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

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(#M40115375) TCS Numerical Ability question numbers Keep an EYE

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

Asked In TCS harish (9 years ago)

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vinit ratan  (9 years ago)

solution is posible in this case also take a=b=5/68 c=29/68 d=1/68 e=1/68 then a+b+c+d=40/68

Is this solution Helpfull? Yes (8) | No   

jeremiah  (9 years ago)

question is wrong
it should be
(68-a)(68-b)(68-c)(68-d)(68-e)
and the ans is 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=232

Is this solution Helpfull? Yes (7) | No   

MATHU  (9 years ago)

how the ans is 232

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Ritesh   (9 years ago)

plzzz
explain this question

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pushkar  (9 years ago)

fisrt take out the factor of 725 by using factorization. the factor is 5,5,29. but we five variables a,b,c,d,e, so take 5 as -5 becoz every no exist in +- form on number line. also arrange in ascending order, then (29,5,-5,1,-1). now compare and find abcde ie a=39,b=63,c=73,d=67,e=69. then find a+b+c+d=39+63+73+67=242. hence 232 is wrong.

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