Latest TCS Aptitude Question SOLUTION: . (40*40*40 – 31*31*31)/(40*40+40*31+31*31)=? a)8 b)9 c)71 d)51 Solution: a3 -b3 =(a-b)*(a2+a*b+b2) so from this formula we will find (a-b) value

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(#M40040505) TCS Numerical Ability question algebra Keep an EYE

. (40*40*40 – 31*31*31)/(40*40+40*31+31*31)=?
a)8
b)9
c)71
d)51
Solution: a3 -b3 =(a-b)*(a2+a*b+b2) so from this formula we will find (a-b) value

Asked In TCS ANUP KUMAR GUPTA (9 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

Krishnendu Dutta  (9 years ago)

Answer is 9.
let a=40,b=31
so, (40^3-31^3)/(40^2+40*31+31^2)
Now as we know, a^3-b^3=(a-b)*(a^2+a*b+b^2)
(40-31)(40^2+40*31+31^2)/(40^2+40*31+31^2)
So, (40-31)=9.

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Tania Chatterjee  (9 years ago)

Let, a=40 and b=31
(a^3-b^3)/(a^2+ab+b^2)=(a-b)=(40-31)=9
Option b)9

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

ANIMESH GORAI  (9 years ago)

let a=40,b=31,
(a^3-b^3)/(a^2+ab+b^2)=(a-b)(a^2+ab+b^2)/(a^2+ab+b^2)=(a-b)=40-31=9
b)9

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Trishla kapoor  (9 years ago)

the solution for this kind of questions is
(a+b) i.e. a-40, b=31
therefore, (40+31)=71
options c is correct

Is this solution Helpfull? Yes | No (1)  

karthiga  (9 years ago)

a^3 - b^3 = (a - b)(a^2 + a*b +b^2)
40^3 - 31^3 = (40 - 31)(40^2 + 40*31 +31^2)/(40^2 + 40*31 +31^2) = 9

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sangram nayak  (9 years ago)

let a=40 and b=31
simply the above formula saatisfies the alzebric formula
so a-b=40-31=9

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TARAK  (9 years ago)

Trick- Guys try to solve these type of question by solving the last digit.
takink the last digit of 40 i.e 0
, likewise last digit of 31 is 1,
Now (0*0*0-1*1*1)/(0*0+0*1+1*1)
=(0-1)/(1)
Now for numerator clearly (0-1) =9 i.e the last digit of the Numerator.
Denominator is 1
So, the ans will will be defenitly will the last digit 9.
Only one option satsfy so ans is B.

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