231

Is this solution Helpfull? Yes (19) | No (11)  

SUMANTA SAMANTA  (9 years ago)

The L.C.M of 725 = 5 * 5 * 29 * 1 * 1
we can form above things like this b/c :-
(68-a)(68-b)(68-c)(68-d)(68-e) => 5 * 5 * 29 * 1 * 1
As,
(68-a)=5 ; so, a=63;
(68-b)=5 ; so, b=63;
(68-c)=29; so, c=39;
(68-d)=1 ; so, d=67;
(68-e)=1 ; so, e=67;
So we can write ,
(a+b+c+d)=(63+63+39+67)= 232 ;

N.B : This Math solution can be different if anyone consider
(68-a) = 1 or 5 or 29
(68-b) = 1 or 5 or 29
(68-c) = 1 or 5 or 29
(68-d) = 1 or 5 or 29
(68-e) = 1 or 5 or 29
That why it will confuse every one .
As it is not given no relation between a,b,c,d,e ( I mean a>b>c>d>e like
this ).

Is this solution Helpfull? Yes (11) | No (2)  

virendra tiwari  (9 years ago)

factor or 725=5*5*29
68-a=5
68-b=5
68-c=29
68-d=1
68-e=1
but sequence can be change so ans will be,,63+63+39+67=232 and 63+39+67+67=236,,,and 63+63+67+67=260,,, go to option

Is this solution Helpfull? Yes (2) | No   

Rohit Kumar  (9 years ago)

this question is not complete. According to given informations it can have multiple answers.

Is this solution Helpfull? Yes (1) | No (3)  

Souradip Nath  (9 years ago)

725 = 29 * 5 * 5 * 1 * 1
68-a=29 => a = 39
68-b=5 => b = 63
68-c=5 => c = 63
68-d=1 => d = 67
68-e=1 => e = 67

a+b+c+d+e = 299

Is this solution Helpfull? Yes (1) | No   

Saikat Maji  (9 years ago)

725 factors are 29x5x5x1x1....
so as we equal the 68-a=29 then a=39
consequently b=63,c=63,d=67,e=67
so a+b+c+d=232

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ANIK KUMAR SAHA  (9 years ago)

we have 725= 5x5x29
hence we can have (68-67)(68-67)(68-63)(68-63)(68-39)=725
hence a+b+c+d+e= 67+67+5+5+29=173

Is this solution Helpfull? Yes | No (1)  

ANIK KUMAR SAHA  (9 years ago)

we have 725= 5x5x29
hence we can have (68-67)(68-67)(68-63)(68-63)(68-39)=725
hence a+b+c+d+e= 67+67+63+63+39=299

Is this solution Helpfull? Yes | No   

sonali sahoo  (9 years ago)

first of all factorize 725- >> 29*5*5*1*1
now try to manipulate all the factors in the form of the equation
(68-63)*(68-63)*(68-39)*(68-67)*(68-67)
a =63
b=63
c=39
d=67
e=67
now a+b+c+d = 232

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haran  (9 years ago)

In actual question, there are 5 elements and they are distinct, i.e. a,b,c,d,e & a!=b!=c!=d!=e, it can be solved like, 725=29*5*(-5)*1*(-1)
which makes the values 63+73+69+67+39 = 311

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hashmi  (9 years ago)

725=5*5*29
68-5=67
so a=b=67
68-29=39 so c=39
d=67,e=67 by using this a+b+c+d=232

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Goldi kumari  (9 years ago)

factorizing of 725,we get prime factors 5,5 and 29 i.e.,
725=5*5*29
(68-a)=5
(68-b)=5
(68-c)=29
(68-d)=1
(68-e)=1
a=63
b=63
c=39
d=67
e=67
a+b+c+d
63+63+39+67=232

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Abhishek Kumar  (9 years ago)

Ans: 311

(68-a)(68-b)(68-c)(68-d)(68-e) = 725
factors of 725=5*5*29
68-a=5 ==>a=63
As a, b, c, d can not have similar value,
so, 68-b=-5 b=73
68-c=29 ==> c=39
68-d=1 ==> d=67
68-e=-1 ==> e=69
Hence sum
a+b+c+d= 63+73+39+67+69=311

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