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(68-a)(68-b)(68-c)(68-d)(68-e)=725.wat is the value of a+b+c+d?
Read Solution (Total 17)
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- factors of 725= 5,5 29
so (68-63)=5 ,(68-63)=5,(68-39)=29, (68-67)=1 ,(69-67)=1
so a+b+c+d= 232 - 10 years agoHelpfull: Yes(28) No(1)
- (68-63)(68-63)(68-39)(68-67)(68-67)=725
=5*5*29*1*1
=725
and a+b+c+d
=63+63+39+67=232 - 10 years agoHelpfull: Yes(15) No(1)
- it is possible to find sum of all a+b+c+d+e but not possible to find sum of a+b+c+d as different people can take different values of a,b,c,d,e
- 10 years agoHelpfull: Yes(13) No(0)
(68-a)(68-b)(68-c)(68-d)(68-e)=5*5*29*1*1
a=63 b=63 c=39 d=68 e=68
a+b+c+d=232- 10 years agoHelpfull: Yes(3) No(0)
- @deepak goswami abcde should b distinct
- 10 years agoHelpfull: Yes(2) No(1)
- see 725 is equal to 5*5*29*1*1 however all the digits should be unique so one of the two duplicats is necessarily negative that is 5*-5*29*1*-1
- 10 years agoHelpfull: Yes(2) No(0)
- ans is 311
- 10 years agoHelpfull: Yes(1) No(5)
- a+b+c+d=232
- 10 years agoHelpfull: Yes(1) No(0)
- 242 - a=39,b=63,c=67,d=73,e=36
- 10 years agoHelpfull: Yes(1) No(1)
- =5*5*29*1*1(factors)
in order to get 5 in first place a=63
same for b ,c,d also
so a=63,b=63,c=39,d=1,d=1 - 10 years agoHelpfull: Yes(1) No(2)
- if i take e value 39, the sum will change
- 10 years agoHelpfull: Yes(1) No(0)
- dummy question
- 9 years agoHelpfull: Yes(1) No(0)
- how it is possable(a=63)?plz explain.
- 10 years agoHelpfull: Yes(0) No(0)
- 725= 5 * 5 * 29
(68-63)(68-63)(68-39)(68-67)(68-67)
63+63+39+67=232 - 10 years agoHelpfull: Yes(0) No(0)
- 63+63+67+67+39=299
- 10 years agoHelpfull: Yes(0) No(0)
- DEEPAK GOSWAMI
how you took once 67
its not possible
say it clear.... - 10 years agoHelpfull: Yes(0) No(0)
- For all the distinct digits...
1*5*29*-1*-5
(68-67)*(68-63)*(68-39)*(68-69)*(68-73)
Thus 67+63+39+69+73=311 - 10 years agoHelpfull: Yes(0) No(0)
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