If x = 3, y = 4, z = 5, then sum of the modulus of the numbers obtained by differences in the man

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15 Dec 2011 number game

If x = 3, y = 4, z = 5,
then sum of the modulus of the numbers obtained by differences in the manner E1 - E2 and E2 - E3
Meaning of operations are
E1 : sum of x,x^2,x^3;
E2 : sum of y,y^2,y^3 and
E3 : sum of z,z^2,z^3

OPtion

1) 96
2) 104
3) 110
4) 116
5) 129
6) 134
7) 140
8) 148
9) 152
10) none of these

Solution

E1 = 39
E2 = 84
E3 = 155
now E1 - E2 = 39 - 84 = -45 its modulus means numerical value = 45
and E2 - E3 = 84 - 155 = -71 its modulus means numerical value = 71
so sum = 45+71 = 116

Correct Option: 4 Best Solution (12)

Sidak   12 years AGO

E1 : sum of x,x^2,x^3 = 39;
E2 : sum of y,y^2,y^3 = 84
and
E3 : sum of z,z^2,z^3 = 155

modulus of E1-E2 = 45
modulus of E2-E3 = 71

sum of modulus = 45+71= 116 .. option 4

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mohan   12 years AGO

E1-3,9,27=39,E2-4,16,64=84, E3-5,25,125=155, MODULES of E1-E2=45,MODULES OF E2-E3=71, sum of this modules =45+71=116

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sonam wasule   12 years AGO

E1=3+9+27=39
E2=4+16+64=45
E3=5+25+125=155
MIOD OF E1-E2=45
MOD OF E2-E3=71
ANS=45=71=116

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GAURAV KUMAR SRIVASTAV   12 years AGO

MOD(E1-E2)+MOD(E2-E3)
(x-y)(1+x+y+x^2+y^2+xy)+(y-z)(1+y+z+y^2+z^2+yz)
MOD(-45)+MOD(-71)
116

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Mullaji Zubair Dawood   12 years AGO

E1=(3*3^2*3^3)=3+9+27=39
E2=(4*4^2*4^3)=4+16+64=84
E3=(5*5^2*5^3)=5+25+125=155
|E1-E2|=45
|E2-E3|=71
.:|E1-E2|+|E2-E3|=116 so Ans4.

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sandeep potnuru   12 years AGO

sum of x,x^2,x^3=39
sum of y,y^2,y^3=84
sum of z,z^2,z^3=155,E1-E2 mod=45,E2-E3 mod=71
sum of these two is 116

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Sudhish Surendran   12 years AGO

E1 = 3+3^2+3^3 = 39
E2 = 4+4^2+4^3 = 84
E3 = 5+5^2+5^3 = 155
|E1 - E2| + |E2 - E3| = |-45|+|-71| = 45+71 = 116
i.e. option 4

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Rakesh kumar   12 years AGO

modulus {(3+9+27)-(4+16+64)}+{(4+16+64)-(5+25+125)}=116

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Swetha.R   12 years AGO

E1=39,E2=84,E3=155
|E1-E2|+|E2-E3|=116

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Manish Kumar   12 years AGO

E1=3+9+27=39
E2=4+16+64=84
E3=5+25+125

NO=|39-84| + |84-125| = 116

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sivakumar   12 years AGO

E1=3+9+27 = 39;
E2=4+16+64 = 84;
E3=5+25+125=155;
MODULUS OF (E1-E2)=45;MODULUS OF (E2-E3)=71;
SUM IS 116;

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Balakumaran N.A   12 years AGO

By, substituting X,Y and Z values:
E1=39; E2=84; E3=155;
|E1-E2|+|E2-E3|=|-45|+|-71| = 45+71= 116.Option 4.

Like? Yes (2) | No (4)