IBM
Company
Numerical Ability
Algebra
What is the value of 44444445*88888885*444444442*44444438/44444448^2
1) 88888883
2) 88888884
3) 88888888
4) 44444443
Read Solution (Total 1)
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- actual question is (44444445*88888885*444444442+44444438)/44444444^2.
then x=44444444
equation becomes
=((x+1)*(2x-3)*(x-2)+(x-6))/x^2
=((2x^2-x-3)(x-2)+(x-6))/x^2
=(2x^3-4x^2-x^2+2x-3x+6+x-6)/x^2
=(2x^3-5x^2)/x^2
=2x-5
=2(44444444)-5
=88888888-5
=88888883 - 10 years agoHelpfull: Yes(63) No(2)
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